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Undergraduate Lecture Notes in Physics

Lev Kantorovich

Mathematics
for Natural
Scientists
Fundamentals and Basics

Undergraduate Lecture Notes in Physics

More information about this series at http://www
...
com/series/8917

Undergraduate Lecture Notes in Physics (ULNP) publishes authoritative texts
covering topics throughout pure and applied physics
...

ULNP titles must provide at least one of the following:
An exceptionally clear and concise treatment of a standard undergraduate subject
...

A novel perspective or an unusual approach to teaching a subject
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The purpose of ULNP is to provide intriguing, absorbing books that will continue
to be the reader’s preferred reference throughout their academic career
...
1007/978-1-4939-2785-2
Library of Congress Control Number: 2015943266
Springer New York Heidelberg Dordrecht London
© Springer Science+Business Media, LLC 2016
This work is subject to copyright
...

The use of general descriptive names, registered names, trademarks, service marks, etc
...

The publisher, the authors and the editors are safe to assume that the advice and information in this book
are believed to be true and accurate at the date of publication
...

Printed on acid-free paper
Springer Science+Business Media LLC New York is part of Springer Science+Business Media (www
...
com)

Preface

The idea to write a mathematics textbook for physics undergraduate students
presented itself only after more than 10 years of teaching a mathematics course
for the second year students at King’s College London
...
Instead, the assumption is made that a baseline understanding of numbers,
functions, numerical sequences, differentiation and integration was adequately
covered in the school curriculum and thus they remain poorly represented at
university
...
Students, for instance, have no understanding of “a
proof”
...
In fact, the whole approach to
teaching mathematics (and physics) at school is largely based on memorising facts
...
This can only be
avoided if students are taught how to think, and part of this learning must come from
understanding the foundations of the most beautiful of all sciences, mathematics,
which is based on a magnificent edifice of logic and rigour
...

Unfortunately, very little exists in the offering today for physics students
...
However, these books in their presentation
lack rigour; most of the important statements (or theorems) are not even mentioned
v

vi

Preface

while others are discussed rather briefly and far too “intuitively”
...
g
...

I myself studied physics and mathematics between 1974 and 1979 at the Latvian
University in Riga (the capital of Latvia, formerly USSR) where a very different
approach to teaching students was adopted
...
The whole first year of studies
was almost completely devoted to mathematics and general physics; the latter was
an introductory course with little mathematics, but covering most of physics from
mechanics to optics
...
There were several
excellent textbooks available to us at the time, e
...
[6, 7], which I believe are still
among the best available today, especially for a student yearning for something
thought provoking
...

In this project, an attempt has been made to build mathematics from numbers up
to functions of a single and then of many variables, including linear algebra and
theory of functions of complex variables, step-by-step
...
Practically everything is accompanied by proofs
with sufficient rigour
...
At the same time, the material must be written in such
a way that it is not intimidating and is easy to read by an average physics student
and thus a decision was made not to overload the text with notations mathematicians
would find commonplace in their literature
...
of mathematical logic are not used although they would have enabled
in many places (e
...
in formulating theorems and during proofs) to shorten the
wording considerably
...
As a result, the material is presented by
a language that is more accessible and hence should be easier to understand but
without compromising the clarity and rigour of the subject
...
Though
adding examples is not that difficult, adding many problems turned out to be a
formidable task requiring considerable time
...
In addition, some of the problems were designed
to illustrate the theoretical material, appearing at the corresponding locations
throughout the text
...
Nevertheless,
there are up to a hundred problems in each chapter, and almost all of them are

Preface

vii

accompanied by answers
...
There are many
other books available now to students, which contain a large number of problems,
e
...
[1–5], and hence students should be able to use them much more easily as an
accompaniment to the explanations given here
...
This one you are holding
now is the first book
...
1), and moves on to functions
and limits (Chap
...
3 and 5), integration (Chaps
...
7) and ordinary differential equations (Chap
...
As we
build up our mathematical tools, examples from physics start illustrating the maths
being developed whenever possible
...
This is a win-win situation as on the other hand, a more in-depth
knowledge of physics may follow by looking more closely at the mathematics side
of a known physics problem
...

In the second book, more advanced material will be presented such as: linear
algebra, Fourier series, integral transforms (Fourier and Laplace), functions of
complex variable, special functions including general theory of orthogonal polynomials, general theory of curvilinear coordinates including their applications in
differential calculus, partial differential equations of mathematical physics, and
finally calculus of variation
...
Both volumes taken together should comprise a comprehensive
collection of mathematical wisdom hopefully presented in a clear, gradual and
convincing manner, with real illustrations from physics
...

I am also convinced that the book should serve well as a reach reference material
for lecturers
...

In preparing the book, I have consulted extensively a number of excellent existing
books [6–9], most of them written originally in Russian
...
To make the reading easier, I do not cite the specific
source I used in the text, so I’d like to acknowledge the above-mentioned books in
general here
...
In addition, I would also like to acknowledge
the invaluable help I have been having along the way from Wikipedia, and therefore
my gratitude goes to all those scholars across the globe who contributed to this
fantastic online resource
...
These are, first of all, Era Lepina
who taught me the foundations of analysis during my first year at University (I still
keep the notes from her excellent lectures and frequently consulted them during
the work on this book!), Michael Belov and Teodors Cirulis (theory of functions
of complex variables and some other advanced topics such as asymptotic series)
...

Finally, I would like to apologise to the reader for possible misprints and errors
in the book which are inevitable for the text of this size in spite of all the efforts
to avoid these
...
kantorovitch@kcl
...
uk)
...
McQuarrie D (2003) Mathematical methods for scientists and engineers
...
ISBN 1-891389-29-7
2
...
Cambridge University Press, Cambridge
...
Boas M (1983) Mathematical methods in the physical sciences, 2nd edn
...
ISBN 0-471-04409-1
4
...
Palgrave, New York
...
Arfken GB, Weber, HJ Harris, FE (2013) Mathematical methods for physicists
...
Academic Press, New York
...
Смирнов ВИ (1974) Курс высшей математики, т
...
Apparently, there is a rather old English translation: Smirnov VI A
course of higher mathematics: Adiwes international series in mathematics, vol
1
...
ISBN 1483120171, 3-Part-1
...
ISBN B00GWQPPMO
...
Фихтенголц ГМ (2001) Курс дифференциального и интегрального
исчисления, Москва, Физматлит (in Russian)
...
Шипачев ВС (1998) Высшая математика, Москва, Высшая Школа (in
Russian)
...
Widder DV (1989) Advanced calculus, 2nd edn
...
For reader’s convenience their names (together with some information
borrowed from their Wikipedia pages) are listed here in the order they first appear
in the text:
René Descartes (Latinized: Renatus Cartesius) (1596–1650) was a French
philosopher, mathematician and writer
...
570 BC – c
...

Tullio Levi-Civita (1873–1941) was an Italian mathematician
...

Gerolamo (or Girolamo, or Geronimo) Cardano (1501–1576) was an Italian
mathematician, physician, astrologer and gambler
...

William Rowan Hamilton (1805–1865) was an Irish mathematician, physicist
and astronomer
...

Abu Bakr ibn Muhammad ibn al Husayn al-Karaji (or al-Karkhi) was a
Persian mathematician and engineer
...

Viktor Yakovych Bunyakovsky (1804–1889) was a Ukrainian mathematician
...

Guido Grandi (1671–1742) was an Italian mathematician
...

Lodovico Ferrari (1522–1565) was an Italian mathematician
...

ix

x

Famous Scientists Mentioned in the Book

Gerolamo Cardano (1501–1576) was an Italian mathematician
...

Niels Henrik Abel (1802–1829) was a Norwegian mathematician
...

Abraham de Moivre (1667–1754) was a French mathematician
...

Bernhard Placidus Johann Nepomuk Bolzano (1781–1848) was a Czech
mathematician, philosopher and theologist
...

Sir William Rowan Hamilton (1805–1865) was an Irish physicist, astronomer
and mathematician
...

Brook Taylor (1685–1731) was an English mathematician who is best known
for Taylor’s theorem and Taylor’s series
...

Colin Maclaurin (1698–1746) was a Scottish mathematician
...
He made significant contributions to the fields of
analysis, number theory, and both classical and celestial mechanics
...

Johannes Diderik van der Waals (1837–1923) was a Dutch theoretical physicist and thermodynamicist
...
He is a co-author of a famous ten volume course of theoretical physics
...

Georg Friedrich Bernhard Riemann (1826–1866) was a German mathematician
...

Henri Léon Lebesgue (1875–1941) was a French mathematician
...

Jean Baptiste Joseph Fourier (1768–1830) was a French mathematician and
physicist, best known for Fourier series and integral
...

He made important discoveries in various fields of mathematics and introduced
much of the modern mathematical terminology and notation
...

Paul Langevin (1872–1946) was a prominent French physicist
...

Lorenzo Romano Amedeo Carlo Avogadro di Quaregna e di Cerreto (1776–
1856) was an Italian scientist
...

Félix Savart (1791–1841) was a French physicist
...

Hermann Ludwig Ferdinand von Helmholtz (1821–1894) was a German
physician and physicist
...

James Clerk Maxwell (1831–1879) was a Scottish mathematical physicist best
known for his unification of electricity and magnetism into a single theory of
electromagnetism
...

Carl Gustav Jacob Jacobi (1804–1851) who was a German mathematician
...

Albert Einstein (1879–1955) was a German-born theoretical physicist famously
known for his relativity and gravity theories
...

Andrey Nikolaevich Kolmogorov (1903–1987) was a Soviet mathematician,
one of the founders of modern probability theory
...

George Green (1793–1841) was a British mathematical physicist
...

Johann Benedict Listing (1808–1882) was a German mathematician
...

Mikhail Vasilyevich Ostrogradsky (1801–1862) was a Russian - Ukrainian
mathematician, mechanician and physicist
...

Archimedes of Syracuse (c
...
212 BC) was a Greek mathematician,
physicist, engineer, inventor and astronomer
...

Hendrik Antoon Lorentz (1853–928) was a Dutch physicist
...


xii

Famous Scientists Mentioned in the Book

Michael Faraday (1791–1867) was a famous English physicist
...

Jean-Baptiste le Rond d’Alembert (1717–1783) was a French mathematician,
physicist, philosopher
...

Douglas Rayner Hartree (1897–1958) was an English mathematician and
physicist
...

Satyendra Nath Bose (1894–1974) was an Indian Bengali physicist specialising
in mathematical physics
...

Charles Hermite (1822–1901) was a French mathematician
...

Pafnuty Lvovich Chebyshev (1821–1894) was a Russian mathematician
...

´
Józef Maria Hoene-Wronski (1776–1853) was a Polish Messianist philosopher
who worked in many fields of knowledge, including mathematics and physics
...
He also contributed in other
fields fo physics such as statistical mechanics and thermodynamics, physics of
dielectrics, colour theory, electrodynamics, general relativity and cosmology
...

Friedrich Wilhelm Bessel (1784–1846) was a German mathematician and
astronomer
...

Konstantin Eduardovich Tsiolkovsky (1857–1935) was a Russian and Soviet
rocket scientist and pioneer of the astronautic theory
...


Contents

Part I

Fundamentals

1 Basic Knowledge
...
1 Logic of Mathematics
...
2 Real Numbers
...
3 Cartesian Coordinates in 2D and 3D Spaces
...
4 Elementary Geometry
...
5 Introduction to Elementary Functions and Trigonometry
...
6 Simple Determinants
...
7 Vectors
...
7
...

1
...
2 N-dimensional Space
...
8 Introduction to Complex Numbers
...
9 Summation of Finite Series
...
10 Binomial Formula
...
11 Combinatorics and Multinomial Theorem
...
12 Some Important Inequalities
...
13 Lines and Planes
...
13
...

1
...
2 Polar and Spherical Coordinates
...
13
...

1
...
4 Planes
...
13
...


3
3
5
10
11
14
23
26
26
35
37
41
44
49
52
56
56
57
58
59
62

2 Functions
...
1 Definition and Main Types of Functions
...
2 Infinite Numerical Sequences
...
2
...

2
...
2 Main Theorems
...
2
...

2
...

2
...
1 Polynomials
...
4

Part II

2
...
2 Rational Functions
...
3
...

2
...
4 Number e
...
3
...

2
...
6 Hyperbolic Functions
...
3
...

2
...
8 Trigonometric Functions
...
3
...

Limit of a Function
...
4
...

2
...
2 Main Theorems
...
4
...

2
...
4 Several Famous Theorems Related to
Continuous Functions
...
4
...

2
...
6 Dealing with Uncertainties
...

3
...

3
...

3
...

3
...

3
...

3
...

3
...

3
...

3
...

3
...


123
123
127
132
136
137
140
146
155
157
160

4 Integral
...
1 Definite Integral: Introduction
...
2 Main Theorems
...
3 Main Theorem of Integration: Indefinite Integrals
...
4 Indefinite Integrals: Main Techniques
...
4
...

4
...
2 Integration by Parts
...
4
...

4
...
4 Integration of Trigonometric Functions
...
4
...

4
...
6 Integration of Irrational Functions
...
5 More on Calculation of Definite Integrals
...
5
...


175
175
181
188
195
195
198
204
209
212
213
220
220

Contents

xv

4
...
2 Integrals Depending on a Parameter
...
5
...

4
...
4 Cauchy Principal Value
...

4
...
1 Length of a Curved Line
...
6
...

4
...
3 Volume of Three-Dimensional Bodies
...
6
...

4
...
5 Simple Applications in Physics
...


223
226
235
237
238
242
245
248
250
259

5 Functions of Many Variables: Differentiation
...
1 Specification of Functions of Many Variables
...
1
...

5
...
2 Ellipsoid
...
1
...

5
...
4 Two-Pole (Two Sheet) Hyperboloid
...
1
...

5
...

5
...

5
...
Tangent Plane
...
5 Exact Differentials
...
6 Derivatives of Composite Functions
...
7 Applications in Thermodynamics
...
8 Directional Derivative and the Gradient of a Scalar Field
...
9 Taylor’s Theorem for Functions of Many Variables
...
10 Introduction to Finding an Extremum of a Function
...
10
...

5
...
2 Characterising Stationary Points: Sufficient Conditions
...
10
...

5
...
4 Method of Lagrange Multipliers
...

6
...

6
...
1 Definition and Intuitive Approach
...
1
...

6
...
3 Improper Integrals
...
1
...

6
...

6
...
1 Definition and Calculation
...
2
...

6
...

6
...
1 Line Integrals for Scalar Fields
...
3
...

6
...
3 Two-Dimensional Case: Green’s Formula
...
3
...


315
315
315
317
323
327
333
333
335
338
338
342
346
351

4
...
7

xvi

Contents

6
...
5

6
...
7

Surface Integrals
...
4
...

6
...
2 Area of a Surface
...
4
...

6
...
4 Surface Integrals for Vector Fields
...
4
...

6
...
6 Three-Dimensional Case: Exact Differentials
...
4
...

Application of Integral Theorems in Physics: Part I
...
5
...

6
...
2 Archimedes Law
...

6
...
1 Divergence of a Vector Field
...
6
...

6
...
3 Vector Fields: Scalar and Vector Potentials
...

6
...
1 Maxwell’s Equations
...
7
...

6
...
3 Hydrodynamic Equations of Ideal Liquid (Gas)
...

7
...

7
...
1 Series with Positive Terms
...
1
...

7
...
3 Alternating Series
...
1
...

7
...

7
...
1 Uniform Convergence
...
2
...

7
...
3 Properties: Integration and Differentiation
...
3 Power Series
...
3
...

7
...
2 Uniform Convergence and Term-by-Term
Differentiation and Integration of Power Series
...
3
...


417
418
420
425
426
429
434
435
437
439
441
442

8 Ordinary Differential Equations
...
1 First Order First Degree Differential Equations
...
1
...

8
...
2 “Exact” Differential Equations
...
1
...

8
...
4 Homogeneous Differential Equations
...
1
...


455
456
456
458
460
462
464

445
446

Contents

8
...
3
8
...
5

xvii

Linear Second Order Differential Equations
...
2
...

8
...
2 Inhomogeneous Linear Differential Equations
...

Series Solution of Linear ODEs
...
4
...

8
...
2 Series Solutions About a Regular Singular Point
...
4
...

Examples in Physics
...
5
...

8
...
2 Falling Water Drop
...
5
...

8
...
4 Distribution of Particles
...
5
...

8
...
6 Defects in a Crystal
...
521

Part I

Fundamentals

Chapter 1

Basic Knowledge

This is an introductory chapter which sets up some basic definitions we shall need
throughout the book
...
Hence the content of
this chapter will enable the reader to work on the forthcoming chapters without the
need of consulting other texts or web pages, and should make the book more or less
self-contained
...
1 Logic of Mathematics
In mathematics (and in other natural sciences, especially in physics) we would like
to establish relations between various properties A, B, C, etc
...
e
...
e
...
This kind of argument is what we will be referring
to as a “proof”
...
For instance, if we would like to
prove that A)B, then we use known properties of A, and via a logical argument,
the property B should follow
...
e
...
e
...
If it was possible to
provide a logical path in both directions, then one can say that B is true “if and only
if” A is true
...
For instance,
A could be the statement that “a number consists of three digits”, while B is that the

© Springer Science+Business Media, LLC 2016
L
...
1007/978-1-4939-2785-2_1

3

4

1 Basic Knowledge

number lies inclusively between 100 and 999
...

Indeed, from A (three digits) obviously B follows (between and including 100 and
999), and, conversely, from B immediately follows A
...
g
...
g
...
e
...
If for B to be true A
is necessarily required (but there are also some additional conditions required for
B to be true as well), then it is said that A is a necessary condition for B
...
Therefore, if both statements are
proven, i
...
that A is both necessary and sufficient for B to be true, then obviously
A and B are equivalent
...
Hence, these two conditions are indeed equivalent
...
e
...
Similarly, all numbers ending with
2 are divisible by 2, this is a sufficient condition; however, there are also other
even numbers divisible by 2
...
e
...

In some cases several conditions are equivalent
...
If, e
...
, we would like to prove the equivalence of three conditions
A, B and C, then we prove that: A)B, then B)C and, finally, C)A
...
If the reverse is to be
proven, the process is repeated, but this time assuming that B is true, and, on the
acceptance of that, a logical path is built towards A
...
Other methods of proof exist in mathematical
logic as well but will not be used in this book
...

In some cases a statement can be proven right by assuming an opposite and then
proving that the assumption was wrong—this is called proving by contradiction
...
Suppose, there is a property which depends on the natural number
n D 1; 2; 3; : : :
...
We perform detailed calculations
for some small values of n, say n D 1; 2; 3; 4, and see a pattern or a rule, and devise

1

The corresponding Latin expression “reductio ad absurdum” is also frequently used
...
2 Real Numbers

5

such a formula for a general n
...
Nevertheless, there is a beautiful logical construction which allows
one to prove (or disprove) our formula
...
g
...
e
...

If the last step is successfully proven, then that means the formula is indeed valid
for any value of n
...

Then, because of the last two steps of the induction, it must also be valid for the next
value of n D 2
...

As an example, consider a trivial case of even natural numbers
...
Indeed, for k D 1 we get the number N1 D 2 1 D 2, which is obviously
even
...

Let us prove that the number associated with the next value of k is also even
...
k C 1/
which is exactly of the required NkC1 form
...
E
...
2
Problem 1
...
Demonstrate that any odd number can be written as Nk D 2k 1,
where k D 1; 2; 3; : : :
...


1
...
For this
we use numbers
...
These numbers

2

Which means (from Latin “quod erat demonstrandum”) “which was to be demonstrated”
...

By infinite here we mean that the set never ends: after each, however big, number
there is always standing the next one bigger by one
...
We also define the
simplest operations on the numbers such as addition, subtraction, multiplication and
division
...

Using the subtraction operation, negative integer numbers and the zero number
are created forming a wider set Z D f0; ˙1; ˙2; ˙3; : : :g of positive and negative
integers (including zero)
...
3 The integer numbers have a property that summation, subtraction or
multiplication of any two members of the set Z is a member of the same set, i
...
it
is an integer number
...
We also define that zero power gives always one: n0 D 1
...
1)

where on the left we wrote an integer consisting of n C 1 digits (with the
corresponding plus or minus sign), and on the right the decimal representation of
the whole number
...
The representation above allows proving
familiar divisibility rules
...

Problem 1
...
Prove that a number is divisible by 3 or 9 if the sum of all its
digits is divisible by 3 or 9, respectively
...
]
Problem 1
...
Prove the following divisibility rule for 7: multiply the first, the
second, the third, etc
...
If the sum is divisible by
7, so is the whole number
...


1
...
These are formed by dividing all possible
numbers m and n from Z, i
...
via m=n with n ¤ 0
...
Any sum,
difference, product (including power) or division of two rational numbers is a
rational number, i
...
it belongs to the same set over all four operations
...

Problem 1
...
Prove that a product or division of two rational numbers is a
rational number
...
5
...
[Hint: construct, e
...
, a set of n equidistant
numbers lying along a linear interpolation between the two numbers and show
that they are all rational
...
For instance, consider z > 0 with 0 < m < n
...
If m > n, then it can always
be written as m D i n C m1 , where m1 is the remainder between 0 and n excluding
n (i
...
0 Ä m1 < n)
...
For instance, 13=6 D
...

In the decimal representation positive rational numbers are represented as a
number with the dot: the integer part is written on the left of the dot, while the
rest of it (which is between 0 and 1) on the right
...
Let us assume that
there exists the smallest positive integer k such that n k D 1 „ƒ‚… D 10l , i
...
it
0 0
l

is represented as 1 following by l zeroes
...
Then,
m1 k
m1 k
m1
D
D
:
n
n k
10l

8

1 Basic Knowledge

Since m1 < n, then m1 k < 10l and hence the number m1 =n is represented by no
more than l digits
...
For instance, 3=8 D
...
8 125/ D 375=1000 D 0:375,
12=5 D 2 C 2=5 D 2 C 4=10 D 2:4
...

It is easy to see that if a positive rational number has an integer part i and the rest
r D m1 =n consists of digits d1 d2 dl , i
...
r D 0:d1 d2 dl , then one can always
write:
zD

m1
d1
m
d2
d3
DiC
DiC 1 C 2 C 3 C
n
n
10
10
10

C

dl
:
10l

(1
...
g
...
2) via a sum of inverse powers of 10 will
contain a finite number of such terms
...
g
...
3/ (which means it is of the period of 3),
1=6 D 0:16666: : : D 0:1
...
3/ D 0C 1 C 2 C 3 C 4 C
3
10
10
10
10

or

1
6
6
6
1
D 0:1
...
The numbers m=n with periods in the decimal representation
arise if it is impossible to find such integer k that n k is represented as 10l with some
positive integer l
...
These are called irrational numbers
...
e
...
1=2/2 D 1=4, or, inversely, 1=2 D 1=4 is expressed via the square root operation
p
: : :
...
g
...
In the decimal representation irrational numbers
are constructed from a sequence of an infinite number of digits which is never
p
repeated; 2 D 1:414213562373095 : : : is an example of an irrational number
...
2 Real Numbers

9

All positive and negative rational and irrational numbers form a complete set of
real numbers R
...

Operations of summation and multiplication are commutative, i
...
they do not
depend on the order of terms (x C y D y C x and xy D yx), and associative
(
...
y C z/ and
...
yz/); they are also distributive:
x
...
The subtraction operation x y is understood as summation of
x and y; division of x and y is understood as a multiplication of x and 1=y
...
1
...
The distance between a point on the axis and zero represents an absolute
value jxj of the number x; its sign is plus if the point is on the right of 0, while the
sign is minus if it is on the left
...
Numbers run in both directions from zero indefinitely,
i
...
for any number x > 0 there is always another number x0 > x on the right of it;
similarly, for any number x < 0 on the left of the zero there can always be found a
number x0 < x on the left of it
...

These are not actual numbers (since for any real number there is always a real
number either larger or smaller), but special symbols indicating unbounded limits of
the sequence of real numbers at both ends
...
1
...
Division of a positive
real number by 0 is defined to be C1, while division of a negative number by 0 is
defined to be 1
...
These subsets are called intervals and could be
either: (i) unbounded, when a < x < b (both ends are open); (ii) semi-bounded, if
either a Ä x < b or a < x Ä b (one end is open and one closed); or (iii) bounded,
if a Ä x Ä b (both ends are closed)
...
3)

Then, jxj < a corresponds to the unbounded interval between a and a, and jxj Ä a
to the same interval with the addition of the upper and lower bounds a and a
(assuming here that a > 0)
...
All real numbers can be
represented as an interval between ˙1, i
...
1 < x < 1
...

Frequently other notations for the intervals are also used: Œa; b is equivalent to
a Ä x Ä b, Œa; b/ is equivalent to a Ä x < b,
...
a; b/ is equivalent to a < x < b
...
g
...
a; b/; however, we shall not use these notations here
in the book
...
3 Cartesian Coordinates in 2D and 3D Spaces
The number axis in Fig
...
1(a) corresponds to a one-dimensional (often denoted as
1D) space R
...

By taking an ordered pair
...
1
...
In this case one draws two perpendicular
number axes, one for x and another for y
...


1
...
This is denoted P
...
The 2D space formed by pairs of real numbers
is denoted R2
...
x; y; z/ of three real numbers, x, y and z, corresponds to a point
in a three-dimensional space R3 (or 3D)
...
1
...
x; y; z/
...


1
...
We shall start from
a circle on the x y plane
...
Every point
...
Such a
circle of unit radius is shown in Fig
...
2(a)
...
We have already mentioned above that two lines
can be “perpendicular”, meaning they make 90 degrees (denoted 90ı ) with each
other
...
Here we shall introduce
general angles properly
...
1
...
Here the point
A is obtained from the point B by moving around the circle in the anticlockwise
direction
...
1
...
Alongside degrees, radians are
also frequently used: 360ı are equivalent to 2 radians, where D 3:1415 : : : is
an irrational number
...
1
...
1
...
Specific angles of 90ı D =2 (blue), 180ı D
270ı D 3 =2 (magenta) and 360ı D 2 (orange) are shown in (b)

12

1 Basic Knowledge

 D † AOB D 360

LAB

...
radians/:
Lcircle

corresponds to the half of the circle, while quarter of
The angle of 180ı D
the circle gives 90ı D =2
...

One also introduces negative angles which correspond to going around the unit
circle in the clockwise direction starting from the positive part of the x axis
...
Then, for instance, † BOD in Fig
...
2(b) is 90ı
(or
=2)
...

Consider now a line L1 in Fig
...
3(a)
...

This is because the opposite angles are always the same
...
Exactly the same
angles will be formed at the crossing with another line L2 parallel to the first one:
If we slide L1 along T keeping it parallel all the way (i
...
keeping the same angles
˛ and ˇ), then we shall arrive at L2 and the crossing will be perfectly identical; this
follows from the fact that the lines are parallel: we call lines parallel if they never
cross
...
1
...

We shall be using a lot of various 2D figures, and amongst them the triangle and
parallelogram play an essential role
...
1
...


Problem 1
...
Prove using Fig
...
3(b) that the sum of three (internal) angles of
a general triangle is 180ı , i
...
˛ C ˇ C D
...
1
...
(b) The sum of the angles of any triangle is 180ı D

1
...
1
...
(b) An
isosceles triangle with two equal sides AC D AB and two equal angles † ACB D † ABC D ˛
...
(d) To the proof of the cosine theorem for a general
triangle ABC
Fig
...
5 Triangles in (a) and
(b) are similar; the heights of
the triangles in (c), AD1 and
AD2 , also differ by the same
factor as the sides

The triangle in Fig
...
4(a) is called a right triangle as one of its angles is 90ı
(or =2)
...
A general triangle
would have its three angles ˛, ˇ and all different, and its three sides AB, BC and
AC different as well
...
It is called equilateral
...
g
...
1
...
This is an isosceles triangle
...

In practice very often it is necessary to consider similar triangles, shown in
Fig
...
5(a) and (b); these are the ones which are scaled up or down with respect
to each other; these have exactly the same angles, and their sides are scaled by the
same factor: A1 B1 =A2 B2 D A1 C1 =A2 C2 D B1 C1 =B2 C2
...
7
...
1
...


1
...
However,
for the purposes of this chapter, some elementary knowledge is necessary
...

We shall start by reminding the reader the main ideas of algebra
...
result in a
single answer in a form of another number and no general rules are seen through
this arithmetic calculation
...
Replacing the letters by numbers in the answer (i
...
in
the resulting formula) gives the required numerical value
...

Moreover, using algebraic manipulations it is often discovered that many terms may
cancel out in the final result yielding a simple final expression
...

However, we need to work out rules for algebraic manipulations
...
Since multiplication of two numbers
does not depend on their order, e
...
5 3 D 3 5, we establish a general rule that
multiplication of two letters, say a and b, is commutative: a b D b a
...
Similarly summation is also commutative,
e
...
3 C 5 D 5 C 3, and so is summation of letters in algebra: a C b D b C a
...
g
...
5 C 6/, may also be performed
as 3 5 C 3 6; we say that this binary operation is distributive
...
b C c/ as ab C ac,
i
...
“open the brackets”
...
g
...
Both summation and multiplication are
also associative, i
...
they do not depend on the order in which they are performed:

...
b C c/ and
...
bc/
...
5 Introduction to Elementary Functions and Trigonometry

15

As an example, let us consider an expression
...
By definition of the power,
this is
...
a C b/
...
a C b/2 D
...
a C b/ C b
...
a C b/2 D a
...
a C b/ D a2 C ab C ba C b2 :
Because a and b commute, ab D ba, we can sum up ab and ba together as abCba D
ab C ab D 2ab, which finally allows us to arrive at the familiar result:

...
8
...
a

b/
...
a ˙ b/3 D a3 ˙ 3a2 b C 3ab2 ˙ b3 I


...
a ˙ b/ a2

ab C b2 :

In the last two cases either upper or lower signs should be consistently used
when ˙ or signs are given
...
If a number a is strictly
larger than b, it is written as a > b; by writing a b, we mean that a could also be
equal to b, i
...
a is not smaller than b
...
Obviously, a > b and b < a are
equivalent, as are a b and b Ä a
...
Firstly, one
may add to both sides of an inequality the same number, i
...
if a < b, then a C c <
bCc
...
e
...
The latter inequality can also be written as b a > 0
...

It is possible to multiply both sides of the given inequality by some number c ¤
0: If c > 0, the sign of the inequality does not change; however, if c < 0, it changes
to the opposite one
...


16

1 Basic Knowledge

However, 5a >
the first one:

5b
...

Problem 1
...
Prove, that if a < b and 0 < c < d (i
...
both c and d are positive
numbers), then ac < bd, i
...
one can “multiply both inequalities” side-by-side
...
10
...

Two inequalities of the same type (i
...
both are either “less than” or “larger than”
types) can be “summed up”, i
...
from a < b and c < d follows aCc < bCd
...
However, since c < d, then we can similarly state
that b C c < b C d, which proves the above made statement:
aCc
and b C c < b C d

H)

aCc
Similarly, from a
b and c
d follows a C c
b C d
...
g
...

Indeed, d < c, so that summing up this one with a < b, we get a C d < c C b or
a c < b d, as required
...
A function y D f
...
There must be a single value of y
for each value of the x
...
(1
...
The other simplest class of functions are polynomials:
f
...
4)

iD0

This is a general way of writing a polynomial of degree n, where a0 , a1 , etc
...
n C 1/ real P
coefficients; also, it is assumed that at least an ¤ 0
...
It is simply a short-cut notation for the sum written explicitly in
Eq
...
4); the index i is called a summation index and it runs here from 0 (shown
underneath the sum symbol) to the value of n (shown on top of it) and it is a matter
of simple exercise to check that the full expression just after the f
...


1
...
1
...
x/ of degree n on another one Qm
...

The simplest polynomials are the first and the second order ones
...
1
...
When plotted, a linear polynomial P1
...
0/ being the point where its graph crosses the y axis,
while a1 shows its slope: if a1 > 0, then the curve goes up from left to right (curves
1 and 2 in Fig
...
6(a)), while when a1 < 0 the line goes down (curves 3 and 4)
...
It happens always as long
as a1 ¤ 0
...

Graphs of various second order polynomials (called parabolas), P2
...
1
...
Depending on the values of the constants a0 , a1 and a2
the parabola may either cross the x axis in 1 (curves 3 and 6) or 2 (curves 2 and 5)
points or do not cross at all (curves 1 and 4)
...
x/ D a2 x C a1 x C a0 D a2 x C 2
2a2
"
Â
Ã
Â
à ! Â
à #
a1
a1 2
a1 2
2
D a2
x C2
C a0
xC
2a2
2a2
2a2
"Â
Ã
Ã#
Â
Ã
Ã
Â
Â
a2
a1 2
a1 2
a1 2
1
D a2
C
C a0 :
xC
C a0 D a2 x C
2a2
2a2
2a2
4a2
(1
...
If a0 a2 =4a2 > 0 and a2 > 0, then the entire graph of this
1
function is above the x axis and there will be no roots (0 crossings of the x axis); if,
however, a2 < 0, then there will be 2 roots
...
The roots can be established
by setting to zero the complete square expression obtained above5:
Â
Ã
Ã
Â
Ã
Â
a1 2
a1 2
a2
a2 4a0 a2
1
a2 x C
xC
C
C a0 D 0 )
D 1
2a2
4a2
2a2
4a2
2
s
s
a2 4a0 a2
a2 4a0 a2
a1
a1
1
1
)xC
D˙
) xD
˙
)
2a2
2a2
4a2
4a2
2
2
q
a1 ˙ a2 4a0 a2
1
xD
:
(1
...
These are the same results as
previously worked out by inspecting the graphs of the parabolas
...
But first, we have
to briefly return to the right triangles, i
...
the ones which have one right angle (so
that the sum of the other two is exactly 90ı ), see Fig
...
4(a)
...
There are many proofs of this theorem, one of the simplest is based
on the construction of four identical triangles depicted in Fig
...
4(c)
...


Problem 1
...
Prove the Pythagorean theorem using the construction in
Fig
...
4(c) made of four identical right triangles of Fig
...
4(a)
...
]
We are now in the position to define the necessary trigonometric functions
...
1
...
7)

Similarly, sin ˇ for the other acute angle is defined as b=c, where b is the length of
the side opposite to the angle ˇ
...
One has to understand
this as two separate equations: one root is obtained with the upper sign (which is “C”), and another
one with the lower sign “ ”
...
5 Introduction to Elementary Functions and Trigonometry

19

Fig
...
7 To the definition of
the trigonometric functions
for arbitrary angles

cos ˛ D

b
b
:
D p
2 C b2
c
a

(1
...
9)

which follows from the Pythagorean theorem
...
However, these are generalised to any angle by means
of the unit radius circle in the x y plane depicted in Fig
...
7
...
1
...
The same holds for
any other angle
...
1
...
In this case cos ˛ is defined as OD and is negative (OD itself is a
positive length) since the point D is on the negative part of the x axis; at the same
time, sin ˛ DOA and is still positive
...
In fact,
sine and cosine functions for arbitrary angle ˛ can be defined as the projection on
the y and x axes of the line drawn from the origin to the circle of unit radius and
making the angle ˛ with the x axis
...
Firstly,
by rotating the radius by 2 we arrive at the same point on the circle, so that both
functions remain the same for angles ˛ and ˛ C 2 ; we say that the functions are
periodic with the period 2 :
sin
...
˛ C 2 / D cos ˛ :

(1
...
1
...
We have:
sin ˛ D sin
...


ˇ/ D

OD D OC D

cos ˇ :

If negative angles are used, the same definitions for the sine and cosine functions
apply, see Fig
...
7(b):
sin
...
11)

and
cos
...
12)

A function is called even with respect to its argument if f
...
x/ for every
value of the x, while it is called odd if f
...
x/
...


Problem 1
...
Now we shall consider a general triangle to derive the cosine
theorem which generalises the Pythagoras theorem to arbitrary triangles
...
1
...
Prove using the additional
construction shown there the cosine theorem, stating that
c2 D a 2 C b 2

2ab cos

:

(1
...
13
...
1
...
14)

[Hint: expressing the height h using either ˇ or will immediately give the first
part of the above identity; the other part is proven similarly by introducing the
appropriate height h0
...
All their properties follow from their definition and the
properties of the sine and cosine functions
...
1
...
15)

1
...
1
...
=6/ and (b) sin
...
1
...
e
...
For cot ˛
this is the other way round
...
Consider
an equilateral 4ABC shown in Fig
...
8(a); it has all three angles equal: ˛ D 60ı
...
Since the lines DE and CB are parallel, the angles
† ADC and † AEB are the right angles
...
The
two triangles 4DAC and 4AEB are equal
...
14
...
1
...
Correspondingly, show that sin 60ı D
p
cos 30ı D 3=2
...
15
...
1
...

In applications it is frequently needed to calculate areas of various objects, and
we shall come across this type of problem very often in this course
...
A square has four sides of the same length a
and all four of its angles are the right angles, Fig
...
9(a)
...

The square is generalised by a rectangle in which case the two adjacent sides are
not the same and equal to a and b, while the four angles are still of 90ı ; its area
Srectang D ab
...
1
...
Its opposite sides are of the same length, but the four
angles are in general different from 90ı , although ˛ C ˇ D 180ı
...
16
...
1
...

Problem 1
...
Show using Fig
...
9(c) that the area of a triangle 4ABD is
equal to S4 D ah=2, where h is the height perpendicular to the side a
...
Coordinates
...
x0 ; y0 / are given by

...
y

y 0 /2 D R 2 :

The simplest generalisation of the circle is an ellipse
...
x; y/ are coordinates of an arbitrary
points on the curve
...
1
...
There are two axes: the one drawn between
points with coordinates
...
a; 0/ along the x axis and the other between
points
...
0; b/ along the y axis
...
If a D b, the ellipse
coincides with a circle of radius R D a D b
...
˙a; 0/ and
...
1
...

The distance to the centre O from any other point on the curve of the ellipse lies
between b and a values (a > b for the ellipse in the figure)
...
1
...
6 Simple Determinants

23

s
eD

1

 Ã2
b
;
a

where it is assumed that b and a are halfs of the minor and major axes, respectively
(a > b)
...
A peculiar property of the ellipse is that the sum of distances F1 P C
F2 P D d1 C d2 from the focal points to any point P on the ellipse is a constant
quantity equal to 2a (obtained by taking the point P to be one of the vertices on the
major axis)
...
18
...
˙f ; 0/
...
x; y/ on the ellipse the distance D D d1 C d2 D
2a is constant if and only if f 2 D a2 b2
...
6 Simple Determinants
In many instances it appears to be extremely convenient to use special objects called
determinants
...

A 2 2 determinant is a scalar (a number) composed from four numbers
a11 , a12 , a21 and a22 (which we distinguished by using unique double indices for
convenience) written on a 2 2 grid between two vertical lines and its value is
calculated as follows:
ˇ
ˇ
ˇ a11 a12 ˇ
ˇ
ˇ
(1
...
a22 a33
Ca13
...
a21 a33

a22 a31 / :

a23 a31 /
(1
...
Note how the elements of the determinants are denoted: they have a double
index, e
...
a13 has the left index 1 and the right index 3 in the 3 3 determinant
...
Indeed, the element a13 can be found at the end
of the first row in the 3 3 determinant
...

When defining the value of the determinant in (1
...
Notice that the 2 2 determinants can be obtained
by removing all elements of the row and column which intersect at the particular
ˇ
ˇ
ˇa a ˇ
element which appears as a pre-factor to them
...
The same is true for the
other two as well as can easily be checked
...

An important property of the determinants that we shall need in the following
is that a determinant is equal to zero if any one of its rows (or columns) is a
linear combination of the other rows (columns)
...
Starting from the 2 2 case, if the second row is linearly dependent
on the first, that means that the elements of the second row are related to the
corresponding elements of the first row via the same scaling factor , i
...
a21 D a11
and a22 D a12
...
16) of the determinant, we have:
ˇ ˇ
ˇ
ˇ
ˇ a11 a12 ˇ ˇ a11 a12 ˇ
ˇDˇ
ˇ
ˇ
ˇ a21 a22 ˇ ˇ a11 a12 ˇ D a11 a12

a12 a11 D 0 ;

as required
...
18)

You can see that elements of the first row are indeed expressed as a linear
combination of the elements from the other two rows taken from the same column
...
Now, let us see explicitly that if we have such a
linear combination (1
...
This can be easily
established from the definition of the determinant
...
6 Simple Determinants

25

Problem 1
...
Show using the definition of the 3 3 determinant that if the
elements of the first row are related to those of the second and the third rows as
given in Eq
...
18), then the determinant is zero:
ˇ
ˇ ˇ
ˇ
ˇ a11 a12 a13 ˇ ˇ a21 C a31 a22 C a32 a23 C a33 ˇ
ˇ
ˇ ˇ
ˇ
ˇ a21 a22 a23 ˇ D ˇ
ˇD0
a21
a22
a23
ˇ
ˇ ˇ
ˇ
ˇa a a ˇ ˇ
ˇ
a31
a32
a33
31 32 33
for any values of

and
...

Again, this is most easily checked by a direct calculation: use the definition (1
...
a12 a33 a13 a32 / a22
...
a11 a32 a12 a31 / :
ˇ
ˇ
ˇa a a ˇ
31 32 33
Rearranging the terms in the right-hand side, we obtain:
ˇ
ˇ
ˇ a21 a22 a23 ˇ
ˇ
ˇ
ˇ a11 a12 a13 ˇD
ˇ
ˇ
ˇa a a ˇ
31 32 33

a11
...
a21 a32

a23 a32 / C a12
...
e
...
The same happens if any two rows or
columns are interchanged
...


Problem 1
...
Show that
ˇ
ˇ
ˇ
ˇ
ˇ1 0 2ˇ
ˇ1 1 0ˇ
ˇ
ˇ
ˇ
ˇ
jAj D ˇ 3 1 0 ˇ D 29 ; jBj D ˇ 0 2 1 ˇ D 4 :
ˇ
ˇ
ˇ
ˇ
ˇ0 5 1ˇ
ˇ3 1 0ˇ

26

1 Basic Knowledge

1
...
7
...

It is convenient to define objects, called vectors, belonging to the space denoted
R3 , such that they can undergo summation and subtraction operations resulting in
the same type of objects (i
...
vectors as well) belonging to the same space R3
...

This notation we shall be rarely using, however; instead, another notation will be
frequently employed whereby a vector is shown by a bold letter, e
...
a
...
xA ; yA ; zA / and
...
xB xA ; yB yA ; zB zA /
...
The length of the vector (or
ˇ !ˇ
ˇ ˇ
its magnitude) is the distance between two points, A and B, and it is denoted ˇABˇ
or jaj
...
g
...
Two vectors a and b are said to be identical, a D b, if they have the
same length and direction
...
In particular, any vector can be translated to the
position in which its starting point is at the centre of the coordinate system with zero
coordinates as shown in Fig
...
1(c)
...
It is easy to see from the same
figure, that the length of the vector,
q
ˇ !ˇ
p
ˇ ˇ
OPˇ D jaj D x2 C y2 C z2 D
...
yB

yA /2 C
...
19)

corresponds to a diagonal of the cuboid with sides x, y and z
...
1
...

A vector of unit length is called unit vector
...
As was mentioned, formula (1
...

Two vectors are collinear if they have the same or opposite directions; in other
words, if translated such that they begin at the centre of the coordinate system, they
lie on the same line
...


1
...
1
...
A vector a if multiplied
by a scalar (a real number) c results in another vector b D ca which is collinear
with a and has the length jbj D c jaj
...
The vector b has the same
direction as a if c > 0, opposite to that of a if c < 0, and is the null vector if c D 0
...

The sum of two vectors a and b is defined as follows: translate the vector a into
!
!
!
the vector AB, vector b into BC, then the sum c D a C b is defined as the vector AC,
see Fig
...
11(a)
...
Similarly, one defines a
difference b D c a of two vectors c and a as a vector b for which c D a C b,
also see Fig
...
11
...
are real numbers
...
The summation operation of vectors is both
commutative and associative, similarly to real numbers; this is because by summing
vectors we perform summation of their coordinates which are real numbers
...
a1 ; a2 ; a3 / and b D
...
a; b/ D a1 b1 C a2 b2 C a3 b3 :

(1
...
a; b/) will be frequently used
...
a; b/, which directly follows from the definitions of the two
...
a; b/ D
...
c; a C b/ D

...
c; b/, which can easily be checked from their definitions; e
...
for the latter:

...
a1 C b1 / C c2
...
a3 C b3 /
D
...
c1 b1 C c2 b2 C c3 b3 / D
...
c; b/ :

28

1 Basic Knowledge

Two vectors a and b are said to be orthogonal, if their dot product is zero
...
Indeed, consider two vectors a and c with the angle  between
them, Fig
...
11(b)
...
a c/2 D
...
a; a/ 2
...
c; c/ D jaj2 2
...
a; c/ ;

where a D jaj and c D jcj
...
1
...

Therefore, from the well-known cosine theorem of Eq
...
13), b2 D ja cj2 D
a2 C c2 2ac cos Â
...
a; c/ D jaj jcj cos  :

(1
...
e
...

It is convenient to introduce three special unit vectors, frequently called unit base
vectors of the Cartesian system,
i D
...
0; 1; 0/ ; k D
...
22)

which run along the Cartesian x, y and z axes
...
i; j/ D
...
j; k/ D 0, and all have the unit length
...
Then, a vector a D
...
1; 0; 0/ C a2
...
0; 0; 1/ D a1 i C a2 j C a3 k :

(1
...
If a D
...
b1 ; b2 ; b3 /,
then one defines the third vector c D a b called their vector product as a vector of
length
jcj D ja

bj D jaj jbj sin Â

(1
...
1
...
Note that the vector c is perpendicular to both
vectors a and b
...
c1 ; c2 ; c3 /, let us first make use of the conditions that the vector c is
perpendicular to both a and b:
c?a

)


...
25)

c?b

)


...
26)

1
...
1
...
27)

Here we assumed that b2 a1 a2 b1 ¤ 0
...
25) or (1
...
28)

The obtained components of c D
...
So, we
must choose it in such a way as to comply with the definition of Eq
...
24)
...
(1
...
a2 b1

2

cos  D a b

b2 C b2 C b2
1
2
3

a1 b2 /2 C
...
a b/2


...
a2 b3

a 3 b 2 /2 :

(1
...
Comparing (1
...
(1
...
28) for c1 and c2 , we see that choosing c3 D ˙
...
To fix the sign here, consider a particular case when a
is a unit vector directed along the x axis, a D i D
...
0; 1; 0/
is a unit vector along the y axis
...
1
...
e
...
Therefore, in c3 D
˙
...
We finally have:
a

b D
...
a3 b1

a1 b3 /j C
...
30)

30

1 Basic Knowledge

The vector product of two vectors can most easily be remembered if written via
a determinant which we introduced in Sect
...
6; specifically, compare Eqs
...
17)
and (1
...
Hence, one can write:
ˇ
ˇ
ˇ i j kˇ
ˇ
ˇ
a b D ˇ a1 a2 a3 ˇ :
(1
...
21
...
30) has perfect sense also in the case
of b2 a1 a2 b1 D 0, i
...
show explicitly that in this case the vector a b is
perpendicular to both vectors a and b, and that its length corresponds to the
definition of Eq
...
24)
...
22
...
1; 0; 1/ and b D
...
a; b/ and a

b
...
23
...
x; x2 ; x3 / and b D
...
a; b/, a b and jaj and jbj
...
x3 ; x4 ; x5 /; 3;
...
]
Another form of the vector product, very useful in practical analytical work, is
based on the so-called Levi-Civita symbol ijk defined as follows: if its indices i; j; k
are all different and form a cyclic order (123, 231 or 312), then it is equal to C1, if
the order is non-cyclic, then it is equal to 1; finally, it is equal to 0 if at least two
indices are the same
...
32)

j;kD1

In the last passage here a single sum symbol was used to simplify the notations; this
is what we shall be frequently doing
...
In particular, a a D 0
...
a

b/

c¤a


...
7 Vectors

31

However, the vector product is distributive:

...
1
...
30) or (1
...
Of course, vector multiplying either of the three vectors
with themselves gives zero
...
24
...
(1
...

Problem 1
...
Using Eq
...
30), prove that the vector product is distributive
...
x; y/ plane; of course, the result should not depend
on the orientation of the Cartesian axes!]
Problem 1
...
Using the distributivity of the vector product, derive formula (1
...

Problem 1
...
Using the explicit expression for the vector product given
above, prove the following identities:
Œa

b Œc

d D
...
b d/


...
b c/ ;

(1
...
a c/ b


...
34)

Œa

b

c D
...
c b/ a ;

(1
...
36)

Problem 1
...
Show that the area of a parallelogram defined by two vectors
a and b, as shown in Fig
...
13(a), is given by the absolute value of their vector
product:
Sparalellogram D ja

bj :

(1
...
0; 1; 3/, v D
...
1; 2; 5/, u

...
u; v/ D 0
...
1
...
1 2

3 1/i C
...
1/

0 2/j C
...
1//k D
...
g
...
Indeed, a vector product of two vectors is a vector which can be dotmultiplied with a third vector giving a scalar:
Œc; a; b D c Œa

b

D
...
a2 b3
D
...
a3 b1
ˇ
c3 ˇ
ˇ
a3 ˇ :
ˇ
b3 ˇ

a3 b2 /i C
...
a1 b2

a1 b3 /j C
...
38)

The last equality trivially follows from formula (1
...
(1
...
(1
...
The triple product has some important symmetry properties listed below
which follow from its determinant representation and the fact, discussed earlier in
Sect
...
6, that a determinant changes sign if two of its rows are exchanged:
Œc; a; b D Œa; b; c D Œb; c; a D Œc; b; a D

Œb; a; c D Œa; c; b ;

(1
...
If two permutations are performed
(this corresponds to the cyclic permutation, e
...
abc ! bac ! bca), then no change
of the sign occurs
...
7 Vectors

33

Problem 1
...
Show that
Œa; a; b D Œa; b; a D Œb; a; a D 0 :

(1
...
38) for the triple product preceding the
determinant
...
30
...
1
...
41)

The immediate consequence of the definition of the triple product is that it is
equal to zero if the three vectors lie in the same plane, i
...
are coplanar
...

The Cartesian basis introduced above, fi; j; kg, is not the only one possible; one
can define any three non-collinear and non-planar vectors fa1 ; a2 ; a3 g to serve as a
new basis and then any vector x from R3 can then be expanded with respect to it:
x D g1 a1 C g2 a2 C g3 a3 :

(1
...
Let us write the new basis via their Cartesian
coordinates: a1 D
...
Here aij corresponds
to the j-th Cartesian component (j D 1; 2; 3) of the vector ai (i D 1; 2; 3)
...
43)

One can see that the “coordinates” fg1 ; g2 ; g3 g in the new basis define uniquely
the Cartesian components fxi g of the vector x; inversely, given the Cartesian
components of x D
...
43)
...


34

1 Basic Knowledge

shown in linear algebra that for the unique solution of these equations to exist it is
necessary that the following condition be satisfied:
ˇ
ˇ
ˇ a11 a21 a31 ˇ
ˇ
ˇ
ˇ a12 a22 a32 ˇ ¤ 0 ;
(1
...
e
...
It must be clear that this condition is satisfied as long as the three vectors
are non-collinear and non-planar (in the latter case one must be a linear combination
of the other two)
...
1
...
2; 1; 1/ in terms of three vectors a1 D

...
1; 1; 0/ and a3 D
...

Solution
...
43),
8
< g1 C g2 C g3 D 2
;
g2 C g3 D 1
:
g3 D 1
we obtain g3 D 1, then g2 D 2 and g1 D 1, i
...
the expansion sought for is
x D a1 C 2a2 a3 , which can now be checked by direct calculation
...
These
basis vectors are defined by the following relations:
b1 D

1
Œa2
vc

a3  ; b2 D

1
Œa3
vc

a1  ; b3 D

1
Œa1
vc

a2  ;

(1
...
Notice that the index of each
reciprocal vector and the two old vectors it is related to form an ordered triple in
each case
...
39) and (1
...
46)

where we introduced the Kronecker symbol which is equal to zero if its two indices
are different, otherwise it is equal to one
...
Indeed,

7

In solid state physics and crystallography the reciprocal lattice is defined with an additional factor
of 2 , so that instead of (1
...


1
...
a2 b2 /
...
a2 b3 /
...
47)

where we made use of (1
...
46) for the four dot products of direct and
reciprocal vectors
...

This is proven in the following problem
...
31
...
(1
...
e
...
(1
...
(1
...
35) for the double vector product
...
32
...
By repeating this cell in all three
directions the whole periodic crystal lattice is constructed
...
Assuming that a1 lies along the x axes, a2 in the x y plane, work
out the Cartesian components of all three lattice vectors
...


1
...
2 N-dimensional Space
Most of the material we considered above is generalised to abstract spaces of
arbitrary N dimensions (N
1)
...
e
...
x1 ; : : : ; xN /
...
x1 ; : : : ; xN / and Y D
...
19) for the 3D space:
v
u N
uX
dXY D t

...
48)
iD1

36

1 Basic Knowledge

!
We can also define an N-dimensional vector XY D a D
...
Similarly to
the 3D case, vectors can be added or subtracted from each other or multiplied by a
number; each such an operation results in a vector in the same space: if there are
two vectors a D
...
b1 ; : : : ; bN /, then we define
a ˙ b D c D
...
Correspondingly, we can also define a dot or scalar
product of two vectors a D
...
b1 ; : : : ; bN / as a direct
generalisation of Eq
...
20):
a b D
...
49)

iD1

As was the case for the 3D space, the square of a vector length is given by the dot
product of the vector with itself: jaj2 D a a
...
The distance between two points X
and Y is then given by
q
p
dXY D
...
x y/
...
1; 0; : : : ; 0/, e2 D
...
, eN D
...
Each of these vectors
consists of N 1 zeros and a single number one; the latter is positioned exactly at the
i-th place in ei
...
e
...
Then, any vector a D
...
g
...
x1 ; : : : ; xN /, when the
coordinates of the vector are listed along a line (a row)
...


1
...
8 Introduction to Complex Numbers
We know that a square of a real number is always positive
...
This
fact causes some problems in solving algebraic equations of power more than one
...
8
These are defined by a pair of two real numbers x (which is called a real part)
and y (a complex part) and a special object i as follows: z D x C iy
...
e
...
It is also a
complex number with zero real part and the complex part equal to one
...

Complex numbers are conveniently shown as points
...
1
...
The distance r D x2 C y2 from the point to the origin of
the coordinate system is called the absolute value of the complex number z, while
the angle (see the figure) the argument or phase
...
e
...
cos

C i sin / :

(1
...
e
...
1
...
x; y/ with coordinates x and y on the
2D plane, the so-called complex plane
...
The complex conjugate number
...
It has the same r but its phase is

8

These were developed further by Rafael Bombelli and William Rowan Hamilton
...
The phase can be obtained from the values of
y D r sin and x D r cos via
r sin
y
D
x
r cos

D tan

;

(1
...
e
...
We shall learn in
Sect
...
3
...


Problem 1
...
Determine the absolute value r and the phase pof the followp
ing complex numbers: 1 C i; 1 i; 3 C i
...
]
The vector from the coordinate origin to the point
...
1
...

In that case r is the vector length and is the angle the vector makes with the x axis
...

If y D 0, the numbers are positioned on the x axis and are purely real, and the
phase is D n with any positive and negative integer n, including zero: even n
corresponds to positive, while odd n corresponds to the negative direction of the x
axis
...
Any arithmetic operation with any two real numbers would result in
a number on the same axis, i
...
within the manifold of real numbers as we discussed
at length above
...
g
...
1/ D 9
1 D 3i
...
x1 C iy1 /C
...
x1 C x2 /Ci
...
x1 C iy1 /


...
x1

x2 / C i
...
52)

(subtraction)
...
53)

These definitions ensure that the sum (difference) of two numbers are equivalent
to a sum (difference) of two vectors corresponding to them
...
g
...
z2 C z3 /
...
e
...
e
...
These conditions are satisfied if we treat a product

1
...
e
...
Hence we define the product as follows:

...
x2 C iy2 / D x1 x2 C iy1 x2 C ix1 y2 C i2 y1 y2
D x1 x2 C i2 y1 y2 C i
...
x1 C iy1 /
...
x1 x2

y1 y2 / C i
...


(1
...
54) that if y1 D y2 D 0, then the imaginary part of the
product is also equal to zero, i
...
we remain within the real numbers and that indeed
we get x1 x2
...
Finally,
the product is commutative, i
...
z1 z2 D z2 z1 , and associative,
...
z2 z3 /,
which guarantee (together with the analogous properties of the sum and difference)
that the algebraic rules, however complex, will be indeed respected
...
This is defined as an operation
inverse to the product, i
...
z1 =z2 D z3 is understood in the sense that z1 D z2 z3
...
34
...
54), prove the following formula for the division:
z1
x1 C iy1
x1 x2 C y1 y2
y1 x2 x1 y2
D
D
Ci 2
z2
x2 C iy2
x2 C y2
x2 C y2
2
2
2

(division)
...
55)

[Hint: write z1 D z2 z D z2
...
]
We shall see later on in Sect
...
3
...

Finally, a special operation characteristic only to complex numbers is also
introduced whereby the signs to all i in the expression are inverted
...
e
...

The complex conjugate number z corresponds to the reflection of the position of
the number z on the complex plane with respect to the x axis, Fig
...
14
...
35
...
36
...
56)

Problem 1
...
The square of the absolute value of a complex number r2 is very
often denoted jzj2 and called the square of the module of the complex number
z
...
57)

Another (much simpler) way of proving the result for the division (1
...
(1
...
x1 C iy1 /
...
38
...
e
...
1 C i/
...
1 C i/
...
1 C i/


...
1 C 5i/ =13; 2 C 3i
...
39
...
]
1 C i, 1 i, i, 1 and 1
...
9 Summation of Finite Series

41

1
...
58)

iD0

containing n C 1 terms constructed via a certain rule, i
...
each term ai is constructed
from its number i (the index)
...

In a geometrical progression ai D a0 qi , i
...
the numbers ai form a finite sequence
a0 , a0 q, a0 q2 ,: : :, a0 qn
...
We would like to calculate the sum
n
X
ai D a0 C a0 q C a0 q2 C C a0 qn :
Sn D

(1
...
60)

iD0

The trick here is to notice that if we multiply Sn by q, we can manipulate the result
into an expression containing Sn again:
Sn q D a0 q C a0 q2 C a0 q3 C

C a0 qnC1 D a0 C a0 q C a0 q2 C

a0 C a0 qnC1 D Sn C a0 qnC1

C a0 qn

1

Solving this equation with respect to Sn , we finally obtain:
Sn D a0

qnC1 1
;
q 1

(1
...
Note that this formula is valid for both q > 1 and q < 1,
but not for q D 1
...
60) are identical
and Sn D
...
We shall see later on in Sect
...
9 that the
formula (1
...
e
...

Although we have derived the sum of the geometrical progression from its
definition, it is also instructive to illustrate using this example how the induction
principle we discussed in Sect
...
1 works
...
61) for the sum of the progression (1
...
First of all, we check that the formula (1
...
Then we assume that (1
...
61) by the
substitution n ! n C 1, as required
...
2
...
This is a geometrical progression with a0 D 32 D 9 and q D 1=3,
containing n C 1 D 13 terms
...
61), we thus get
Â
Ã
27
1

...
40
...
a/ 1 C 2 C 4 C 8 C
[Answers: 211

C 1024 I

1; 33 C 3

10


...
]

Problem 1
...
Show that
Ã
N
XÂ
1 2
n
a C n
D 2
...
a2 1/

In an arithmetic progression ai D a0 C ir, i
...
each term is obtained from the
previous one by adding a constant r:
aiC1 D ai C r ; where i D 0; 1; 2; : : : ;

(1
...
n

1/r C Œa0 C nr :

iD0

(1
...
9 Summation of Finite Series

43

The trick here which will enable us to derive a simple expression for the sum Sn
is to notice that each term in the series can also be written via the last term: ak D
a0 C kr D an
...
Therefore, the sum (1
...
n

1/r C Œan


...
64)

Here we have exactly the same terms containing r as in (1
...
Therefore, all these terms cancel out if we sum
the two equalities (1
...
64):
2Sn D
...
n C 1/an ) Sn D

a0 C an

...
65)

where the factor of
...
The final result is very simple: you take the average of
the first and the last terms which is to be multiplied by the total number of terms in
the finite series (which is n C 1)
...
42
...
[Answer:
500455
...


Problem 1
...
Use the mathematical induction to prove the formulae given
below:
n
X

i2 D 1 C 22 C 32 C

C n2 D

iD1
n
X

i3 D 1 C 23 C 33 C

1
n
...
2n C 1/ I
6

(1
...
n C 1/2 I
4

(1
...
a0 Cir/qi D a0

1

1
i2

1

D

3
4

2n C 1
I
2n
...
1 q/2


...
68)

(1
...


44

1 Basic Knowledge

Problem 1
...
Prove that
n
X
iD1

1
D1
i
...
i C 1/ D 1=i

(1
...
i C 1/
...
10 Binomial Formula
There are numerous examples when one needs to expand an expression like
...
We can easily multiply the brackets for small values of the n
to get:

...
a C b/2 D a2 C 2ab C b2 D a2 b0 C 2a1 b1 C a0 b2 ;

...
a C b/4 Da4 C4a3 bC6a2 b2 C4ab3 Cb4 D a4 b0 C4a3 b1 C6a2 b2 C4a1 b3 Ca0 b4 :
This can be continued, but the expressions get much longer
...
a C b/n from the particular cases considered
above
...
n i/ D
n
...
1; 1/ for n D 1,
...
1; 3; 3; 1/ for n D 3,
...
One may
notice that a set of coefficients for the power n C 1 can actually be obtained from
the coefficients corresponding to the previous power n
...
1
...
Consider, for instance, the next to n D 4
set of coefficients, i
...
for n D 5
...
1; 5; 10; 10; 5; 1/, and they are constructed following two simple rules:
(i) the first and the last coefficients (for the terms a5 b0 D a5 and a0 b5 D b5 ) are 1;
(ii) any other coefficient is obtained as a sum of the two coefficients directly above
it from the previous line (corresponding to n D 4 in our case)
...


1
...
1
...
a C b/n analytically for a general integer n
...
a C b/ D
n

n
XÂnÃ
iD0

i

ai bn

i

;

(1
...
n i/Š

(1
...
Here nŠ D 1 2 3 : : : n is a short-hand notation for
a product of all integers between 1 and n, called n-factorial
...
71) can
also formally be written as follows:


...
73)

k1 Ck2 Dn

10

The formula and the triangle were described by Blaise Pascal, but apparently were known before
him; in particular, a Persian mathematician and engineer Abu Bakr ibn Muhammad ibn al Husayn
al-Karaji (or al-Karkhi) wrote about them in the tenth century as well; he has also proved the
formula by introducing the principle of mathematical induction
...
The sum in this formula is taken over all possible indices
where
k1
k1 ; k2
k1 and k2 between 0 and n such that k1 C k2 D n
...
(1
...
73) becomes a single sum over i between 0 and n
...
74)

i
...
the first and the last coefficients (which are by the highest powers of a and b)
are, for any n, equal to 1
...
The second
property of Pascal’s triangle is established by the following identity:
Â

Ã

n
i

1

C

 Ã
n
D
i

...
i

nŠ
nŠ
C
1/Š
...
n i/Š
Ä
1
1
nŠ
C
1/Š
...
nC1/Š
nC1
nŠ
D
D

...
n i/Š i
...
nC1 i/Š

Â

Ã
nC1
; (1
...
m 1/Šm for any positive integer
m
...
i 1/-th and i-th
terms of the expansion of
...
a C b/nC1
expansion
...
45
...
76)

which basically means that the binomial coefficients are the same in the
binomial formula whether one counts from the beginning to the end or vice
versa
...
74) is a particular case of this symmetry written for the first
and the last coefficients
...
46
...
10 Binomial Formula

47

These properties do not yet prove the binomial formula
...
Consider the case of n D 5
...
e
...

Now we shall prove the formula using induction
...
Indeed, since
1
0

...
Next, we assume that the
formula (1
...
n C 1/:

...
a C b/
...
aCb/
n

n
XÂnÃ

ai bn

i

Ã
n i nC1
ab
i

i

iD0

D

n
XÂ
iD0

Ã

n iC1 n i
a b C
i

n
XÂ
iD0

i

:

In the first sum we separate out the term with i D n, while in the second the term
with i D 0:

...
n 1/ in the first sum and starts
from i D 1 in the second
...
74), the first
and the last binomial coefficients in the last expression are both equal to one
...
n 1/ we introduce k D i C 1 changing
between 1 and n:
n 1
XÂnÃ
iD0

i

a

iC1 n i

b

n
XÂ n Ã
ak bn
D
k 1
kD1

kC1

n
XÂ n Ã
ai bn
D
i 1
iD1

iC1

;

48

1 Basic Knowledge

where in the last passage we replaced k back with i for convenience
...
a C b/nC1 D anC1 b0 C

n
X Ä n à  n Ã
C
ai bn
i 1
i

iC1

C a0 bnC1 :

(1
...
74) and the sum of two binomial
Ã
Â
nC1
because of (1
...
aCb/nC1 in (1
...
71)
with
...
This completes the proof by induction of the binomial
formula
...
3
...
consider the binomial expansion of order n for a D b D 1:
 à  Ã
n
n

...
78)

which shows that the sum is equal to 2n
...
47
...
1 C 1/n

11

D 2n

D 2n
1

1

1

1 ; if n D odd,

(1
...


(1
...
(1
...
]

After all, the summation index is a dummy index which is only used to indicate how the
summation is made; any letter can be used for it
...
11 Combinatorics and Multinomial Theorem

49

Problem 1
...
Similarly, prove:
 à  à  Ã
n
n
n
C
C
C
5
3
1

Â
C

 à  à  Ã
n
n
n
C
C
C
5
3
1

Ã

n
Â

C

D 2n

2

n

Ã

n
n

1

1

D 2n

1

1 ; if n D odd,

(1
...


(1
...
49
...
83)

[Hint: expand
...
1 C x/n in two different ways and compare the
coefficients by xn
...
76)
...


1
...
This may seem
only to deserve some attention of a mathematician as a purely abstract problem,
which is entirely irrelevant to a physicist or engineer
...

There are several questions one may ask, the simplest of them being as follows:
in how many combinations can one place n different objects? For instance, consider
different numbers: there will be exactly two combinations,
...
ba/, in
arranging two different numbers a and b; if there are three numbers, a, b and c,
however, the number of combinations is six, namely:
...
acb/,
...
bca/,

...
cba/
...
For instance,
choosing c at the first position, a and b are left giving us two possibilities: c
...
ba/
...

Problem 1
...
Use the method of induction to prove that n distinct objects can
be arranged exactly in nŠ different combinations
...
If the balls were all different, then there would
have been 5Š D 120 such combinations
...
Within the
5Š D 120 combinations, for any given arrangement of white balls in 5 places, there
will exactly be 3Š D 6 equivalent combinations of the black ones permuted between
themselves; similarly, for any given arrangement of 3 black balls in 5 positions,
there will be 2Š D 2 identical combinations associated with the permutation of
the two white balls between themselves
...
2Š 3Š/ D 120=
...
1
...
The same argument can be extended to n black and m white balls, in
which case the total number of combinations à be
...
nŠ mŠ/
...
(1
...

n
m

Fig
...
16 There are ten combinations in arranging three black and two white balls

1
...
71), each term in the sum
contains i-th power of a and
...
However, the total number of combinations in which
we arrange the numbers in the product is equal to the total number of arrangements
 Ã
n
, the binomial coefficient in Eq
...
71)
...
e
...
12 Since the total number of balls is k1 C k2 C C kn D m, the number
of different (unique) combinations will be
Â

m
k1 ; k2 ; : : : ; kn

Ã
D

Pn

...
84)

Q
where the symbol
means “a product of”, i
...
in our case above this is simply
the product of k1 Š, k2 Š, etc
...
73)
...
However, this number is to be reduced by the
numbers k1 Š, k2 Š, and so on of permutations of all identically coloured balls, so we
divide mŠ by the product of k1 Š, k2 Š, etc
...
a1 C a2 C
„

C an /
...
a1 C a2 C


...


12

Of course, many other interpretations of the same result exist
...

and so on, k1 C k2 C

52

1 Basic Knowledge

Problem 1
...
As an example, consider
...
First, expand the
cube without combining identical terms ai aj ak (where i, j, k are 1, 2 or 3);
verify that there will be all possible combinations of the indices
...
Then,
combine identical terms having the same powers for a1 , a2 and a3
...
i; j; k/ corresponds to the number of ways the indices can
be permuted giving different sequences
...
(1
...

Returning to the general case, we can combine together all combinations which
contain a1 exactly k1 times, a2 exactly k2 times, and so on, as they give an identical
contribution of ak1 ak2 : : : akn
...
How many such identical terms there will be? The
answer is given by the same Eq
...
84)! Therefore, combining all identical terms,
we arrive at an expansion in which a numerical pre-factor (1
...
e
...
a1 C a2 C C an / D
a k1 a k2 : : : a kn ;
(1
...
The sum here is run over all possible
indices k1 , k2 , etc
...

Problem 1
...
Prove Eq
...
85) using the method of induction
...
a1 C a2 C C an C anC1 /m into a single number
bn D an C anC1 , apply the hypothesis and use the binomial theorem (1
...
]
Problem 1
...
Consider 2n objects
...
2n/Š
:
2n nŠ

(1
...
12 Some Important Inequalities
There is one famous result in mathematics known as Cauchy–Bunyakovsky–Schwarz
inequality
...
12 Some Important Inequalities
n
X

!2
xi yi

Ä

iD1

n
X

53

!

n
X

x2
i

iD1

!
y2
i

n
X

or

iD1

v
v
u n
u n
uX uX
xi yi Ä t
x2 t
y2 :
i
i

iD1

iD1

iD1

(1
...
t/ D

...
This means that the determinant of this square polynomial D D
4b2 4ac is either equal to zero (one root: the parabola just touches the t axis at a
single point) or is negative (no roots: the parabola is lifted above the t axis), therefore
2
!2
! n
!3
n
n
X
X
X
D D 4 b2 ac D 4 4
xi yi
x2
y2 5 Ä 0 ;
i
i
iD1

iD1

iD1

which is exactly what we wanted to prove
...
Indeed, consider two vectors xD
...
y1 ; y2 ; y3 /
...
88)

v
u 3
sX
uX
X
X
2
t

...
87) for the third term inside the square root,
which gives:
v
sX sX
sX
uX
X
X
X
u
2
2
xi C
yi C 2
xi yi Ä t
x2 C
y2 C 2
x2
y2
jx C yj D
i
i
i
i
i

i

i

i

i

i

i

v
12
u0
sX
sX
sX
u sX
u
2
2A
2
t@
D
xi C
yi
D
xi C
y2 D jxj C jyj ;
i
i

i

i

i

which is the desired result
...


54

1 Basic Knowledge

Problem 1
...
Prove similarly that for any two complex numbers z1 and z2 ,
we have:
jz1 C z2 j Ä jz1 j C jz2 j :

(1
...
55
...
90)
jzi j :
ˇ
ˇ
ˇ
i

i

Problem 1
...
Prove that for any two complex numbers z1 and z2 :
jz1

z2 j

jz1 j

jz2 j :

(1
...
(1
...
]
Problem 1
...
Prove that one side of a triangle is always shorter than the sum
of two other sides
...
92)

iD1

i
...
the absolute value of a sum of real numbers is always less than or equal to the
sum of their absolute values
...

Let us first consider the case of n D 2
...
x1 C x2 /2 D x2 C x2 C 2x1 x2 D
...
jx1 j C jx2 j/2 C 2
...
Hence, 2
...
12 Some Important Inequalities

55

Fig
...
17 To the proof of
inequality (1
...
jx1 j C jx2 j/2 C 2
...
jx1 j C jx2 j/2 D jx1 j C jx2 j ;

as required
...
x2 C x3 /j Ä jx1 j C jx2 C x3 j Ä jx1 j C jx2 j C jx3 j ;
and so on
...
It is used in
various estimates and proofs:
sin ˛ < ˛ < tan ˛ :

(1
...
1
...
Obviously, the area of
the 4OAC is strictly smaller than the area of the sector, which in turn is smaller
than the area of the 4ODC
...
The areas of the two triangles are easy to calculate as
S4OAC D 1 OC AB D 1 sin ˛ and S4ODC D 1 OC DC D 1 tan ˛
...
Comparing the three and
area R2 D
of the whole circle14 : 2˛
2
cancelling out on the factor of 1=2, we arrive at (1
...

Problem 1
...
Prove the same result (1
...
15

14

The area of a circle of radius R is R2 , this will be established later on in Sect
...
6
...


15

The length of the circle circumference is 2 R, where R is the circle radius (see Sect
...
6
...


56

1 Basic Knowledge

1
...
We shall start from lines,
then turn to surfaces, after which we shall consider various geometrical problems
related to both of them
...
13
...
xA ; yA ; zA / and
B
...
xA ; yA ; zA / and rB D

...
If the vector r D
...
1
...
rB

rA / ;

(1
...
This is a parametric equation of the line,
specified either via two points A and B or via a vector a (it is frequently convenient
for it to be a unit vector) and a point A
...
59
...
94) is equivalent to three scalar ones
for the three Cartesian components of r
...
95)
xB xA
yB yA
zB zA

Fig
...
18 An illustration to the derivation of the equations for a line (a) and a plane (b)

1
...
60
...
96)

Note that this form is more general than the familiar y D kx C c (with k and
c being constants) since it contains also lines parallel to the y axis (Ax D C)
which are not contained in the other form
...
e
...
Note,
however, that in the second case the lines may not necessarily cross
...
1
...
5
...
13
...

In polar coordinates the length r D jrj of the vector r and its angle with the x
axis (0 Ä Ä 2 ) are used instead of the Cartesian coordinates
...
1
...
As follows from the figure, the relationship between the two types of
coordinates is provided by the following connection relations:
x D r cos

;

y D r sin

:

(1
...
1
...
The
components of the vector r are its projections x D OA, y D OB and z D OC on the Cartesian axes

58

1 Basic Knowledge

In 3D space polar coordinates are generalised to cylindrical coordinates by
adding the z coordinate as shown in Fig
...
19(b); in this case,
...
r; ; z/ as follows:
x D r cos

;

y D r sin

;

zDz:

(1
...

Finally, we shall also introduce another very useful coordinate system—
spherical
...
1
...
Here the coordinates used are
...
The connection relations in this case are:
x D r sin  cos ;

y D r sin  sin ;

z D r cos Â;

(1
...

The special coordinate systems introduced above are frequently used in the
calculus to simplify calculations, and we shall come across many such examples
throughout the course
...
Many other so-called curvilinear coordinates also exist
...
13
...
The idea is based on writing r as a function of , e
...

r D r
...
/ D r
...
/ D r
...
100)

as parametric equations of curved lines via the angle serving as a single parameter
...
/ may be even negative, i
...
only its absolute value
corresponds to the distance of the
...

As an example, consider a circle of radius R centred at the origin
...
e
...
In that case
from (1
...


1
...
61
...
0; R/
...
x; y/ D 0 for the curve
...
62
...
1
...
Convince yourself that these geometrical figures correspond to the equations given
...
16
Problem 1
...
Consider a family of curves specified by r D a cos
...
How does the number of petals depend on the value of k?

1
...
4 Planes
A plane can be specified by either three points A, B, C or by a normal vector n and a
single point A, see Fig
...
18(b)
...
rB rC /
...
So, it is sufficient to consider the specification of the plane
via a normal and a point, say the point A
...
r

rA / D 0

H)

r n D rA n ;

(1
...
One can see the point rA lies on the plane
(as r D rA also satisfies the equation), and that the normal does not need to be a unit
vector
...

Another form of the equation for the plane follows immediately from (1
...
102)

It follows from this discussion that the coefficients
...
Note that if h ¤ 0, one may divide both
sides of the equation by it, resulting in a simpler form a0 x C b0 y C c0 z D 1, however,
for some surfaces h D 0 (e
...
the x y plane has the equation z D 0), so that the
form above is the most general
...


60

1 Basic Knowledge

Fig
...
20 Examples of various famous curves specified in polar coordinates: (a) a three-leaf rose
(or rhodonea curve) specified as r D a cos
...
a; b/; note how the curves
continuously change their shape when going from a < b to a > b across a D b; (c) Archimedean
spiral r D a (plotted for a D 1); (d) spiral r D ae ˛ (plotted for a D 1 and ˛ D 0:1)
...

We shall now derive a parametric equation for a plane
...
(1
...
In
the case of a plane there must be two such parameters as the plane is equivalent to
a 2D space
...
Now, consider two vectors a1 and a2 lying in the plane (it is usually convenient
to choose them of unit length), and the vector rA pointing to a point A on the plane
...
13 Lines and Planes

61

Then, evidently, any point P on the plane would have the vector
r D rA C t1 a1 C t2 a2

(1
...

Two planes are parallel if their normal vectors n1 and n2 are parallel, i
...
if n1
n2 D 0; two planes are perpendicular to each other if their normals are orthogonal,
i
...
if n1 n2 D 0
...
Let us derive the equation
of that line
...
(1
...
e
...
The only other unknown
to find is a point A with vector rA lying on that crossing, as after that the parametric
equation of the line is completely determined
...
e
...
101) or (1
...
This step will result in two algebraic equations for three
unknowns x, y and z, so that one can only express any two of them via the third; we
choose that third coordinate arbitrarily (our particular choice will result in a certain
position of the point A along the line) and then would be able to determine the other
two
...

Example 1
...
I Determine the parametric equation of the line at which two
surfaces x C y C z D 1 and x D 0 cross
...
the normal of the first surface n1 D
...
1; 0; 0/ for the
second
...
0; 1; 1/ :

Considering the two equations simultaneously, we get y C z D 1
...
0; 0; 1/ on the crossing line
...
e
...


62

1 Basic Knowledge

Fig
...
21 Typical problems related to lines and planes
...
Then, we have x D 0, y D t and
z D 1 y D 1 t
...
J
Problem 1
...
Prove that this method will always give the correct solution,
i
...
points on the line constructed in this manner will satisfy Eq
...
101) for
both planes
...
65
...
[Answer: r D
...
2; 3; 1/
...
13
...

Distance from a Point to a Line or Plane
...
What
we would like to calculate is the distance between them; this will be the length of
the perpendicular drawn from the point P to the line as shown in Fig
...
21(a)
...
13 Lines and Planes

63

!
!
AB k a, and hence the angle can be obtained from the dot product of a and AP, i
...

ˇ !ˇ
!
ˇ ˇ
AP a D jaj ˇAPˇ cos
...
Then, the
ˇ !ˇ
ˇ ˇ
required distance BP D ˇAPˇ sin
...
Let point A be
...
e
...
The point P has coordinates
...
Then, by considering two points
A D
...
xP ; kxP C b/ on the line, we can calculate its direction vector as
a D
...
xP ; yP
cos

0/ i C Œ
...
Therefore,

!
AP a
xP C k
...
yP b/
P
D ˇ !ˇ
Dq
D q
p
p
ˇ ˇ
2
ˇAPˇ jaj
x2 C
...
yP b/2 1 C k2
P
P

and hence
sin2

D1

Dh

cos2

D1


...
yP
P

x2 C 2kxP
...
yP
P

b/ C k2
...
1 C k2 /


...
1 C k2 /

...
104)

Problem 1
...
Show that the distance between a point P
...
1
...
105)

where n is the unit normal to the plane
...
Indeed, manipulate this
expression into
dD

axP C byP C czP h
p
:
a 2 C b 2 C c2

(1
...
The necessary conditions for crossing of two lines can be
established as follows
...
The lines will cross if and only if
there exists a plane to which they both belong
...
This in turn means that the vector r1 r2 can be
represented as a linear combination s1 a1 C s2 a2 of the vectors a1 and a2 with some
numbers s1 and s2 since any vector within the plane can be represented like this
unless the two vectors a1 ; a2 are parallel which we assume is not the case
...
e
...
The simplest way of
writing this condition down is by using the triple product:
ˇ
ˇ
ˇ x1 x2 y1 y2 z1 z2 ˇ
ˇ
ˇ
(1
...
a1 /x
...
a1 /z ˇ D 0 ;
ˇ
ˇ
ˇ
ˇ
...
a2 /y
...
e
...
Indeed,
as we know from the properties of the determinants mentioned above, it is equal
to zero if any of its rows (or columns) is a linear combination of any other rows
(columns)
...
1
...
Eq
...
38)
...
5
...
1; 0; 0/ C t1
...
There is
also a family of lines specified via r D
...
0; cos '; sin '/; each line in the
family corresponds to a particular value of the angle '
...
e
...

Solution
...
(1
...
1; 1; 1/, a1 D
...
0; cos '; sin '/; hence, building up the determinant and calculating it,
we get:
7
6
61 1
1 7
7
6
41 0
0 5D1
...
1/
...
cos ' 0 0/
0 cos ' sin '
D sin '

cos ' D 0 ;

i
...
tan ' D 1 and Á D =4, so the equation of the second line is r D
...
The point of crossing is then obtained by solving the vector
equation

Note that the case of a1 k a2 is not interesting here as in this case the lines do not cross—unless
they are the same, of course, which is of even less interest!

17

1
...
1; 0; 0/ C t1
...
0; 1; 1/ C t2 0; p ; p
2
2
with respect to t1 and t2 in components
...
107) we just used, one of the equations should be equivalent to the
other two, so we just have to consider two independent equations
...
1; 0; 0/ C t1
...
0; 0; 0/ is the centre of the coordinate system (of course, the same result is obtained
by using the equation of the second line and the value of t2 )
...
J
Crossing Point Between a Line and Plane
...

The line will definitely cross the plane at some point A, Fig
...
21(c), if the vector a
is not perpendicular to the line normal: a n ¤ 0
...
rC rB /
n
...
108)

where rB is a point on the line, while rC is a point on the plane
...
Consider two lines,
1 and 2 , which are not parallel (their vectors a1 and a2 are not parallel, a1
a2 ¤ 0) and do not cross, Fig
...
21(d)
...
e
...
It is easy to see that
the line connecting these two points will be perpendicular to both lines
...
Let us first draw the
vector a2 from the point A on the first line, and the vector a1 from the point B on the
second, see Fig
...
21(d)
...
Since the two normal vectors are collinear,
the two planes are parallel
...
The distance
between these planes, which is given by the perpendicular line drawn from one plane
to the other, is exactly what is needed—the minimal distance between the points of
the two lines, shown as a blue dotted line in Fig
...
21(d)
...
For this task to be accomplished, we,
however, require a derivative which we have not yet considered
...
The vectors BA D rA rB , a1 and a2 form a
parallelepiped shown in the figure
...
On the other hand, its volume is
also given by the area of the parallelogram at the base of the parallelepiped and
formed by a1 and a2 (which is equal to the absolute value of the vector product
ja1 a2 j), multiplied by its height which is h
...
rB rA / Œa1
D
ja1 a2 j
ja1 a2 j

a2 j

;

(1
...

Distance Between Two Parallel Planes
...
Consider two planes 1 and 2 with collinear
normal vectors n1 and n2 , given by the equations n1 r D n1 rA and n2 r D n2 rB
...
67
...
Fig
...
21(a,
b), that the distance between two parallel planes as specified above can be
calculated via
hD


...
110)

Crossing of Two Planes
...

Two planes are parallel if their corresponding normal vectors are collinear, i
...
their
vector product is equal to zero
...
In general, one can define an angle ˛ between
two planes as shown in Fig
...
21(e); if ˛ D 0, the two planes are parallel, while if
˛ D =2, the two planes are perpendicular
...
68
...
1
...
Correspondingly, demonstrate that the angle ˛ between two planes given by equations
a1 x C b1 y C c1 z D d1 and a2 x C b2 y C c2 z D d2 is given by:
cos ˛ D q

a 1 a 2 C b 1 b 2 C c1 c2
q
a 2 C b 2 C c2 a 2 C b 2 C c2
1
1
1
2
2
2

(1
...
In particular, verify the above criteria for
the two planes to be parallel or perpendicular to each other
...
1 Definition and Main Types of Functions
As was mentioned already in Sect
...
5, a function y D f
...
It is
not an essential condition for the function to exist that the values of x and/or y
form continuous intervals
...
It makes a correspondence between all integer numbers (serving as
the values of x) and some specific set of integer numbers (1, 2, 6, 24, etc
...
It is essential, however, that to a single value of x there is always one and
only one value of y
...

Indeed, consider, e
...
, the function y D x2 for which two values of x D ˙1 result in
the same value of y D 1
...
x/ and g
...
x/ ˙ g
...
x/g
...
x/=g
...
x/ D 0)
...
x/ and g
...
w/ and w D f
...
e
...
f
...
For instance, y D sin2
...
x/
...
g
...
f
...
x///
...
g
...
x/ D

1;
0;

if x 0
;
if x < 0

© Springer Science+Business Media, LLC 2016
L
...
1007/978-1-4939-2785-2_2

(2
...
2
...
(2
...
2)
...
0/ D 1
...
2
...
Other definitions of H
...
Another useful example of such a definition is the sign
function which is defined similarly as:
8
< 1; if x < 0
sgn
...
2)

shown in Fig
...
1(b)
...
x/ D 2H
...
However, if H
...
0/ D 1=2,
then the mentioned relationship would hold even for x D 0
...
x/ at x D 0 is sometimes useful
...
1
...
x/ D 2H
...
x/

H
...
x/ are displayed on the x y plane
...
g
...
g
...
However, when discussing functions, we shall
mostly be dealing with continuous functions which are defined on a continuous
interval of the values of x in R
...
A function can be limited
from above on the given interval of x if for all values of the x within it there exists a
real M such that f
...
For instance, y D x2 is limited from above, y < M D 0,
for all x in R, i
...
for any 1 < x < 1
...
x/ m, then the function y D f
...
g
...
A function can also be limited both
from above and below if there exists such positive M that jf
...
An obvious
example is y D sin
...
x/ Ä 1, i
...
for the sine function M D 1
...
1 Definition and Main Types of Functions

69

since it is impossible to find a single constant M > 0 such that 1=x2 Ä M for
all positive x: whatever value of M we take, there always be values of x such that
p
1=x2 > M in the interval 0 < x < 1= M
...
x/ is strictly increasing monotonically within an interval, if
for any x1 < x2 follows f
...
x2 /, while the function is strictly decreasing
monotonically if f
...
x2 / follows instead
...
Strictly decreasing or increasing
functions are called monotonic
...
x C T 0 / D f
...
The smallest such number T is called the
period
...
x/ is a periodic function since sin
...
x/
for any integer n D 1; 2; : : :, i
...
the values of T 0 form a sequence 2 , 4 , etc
...

Above we have defined a direct function whereby each value of x is put in the
direct correspondence with a single value of y via y D f
...
One can, however,
solve this equation with respect to x and write it as a function of y
...
x/ and it is called (if exists, see below) the inverse function
x D f 1
...
Conventionally, however, we would like the x to be the argument and y
to be the function
...
x/
...
1
...

Solution
...
e
...
x/ D 5x is f 1
...
J
5
Example 2
...
I Obtain the inverse function to y D x3
...
Solving for x and interchanging x and y, we obtain y D f
which is the cube root function (see Sect
...
3
...
J

1


...
x/
...
x1 / ¤ y2 D f
...
This is because for each value of the x there must be a unique
value of y D f
...
y/
...
x1 / < y2 D f
...
Similarly, the inverse
function f 1
...
x/ is monotonically decreasing
...
2
...
x/ and f 1
...
2
...
x/ is an increasing (decreasing) function, then so
is its inverse (if it exists)
...
3
...
x/ whose inverse is f 1
...
Prove
that the inverse to f 1
...
x/, i
...
the two functions are mutually inverse to
each other
...
x/ and y D f 1
...
2
...
Indeed, consider a point A on the curve of
the function f
...
x1 ; y1 D f
...
Consequently, x1 D f 1
...
However, conventionally, we write this as
y2 D f 1
...
e
...
It is seen that the point A on f
...
x/ with the coordinates
...
It is now obvious from
the drawing in Fig
...
2 that the two points A and B are located symmetrically with
respect to the dashed brown diagonal line
...
x/, the reasoning given would correspond to any point on f
...
e
...
x/ and f 1
...

It follows from that simple discussion that the graph of f 1
...
x/ by reflecting the latter across the diagonal of the quadrants I
and III
...
Consider, for
instance, a parabola y D x2 , shown in Fig
...
3 with the red line
...
In other
words, in this case the two curves (the red and blue) are rotated by 90ı with respect
to each other
...
This is of course because while solving the equation y D x2 with respect

2
...
2
...
It follows from this discussion that,
strictly speaking, the function y D x2 does not have an inverse
...
This way we obtain the function
and
y D x > 0 drawn in Fig
...
3 with the solid blue line
...

The reduction operation considered above for the parabola was necessary to
define properly the inverse function to it, and the same operation is applied in some
other cases as will be demonstrated in the next section
...
2 Infinite Numerical Sequences
2
...
1 Definitions
To continue our gradual approach to functions, we have to consider now numerical
sequences and their limit, otherwise it would be difficult for us to define power and
exponential functions which are the matter of the next section
...

In other words, the numbers xn D f
...
For instance, the set 1; 1; 1; 1; 1; : : : can be obtained by
calculating
...
Another set
1;

1 1 1 1 1 1
; ; ; ; ; ; :::
2 3 4 5 6 7

(2
...
If the former set
...
One may say that if the
former sequence does not have a definite limit when n becomes indefinitely large

72

2 Functions

(we say “n tends to infinity”, which is denoted as 1, and write it as n ! 1),
the latter sequence does: 1=n ! 0 as n ! 1
...
To formulate the required definition, let us assume that a sequence
x0 ; x1 ; x2 ; x3 ; : : : ; xn ; : : : approaches a
...
e
...
g
...
1/n =n)
...
e
...
This
is written as follows:
lim xn D a:

n!1

(2
...
Let us take a small positive number 4 D 0:01
...
e
...
Indeed, from the inequality 1=n < 4
we get n > 1=4 D 100; in other words, there exist such number N D 100 so
1
1
that any member of the sequence x101 D 101 ' 0:0099, x102 D 102 ' 0:0098,
1
x103 D 103 ' 0:0097, etc
...
If we
take even smaller value of 4 D 0:001, then solving for n in 1=n < 4 will result in
N D 1=4 D 1000, i
...
for all n > 1000 all members of the sequence xn are smaller
than this smaller value of 4
...
This means that indeed, according to the given above
definition, the sequence xn D 1=n tends to zero
...
e
...
e
...

Problem 2
...
Prove that the infinite numerical sequence xn D 1=n2 tends as
well to zero, i
...
its limit is equal to zero
...
2 Infinite Numerical Sequences

73

Problem 2
...
Prove, using the definition, that the infinite numerical sequence
xn D
...

Problem 2
...
Prove that the sequence xn D qn does not have a limit if jqj > 1
...
Consider again Problem 2
...
Each element of the
sequence can alternatively be written as
xn D

1
C 2;
n

Then, considering the original sequence xn D yn C zn as a sum of two sequences,
yn D 1=n and zn D 2, we may apply the limit to each sequence separately to get
yn D 1=n ! 0, zn ! 2, and hence xn ! 2 as n ! 1
...
3
...
1/n
:
3n2 4n C 5

Solution
...
1/n =n2
2
! ;
2
3 4=n C 5=n
3

as all other terms tend to zero as n ! 1
...
2
...
This is accomplished by the following simple theorems which
we shall now discuss
...
1 (Addition Theorem)
...


74

2 Functions

Proof
...
e
...
e
...
Similarly, for the second sequence: for any y D =2 there would
exist such Ny , so that for any n > Ny we have jyn Yj < y
...
X C Y/j D j
...
yn

Y/j Ä jxn

Xj C jyn

Yj < x C y D :
(2
...
92) stating that an absolute value of a sum of numbers
is always less than or equal to the sum of their absolute values
...
5), e
...
N D max Nx ; Ny
...
Q
...
D
...
2 (Product Theorem)
...


Proof
...
xn
Ä jxn

X/
...
yn

Y/ C Y
...
e
...
e
...
Q
...
D
...
3 (Division Theorem)
...


2
...
Again, this is proven by the following manipulation:
ˇ
ˇ xn
ˇ
ˇy
n

ˇ
Xˇ
ˇD
Yˇ
Ä

ˇ ˇ
ˇ
ˇ xn Y yn X ˇ ˇ
...
yn
Yyn
Yj jXj

<

ˇ
Y/ X ˇ
ˇ
ˇ

x jYj C y jXj
:
jYj jyn j

Since yn ¤ 0 for any n, then it should be possible to find such ı > 0 that jyn j
for any n
...
Q
...
D
...
1/n
2 C 3=n C
...
3=n/ C limn!1
...
4=n/ C limn!1
...
g
...
g
...
3)
...
This is stated by the following

Theorem 2
...
If a sequence xn (with indices n
taking all integer values Nat) tends to a limit X, then its arbitrary subsequence
yk (where indices k take on only a subset of values from the original set Nat
and for these indices yk D xk ) tends to the same limit
...
This follows directly from the definition of the limit
...
Take now a subset of indices, let us call them k
...
Q
...
D
...
2n/ containing only
inverse even integers
...

Problem 2
...
Prove that the sequence xn D
...

If a sequence has a limit, it must be unique, it is impossible for a sequence to
have two different limits
...
5 (Uniqueness)
...


Proof
...
Assume that the sequence xn
has two different limits, X1 and X2 > X1
...
X2 X1 / =2 > 0
...
e
...
Similarly, since X2 is also a
limit, there must be such N2 so that for any n > N2 we have jxn X2 j <  with the
same 
...
X2 xn / C
...
Note that we have the strict “lesser than” sign here!
Hence, our assumption is wrong
...
Q
...
D
...
We shall consider them next
...
6
...
e
...


Proof
...
Assume that X < A
...
This means that 4 < xn X < 
...
2 Infinite Numerical Sequences

77

xn <  C X
...
Our initial assumption is therefore wrong
...
E
...


Theorem 2
...
If all elements of the sequence, starting from some particular
number n0 , satisfy xn Ä A and the sequence has the limit X, then X Ä A
...
8
...


Theorem 2
...
If elements of two sequences xn and yn , starting from some
number n0 , satisfy xn Ä yn and their limits exist and equal to X and Y,
respectively, then X Ä Y
...
Consider a sequence zn D yn xn
...
Therefore, according
to Theorem 2
...
Q
...
D
...
e
...
In particular, if Z D X, then Y D X D Z
...
2
...
Sect
...
9)
SN D

N
X

xn

nD0

of its first N members
...
It is instructive to explicitly state here the definition of the convergence of the
infinite series which is nothing but a rephrased definition (for SN in this P
case) of
the limit given above in Sect
...
2
...
e
...


78

2 Functions

As an example of an infinite numerical series, consider an infinite geometrical
progression [cf
...
(1
...
6)
where we used the formula (1
...
Now, if jqj > 1, then qNC1 ! 1 as N ! 1 (obviously, qN
increases with N), and hence the series does not have a finite limit
...
If
q D 1, then the series contains alternating a0 and a0 terms, and hence (similarly
to the case of xn D
...
Therefore, it
only makes sense to consider the case of jqj < 1
...
6)
an important result:
Sgeom D a0 q0 C a0 q1 C a0 q2 C a0 q3 C

D lim SN D lim

a0 1 qNC1
a0
D
:
N!1
1 q
1 q

Sgeom D lim

(2
...
4
...
123/, which contains a periodic sequence of digits in its decimal representation
...
we write:
r D 0:9 C 0:0123 C 0:0000123 C

D 0:9 C 0:0123 1 C 10

3

C 10

6

C

:

The expression in the brackets is nothing but the infinite geometric progression
...
7) for its sum, we finally obtain:
r D 0:9 C

123
9
123
9114
1
0:0123
9
C
D
C
D
:J
D
3
1 10
10
10000 0:999
10
9990
9990

Problem 2
...
Prove that any number r < 1 specified via a periodic sequence
of digits in the decimal representation is a rational number, i
...
it can always
be written as a ratio of two integers
...
3 Elementary Functions
Here we shall introduce all elementary functions and their main properties
...


2
...
3
...
Its graph
is represented by a horizontal line crossing the y axis exactly at the value of C
...

A linear combination of x in integer powers, from zero to n, with real coefficients,
as was already mentioned in Sect
...
5, gives rise to a function called a polynomial
of degree n:
f
...
8)

iD0

It is of special utility to know the roots of the polynomial, i
...
the values of x at
which f
...
The reader must be familiar with the methods of finding the roots
of the polynomials of the degree 1 (linear) and 2 (quadratic), see Sect
...
5
...
e
...
In fact, it was later proven2 that it is not possible
to express the roots of general polynomial equations with degree higher than 4 in
radicals, so that in practice numerical solutions (or analytical ones in simple cases)
are usually made
...
x/ D 0
always has at least one solution, and the maximum possible number of different
roots is n
...
However, in that case they will always
come in complex conjugate pairs, a C ib and a ib
...
This fact can be simply put by writing
the function f
...
x/ D an
...
x

x 2 /k 2


...
9)

which states explicitly that f
...
Multiplying the brackets
in the above equation, a polynomial of degree n is obtained; comparing the same
powers of x in it with the definition in Eq
...
8), algebraic relationship between the
roots fxi g and the coefficients of the polynomial fai g can be established
...
x x1 /
...
x1 Cx2 / xCx1 x2

H)

a1
D
a2

x1 x2 ;

a0
D x1 x2 :
a2

1

This was done by Italian mathematicians Lodovico Ferrari, Niccolo Fontana Tartaglia and
Gerolamo Cardano
...


80

2 Functions

Problem 2
...
Show that such a relationship for the cubic polynomial, assuming all roots are different and also that a3 D 1, is:
a 2 D x1

x3 I

x2

a 1 D x1 x2 C x1 x3 C x2 x3

and a0 D

x1 x2 x3 :

Problem 2
...
The same for the quartic polynomial:
a 3 D x1

x2

x3

a 1 D x1 x2 x3

x4 I

a 2 D x1 x2 C x1 x3 C x1 x4 C x2 x3 C x2 x4 C x3 x4 I

x1 x2 x4

x1 x3 x4

x2 x3 x4 I

a 0 D x1 x2 x3 x4 :

Note that all coefficients are fully symmetric with respect to an arbitrary
permutation of the roots
...
For instance, consider the quartic case
...
2Š2Š/ D 6 combinations, while a3 contains
sum of 6 terms due to
2
 Ã
 Ã
4
4
D 4Š=3Š D 4 terms and a0 contains only
D 1
...
9) can also be written entirely via real factors
...
x

a

ib/
...
Therefore, in general, a real polynomial is represented in a
factorised form as
f
...
x

x 1 /k 1


...
10)

Here we first have m simple factors x xi corresponding to real roots of f
...

Graphs of several integer power functions y D xn are shown in Fig
...
4(a)
...
e
...
x/n D xn
...
3
...
In many applications it is
necessary to decompose such a function into simpler functions
...
3 Elementary Functions

81

Fig
...
4 (a) Power functions y D xn for selected values of n; (b) exponential functions y D ax for
several values of the base a

down a desired decomposition with undetermined coefficients and then finding the
coefficients by comparing both sides (as the equality should hold for any value of x)
...
x/ D
Since x2
follows:

1 can be factored as
...
x C 1/, we attempt to decompose f
...
Summing up the two terms
above in the right-hand side, we obtain a fraction with the denominator identical to
the original one, x2 1, standing on the left
...
x C 1/ C B
...
11)

Since this equation is to hold for any x, we may, for instance, put two values of x
to obtain two independent equations for the two constants
...
12)

Another way of approaching the problem of solving Eq
...
11) is to rearrange the
right-hand side first in a form of a polynomial:

82

2 Functions

1 D
...
A C B/ x:

In the left-hand side the polynomial is equal to just a constant, i
...
when we compare
the coefficients in both sides to the same powers of x, we obtain two equations
1DA B
0DACB

A D 1=2
;
B D 1=2

H)

resulting in the same solution
...
11) must hold for any x
...
12
...
x

1
1/
...
x

2/

:

Generally, if the function f
...
x/=Qm
...
The latter case can always be manipulated into the first one by singling
out the denominator in the numerator at its highest power term(s), as, e
...
, done in
the following example:
3x x2 C1
3x x2 C1
1 C2x C 1
xC1
3x3 C2xC1
D
D
D 3x
2 C1
2 C1
2 C1
x
x
x
This way one can always represent Pn =Qm with n

x 1
:
x2 C 1
(2
...
14)

where Rr and Gl are polynomials and l < m
...
13
...
(2
...
15)
12x 1
:
(2
...
3 Elementary Functions

83

The rational function Pn =Qm with n < m can then be decomposed further, as we
have seen above, into a sum of simpler fractions
...
x/ is decomposed in a
product of simple factors as in Eq
...
10), then the fraction Pn =Qm (n < m) can be
written as a sum of simpler fractions using the following rules:
• each single root a which is repeated r times gives rise to the following sum of
terms:
A1
A2
Ar
C
C C
2
x a

...
x a/
with r arbitrary constants A1 ; : : : ; Ar ;
• each pair of complex roots z D a Cib and z D a ib corresponding to the square
polynomial x2 C px C q results in the sum of terms
B1 x C C1
B2 x C C2
C
C
C px C q

...
x2 C px C q/l

with 2l arbitrary constants B1 ; C1 ; : : : ; Bl ; Cl , where l is the repetition number for
the given pair of roots
...
5
...
x

1/

2

1


...
Using the rules, we write:
RD

A1
...
BxCC/
...
x 1/2 x C1

...
x2 C1/

Opening up the brackets in the numerator above, grouping terms with the same
powers of x and comparing them with the corresponding coefficients of the original
fraction, we obtain the following four equations for the four unknown coefficients
to x3 , x2 , x1 and x0 , respectively:
8 3
ˆx
ˆ 2
<
x
ˆ x1
ˆ
: 0
x

W
W
W
W

A1 C B D 0
A1 C A2 C C 2B D 3
A1 2C C B D 1
A1 C A2 C C D 1

H)

8
ˆ A1 D 2
ˆ
<
A2 D 3=2
;
ˆ BD 2
ˆ
:
C D 1=2

i
...
we finally obtain the decomposition sought for:
3x2 C x

...
x2

1
C 1/

D

2
x

1

C

3
2
...
x2 C 1/

(2
...
14
...
2x 1/
...
x C 2/2
Ä
x3 C x2 C 1
899
137
189x C 237
1
C
I
C
D
2
2
2 C 3/
2523 3x 1
2x2 C 3

...
2x

...
x2 C 1/2
...
x2 C 1/2
[Hint: note that in the first fraction x2 C 4x C 4 should be factored first
...
15
...
Of course, the four
roots of x4 C 1 D 0 can also be formally found and then this expression be
fully factored; but the roots are all complex and this method requires a better
knowledge of complex numbers
...
3
...
Definition of the power function to any real power is done in stages
...
e
...
In addition, we define that x0 D 1 for any x
...
Indeed, consider first n D 2
...
e
...
Similarly, that property can be demonstrated for any integer n
...
3 Elementary Functions

85

Then, we define the power being an inverse integer, ˛ D 1=n, as the solution for
p
y of the equation yn D x, which we write y D x1=n or y D n x
...
Note that in the cases of
even values of n we only accept positive solutions for y and x
0; this method
would prevent us from arriving at multi-valued functions, as was discussed at the
end of Sect
...
1 using parabola as an example
...

At the next step we define the power being a general rational number ˛ D n=m
via:
y D xn=m D x1=m

n

This is an example of the composition of two functions mentioned at the beginning
of Sect
...
1: indeed, we combined two functions, y D wn and w D x1=m , into one:
n
y D wn D x1=m
...
Hence, the function y D x˛ for any rational ˛ increases
monotonically
...
Since we already covered general rational numbers (which are
represented by either finite or periodic decimal fractions, Sect
...
2), we only need to
consider general irrational values of ˛ which are represented by infinite non-periodic
decimal fractions
...

p
For instance, consider ˛ being 2 D 1:1414213562 : : :
...
g
...
Then, since the
function y D xr with rational r is increasing monotonically, then xrlower < xrupper
...
Of course, these two bounds are only approximations to the
x
actual value of p2, so we could do much betterp enclosing it in tighter bounds,
by
e
...
x1:14142 < x 2 < x1:14143 , then x1:141421 < x 2 < x1:141422 , and so on, i
...
this
process can be continued by moving the boundaries on the left and on the right
p
of 2 closer to it simultaneously from both directions
...
1414, 1
...
141421, etc
...
1415, 1
...
141422, etc
...
The sequence on the right is
composed also by adding an extra digit at the same position as in the first sequence;
however, every time that digit is larger by 1 than in the previous case
...
These two sequences
p
which bound x 2 on both sides converge to the same limit which, by definition, is
p
p
on
taken as the exact value of x 2
...


86

2 Functions

Finally, a real negative power is defined as follows: y D x ˛ D 1=x˛ with ˛ > 0
...

Let us now consider properties of the power function
...
xn /m D xnm , due to the definition of the integer
1=m
D x1=
...
For the inverse integer powers, we also have x1=n
is the inverse function of the integer power
...
yn /m D
ynm D x follows that y D x1=
...
On the other hand, from
...
This means that x1=n
D x1=
...
Now we can consider the case of both powers being arbitrary rational
numbers (n, m, p and q are positive integers):
xn=m

p=q

D

n

x1=m

o
n p 1=q

D

x1=m

np 1=q

i1=q
h
D
...
xnp /1=
...
mq/ ;

as required
...
mq/ xpm=
...
mq/

nq

x1=
...
mq/

nqCpm

D x
...
mq/ D x
...
p=q/ :
Extension of these properties to arbitrary positive real numbers is achieved using the
method of bounding rational sequences as already has been demonstrated above
...
16
...

Hence, we conclude that for any real ˛ and ˇ:
x˛ xˇ D x˛ˇ
where generally we must consider x

and


...
18)

0
...
3
...
We shall show that
n
it tends to a limit e D 2:71828182845 : : : (the base of the natural logarithm)
...
71):
3

This is actually an irrational number although we won’t be able to demonstrate this here
...
3 Elementary Functions

87

Ã
Â
n  Ã
1 n X n n
1
xn D 1 C
D
k
n
kD0

k

 Ãk X  Ã
n
1
n 1
D
k nk
n
kD0

1
n
...
n 2/ 1
n
...
n 1/
...
n k C 1/ 1
C C
;
C
k
kŠ
n
nŠ nn

(2
...
72) can be written as
 Ã
n
...
n k/Š

1/
...
n

k C 1/

for any k
1
...
19) may be
written as follows for any k 1:
!
n
k

ÃÂ
à Â
Ã
Â
ÃÂ
à Â
Ã
Â
n 2
n kC1
1
1 nÁ n 1
1
1
2
k 1
D
:::
D
1
1
::: 1
;
kŠ n
n
n
n
kŠ
n
n
n
nk

so that
Â
Ã
Â
ÃÂ
Ã
Â
ÃÂ
ÃÂ
Ã
1
1
1
1
2
1
1
2
3
1
C
1
1
C
1
1
1
C
2Š
n
3Š
n
n
4Š
n
n
n
Â
ÃÂ
à Â
Ã
Â
ÃÂ
à Â
Ã
1
1
2
k 1
1
1
2
n 1
C
1
1
::: 1
C C
1
1
::: 1
;
kŠ
n
n
n
nŠ
n
n
n

xn D 1C1C

(2
...
19) also in the same form as the
others
...
20) of xn
...
20):
xnC1 D 1 C 1 C
Â
1
1
4Š
Â
1
C
1
kŠ

Â
1
1
2Š

C

C

1

...
21)

88

2 Functions

This expansion of xnC1 has n C 2 terms
...
It is
clear that for any k 1
1

k
:
n

k
>1
nC1

Therefore, the third term of xnC1 is greater than the third term of xn , the fourth term
of xnC1 is greater than the fourth term of xn , and so on
...

Therefore, we strictly have xnC1 > xn for any value of n
...

At the same time, the sequence is limited from above
...
However, the infinite sum represents an infinite geometric progression
which can be calculated as we discussed above by virtue of Eq
...
7):
1
X
kD0

2

k

D1C

1
1
C 2C
2
2

C

1
C
2n 1

D

1
1

1
2

D2;

so that
xn < 1 C

1
X

2

k

D 1 C 2 D 3:

kD1

Hence, we showed that 2 < xn < 3 and xn increases with n
...
Hence, we can state that
Ã
Â
1 n
lim 1 C
D e:
n!1
n

4

(2
...
2
...
2
...
3 Elementary Functions

89

From this it is easy to see that, for any positive integer m,
lim

n!1

"Â
Ã
à #m
Â
1 n=m
m Án
1 n
1C
1C
D lim 1 C
D lim
:
n!1
n!1
n
n=m
n=m

Here n=m may be either integer or generally rational, and we have made use of the
fact that
...
However, when we take the limit n ! 1, we can run n only
over a subset of integer numbers which are divisible by m
...
4)
...
23)

Ã

1
m

1

1
...
Therefore, we obtain that
Ã
Â
1 n
lim 1 C
D e:
n! 1
n

(2
...
17
...
25)

90

2 Functions

2
...
5 Exponential Function
Consider a general definition of y D ax for any real positive a (and avoiding
the trivial case of a D 1) and arbitrary x from R
...
It is important in this respect to establish first that the exponential
function is monotonic
...
First we prove that a > 1 for any positive
...
We also have that a1=m > 1 for
any positive integer m
...
e
...
e
...
1 ˛/m
...
e
...
1 ˛/m < 1m D 1 as 1 ˛ < 1; hence, we
arrive at a < 1 which contradicts our assumption of a > 1, and that proves the
statement made above
...

Indeed, an=m D
...
The proof for any positive real is
extended by applying the bounding numerical sequences
...
ax2

x1

1/ > 0;

since a > 1 for any positive as was shown above
...

Problem 2
...
Prove that the exponential function for 0 < a < 1 is
monotonically decreasing
...
2
...

Due to the properties of powers, Eq
...
18), we also have:
ax1 ax2 D ax1 Cx2

and


...
26)

Exponential functions y D ax for selected values of the base a are shown in
Fig
...
4(b)
...
0
when it becomes indeed horizontal (and hence a constant); then for a > 1, the
function monotonically increases; the larger the value of a, the steeper the increase
...
3 Elementary Functions

91

2
...
6 Hyperbolic Functions
These are combinations of exponential functions:
sinh
...
e
2

e x/ I

(2
...
x/ D

1 x

...
28)

tanh
...
x/
D x
cosh
...
x/ D

cosh
...
x/
e
e

x
x
x
x

I

(2
...
30)

It can be shown that these functions are closely related to trigonometric functions
sin
...
x/, tan
...
x/, and are called, respectively, hyperbolic sine, cosine,
tangent and cotangent functions
...
19
...
x/

sinh2
...
31)

sinh
...
x/ cosh
...
32)

cosh
...
x/ C sinh2
...
33)

sinh
...
x/ cosh
...
x/ sinh
...
34)

These identities have close analogues in trigonometric functions; for instance,
the first one, Eq
...
31), is analogous to (1
...

One can also define inverse hyperbolic functions
...


2
...
7 Logarithmic Function
The logarithmic function is defined as an inverse to the exponential function: from
y D ax we write x D loga y, or using conventional notations, y D loga x
...
From the properties of the exponential

92

2 Functions

function, we can easily deduce the properties of the logarithmic function
...
y1 y2 / D x1 Cx2 D loga y1 C loga y2 :
(2
...
36)

Raise both sides of this equality into power z:
aloga x

z

D az loga x D xz

)

z loga x D loga
...
37)

Finally, we write
x D aloga x D blogb a

loga x

D bloga x logb a ;

from which it follows that
logb x D loga x logb a;

(2
...

If a D e is used as the base, then the logarithmic function is denoted simply
ln x D loge x and is called natural logarithm
...
The
natural logarithm y D ln x and the exponential function y D ex , which is the inverse
to it, are compared in Fig
...
5
...


Problem 2
...
Show that the inverse hyperbolic cosine function accepts two
branches given by:
cosh 1
...
39)

Normally the plus sign is used leading to positive values of the function for any
x 1
...
Sketch both branches
of the function
...
x/ D ln x C

Á
p
x2 C 1 ;

1 < x < 1;

(2
...
3 Elementary Functions

93

Fig
...
5 Mutually inverse y D ex and y D ln x functions

Problem 2
...
x/ D

1 1Cx
ln
;
2 1 x

jxj < 1:

(2
...
42)
2 x 1
Using the fact that the logarithmic function is only defined for its positive
argument, explain the chosen x intervals in each case
...
x/ D

2
...
8 Trigonometric Functions
We have already defined sine and cosine functions in Sect
...
5 and considered some
of their properties
...
The two functions are plotted in Fig
...
6(a)
...
Both functions are
periodic with the period of 2
...
2
...
They both have the period of and are singular
at some periodically repeated points: tan
...


94

2 Functions

Fig
...
6 (a) Functions y D sin
...
x/ with the period of 2 ; (b) functions y D tan x
and y D cot x with the period of
...
2
...

First, we shall prove double angle identities
...
2
...
The triangle 4AOC is an isosceles triangle since it has two
its sides, OC and OA, equal (in fact, equal to 1), hence two angles †OAC D ˇ and
†OCA are also equal, and since all three angles of the triangle sum up to , we have
2ˇ C ˛ D
...
Then, draw the bisector
of the †AOC which crosses AC at the point D; then, AC D 2DC D 2 sin ˛
...
2˛/ D 2 sin ˛ cos ˛:

(2
...
32) we discussed above for
the hyperbolic functions
...
3 Elementary Functions

95

Similarly, looking again at Fig
...
7(a), we can write:
OB D cos ˛ D 1

BC D 1
D1

AC cos ˇ D 1
sin ˛
˛
sin D 1
cos ˛
2
2

sin ˛
cos
cos ˛
2
2
2 sin2

˛Á
2

˛
;
2

where we decomposed sin ˛ using (2
...
Written in a more conventional form:
2 sin2 ˛ D cos2 ˛

cos
...
44)

where we have also used Eq
...
9)
...
Formula (2
...

Next, we consider a more general formula for a sum of two angles
...
2
...
The coordinates
of points A and D obviously are: A
...
˛ C ˇ/ ; sin
...
cos ˛; sin ˛/,
and hence
AD2 D
...
˛ C ˇ/
D2

cos ˛/2 C
...
˛ C ˇ/

sin ˛/2

2 Œcos
...
˛ C ˇ/ sin ˛

after opening the squares and using Eq
...
9)
...
1 cos ˇ/ ;
2 2 Œcos
...
˛Cˇ/ sin ˛ D 2 sin
2
2
after transforming the right-hand side using the double-angle formula (2
...

Opening the brackets and replacing D ˛ C ˇ, we obtain:
cos
...
45)

Consequently, replacing ˛ with ˛ in the above formula and recalling that sine and
cosine functions are odd and even functions, respectively, we obtain:
cos
...
46)

96

2 Functions

Problem 2
...
Using one of the equations above, prove similar identities for
the sine function as well:
sin
...
47)

Problem 2
...
By adding (subtracting) Eqs
...
45)–(2
...
49)

Ã
Â
Ã
˛Cˇ
˛ ˇ
cos
I
2
2

(2
...
51)

Â
cos ˛

Ã

(2
...
23
...
˛
2

sin ˛ cos ˇ D

1
Œsin
...
˛
2

ˇ/ I

(2
...
˛ C ˇ/
2

ˇ/ I

(2
...
˛ C ˇ/ :

(2
...
˛
2

ˇ/

cos
...
˛

(2
...
This may be useful in calculating integrals as
demonstrated in later chapters
...
(2
...
46), as well
as the particular cases of these for equal angles D ˛, we obtain:

2
...
3˛/ D sin
...
2˛/ cos ˛ C cos
...
56)

cos
...
2˛ C ˛/ D cos
...
2˛/ sin ˛

2 sin2 ˛ cos ˛ D cos3 ˛

3 sin2 ˛ cos ˛:
(2
...
24
...
4˛/ D 4 sin ˛ cos3 ˛
cos
...
58)

6 sin2 ˛ cos2 ˛ C sin4 ˛I

(2
...
]
sin
...
60)

cos
...
61)

Problem 2
...
Using the following identity,
sin
...
62)

which follows from Eq
...
51), derive the formula below for the finite sum of
sine functions:
n
X

sin
...
n C 1/ 
sin
:
sin

2
2
sin 2

(2
...
62) for the sine functions in the sum and then observe that only
the first and the last terms remain, all other terms cancel out exactly
...
26
...
k/ D

1
2 sin


2

Ä Â Â
ÃÃ
1
sin  n C
2

sin


...
64)

98

2 Functions

Problem 2
...
When studying integrals containing a rational function of sine
and cosine, a special substitution replacing both of them with a tangent function
is frequently found useful
...
65)

Problem 2
...
Prove the following identity:
cot
...
x

y/ D

2 sin 2y
:
cos 2x cos 2y

(2
...
29
...
50) that when we multiply two complex
numbers, their absolute values multiply and the phases sum up
...
30
...

Problem 2
...
Correspondingly, derive the following de Moivre’s formula:

...
n / C i sin
...
3
...
These functions have the meaning inverse to that
of the direct functions
...
e
...

Because of the periodicity of the direct functions, there are different values
of y (the angles) possible that give the same value of the functions x
...
g
...
Then, for all angles given by
y D y0 C 2 n with n from all integer numbers Z the same value of the sine function
x D sin y D sin y0 is obtained
...
This difficulty is similar to the one we came across before in
Sect
...
1 when considering the inverse of the y D x2 function, and we realised there
that by restricting the values of the argument of the direct function, y, the one-to-one
correspondence can be established
...


2
...
2
...
2
...
The graphs of both
functions are shown in Fig
...
8(a)
...
2
...

Similarly, since the cosine function varies between 1 and 1 when the angle
changes between 0 and , the arccosine function and its inverse are well defined
within these intervals
...
Both functions are drawn in Fig
...
8(b)
...
The graphs of these two
functions together with the corresponding direct functions are shown in Fig
...
9
...
For instance, equation sin x D a (with 1 Ä a Ä 1) has
the general solution x D arcsin a C 2 n, where n D 0; ˙1; ˙2; : : : is from Z
...

Obviously, it follows from their definitions that

100

2 Functions

sin
...
arccos x/ D x and tan
...
67)

Other properties are all derived from their definitions, i
...
from the properties of
the corresponding direct functions
...
x/ D sin x, from where it follows that, on the one hand, x D arcsin y, but
on the other, arcsin
...
This means that the arcsine function is antisymmetric
(odd):
arcsin
...
68)

Problem 2
...
Using the properties of the trigonometric functions, prove the
following properties of the inverse trigonometric functions:
arctan
...
69)

arccot
...
70)

arccos
...
71)

arcsin x C arccos x D
arctan x C arccot x D

2
2

sin
...
arcsin x/ D

;

(2
...
73)

p
1

x2 ;

(2
...
75)

1
cos
...
76)

2
...
4
...
Several
different cases need to be considered
...
4 Limit of a Function

101

Fig
...
10 Illustration of two equivalent definitions of the limit of a function y D f
...
4
...
1 Limit of a Function at a Finite Point
We shall start by considering two definitions of a finite limit of a function at a point
1 < x < 1
...

Let f
...

Definition 2
...
of the limit of a function f
...
5
If we take a numerical sequence of numbers xn which is different from x0 and
converges to x0 , then it generates another sequence, a sequence of the values of
the function,
f
...
x2 /; f
...
xn /; : : : ;

(2
...
2
...
Then A is the limit of the function f
...
e
...
x/ D A;

x!x0

if arbitrary sequence of numbers converging at x0 generates the functional
sequence (2
...

Note that each sequence, if it has a limit, will converge to a unique limit
(Sect
...
2); however, what we are saying here is that if any sequence, converging
at x0 , results in the same limit A of the generated functional sequence, only then the
function f
...
It follows from here that the function f
...


5

Due to Heinrich Eduard Heine
...
x/ D x
...
77) would coincide with the sequence xn since f
...

Take now the parabolic function f
...
The corresponding functional
sequence becomes x2 , x2 , x2 , etc
...
2 in Sect
...
2),
0
where x0 is the limit of the sequence xn
...
Hence, the
0
function f
...

0
For our next example we consider the function f
...
1=x/ which is defined
everywhere except x D 0
...
n /,
where n D 1; 2; 3; : : :
...
xn / D sin
...
One may think then that our function f
...
However, this function in fact does not have a limit, and to prove that it is
sufficient to build at least one example of a sequence xn , also converging to zero,
which generates the corresponding functional sequence converging to a different
limit
...
Consider the second sequence, xn D
1=
...
However, generated
by it functional sequence f
...
=2 C 2 n/ converges to 1 instead
...
x/ D sin
...

Now we shall give another, equivalent, definition of the limit of a function
...
2
...
x/ at x0
...
x/
is A when x ! x0 , if for any > 0 one can indicate a vicinity of x0 such that for
any x from that vicinity f
...
Rephrased in a
more formal way: for any > 0 one can find such ı > 0, so that for any x satisfying
jx x0 j < ı it follows jf
...
This is still exactly the same formulation:
jx x0 j < ı means that x belongs to some ı-vicinity of x0 , i
...
x0 ı < x < x0 C ı,
and jf
...
x/ is in the vicinity of A by no more than
as illustrated in Fig
...
10(b)
...
x/ to A, we shall always find the corresponding values of x close to x0 for
which this condition will be satisfied
...

Note that ı D ı
...
e
...

As we said before, the two definitions are equivalent
...


6

Due to Augustin-Louis Cauchy
...


2
...
x/ D x at x0
...
x/ x0 j D jx x0 j < ı D , i
...
f
...

Consider now the limit of f
...
Let us prove that the limit exists and
it is equal to x2
...
Then, for all x in
0
that vicinity, we can estimate:
ˇ
ˇ ˇ
ˇ
ˇf
...
x x0 / C2x0 j Ä ı
...
ıC2 jx0 j/ :
0
0
If we now take ı as a solution of the equation D ı
...
g
...
So, f
...

0
Next, let us consider f
...
We would like to prove that it tends to sin x0
when x ! x0
...
We
can write:
x C x0
x x0
cos
;
sin x sin x0 D 2 sin
2
2
so that

ˇ
ˇ
ˇ
ˇ x x ˇ ˇ
ˇ ˇ
ˇ
ˇ
xCx0 ˇ
0ˇ ˇ
ˇ
ˇ < 2 ˇ x x0 ˇ ˇcos xCx0 ˇ < jx x0 j < ı;
jsin x sin x0 j D 2 ˇsin
ˇ
ˇ ˇcos
ˇ ˇ
ˇ
ˇ
2
2
2
2 ˇ

where we used the first part of the inequality (1
...
The result we have just obtained clearly
shows that for any there exists such ı D , so that from jx x0 j < ı immediately
follows jsin x sin x0 j < , which is exactly what was required
...
e
...
78)
1j < , or that
(2
...
Then, as
the logarithm is a monotonically increasing function, the inequality (2
...
1
/ < x < loga
...
1
/, is negative (since 0 < 1
< 1), while the right one,
loga
...
We see that if we take ı as the smallest absolute
value of the two boundaries,
ı D min fjloga
...
1 C /jg ;

then we obtain ı < x < ı, i
...
ı D ı
...

This proves the limit (2
...


104

2 Functions

Fig
...
11 Heaviside (a) and sign (b) functions experience a jump at x D 0
...
The
arrows on some lines indicate that the lines just approach the point x D 0, but never reach it
...
(2
...
2)

Problem 2
...
Prove, using Eq
...
78), that limx!x0 ax D ax0
...
34
...


p
p
x D x0

2
...
1
...
For instance, the Heaviside
function H
...
(2
...
2
...
Similarly, the sign function, Eq
...
2), experiences a jump as well, see
Fig
...
11(b)
...
Both definitions of the limit are easily extended to this
0
case
...
x/ at x0 is equal to A (written: limx!x0 0 f
...
x/ Aj <
...
35
...
x/DA
...
2
...
It is clear that if the limit
of y D f
...
Inversely, if both the left
and right limits exist and both are equal to A, then there exists the limit of f
...
If, however, at point x the left and right limits exist, but are different,

2
...
2
...
x/ at x D x0 using the language of
intervals

then at this point the limit of the function f
...
The functions H
...
x/ shown in Fig
...
11 do not have a limit at x D 0, although both their left
and right limits exist
...
This function is only defined
for x 0 and hence we can only consider the right limit
...
It is clear that this inequality is indeed fulfilled if ı D 2
...
36
...

Problem 2
...
Prove that the left and right limits of the sign function around
x D 0 are 1 and 1, respectively
...
38
...


2
...
2 Main Theorems
Similarly to the numerical sequences, one can operate almost algebraically with
the limits of functions
...
2
...
Therefore, similar to
sequences, analogous theorems exist for the limits of the functions
...


106

2 Functions

Theorem 2
...
If the function y D f
...


It is proven by contradiction
...
10
...
x/ and g
...
x/ C g
...
x/g
...
x/=g
...


These are easily proven using the language of intervals
...
11
...
x/ D A and f
...


F (or f
...


Theorem 2
...
If in some vicinity of x0 the two functions f
...
x/, and
both have definite limits at x0 , i
...
limx!x0 f
...
x/ D G,
then F Ä G
...
x/ D f
...
x/ Ä 0 and
limx!x0 h
...

It then follows from the last two theorems that if f
...
x/ Ä f
...
x/, and both functions
...
x/ tend to the same
limit A at x ! x0 , so is the function f
...

All these theorems can be proven similarly to those for numerical sequences
...
Since the two functions have
definite limits F and G, then for any positive 1 and 2 one can always find such
positive ı1 and ı2 , so that from jx x0 j < ı1 follows jf
...
x/ Gj < 2
...
We can then write:
jf
...
x/

FGj D j
...
x/
<

F/
...
x/
1 2

C jFj

G/ C F
...
x/
2

C jGj

G/ C G
...
x/

F/j

1;

which gives an expression for the in the right-hand side above, as required
...
4 Limit of a Function

107

Problem 2
...
Prove all the above theorems using the “

ı” language
...


Theorem 2
...
Consider a function f
...
e
...
x/j Ä M
...
x/ D 0, then limx!x0 Œf
...
x/ D
0 as well
...
x/ may not even have a definite limit at x0 :

Proof
...
x/ has the zero limit, then for any x within jx
jg
...
x/j < 0
...
x/g
...
e
...
x/g
...
x/j < M 0 ;

D M 0
...
E
...


As an example, consider limx!0 Œx sin
...
The sine function is not well
defined at x D 0 (in fact, it does not have a definite limit there, recall a discussion
on sequences in Sect
...
2); however, it is bounded from below and above, i
...
it is
limited: jsin
...
Therefore, since the limit of x at zero is zero, so is the limit
of its product with the sine function: limx!0 Œx sin
...

At variance with the sequences, two more operations exist with functions: the
composition and inverse; hence, two more limit theorems
...
14 (Limit of Composition)
...
t/
and t D g
...
x/ D f
...
x//
...
x/ D t0 and limt!t0 f
...
x/ D F
...
since g
...
x/ t0 j < 1
...
t/ also has a definite limit at t0 ,
i
...
jt t0 j < ı1 implies jf
...
However, since t D g
...
x/ t0 j < ı1
...
x/

t0 j D jt

proving the theorem, Q
...
D
...
t/

Fj D jf
...
x//

Fj < ;

108

2 Functions

This result can be used to find limits of complex functions by considering them
as compositions of simpler ones, for instance,
Â
Ã2
lim sin2 x D lim sin x D sin2 x0 :
x!x0

x!x0

Theorem 2
...
Let f 1
...
x/; the latter has a limit at x0 equal to F, i
...
limx!x0 f
...

Then, limx!F f 1
...

Proof
...
x/ is the direct function, then x D f 1
...
Since the direct
function has the limit, then for each we can find ı such that jx x0 j < ı implies
jf
...
We need to prove that for any 1 there exists ı1 such that
for any y where jy Fj < ı1 we may write jx x0 j < 1
...
E
...


2
...
3 Continuous Functions
Intuitively, it seems that if one can draw a function y D f
...
A more
rigorous consideration, however, requires some care
...
x/ within an interval a <
x < b is that its limit at any point x0 within this interval coincides with the value of
the function there:
lim f
...
x0 / :

x!x0

(2
...
x/ to have a definite limit for any x within
some interval to be continuous there
...
(2
...
We shall prove this by contradiction: assume the
opposite that the limit of f
...
x0 /
...
x/ D A > f
...
This means that for any > 0 there exists ı > 0 such
that jx x0 j < ı implies jf
...
Let us choose the positive as follows:
D 1
...
x0 //
...
x/

Aj <

D

1

...
x0 // ;

2
...
x0 /

Aj D A

f
...
A
2

f
...
e
...
x0 / is
incorrect
...
x0 / which also is easily
shown to be wrong
...
x0 /, Q
...
D
...
x/ is continuous within some interval of x if it has
well-defined limits at any point x belonging to this interval
...

Yet another feeling for the continuity of a function one can get if we rewrite
Eq
...
80) in a slightly different form
...
x/ D
f
...
(2
...
x0 C x/

x!0

f
...
x0 / D 0:
x!0

(2
...
x0 / D f
...
x0 / is the corresponding change of the function,
see Fig
...
13
...
x/ which
corresponds to the coordinates x0 C x and f
...
x0 /; as x becomes smaller, so is
y D f
...
x0 /
...
81) exists for any point x (within
some interval), this kind of continuity exists at each such point; as we change x
continuously between two values x1 and x2 > x1 , the function y D f
...
x1 / and f
...

Since a continuous function has well-defined limits for all values of x within
some interval, the function cannot jump there, i
...
it cannot be discontinuous
...
If this condition is not satisfied at
some point x0 , the function f
...
There are two types
Fig
...
13 Change of a
function y tends to zero as
change x of x tends to zero

110

2 Functions

Fig
...
14 The hyperbolic
function y D 1=x

of discontinuities
...
x/ has well-defined left and right limits at point
x0 , which are not equal to each other,
lim f
...
x0 C 0/ ¤ lim f
...
x0

x!xC
0

x!x0

0/;

(2
...
All other discontinuities
are of the second kind
...
x/ and sgn
...
On the other
hand, the function y D 1=x has infinite limits on the left (equal to 1) and on
the right (equal to C1) of the point x D 0, and hence has the discontinuity of the
second kind there, see Fig
...
14
...
Since this
property is based on the functions having a well defined limit at each point, the limit
theorems of Sect
...
4
...
x/ and g
...
x/ ˙ g
...
x/g
...
x/=g
...
x/ D 0 in the latter case)
...
g
...
Moreover, if the function f
...

Let us now look at the elementary functions and establish their continuity
...
Consider first a constant function y D C
...
Then, let us consider an integer power function
y D xn
...
x C x/n xn D

n
XÂnÃ
iD0

i


...
x/i xn i ;

(2
...
4 Limit of a Function

111

1 for any n (see Sect
...
10); this means that xn cancels out exactly
...

This means that the integer power function is continuous
...
e
...
Since arithmetic operations involving
continuous functions result in a continuous function, any polynomial is a continuous
function
...
Consider now y D x˛ , a general power function, where
˛ is an arbitrary real number
...
2
...
We first
consider ˛ D 1=n with n being positive integer
...
The function y D xn=m (with ˛ D n=m being
a positive rational number) is a composition of two functions just considered, y D
n
x1=m , and hence is also a continuous function
...
Following the same method as in Sect
...
3, we consider two numerical
sequences ˛i0 and ˛i00 (i D 1; 2; 3; : : :) of rational numbers bracketing the irrational ˛
for each i (i
...
˛i0 < ˛ < ˛i00 ) and converging to ˛ from both sides
...
Consider now the x ! x0 limit
...
e
...
Taking now i ! 1 and using the fact
that both sequences on the left and right converge to the same limit x˛ , we arrive at
0
the conclusion that limx!x0 x˛ D x˛ , i
...
any positive real power function is indeed
0
continuous
...
x C x/
x
D
x!0 x
...
We finally conclude that the power function
with any real power ˛ is continuous
...
We have already considered in Sect
...
4
...
1 the limit of
the sine function, y D sin x, and proved that its limit at x ! x0 is equal to sin x0 , i
...

to the value of the sine function at this point
...
The cosine function y D cos x can be composed out of the sine function,
e
...
y D sin
...
The tangent and cotangent are
obtained via division of two continuous functions, sine and cosine, and hence are
also continuous
...
These are also continuous since they are
inverse of the corresponding trigonometric functions which are continuous
...
For any positive a > 0 we can write
Â
lim axCx

x!0

ax D lim ax ax
x!0

1 D ax lim ax
x!0

1 D ax

lim ax

x!0

Ã
1 D 0;

since, as we have seen above in Eq
...
78), limx!0 ax D 1
...

Logarithmic Function
...

Hyperbolic Functions
...


Problem 2
...
Prove that the sine and cosine functions are continuous also by
showing that their change y ! 0 as the change of the variable x ! 0
...
41
...

Problem 2
...
Prove by direct calculation of y that the hyperbolic sine,
cosine and tangent functions are continuous
...


2
...
4 Several Famous Theorems Related to Continuous
Functions
There are a number of (actually quite famous) theorems which clarify the meaning
of continuity of functions
...


Theorem 2
...
If a function f
...
x/ has
the same sign as f
...


2
...
If the function is continuous, then for any x from a vicinity of x0 we have
limx!x0 f
...
x0 /
...
x0 / < f
...
x0 /
...
x/ f
...
x0 / > 0
...
x0 / one obtains
0 < f
...
x0 /, i
...
within (possibly) small vicinity jx x0 j < ı of x0 we have
f
...
If, however, f
...
x0 / > 0 yielding
2f
...
x/ < 0, i
...
within some vicinity of x0 the function f
...

Q
...
D
...
17 (The First Theorem of Bolzano-Cauchy—Existence of a
Root)
...
x/ has different signs at the boundaries
c and d of the interval c Ä x Ä d, then there exists a point x0 belonging to the
same interval where f
...


Proof
...
c/ < 0 and f
...
2
...
The idea
of this “constructive” proof is to build a sequence of smaller and smaller intervals
around x0 such that it converges to the point x0 itself after an infinite number of steps
(if not earlier)
...
c C d/=2
...
2
...
We start with the
original interval X0 between points c and d
...
e
...
c C d/=2 and d1 D d
...
c1 C d1 /=2
...
x/
...
x1 / D 0, then x0 D x1 and we stop here
...
x1 / < 0, then we accept x1 as the new
left boundary c1 of the interval; if f
...
This way we arrive at a new interval c1 Ä x Ä d1 which is two times
shorter, i
...
its length is l1 D
...
Then, we repeat the previous step: choose
the middle point x2 D
...
The length l2 D l1 =2 D
...
Repeating this process, we construct a (possibly)
infinite sequence of intervals cn Ä x Ä dn of reducing lengths ln D
...
It
is essential that, by construction, all of them have their boundaries of different signs
and their length becomes smaller and smaller tending to zero as 1=2n
...
The fact that such a point exists follows from the previous theorem of
conservation of sign of a function: if we assume that f
...
x0 / D 0
...
E
...


Theorem 2
...
Suppose a continuous function f
...
c/ D C and f
...
If A is
any number between C and D, then there exists a point a within the interval,
c Ä a Ä d, where f
...


Problem 2
...
Prove this theorem by defining a new function g
...
x/
and then use the “existence of root” theorem
...
x/ takes on all intermediate
values between C and D within the interval
...


Theorem 2
...
Consider a function f
...
e
...
x1 / < f
...
If the function is limited from above within the interval,
f
...


2
...
2
...
19: the
monotonically increasing
function f
...
(see Fig
...
16) Let us choose some positive
...
x/
increases monotonically, one can always find such value x1 that f
...

This value of x1 will be by some ı on the left from the boundary value b, i
...

x1 D b ı
...
e
...
x/ > f
...
However, on the other hand, f
...
x/ < M C

H)

jf
...
e
...
Q
...
D
...
44
...
e
...
x/ > M > 0), as well as for a
monotonically decreasing function at either of its boundaries
...
4
...
If for any > 0 there exists
ı > 0 such that for any x within jx x0 j < ı one has9 f
...
x/ tends to C1 as x ! x0
...
x/ larger than any given (however large), then
the function f
...


9

Note that here is not longer assumed “small”, it is “large”
...
45
...


We can also consider behaviour of a function at both infinities, i
...
when x0
is either C1 or 1
...
The function f
...
x/ Fj <
...
46
...
x/ to have: (i) a finite
limit F when x ! 1, (ii) an infinite limit F D C1 (or 1) when x ! C1
(or 1)
...
47
...
19 of the previous subsection
for the limits at C1 and 1 for both monotonically increasing and decreasing functions
...
x/ D 1=x2 at x ! 0
...
x/ D 1=x2 > 1=ı 2 D
0j
, i
...
one should take ı D 1=
in this case
...

Now, if f
...
Indeed, for any
p
x > ı we have f
...
e
...

Similarly one can formulate infinite limits for a function at x ! ˙1
...
48
...

Problem 2
...
Prove using the “

ı” language that limx!C1 e

x2

=x D 0
...
84)

This is a generalisation to any real x of the special cases we considered above in
Sect
...
3
...
To prove this result,
we note that any positive x can be placed between two consecutive integers: n Ä

10

ı is assumed to be indefinitely “large” here
...
4 Limit of a Function

117

x Ä n C 1
...
However, from
Eq
...
22) it follows that both limits on the left and right of the last inequality tend
to the same limit e:
1
1C nC1

nC1

1
1C nC1

Ã
à Â
Ã
Â
Ã
Â
Â
1 n
1
1 nC1
1 n
! 1C
1C
! 1C
! e and
1C
! e;
nC1
n
n
n

so that, the function 1 C

1 x
x

will tend to the same limit
...
E
...


Problem 2
...
Prove that
lim

x!C1

1C

y Áx
D ey ;
x

(2
...


2
...
6 Dealing with Uncertainties
The case for the number e considered above is an example of an uncertainty
...
Several other types of uncertainties
exist and often happen in applications, so we shall consider them here
...
86)

lim

There is an uncertainty: indeed, both sin x and x tend to zero as x tends to zero,
so that we encounter the case of 0=0 here
...
The starting point is the inequality (1
...
e
...
Q
...
D
...

Uncertainty 1=1:
2x2 1
2 1=x2
2 0
D lim
D 2:
D
x!C1 x2 C 2x C 1
x!C1 1 C 2=x C 1=x2
1C0C0
lim

Uncertainty 1
lim

x!C1

1:

Á
p
p
x2 1
xC1 D lim

x!C1

"

Á px2
p
1
xC1 p
x2

p
x2

#
p
1C xC1
p
1C xC1

x2 1

...
1 C x/
ˇ
D lim
ln
...
1 C x/1=x D ˇ
ˇ y D 1=x ˇ
x!0
x!0 x
x!0
x
Ã
Ã
Â
Â
Ä
1 y
1 y
D ln e D 1:
D ln lim 1 C
D lim ln 1 C
y!1
y!1
y
y
lim

2
...

Uncertainty 0 1:
lim x cot x D lim x

x!0

x!0

cos x
D
sin x

Â

x
x!0 sin x

ÃÂ

lim

Ã
à Â
sin x
lim cos x D lim
x!0
x!0 x

1

1 D 1:

Problem 2
...
Prove the following limits:
Ã
Â
xC3
3
5x 5 x
D C1I
D 1I
lim
x!C1 5x C 5 x
x!1C0 x
1 x2 1
Ã
Â
Á
p
p
xC3
3
lim
D 1I lim
x
x C 2 D 0I
x!1 0 x
x!C1
1 x2 1
lim

lim

x!0

1

cos x
1
D I
x2
2

lim

x!C1

p
x
x D 1:

[Hint: in the latter case transform first the expression using an exponential
function
...
1 Definition of the Derivative
Very often we would like to know an instantaneous rate of change of something
...
t/
...
0/, then at time t it arrives
at the point with the coordinate x
...
If the particle moved with a constant velocity
vc , then its coordinate would change according to x
...
This means
that in this case the rate of change of x
...
e
...
x x0 / =t
...
e
...
x x0 / =t
...

For instance, we can split the whole journey time t into N equal slices of length t,
i
...
t0 D 0, t1 D t, t2 D 2t, etc
...
e
...
tiC1 / x
...
ti C t/
xiC1 xi
D
D
t
t
t

x
...
1)

for each time interval t D tiC1 ti
...
This simple minded exercise
brings us to a method which would be able to determine the instantaneous velocity
of the particle at any time along the journey: one has to tend the number of slices
© Springer Science+Business Media, LLC 2016
L
...
1007/978-1-4939-2785-2_3

123

124

3 Derivatives

N to infinity (N ! 1, which is equivalent to taking the time length of each slice
t ! 0)
...
t/ at some time t along the journey, we need to consider
the positions of the particle x
...
t C t) at the times t and t C t, and then
calculate
v
...
t C t/
t

x
...
(3
...
This expression becomes more and more precise as t gets
smaller; it becomes exact in the limit of t ! 0
...
Consider
electrons flowing in a thin wire with cross section S
...
This way we can approximately calculate the
electric current passing through the wire at time ti as Ji ' eNi =t (e is the electron
charge)
...
If N
...
t C t/ N
...
t/ '

N
...
t/

;

which becomes more and more exact as the time slices t get smaller and smaller;
in the limit of t ! 0, this formula becomes exact
...
xi C x/
mi
D
x
x

m
...
x/ is the total mass of the rod between points x D 0 and x > 0
...
Exactly in the same way one can define the linear charge
density of a charged rod via the limit of its total charge within a thin slice divided
by its length
...
x/ and we are interested in its rate of change at
the point x, we calculate the limit
y0
...
x C x/
D lim
x!0
x
x

y
...
2)

3
...
x/ at point x
...
x C x/ y
...
x/ over the change x of its argument from x to x Cx
...

Example 3
...
I Calculate the derivative of the following functions: y D c (a
constant), y D x and y D x2
...
In the first case y D c c D 0 and hence the derivative is zero, i
...


...
In the second case y D y
...
x/ D
...
x/0 D 1 for any value of the x
...
x C x/

y
...
x C x/2

x2 D 2xx C
...
x/2
D lim
D lim
...
3)

Note that the change of the function due to the change x of its argument in this
case,
y D 2xx C
...
x/x C ˛
...
4)

is composed of a linear part containing the derivative and an additional term which
tends to zero faster than linearly (in fact, as the square of x in this case) since
˛
...

Problem 3
...
Show using the binomial expansion (see Sect
...
10) that

...
5)

1
...

Consider a function y
...
3
...
Let us try to find an equation
y D a C kx for the tangent line (shown in dashed green) at the point A with
coordinates x and y
...
Our interest here is in obtaining the slope of the curve
y D y
...
The constant a can then be easily determined using the known coordinates of
the point A
...
This line may have a very different direction to the one we are seeking;
however, its slope given by the tangent of AB1 to the x axis may already serve as
the first approximation to the contstant k
...
3
...
x/ at point A

of the function which is somewhere between the points B1 and A; it is now closer to
the point of interest
...
g
...
This process
can be continued: we take a point B3 on the curve lying between A and B2 , then a
point B4 between A and B3 , etc
...
In the limit when the point BN tends to the point A, our procedure would give
the correct tangent line at the point A
...
x/, see Fig
...
1(b), and moving it
towards A
...
x C x/
x

y
...
So, in the limit, we shall get the exact value of the
slope k of the tangent line at point A
...
x/ at point x, i
...
k D y0
...

Similarly to the right and left limits defined in the previous Chapter, we can also
define left and right derivatives, denoted y0
...
x 0/ and y0
...
x C 0/,
C
respectively
...
The
left derivative is defined only via points on the left of x as follows:
y0
...
x/

y
...
6)

3
...
This expression can be written in a form very similar to that of the
right derivative in Eq
...
2) by assuming that x < 0 and tends to zero always being
negative (this is denoted as x ! 0 0 or simply x ! 0):
y0
...
x/

jxj!0

y
...
x C x/
D lim
x! 0
x
jxj

y
...
7)

If the derivative of y
...

Problem 3
...
Consider a line tangent to a curve of the function y D y
...
x x0 / C y0 , with k D
y0
...
x0 /
...
x0 ; y0 / is given by the equation
y D k?
...
x0 /
...
2 Main Theorems
The notion of the derivative of a function is closely related to its continuity
...
x/ at
point x due to change x of its argument in a bit more detail
...
x/
exists (in other words, if the limit (3
...
x/x C ˛x;

(3
...
(3
...
It is essential that ˛ tends
to zero as x ! 0
...
x/ C lim ˛ D y0
...
8) states, quite importantly, that for any x the change of the
function is in the first approximation given by the product of its derivative and x;
the correction term, equal to ˛x, is of a higher order with respect to x, and for
small x may be neglected
...
x/x
...
x/ and is denoted dy
...
x/0 x D x and, since y D x, we have
dx D x as well), we can write:
dy D y0
...
9)

2 0

For instance, if y D x2 , then dy D x dx D 2xdx or simply d x2 D 2xdx
...
x/ D

dy
:
dx

(3
...
x/ as using differentials already implies performing the limit x ! 0
...

Example 3
...
I Derive an explicit expression for the function ˛
...

Solution
...
(2
...
x C x/n

xn D

n
XÂnÃ
kD0

k

xn

k


...
x/k

 Ã
n
XÂnÃ
n n 1
D
x xC
xn k
...
x/
k
kD2

which is of the required form (3
...
xn /0 D nxn 1 , see Eq
...
6)
...
x/ we seek for; it is
clearly seen that it tends to zero as x ! 0 (the sum starts from k D 2)
...
1
...
x/ has a derivative at point x, it is
continuous there
...
Indeed, if y0
...
8) for y
...
2
...
3
...
E
...


3
...
Indeed, consider the function y D jxj
...
At
the same time, the derivative at x D 0 does not exist
...
x/0 D 1
...
x/0 D

Œ
...
x/

D 1:

We see that the derivatives from left and right around x D 0 are different, and hence
the derivative at x D 0 does not exist
...
2 (Sum Rule)
...
x/ D u
...
x/, then y0
...
x/ ˙
v 0
...
e
...
x/ ˙ v
...
x/ ˙ v 0
...
11)

Proof
...
x/ which can be written as a sum of two functions
u
...
x/
...
2), we can write:
y
1
D
Œu
...
x C x/ u
...
x/
x
x
u
...
x/
v
...
x/
u
v
D
C
D
C
:
x
x
x
x
After taking the limit asx ! 0, we obtain the required result
...
u v/0 D u0 v 0
...
E
...

Theorem 3
...
If y
...
x/v
...
x/ D Œu
...
x/0 D u0
...
x/ C u
...
x/:

(3
...
We need to calculate the change of the function y
...
x C x/ v
...
x C x/

u
...
x/

u
...
x C x/ C u
...
x C x/

D
...
x C x/ C u
...
x/

130

3 Derivatives

Dividing y by x and taking the limit, we obtain the required result:
Ä
Ä
v
u
y
D lim
v
...
x/
x!0 x
x!0 x
x!0
x
lim

D lim

x!0

u
v
lim v
...
x/ lim
D u0
...
x/ C u
...
x/
x!0 x
x x!0

Q
...
D
...
3
...
e
...
x/ C ˇv
...
x/ C ˇv 0
...
13)

Problem 3
...
Generalise the product rule for more than two functions:
Œu1
...
x/ : : : un
...
14)

jD1;j¤i

where the product indicated by the symbol
between 1 and n except for j D i
...
4 (Quotient Rule)
...
x/ D u
...
x/ and v
...
x/ 0
0
(3
...
x/ D
v
...
Here we write
y D
D

u
...
x C x/

u
...
x C x/ v
...
x/ v
...
x/
v
...
x/

v
...
xCx/ u
...
x/ Œv
...
x C x/ v
...
x/

D

v
...
x/ v
;
v
...
x/

which, after dividing by x and taking the limit, gives:
v
...
x/ limx!0
y
x
D
x!0 x
limx!0 Œv
...
x/
lim

v
x

D

vu0

uv 0
v2

; Q
...
D
...
2 Main Theorems

131

Problem 3
...
Show that the formula (3
...
e
...
xn /0 D nxn 1
for any n from Z (i
...
n D 0; ˙1; ˙2; ˙3; : : :)
...
5 (Chain Rule)
...
v/ and
v D v
...
e
...
v
...
Then, its derivative is obtained as follows:
Œu
...
x//0 D u0 v 0 ;
v

(3
...
v/ with respect to v
v
(not x)
...
Again, we need to calculate the change y of the function y
...
v
...
v
...
x C x/ D v C v
...
v C v/ D v u C
u
...
v/ when v changes from v to
v C v
...
v
...
v
...
v C v/
y
D
D
x
x
x
v u v
u
...
v/ v
D
;
D
v
x
v x

u
...
(3
...
Q
...
D
...
6
...
x/ 1 D u 2 u0
...
15)? Then, using this result and the
product rule, prove the quotient rule in a general case
...
7
...
x/=dx,
P
where x
...
x/ is its potential energy, p D mP its
x
momentum, and dots above x and p denote the time derivative
...
x/ is conserved in time
...
) [Hint: assume that both p and x are some
functions of time which would make H a composite function, so that you may
use the chain rule
...
6 (Inverse function rule)
...
x/ is the inverse of x D x
...
x/0 D

x0

1
:

...
17)

Proof
...
y/ is continuous, x ! 0 corresponds to y ! 0
as well, so that the limit in the denominator of the above formula can be replaced
with y ! 0, which would give the derivative x0
...
y/ with respect to y,
y
which proves the theorem
...
E
...

As an example, let us derive a formula for the derivative of the square root
p
function y D x
...
x/ is the inverse of the function x
...

Therefore, x0 D 2y
...
5) for the derivative of the integer power
y
p
function (n D 2 in our case)
...
2y/ D 1= 2 x
...
(3
...

Problem 3
...
Show, using a combination of various rules for differentiation
discussed above, that the formula (3
...
e
...
x˛ /0 D ˛x˛ 1 , where
˛ D p=q is a rational number (p and q are positive and/or negative integers,
q ¤ 0)
...
Prove first, using the inverse function rule, that
0
x1=q D 1 x1=q 1
...
(3
...
]

3
...
But first we need to derive formulae
for the derivatives of all elementary functions
...
y D x˛ (˛ is a real number)
...
3 Derivatives of Elementary Functions

133

We have already considered positive and negative values of ˛ and proved that
in these cases
...
Using the definition of the inverse function and the
inverse function rule, we also discussed that this formula is valid for root powers
1=q (q ¤ 0) as well, and then, employing the chain rule, we extended this result for
any rational power ˛ D p=q
...
2
...
4
...
So, for any real ˛,

...
18)

Problem 3
...
Prove that
n2

n 1

n
X ÂnÃ
:
D
k
k
kD1

[Hint: differentiate the binomial expansion of
...
]
Logarithmic Functions
...

We write:
1
1
x C x
Œloga
...
loga x/0 D lim

x!0

D 1=x, which tends to infinity as x ! 0, we can rewrite the limit as
"
Ã
à #
Â
Â
1=x
1=x
0

...
(2
...
Therefore,

...
19)

In the last step we made use of Eq
...
38) to express loga e via the logarithm with
respect to the base e, i
...
ln e D loga e ln a, which immediately gives the required
relationship due to the fact that ln e D 1
...
ln x/0 D

1
:
x

(3
...
y D ax and y D ex
...
Therefore, using
Eq
...
17), we immediately have:

...
y ln a/

...
21)

In particular, if a D e, we obtain a very important result:

...
22)

i
...
the derivative of the ex function is equal to itself
...

Problem 3
...
Prove formula (3
...
x / D e˛ ln x :
This is another way of proving the formula for the derivative of the power
function with arbitrary real power ˛
...
Let us first consider the sine function:
Â
Ã
x
x
y D sin
...
(2
...
Therefore,
Ã
Ä Â
x sin
...
sin x/0 D lim
x!0 x
x!0
2
x=2
Ä
à Ä
Ä
Â
x
sin
...
The last limit we have already seen before in Eq
...
86), and it is
equal to 1
...
sin x/0 D cos x:

(3
...
Indeed,
cos x D sin
...
e
...
cos v/
...
cos x/0 D
...
24)

3
...
11
...
51)
...
12
...
tan x/0 D

1
cos2 x

(3
...
26)

and

...

Inverse Trigonometric Functions
...
It is the
inverse function to x D sin y
...
(3
...
arcsin x/0 D

1
1
1
1
1
D
Dp
D
D
;
x0
cos y
cos
...
sin y/0
1 x2
y
y

(3
...
(2
...


Problem 3
...
Prove that

...
arctan x/0 D

p

1
1

x2

;

1
:
1 C x2

(3
...
29)

You may need identities (2
...
76) here
...
These functions are composed of exponential functions,
and one can use the rules of differentiation to calculate the required derivatives
...
x/:
0

Œsinh
...
e
2

0

e x/

D

1 x 0

...
e x /0 D

1 x

...
x/:
2
(3
...
14
...
x/0 D sinh
...
31)

1
;
cosh2
...
x/0 D

(3
...
x/

Œcoth
...
33)

Problem 3
...
Prove the following formulae for the inverse hyperbolic functions:
0

sinh 1
...
x/ D p

1
x2

C1

1
x2

1

0

;
;
1

0

tanh 1
...
x/ D

1

x2

:

By comparing these equations with the corresponding formulae for the trigonometric functions, one can appreciate their similarity
...


3
...
9) we can derive a number of very useful approximate representations of functions
...

As an example, let us show that for small x
sin x ' x:
Indeed, consider the change of the function y D sin x when x changes by x:
dy
x D
...
x C x/ ' sin x C
...
x C x/
)

sin x '

Taking x D 0 in the last formula, we obtain sin x ' x as required
...
34)

3
...
16
...
35)

x;

(3
...
37)

ex ' 1 C x;

(3
...
1 C x/ ' x:

(3
...
34)–(3
...


3
...
x/ is not the only one possible
...
x/ can also be specified parametrically, e
...

x D x
...
t/

(3
...
If we are interested in
the derivative y0
...
t C t/
D lim
y
...
t C t/
0

D

y
...
t/

y
...
t/
t
limt!0 x
...
t/
t

limt!0

D

y0
...
t/

y
...
t/
t
x
...
t/
t

(3
...
t C t/ y
...
t C t/ x
...
At the second step we changed the limit from x ! 0
to t ! 0 which is possible to do because we assume that the functions y
...
t/ are continuous (recall one of the definitions of the continuity of a function!)
...


Problem 3
...
Prove the same result (3
...
x/ of the function x D x
...
e
...
t/ D
y
...
x// is a composite function
...
16) and Eq
...
17)
for the derivative of an inverse function
...
3
...
t/ D et cos t and y
...
Determine the
slope of the tangent line to the trajectory as a function of time
...
The slope to the trajectory will be given by the derivative y0
...
Therefore,
we obtain:
y0
...
t/
D t
D
0
...
18
...
x/ of the function specified parametrically as y D t2 e 2t and x D 1 C t2 e 2t
...
1 t/=
...
]
A function y D y
...
x; y/ D
0 that cannot be solved with respect to y (and hence the direct calculation of the
derivative is impossible)
...
x/ can be performed
even in this case (although it is not guaranteed that the derivative can be obtained in
a closed form)
...
4
...
x/ be specified via the equation x2 Cy2 D R2
(a circle of radius R and centred at the origin)
...
x/
...
We differentiate both sides of the equation with respect to x, treating
the y2 D
...
x//2 as a composite function and hence using the chain rule
...
x/, i
...
we obtain an equation
2x C 2yy0
...
x/ D x=y
...
It
is easy to see that this is the correct result
...
x/ explicitly as y
...
This expression can

3
...
2x/ D

x
p
D
2
R
x2

x
p
˙ R2

x2

D

x
;
y

i
...
the same result
...
19
...
x/ of the function specified via equation
y4 C 2xy2 C e x y D 0
...
e x 2y/ = 4y3 C 4xy C e x
...
e
...
x/ D u
...
x/
...

For instance, let us calculate the derivative of y D xx
...
Differentiating both sides
(remember to use the chain rule when differentiating the left-hand side as y D y
...
xx /0 D y
...
1 C ln x/ :

Problem 3
...
Derive the general formula
0

u
...
x/ D v 0 uv ln u C u0 uv 1 v
for the derivative of the function y D u
...
x/
...
21
...
x/ D
u
...
x/ as an exponential
...
22
...
x/ of the functions specified in the
following way: (a) y D ln t, x D e 2t sinh t; (b) y 3 C 2e 2xy D 1; (c)
q
p
p
3 3x2
y D 1 C 2x2
;
...
[Answers: (a) e2t = Œt cosh
...
t/;
(b) y= x C
...
2g C 1/ , and
(d) A=B, where B D 1 8yg x C y, A D
2
gDy
x
...
6 Higher Order Derivatives
The derivative y0
...
y0 /0 (“double prime”)
...
Obviously, this
process can be continued and one can talk about derivatives of the n-th order,
denoted y
...
x/ (n is enclosed in brackets to distinguish it from the n-th power)
...
nC1/ D y
...
Similar notations exist for the higher order derivatives
based on the symbols of differentials:
y00 D

d2 y
d3 y
dn y
; y
...
n/ D n :
dx2
dx
dx

Higher order derivatives are frequently used in applications
...
In mechanics acceleration a
...
t/ is the rate of change of
the velocity; at the same time, since the velocity v
...
t/ relates to the rate of
change of the position, the acceleration can also be considered as the second order
derivative of the position: a
...
t/ D
...
t//0 D x00
...

Another important application of first and second order derivatives is in analysing
functions
...
3
...

Example 3
...
I Calculate the 5-th order derivative of y D cos x
...

Solution
...
cos x/0 D sin x;
...
sin x/0 D cos x;
...
cos x/0 D sin x, then
...
4/ D
...
cos x/
...
cos x/0 D sin x
...
A general
formula can be derived if we notice that
...
x C =2/,

...
x C 2 =2/,
...
3/ D sin x D cos
...

We observe that each time a phase of =2 is added to the cosine function
...
cos x/
...
42)

Indeed, the formula works for the cases of n D 1; 2; 3
...
Let us check if it would work for the next value n C 1
...
cos x/
...
cos x/
...
n C 1/ ;
D sin x C n D cos x C n C
2
2
2
2
which is exactly the desired result, i
...
(3
...
The formula is
proven
...
6 Higher Order Derivatives

141

Problem 3
...
Prove the following formulae:

...
n/ D sin x C

2

Á
n ;

(3
...
e˛x /
...
44)

and, assuming a general real ˛,

...
n/ D ˛
...
˛

n C 1/ x˛ n :


...
45)

If ˛ D m is a positive integer, then, in particular,

...
n/ D

mŠ

...
xm /
...
24
...
1/

n


...
46)

Very often it is necessary to calculate the n-th derivative of a product of two
functions
...
This result is especially useful if one of the functions in
the product is a finite order polynomial as will become apparent later on
...
x/ D u
...
x/
...
uv/
n 0
...
n 2/
n
...
n k/
uv
u v
u v
D
...
n/ D uv
...
n/
C Cu v D
(3
...
k/ v
...
nnŠ k/Š are the binomial coefficients which we met in Sect
...
10, and
where
k
we implied that u
...
0/ D v are the functions themselves
...
71) and
therefore is sometimes written in the following symbolic form:
d n
...
uv/
...
u C v/
...
48)

Let us first prove this result using a method very similar to the one we used in
Sect
...
10 when proving the binomial theorem
...
Firstly, we

142

3 Derivatives

check that the formula works for some initial values of n D 1; 2; 3 (in fact, only one
value, e
...
n D 1, would suffice):

...
1/ v C uv
...
1/ v
...
0/ v
...
i
...
familiar u0 v C uv 0 /;


...
2/ v
...
1/ v
...
0/ v
...
uv/000 D u
...
0/ C u
...
1/ C 2 u
...
1/ C u
...
2/ C u
...
2/ C u
...
3/
D u
...
0/ C 3u
...
1/ C 3u
...
2/ C u
...
3/ ;
which are all consistent with Eqs
...
47) or (3
...
Next we assume that the formula (3
...
We should now prove from it that Eq
...
47) holds also for the next
value of n, i
...
for n ! n C 1
...
(3
...
uv/


...
k/
...
kC1/
...
kC1/v
...
k/
...
k/ v
...
uv/
...
k/v
...
k/ v
...
uv/
...
nC1/
...
k/ v
...
0/
...
49)

3
...
(1
...
75), we can finally rewrite Eq
...
49) as follows:

...
nC1/

Â
D

Ã
Ã
n Â
n C 1
...
nC1/ X n C 1
...
nC1
u v
u v
C
k
0

k/

Ã
nC1 Â
n C 1
...
0/ X n C 1
...
nC1
u v
u
v D
k
nC1

k/

kD1

Â
C

Ã

;

kD0

which is the desired result as it looks exactly as Eq
...
47) written for n ! n C 1
...
Differentiation of a function, f 0
...
x/, i
...

f 0
...
x/:
dx
dx

The n-th derivative, f
...
x/, can then be regarded as an action of the same operator
n times:
 Ãn
d

...
x/:
f
...
Let us introduce
d
d
two operators, Du D dx u and Dv D dx v , which differentiate only the functions
u
...
x/, respectively
...
uv/0 D
...
Du u/ v C u
...
1/ v
...
0/ v
...
uv/00 D
...
Du C Dv / uv D
...

Thus, we obtain:

...
Du u/
...
2/ v
...
1/v
...
0/ v
...
Du C Dv /n on the product uv
...
Du C Dv /n formally using
the binomial formula
...
k/ and similarly Dk v with v
...
uv/
...
Du C Dv /n uv D

n
XÂnÃ
kD0

D

n
XÂ
kD0

n
k

k

Ã
Dk u
u

n
Dv k v

D

n
Dk Dv k uv
u
n
XÂnÃ

k

kD0

u
...
n

k/

;

which is exactly the correct expression (3
...
This particular proof shows quite
clearly why the Leibniz formula is so similar to the binomial formula
...
a C b/n we have powers and hence a0 and b0 are both
equal to one and therefore do not need to be written explicitly, in the Leibniz formula
for
...
n/ the “powers” in u
...
n k/ correspond to the order of the derivatives
and thus u
...
0/ coincide with the functions u
...
x/ themselves and are
not equal to one; these should be written explicitly whenever they appear
...
6
...
x/ D x2 C 1 v
...

Solution
...
x/ D x2 C 1 and hence can only be differentiated twice: u
...
2/ D 2, while all higher derivatives give zero
...
n/

 Ã
 Ã
n
...
n/
n
...
n

...
n/ C 2nxv
...
2/
...
n

C n
...
n 1/ =2
...
25
...
x/ D
...
x/ can be written in the following form:
'
...
x/ D a0
...
n/ C a1 x2 f
...
x/
...
n

2/

C a3 f
...
47), that the numerical values of the
coefficients are: a0 D 1, a1 D 3n, a2 D 3n
...
n 2/
...
6 Higher Order Derivatives

145

Problem 3
...
Show that the n-th derivative of the function f
...
2x/,
assuming n is even, is:
f
...
x/ D
...
n 1/ sin 2x
«
1/
...
n

Cn

In some cases the calculation of the n-th derivative could involve some tricks
...
x/ function
...
The calculation of the second derivative
would result in two terms as one has to use the chain rule; more terms arise as we
attempt to calculate more derivatives and there seems to be no general rule as how
to do the n-th derivative (try this!)
...
The solution to this problem appears to be possible if
we notice that
y0
...
1

x/

1=2


...
x/v
...
nC1/


...
n/

D
...
n/

D

n
XÂnÃ
kD0

k

u
...
n

k/

;

where (e
...
by induction)
h
u
...
1
D

x/


...
k/


...
1
2 2
2

...
2kC1/=2

:

Here the double factorial means a product of all odd integers, i
...
2k 1/ŠŠ D
1 3 5 : : :
...
We also adopt that
...
k/ also formally valid for k D 0, i
...
u
...
1 x/ 1=2 D u
...
The double
factorial can always be expressed via single factorials and powers of 2 by inserting
between odd integers the corresponding even ones:

...
2k
D

1/

1 2 3 4 5 6 : : :
...
2k/

...
2k/
2 kŠ

(3
...
k/ D


...
1
4k kŠ

x/


...
2k/Š

...
k/

h

1=2

D
...
1/k


...
k/

D

1/ŠŠ
2k

1
2


...
2kC1/=2

Ã

Â
:::

2k

1

Ã

2


...
1/k
...
1 C x/
Dp
1 C x 4k kŠ

k


...
n

k/


...
2n 2k/Š

...
n k/Š

nCk

:

Combining the obtained derivatives of u
...
x/ in the Leibniz formula, we
finally obtain:
Œarcsin
...
nC1/

D p

1
1

n
X Â n Ã
...
2n 2k/Š
x2 kD0 k 4k kŠ 4n k
...
1/n

k


...
1 C x/


...
2n/Š X
D p

...
51)

where we expressed all factorials via the binomial coefficients
...
7 Taylor’s Formula
Taylor’s formula1 allows representing a function as a finite degree polynomial and
a remainder term to be considered as “small”
...
It is used in numerous analytical proofs and
estimates
...

Let us first do some preliminary work which will help later on
...
x/ which is a polynomial of degree n:
f
...


C ˛n xn :

(3
...
7 Taylor’s Formula

147

This polynomial contains powers of x, but can alternatively be written via powers of

...
x/ D A0 C A1
...
x

C An
...
53)

Problem 3
...
Using the binomial formula for each of the powers
...
x/ written via powers of x again
...
(3
...
Use this method to show that
˛k D

n
X

 Ã
i

...
28
...
(3
...
a/ D A0 , then
f 0
...
x

a/2 C

a/ C 3A3
...
x/ D 2A2 C 3 2A3
...
x

f 000
...
x

H) f 0
...
x

H) f 00
...
a/ D 3 2A3 D 6A3 ;
etc
...
k/
...
k C 1/Š
AkC1
...
k C 2/Š
AkC2
...
k/
...
a/ D

f 0
...
a/
f
...
a/
f
...
a/
; A2 D
; : : : ; Ak D
; : : : ; An D
:
1Š
2Š
kŠ
nŠ
(3
...
29
...
x/ in Eq
...
52) to express derivatives of f
...
Hence, show that the coefficients Ak can be
expressed via ˛i as follows:
Ak D

n
X
iDk

˛i

 Ã
i
ai k :
k

148

3 Derivatives

It follows from Eq
...
54) that any polynomial of degree n can be actually written
as a sum of powers of
...
x/ D f
...
a/

...
k/
...
x

a/ C

f 00
...
x
2Š

a/2 C

C

f
...
a/

...
x/ is not a polynomial, it can never be written via a finite sum of
powers of
...
Let us define the following n-th degree polynomial (Taylor’s
polynomial)
Fn
...
a/ C
D

f 0
...
x
1Š

n
X f
...
a/
kD0

kŠ


...
a/

...
n/
...
x
nŠ

a/n
(3
...
a; a/ D f
...
Then it is clear that formally any function f
...
x/ D Fn
...
56)

where RnC1 , called the remainder term, accounts for the full error we are making
by replacing f
...
(3
...
Taylor’s theorem to be discussed below enables
one to write an explicit formula for the remainder term for any f
...

But before formulating and proving Taylor’s theorem, we need one more
statement to be made, which, we hope, is intuitively obvious
...
x/ within the interval a Ä x Ä b, and let the function have
identical values at both ends: f
...
b/
...
x/ is not
a constant (equal to its edge value)
...
3
...
Indeed, for instance, if the function goes down
from the point x D a, it must eventually go up towards the point x D b as its values
at the edges are the same
...
/ D 0, hence the slope of the
tangent line at that point is zero (it is horizontal)
...
3
...
To prove this
statement, let us suppose that the function has a maximum at point x D
...
/
f
...
/
f
...


3
...
3
...
x/ which has equal
values at both ends of the interval a Ä x Ä b will experience at least one (may be more) extremum
within that interval
...
The extrema (minima and/or maxima) of the functions
are indicated by green dots; the corresponding tangent lines at these points are also shown by
horizontal green lines

i
...
at the neighbouring points on the right and on the left of the point x D the
function f
...
Therefore, we have:
f
...
/

Ä 0 and

f
...


/

0:

Note that the first expression is non-positive, while the second is non-negative
...
/ from
the right and left
...
x/ is continuous, these two derivatives should
be equal to each other, and that means that they should be equal to zero as otherwise
it is impossible to satisfy the two inequalities written above
...
Note that if the function f
...

What is essential for us right now is that if a continuous function f
...
/ D 0
...
t/ D f
...
x; t/


...
x

a/nC1

Œf
...
x; a/ :

Let us first calculate this function at the points t D a and t D x:

...
x/

Fn
...
x/

Fn
...
x

a/nC1


...
x/

Œf
...
x; a/

Fn
...
57)

150

3 Derivatives

and

...
x/

Fn
...
x

x/nC1


...
x/

Fn
...
x/

Fn
...
x; x/ D f
...
55), as was already mentioned
...
a/ D
...
Next, we calculate the derivative of
...
x t/n
dFn
...
n C 1/
Œf
...
x a/nC1


...
k/
...
x; t/
D

...
t/ C
dt
dt
kŠ
kD1
0

D f
...
kC1/
...
t/ C

n
X Ä f
...
t/

kŠ

kD1

D f 0
...
t/
...
x

#
t/

k

Fn
...
58)

n
X d Ä f
...
t/
D f
...
x
dt
kŠ
kD1
0

t/k C

f
...
t/

...
x
kŠ

t/k

1

f
...
t/

...
k 1/Š
Ä
1 000
0
f
...
t/
...
t/
...
x

t/k

t/k

1
...
t/
...
t/
...
nC1/
...
x
nŠ

t/n

f
...
t/

...
n 1/Š

t/

t/2 C
t/n

1

:

It is seen that the first term here is cancelled out with the third (the second term in
the first square bracket), the second with the fifth, the fourth with the seventh, and
so on; all terms cancel out this way apart from the first term in the last bracket, i
...

dFn
...
nC1/
...
x
dt
nŠ

t/n ;

and hence from (3
...
t/ D

f
...
t/

...
n C 1/

D

0

f
...
t/

...
n C 1/


...
x

...
x

t/n
a/nC1
t/n
a/nC1

Œf
...
x; a/
(3
...
x/ and the polynomial
Fn
...
(3
...


3
...

Theorem 3
...
Let y D f
...
e
...
k/
...
Then this function
at some point x can be written as the following
...
a/
f
...
a/ C

...
a/

...
n/
...
x
nŠ

C RnC1
...
x; a/ C RnC1
...
60)

where
RnC1
...
nC1/
...
x

...
61)

is the remainder term with being some point between x and a; of course, the
point generally depends on the choice of n
...
Let us choose a point x > a in the vicinity of a (the case of x < a is
considered similarly) and let Fn
...
Then, consider a function
...
(3
...
Since the
function
...
e
...
/ D 0
...
59) at t D to zero, we obtain:
0


...
nC1/
...
x
nŠ
H)

RnC1 D

/n C
...
nC1/
...
x

...
x

...
(3
...
Q
...
D
...
x

a/ ;

where 0 < # < 1:

This form guarantees that lies between a and x
...

There are several important particular cases of Taylor’s formula worth mentioning
...
x/ D f
...
0/ 2 f 00
...
0/
xC
x C
x C
1Š
2Š
3Š

C

f
...
0/ n
x C RnC1
...
62)

which is called Maclaurin’s formula
...
x/ D

f
...
/ nC1
f
...
#x/ nC1
x
x ;
D

...
n C 1/Š

(3
...
The other important formula (due to Lagrange) is obtained from
the Taylor’s formula in the case of n D 0
...
x/ D f 0
...
x/ D f
...
x
a<

a/ D f
...
a C #
...
x

a/ ;

< x ; 0 < # < 1:

(3
...

Let us now apply the Maclaurin formula to some elementary functions
...
7
...

Solution
...
ex /0 D ex , it is clear that
...
i/ D ex for any i D 1; 2; : : : ; n
...
62) and noting that e0 D 1, we obtain:
ex D 1CxC

X xk e#x xnC1
xn e#x xnC1
C
D
C
:J
nŠ
...
n C 1/Š
kD0
n

x2 x3 x4
C C C
2Š 3Š 4Š

C

(3
...
30
...
1 C x/ D x

X
xn
xk
C RnC1 D

...
66)
n

x2 x3
C
2
3

C
...
1/k

...
1/n
I
n C 1
...
1/nC1 x2n

...
67)
(continued)

3
...
30 (continued)
with
R2nC1 D
cos x D 1
D

x2
x4
C
2Š
4Š

C


...
#x/ 2nC1
x
I

...
1/n x2n
C R2nC2

...
1/k x2k
C R2nC2

...
68)

with
R2nC2 D


...
#x/ 2nC2
x
I

...
˛ 1/ 2
xC
2
n
X²Ã
xk C RnC1 ;
D
k


...
˛ 1/
...
˛ nC1/

xn C RnC1
(3
...
˛
˛
D
k

1/
...
˛

k C 1/

(3
...
1 C #x/˛ n 1 xnC1 :
nC1
 Ã
˛
The notation
for the coefficients is not accidental
...
69) becomes the binomial formula
...
72), and second, verify that the remainder term is zero in this
case, i
...
the expansion contains a finite number of terms
...
x/
...
(3
...
The trick is in noticing that

154

y0 D p

3 Derivatives

1
x2

1

and y00 D

x
x2 /3=2


...
31
...
nC2/ D
...
nC1/ C n2 y
...
e
...

Now apply x D 0 in the above recurrence relation to get y
...
0/ D n2 y
...
0/,
which can be used to derive recursively all the necessary derivatives at x D 0 needed
to derive the Maclaurin formula for the arcsine function
...
2/
...
4/
...
2/
...
, i
...
all the even order
derivatives are equal to zero
...

1=2
; next, y
...
0/ D 12 y
...
0/ D 12 ,
We know that y
...
0/ D 1 from y0 D 1 x2

...
3/
2 2

...
5/
y
...
0/ D 3 1 , y
...
0/ D 52 32 12
...
2k/Š 2
y
...
0/ D Œ
...
50) for the double factorial
...
x/ D

n
X

X

...
2k C 1/
...
2k C 1/Š
4
kD0
(3
...
2kC1/
...
0/ D 0
...
It can be worked out
using Eq
...
51) if desired
...
32
...
x/ and prove the Maclaurin formula for it:
arccos
...
2k/Š
4k


...
kŠ/2

x2kC1 C R2kC2 :

(3
...
8 Approximate Calculations of Functions

155

Problem 3
...
Apply a similar method for obtaining the Maclaurin series for
the arctangent function:
arctan
...
1/k
x2kC1 C R2kC2 :
2k C 1
kD0

(3
...
x/0 D 1 C x2
does not contain radicals, it is
sufficient to consider the identity 1 C x2 y0 D 1 and then differentiate both
sides of it n times by means of the Leibniz formula (3
...
]
Compare the two Maclaurin formulae for the arcsine and arccosine with the
identity (2
...

The apparent contradiction will be resolved in Sect
...
3
...
72) is satisfied exactly
...
8 Approximate Calculations of Functions
Taylor’s formula is frequently used for approximate calculations of various functions
...
55)) and omitting the remainder term
...
e
...

As an illustration of this method, we shall consider a numerical calculation of the
numbers and e
...
65) at x D 1:
1
1
1
1
C C C C
2Š
3Š
4Š
nŠ
n
n
X 1
X
ek
1
D2C
D
:
D
ek with ekC1 D
kŠ

...
74)

The recurrent formula for the terms in the sum given above allows calculating every
next term, ekC1 , from the previous one, ek , and is much more convenient for the
practical calculation on a computer than the direct calculation from their definition
via the factorial
...
3
...
It is really surprising that only seven terms in the sum are already
sufficient to get the number e ' 2:718281828459 with the precision corresponding
to 12 digits
...
3
...
74) and (3
...
If the convergence of the sum for the number e is very fast,
this is not so in the case of the arcsine Maclaurin series for the number
...
71) for the
arcsine function at x D 1 since arcsin
...
2k/Š

kD0

D 2 arcsin
...
2k C 1/
...
2k C 1/2
:
2
...
2k C 3/

(3
...
However, as it follows from the numerical
calculation shown in Fig
...
3(b), this time the series converges much slower; even
with the 105 terms kept in the sum the obtained approximation 3:1303099 is still
with the considerable error of the order of 10 2
...

Taylor’s formula is frequently used for obtaining approximate representations
for the functions in the vicinity of a point of interest
...
x/ around point a, then the Taylor
polynomial could be a good starting point
...
3
...
Its Taylor polynomial around x D 0 is given
by (3
...
It is clearly seen from the plotted graphs in
(a) that a reasonable representation of this function can indeed be achieved using
a small number of terms (a low order polynomial)
...
This
is illustrated with the zoom-in image in (b) where the errors are better seen
...

Generally, the higher the order of the polynomial, the better representation of
the function is obtained
...
9 Calculating Limits of Functions in Difficult Cases

157

Fig
...
4 Approximate representations of the exponential function y D ex using n-order Taylor’s
polynomials (3
...
e
...
The central area of the graph
in (a) is shown in (b) in more detail

function, as this would require a smaller number of terms to achieve the same
precision
...

Problem 3
...
Consider a mass m in the gravitational field of a very large
object (e
...
the Earth) of mass M and radius R
...
d/ D GmM=d,
where G is the gravitational constant
...
z/ ' mgz if z
R, and derive the
expression for the constant g
...
]

3
...
2
...
6 we looked at some simple cases of uncertainties when taking a limit
...
Taylor’s formula is
especially useful in calculating limits of functions in difficult cases of uncertainties
0=0, 1=1 or 0 1
...
86) by expanding the sine function in the
Maclaurin series:
Ä
Ä
˛
...
x/ D lim 1
C
x
D 1;
x!0 x
x!0 x
x!0
3Š
3Š
x
where ˛
...
x/ tends to zero at least as x5 when x ! 0
...
2
...
6
...
35
...
2
...
6)
...
x/=g
...
Expanding
both functions using Taylor’s formula up to the n D 1 term and using the fact that
f
...
x0 / D 0, we have
lim

x!x0

f
...
x0 / C f 0
...
x
D lim
x!x0 g
...
x /
...
x/
0
0
D lim

x!x0

D lim

x!x0

x0 / C
...
x

x 0 /2

x0 / C
...
x

x 0 /2

f 0
...
x

x0 / C
...
x

x0 /2

g0
...
x

x0 / C
...
x

x 0 /2

f 0
...
x/
...
x0 / C
...
x

f 0
...
x0 /

(3
...
x/ and
...
x/ and g
...
The obtained result is known as L’Hôpital’s rule
...
This happens if the functions
...
x/ used above for n D 1 are zeroes at x0
...
x/ and g
...

Problem 3
...
Let the functions f
...
x/ together with their first k
derivatives be equal to zero at x D x0
...
kC1/
...
x/
D
...
x/
g

...
In practice,
however, it is more convenient to apply the rule (3
...


3
...
8
...
Apply L’Hôpital’s rule
L D lim

x!0

sin3 x

0

3
3 sin2 x cos x
sin2 x
D lim
:
x!0
2x
2 x!0 x

D lim


...
x/0

0

D

3
2 sin x cos x
3
lim
D
2 lim sin x D 0: J
x!0
2 x!0
1
2

Above we considered the 0=0 case in applying L’Hôpital’s rule
...
For instance, in the 1=1 case,
when f
...
x0 / D 1, we write:
lim

x!x0

f
...
x/
D lim
:
x!x0 1=f
...
x/

The functions 1=f
...
x/ tend to zero values when x ! x0 , and hence
L’Hôpital’s rule can be applied directly to them:
ˇ
f
...
x/0
Œ1=g
...
x/
x!x0 Œ1=f
...
x/0 ˇxDx0
In the same way the 0 1 uncertainty is considered: if f
...
x0 / D 1,
then in this case
ˇ
f 0
...
x/
ˇ
D
lim f
...
x/ D lim
:
x!x0
x!x0 1=g
...
x/0 ˇxDx0
L’Hôpital’s rule is also useful in finding the limits at ˙1 in any of the cases of
uncertainties
...
x/ and g
...
x/=g
...
This
case can be transformed into the familiar case of the limit at zero if we introduce
a new variable t D 1=x
...
x/
f
...
t//
F
...
t/0
D lim
D lim
D lim
0
x!C1 g
...
x
...
t/
t!0 ŒG
...
x/ x0
...
x/
f 0
...
x/ x0
...
x/
x!C1 g
...
77)

160

3 Derivatives

where F
...
x
...
t/ D g
...
t// are functions of t, and we used the chain
rule when differentiating them with respect to t
...


Problem 3
...
Prove that formula (1
...
That would mean that it is actually valid for any q
...
38
...
1 C x/
D 0;
lim
x!C1
x
p
lim x x D 1 ;
(f)
x!0

D

1
;
2

(b)

lim

x!0

ln
...
]

3
...
x/: one can quickly see in which intervals the function increases or decreases,
or where it has extrema (minima or maxima) and saddle points
...
It seems obvious that if a continuous function
has a positive derivative in some interval a Ä x Ä b, then it increases there
...

Problem 3
...
Show using the Lagrange equation (3
...
x/ > 0 for
all x within the interval a Ä x Ä b, then f
...

[Hint: prove that f
...
x/ for any x > 0:] Similarly show that if the
first derivative is negative within some interval, the function f
...


3
...
3
...
x/ which has two maxima at points A and C, one
minimum at point B and a saddle point at D
...
x/ (red) and y00
...
Note that the tangent lines (dashed purple) at minima,
maxima and saddle points are horizontal, i
...
have zero slope

These statements have a very simple geometrical interpretation: if the function
increases, then the tangent line at any point x will make a positive angle with the x
axis; if it decreases, then this angle will be negative
...
At the beginning, let us
consider an example of a function sketched in Fig
...
5
...
At all these points the slope
of the tangent lines is zero, i
...
the tangent lines are horizontal
...
How can we
define and distinguish different types of such points?
Let us first analyse the behaviour of a function around point A (with x D xa )
where it has a maximum
...
3
...
When moving from some x > xa to the
left towards xa , our function increases and its slope is non-positive, i
...
y0
...
x C dx/ y
...
If we now move towards the point xa from the left (i
...
from some x < xa ),
then y
...
x/ D
Œy
...
x/ =dx
0
...
x/)
changes sign when crossing the point xa of the maximum being exactly equal to zero
at that point; at the point, xa where the first derivative is equal to zero the tangent
line (with the zero slope) is parallel to the x axis
...

Importantly, if you look at the graph of y0
...
3
...
In other words, the derivative of y0
...
e
...
x/ of the original function y
...
e
...
xa / < 0
...


162

3 Derivatives

Note that the same result is valid for the other maximum at point xc : the first
derivative y0
...
xc / < 0 is strictly
negative
...


Problem 3
...
Perform similarly a general analysis of the behaviour of a
function y
...
xb / D 0 and
(ii) y00
...

We see that minima and maxima of a function (i
...
its extrema) can be easily
found by simply solving the equation y0
...
This would all be
good, but unfortunately for some functions this method does not work as the second
derivative at the extremum point of a function is zero
...

Consider the function y D x2
...
x/ D 2x D 0)
the second derivative y00
...
Then, consider the function y D x4 shown in Fig
...
6(b) together with
its first two derivatives
...
x/ D 4x3 D 0 gives x D 0 as the point of
the extremum, however, the second derivative y00
...
0/ if x D 0 corresponds to the
minimum or maximum of the function
...
Indeed, consider two points very close to the point x D 0 of
the extremum lying on both sides of it: x D ˙ (with ! C0)
...
3
...
In the latter case the first (y0 D 4x3 ) and second
(y00 D 12x2 ) derivatives are also shown

3
...
/ D 4 3 < 0 and y0
...
At the same time, the second derivatives at
these points y00
...

Care is needed in applying the condition y0
...
This is because, firstly, the derivative may not exist
...
g
...
3
...
Secondly, there could also be another type of critical points which also
have a horizontal tangent line (of zero slope)
...
3
...
As the slope is horizontal, the first derivative of the function at the saddle
point y0
...
However, the first derivative does not change its
sign when passing this point; in addition, the second derivative at this point is zero
as well: y
...
e
...

Hence, the second derivative is negative on the left of xd and positive on the right
of it, and passes through zero at the saddle point
...
x/ D 0 may
give any type of the critical points (minima, maxima and the saddle), and additional
consideration is necessary to characterise them explicitly
...
We know intuitively that around a
minimum any function is convex, and around a maximum—concave
...
g
...
Investigating the convexity of functions appears to
be quite useful when analysing their general behaviour prior to plotting/sketching
their graph (see below)
...
x/ is convex, its negative, g
...
x/, is concave
...

We start by giving definitions
...
3
...
If we plot a line AB
crossing the convex function at two points,
...
a// and
...
b//, then all points
of the function between a and b would lie not higher than the corresponding
values on the crossing line: f
...
x/ for any a Ä x Ä b
...
The crossing line passes through two points A and B
and hence has the following analytical form:
fl
...
a/ C
h
x
D 1
b

f
...
a/

...
a/ C
a
b

a/
a
f
...
1
a

/ f
...
b/;

(3
...
3
...
The tangent lines at point
C are shown with magenta, while lines AB crossing the function f
...
x a/ =
...
Since x D
...
x/ Ä fl
...
1

/ a C b/ Ä
...
a/ C f
...
79)

D 1=2:
Â
f

aCb
2

Ã
Ä

1
Œf
...
b/ :
2

(3
...

The sign in these inequalities is to be reversed for the concave functions in the
given interval a Ä x Ä b
...
Let us look at the convex function sketched in Fig
...
7(a)
one more time: if we plot a tangent line to any point within the interval a Ä x Ä b,
then it is clear that the tangent line cannot be above the graph of the function itself
...
xc ; f
...
x/ itself
...
x/ ft
...
x/ D f
...
xc /
...
81)

is the tangent line at the point xc (it passes through the point C and has the slope
equal to f 0
...
Similarly, the function is concave within some interval if for any
point x we have f
...
x/, and this situation is shown schematically in Fig
...
7(b)
...
10 Analysing Behaviour of Functions

165

To work out the required criterion for the function to be convex, we shall use the
n D 1 Taylor’s formula (3
...
x/ D f
...
xc /
...
x/ C R2
...
x/ D

1 00
f
...
61) for the remainder term in this case and is somewhere
between x and xc
...
x/ ft
...
x/

ft
...
x/ D

1 00
f
...
/

0;

(3
...
e
...
Since we used arbitrary points xc
and x, it is clear that at any point within the convex interval the second derivative of
the function is not negative
...
These are the required criteria we have been looking for
...

The logarithmic function y D ln x is everywhere3 concave since y00 D
...
1=x/0 D 1=x2 < 0
...
Recall that y D
x4 has the second derivative equal to zero at x D 0, but has the minimum there, not
the inflection point
...

As an example, consider the function
y D 4x3 C 3x2

6x C 1:

(3
...
2x

1/
...
84)

suggests that there are critical points at x1 D 1=2 and x2 D 1
...
At the
point x1 we have a minimum since y00
...
1/ D 18 is negative
...
The point x3 D 1=4
is therefore the inflection point
...
e
...


166

3 Derivatives

x ! C1 it tends to the C1
...
(3
...
Moving from 1 to
the right towards C1, it first increases up to the point x D 1, where it experiences
a maximum; after that point, it decreases up to the point x D 1=2 where it has a
minimum; moving further, it increases again to C1
...
The function is plotted in Fig
...
8
...
3
...
83)
...
10 Analysing Behaviour of Functions

167

Fig
...
9 Functions (a) y D
...
x C 1/ and (b) y D
asymptotes (green)

2x2 C 1 =x (blue) and their

Asymptotes of a Function
...
x/ at x D ˙1 and
also at points x where y
...
Before giving
the corresponding definition, let us first consider two functions:
y1
...
x/ D

1
2x2 C 1
D 2x C
x
x

to illustrate the point
...
x/; its graphs is shown in Fig
...
9(a)
...
e
...
As x gets closer to the value of 1, the graph of the
function gets closer as well to the green vertical line
...
x/
...
x C 1/ which will guide us in this discussion
...
e
...
The green horizontal line y D 1 serves as a useful
guide for this asymptotic behaviour at ˙1 and is called the horizontal asymptote
...
This is illustrated by our second function y2
...
3
...
This function is not defined at x D 0 and hence the vertical
line x D 0 serves as its vertical asymptote
...
x/

168

3 Derivatives

is written, it is clear that as x ! ˙1, the term 1=x ! 0 and the behaviour of the
function is entirely determined by the first part 2x, which is a straight line shown in
green in the figure
...
x/ approaches this straight line either
from above (when x ! C1) or below (when x ! 1)
...
The horizontal asymptote can be considered as a particular case
of the oblique one with the zero slope, while the vertical one—with the 90ı slope
...
The vertical
asymptote is a vertical line x D x0 with x0 being a point at which the function y
...
The function y
...
e
...
A horizontal or oblique
asymptote yas D kx C b is a straight line which approaches the curve of the function
y
...
We
shall now derive a formula for the coefficients k and b of the asymptote (obviously,
k D 0 for the horizontal asymptote) in the case of x ! C1
...
3
...
We need to calculate the distance
AB between the function y
...
x/ D kx C b at this point
...
1
...
5 in Eq
...
104), and the distance is
given by
dD

jkx

p

y
...
x/ which tends to zero at this limit:
lim Œkx

x!C1

H)

Fig
...
10 To the derivation
of the oblique asymptote for
the x ! C1 case

y
...
x/ C b must be equal to some

˛
...
x/

b C ˛
...
x/
D lim
;
x!C1 x
x

(3
...
10 Analysing Behaviour of Functions

169

which is the required formula for k
...
x/ C b

˛
...
x/ C ˛
...
x/
x!C1

x!C1

kx :

(3
...

Example 3
...
I Work out the asymptotes of the function
yD

2x2

3x C 6
:
xC1

(3
...
First we note that x D 1 is the vertical asymptote of the function (note
that the numerator 2x2 3x C 6 is not equal to zero at x D 1)
...
(3
...
86) at
C1 and 1:
2x2 3x C 6
2 3=x C 6=x2
D lim
D 2;
x!C1
x!C1
x
...
x C 1/
2x D lim
D lim
x!C1
x!C1
xC1
xC1

kC1 D lim
bC1

D lim

x!C1

5x C 6
5 C 6=x
D lim
D
x!C1 1 C 1=x
xC1

5;

and the same limits are obviously obtained at x ! 1, i
...
one and the same
asymptote yas D 2x 5 serves at both infinities
...
The function together with its
asymptotes is plotted in Fig
...
11
...
g
...

A usual procedure consists of the following steps:
find the domain of values of x where the function y D y
...
g
...
g
...
x/ and solve the equation y0
...
3
...
x/ of Eq
...
87) (blue) and its asymptotes (green)

• calculate y00
...
x/ D 0);
• make a sketch of the function
...
41
...
x/ D 3
...
x C 1/
...
Check your findings
by plotting the function using any available software
...
]
Problem 3
...
The same for y
...
x C 1/
...
]
Problem 3
...
The same for y
...
x 2/3 =
...
[Answer: maximum at
x D 10, inflection point at x D 2, asymptotes x D 2 and y D x 10
...
44
...
[Answer: critical points at x D
4 ˙ n, where minima are for n D 1; 3; : : :, while maxima for n D 0; 2; 4; : : :,
inflection points are at x D 2 ˙ n, where n D 0; 1; 2; : : :; no asymptotes
...
10 Analysing Behaviour of Functions

171

Problem 3
...
Consider the van der Waals equation of state of a gas
PC

aÁ

...
We
shall consider the case of constant T to investigate the dependence P
...
e
...

(a) Show that the derivative
dP
2a
f
...
v b/2

where

f
...
v

b/2
v3

RT
:
2a

(b) Show that the derivative f 0
...
3b/ < 0 if 8a=27b < RT and f
...

(c) Then show that in either case f
...

(d) Therefore, argue that in the first case, when 8a=27b < RT, pressure P
always decreases with the increase of the volume v
...
v/ D 0 (and hence dP=dv D 0/, where b < v1 < 3b
and 3b < v2 < 1 (a cubic equation needs to be solved to find them, but
the good thing is that our general analysis does not require knowing these
two critical volumes explicitly)
...

(g) The point at which two roots v1 and v2 coincide, i
...
when 8a=27b D RT,
is called the critical point of the van der Waals equation
...
v/ has an inflection point at v D 3b with the pressure
P D a=27b2
...
g
...


172

3 Derivatives

Problem 3
...
A second order phase transition in a ferromagnetic material
between paramagnetic and ferromagnetic phases can be described, within
the so-called Landau–Ginzburg theory, using the following phenomenological
expression for the free energy density of the material:
f
...
T
2

Tc / m2 C f4 m4 ;

(3
...

(a) Show that for T > Tc the free energy has a single stable state (i
...
a single
minimum) with zero magnetisation only
...
Tc T/ =4f4 become stable (f is minimum there)
...
This is characteristic for
the second order phase transition
...
Tc T/ =12f4 D ˙ jm˙ j = 3
...
47
...
The free energy is identical to
Eq
...
88) of a ferromagnetic if one replaces m with P
...
Here we shall consider
a more general case of the material in an external electric field E:
f
...
T
2

Tc / P2 C f4 P4

EP:

(3
...
T Tc / 1 , while in the ferroelectric phase (T < Tc ) we have
D Œ2˛
...
e
...


3
...
48
...
In a supersaturated solution of particles (e
...
,
atoms) a nucleus of a crystalline phase may appear which will then may
grow into the bulk phase
...
This term favours formation of the crystalline
(solid) phase
...
It works against the formation of the nucleus
...
R/ has two extrema: at R D 0 and R D 2 = B j j
...

(c) Show that at R > 3R =2 the free energy becomes negative
...
R/
...


Chapter 4

Integral

A notion of the integral is one of the main ideas of mathematical analysis (or simply
“analysis”)
...
In this chapter we shall use a much clearer and more rigorous route based on
defining the definite integral first as a limit of an integral sum
...


4
...
x/ 0 specified in the interval a Ä x Ä
b as sketched in Fig
...
1
...
If
the function was a constant, f
...
b a/
...
x/ D f
...
b/
b

f
...
x
a

a/ ;

passing through the points
...
a// and
...
b//, in which case we have a trapezoid,
and hence its area A D Œf
...
b/
...

How can one calculate the area under a general curve f
...
First,
we divide the interval a Ä x Ä b into n subintervals by points x1 , x2 , : : :, xn 1
such that
a D x0 < x1 < x2 <

< xn

2

< xn

1

< xn D b;

© Springer Science+Business Media, LLC 2016
L
...
1007/978-1-4939-2785-2_4

175

176

4 Integral

Fig
...
1 Definition of a definite integral between a and b
...
k / and the width xk , as shown in (b)

where for convenience we included the boundary points a and b into the set fxk g
as x0 and xn , respectively
...
Altogether n points k are chosen such that xk 1 Ä k < xk for all k D 1; 2; : : : ; n
as shown in Fig
...
1(a)
...
x/
into n strips (shown in different colours in the figure), and hence we may calculate
the whole area by summing up areas of every such strip
...
k / at the chosen point k within the strip as the strip height,
see Fig
...
1(b), i
...
by writing Ak ' f
...
xk xk 1 / D f
...
This way we replace the actual strip k having a (generally)
curved line at the top with a rectangular of height f
...
The error we are making
with this may be not so large if the strip is narrow enough, i
...
its width xk is
relatively small (compared to the width of the whole interval  D b a)
...
k / xk :

(4
...
It is clear that if we
make more strips, they would become narrower, the function f
...
k /, less severe
...
An important point
is that the exact result can be obtained by making an infinite number of strips (by
taking all widths of the strips xk to zero at the same time which would require
the number of strips n ! 1), or ensuring that the widest of the intervals, i
...

max fjxk jg, goes to zero:
!
n
X
(4
...
k /xk :
AD
lim
maxfjxk jg!0

kD1

4
...
x/dx;

(4
...
Note that b > a, i
...
the top limit is
larger than the bottom one
...
x/
is said to be integrable
...

The definition for the definite integral (4
...
In this case the same
formula (4
...

Moreover, the integral we considered above had a “direction” from a to b (where
b > a), i
...
from left to right
...
In this case the points xk are numbered from
right to left instead and hence the differences xk D xk xk 1 become all negative
...
x/dx;
b

i
...
it looks the same apart from the fact that the top and the bottom limits changed
places: now the bottom limit b is actually larger than the top limit a since b > a
...
x/ is positive
...
Therefore, by
summing up all the contributions in the integral sum we should arrive at the same
numerical value of the integral as when calculating it from a to b, only with the
negative sign
...
e
...
x/dx D

a

f
...
4)

b

Consider now a point c somewhere between a and b, and let us choose one of the
division points xk exactly at c
...
It is clear that the integral sum will split into two, one from a and c, and
another from c and b; at the same time, this must be still the integral from a to b
...
Hence, we obtain the second property of the definite integral
(additivity):
ˆ b
ˆ c
ˆ b
f
...
x/dx C
f
...
5)
a

a

c

Problem 4
...
Prove that this identity still holds even if c > b or c < a
...
x/ be a linear combination of two other functions, u
...
x/, i
...
f
...
x/ C ˇv
...
Then,
" n
#
( n
)
X
X
lim
f
...
k / C ˇv
...
k /xk C

kD1

" n
X

lim

maxfjxk jg!0

kD1

lim

" n
X

maxfjxk jg!0

u
...
e
...
x/ C ˇv
...
x/dx C ˇ
a

a

ˇv
...
k /xk ;

kD1

v
...
6)

a

Of course, the proven linearity property holds for a linear combination of any
number of functions
...

Finally, the integral between two identical points is obviously equal to zero:
ˆ a
f
...
2
...
x/ D f
...
x/ D g
...
Then demonstrate, using the definition of the integral,
that the following identities are valid when these functions are integrated over
a symmetric interval a Ä x Ä a:
ˆ

ˆ

a

f
...
x/dx ; if f
...
x/;

(4
...
x/dx D 0 ; if g
...
x/:
a

(4
...
1 Definite Integral: Introduction

179

It is sometimes useful to remember that any function f
...
x/ D feven
...
x/ ; feven
...
x/ C f
...
x/ D Œf
...
x/:
(4
...
x/ and the interval
boundaries a and b
...
It might also be important how the
points k (at which the function is calculated) are chosen within each subinterval
...
However, before we start
looking into these subtle complications, it is instructive to consider some examples
first
...
1
...


chosen at the end of each

k

Solution
...
x/ D 1 is constant, so for any division
of the interval 0 Ä x Ä 1 into subintervals we have
" n
#
X
xk D
x D x D 1;
lim
JD
lim
maxfjxk jg!0

maxfjxk jg!0

kD1

where x D 1 is the whole integration interval
...
)
In the second case f
...
We divide the interval 0 Ä x Ä 1 into n subintervals
of the same length  D 1=n with points xk D k (k D 1; 2; : : : ; n), and within each
one of them we choose the point k D xk right at the end of it
...
k / D

n
X

xk  D

kD1

n
X

k2 D 2

kD1

n
X
kD1

k D 2

1Cn
n;
2

where in the last passage we have used the formula for a sum of an arithmetic
progression (Sect
...
9, Eq
...
65))
...
Taking this limit in the calculated integral sum, In , we finally obtain
I D 1=2
...
3
...
k 1/ , or in the
middle, k D
...
Make sure that you get the
same result after taking the n ! 1 limit, in spite of the fact that the integral
sums for the given value of n are different in each case
...
4
...
10)

Problem 4
...
Using the same method (subintervals of the same length and a
particular, but simple choice of the points k within them), show that the integral
ˆ

1

x2 dx D

0

1
:
3

[Hint: you will need expression (1
...
] Try several different choices of the points k within the subintervals,
and make sure that the result is always the same in the  ! 0 limit!
Problem 4
...
Using the subintervals of the same length, prove the following
identity:
ˆ

1

e˛x dx D

1

1 ˛

...
61), and
calculate a limit at n ! 1, which contains 0=0 uncertainty
...
7
...
(2
...
]

sin xdx D 2:

4
...
2 Main Theorems
Of course, the above direct method of calculating integrals seems to be extremely
difficult
...
Fortunately, there is a much
easier way whose essence is in relating the integral to the derivative which manifests
itself in the main theorem of calculus to be discussed later on
...

We start from a more rigorous definition
...
g
...
By adopting a particular way of dividing the original interval a Ä
x Ä b into n subintervals and choosing the points k within each of them we shall
obtain a particular value of the integral sum
A
...
n/!0

n
X

f
...
n/ D max fk g is the maximum width of all subintervals for the given
n (assuming for definiteness that any k > 0)
...
1/; A
...
n/; : : :, of the values of the integral sums may tend to some limit
A as n ! 1
...
2
...
x/ on the
interval a Ä x Ä b exists (or the function f
...
n/g of the integral sums obtained by choosing various ways of dividing
the original interval into n subintervals, such that limn!1 
...
This
definition allows us to adopt immediately all main theorems related to the limit of
numerical sequences of Sect
...
2
...

It is possible to give an appropriate definition of the integral using the “
ı”
language as well: the integral on the interval a Ä x Ä b exists and equal A (or the
function f
...
n/ < ı
(i
...
any of them is shorter than ı) we have
ˇ n
ˇX
ˇ
f
...
11)

182

4 Integral

for arbitrary selection of points k within the subintervals
...
e
...
k / k < A C :

(4
...
x/ and the numbers a and b
...

Theorem 4
...
To be integrable, the function f
...
e
...
x/ Ä M
...

Proof (by contradiction):
...
x/ is unlimited from above
(i
...
it can be as large as desired) within the given interval
...
Then, when we divide the interval
into subinterval
...
k / there as large as we like
...
k / k of that subinterval into the integral
sum as large as necessary to disobey the inequality (4
...
Q
...
D
...
Consider a specific division of the interval a Ä x Ä b into n
subintervals, and let mk and Mk be the smallest and largest values of the function f
...
e
...
x/ Ä Mk
...
n/ D

n
X

mk  k ;

(4
...
14)

kD1

Aup
...
These sums correspond to a
specific choice of the points k within each subinterval as illustrated in Fig
...
2: for
the lower sum each point k is chosen where the function f
...
2 Main Theorems

183

Fig
...
2 To the definition of the upper (a) and lower (b) Darboux sums

value within each subinterval, and similarly for the upper Darboux sum
...
n/ will be not
lower than Alow
...
n/:
Alow
...
n/ Ä Aup
...
15)

Note that the Darboux sums only depend on the function f
...
Now let us add one more division point x0 between the points xk 1 and
k
xk
...
The values of mi and Mi for all i
except for i D k will be the same, however, the k-th subinterval now consists of
two, xk 1 Ä x < x0 and x0 Ä x < xk , with the lowest values of the function m0 and
k
k
k
0
00
m00 and its largest values Mk and Mk , respectively
...

For instance, if say mk is taken by the function f
...
0 /
...
This should remain true if we add an arbitrary
number of new divisions
...
2
...


We also need another theorem
...
3
...


184

4 Integral

Proof
...
Consider
1
next a completely different division in which some or all division points are different
from those in the first division
...
Let us construct the third division by adding all new
2
points from the second division to the first
...
From Theorem 4
...
However, exactly
3
1
the same combined division can be constructed by adding all new division points
of the first division to the second one
...
2, we can also write:
up
up
Alow Alow and A3 Ä A2
...
Therefore, we have:
Alow Ä Alow
1
3

up

and Alow Ä A3
3

up

up

and A3 Ä A2

H)

up

Alow Ä A2 :
1
(4
...
Similarly,
up

A1

up

A3

up

and A3

Alow
3

and Alow
3

Alow
2

H)

up

A1

Alow ;
2
(4
...
e
...
Q
...
D
...
This means
that all possible (i
...
obtained using various divisions) lower sums are limited from
above, and all possible upper sums are limited from below
...
These are called lower and upper Darboux integrals, respectively
...
18)

Note that Alow and Aup may correspond to two completely different divisions of the
interval
...
4 (Existence of the Definite Integral)
...
19)

i
...
for any positive there exists such positive ı that for any divisions
satisfying  D max fk g < ı we have 0 < Aup Alow <
...
2 Main Theorems

185

Proof
...
e
...
To prove the necessary condition, we should assume that
the integral exists and hence the condition (4
...
If the
integral exists and equals A than for any division satisfying  < ı follows
A

4

<˛
4

H)

˛

A<

4

˛<

and A

4

;

(4
...
Note that this inequality holds for any choice of
internal points k inside the divisions (subintervals)
...
Still, for the given , we can select the internal points in such a way that the
corresponding integral sum ˛ 0 would satisfy
˛ 0 < Alow C

4

H)

˛0

Alow <

4

:

(4
...
22)

We mentioned that inequalities (4
...

In particular, they are valid for the two internal point selections we have made above,
hence one can also write for them:
˛ 00

A<

4

˛0 <

and A

4

:

(4
...
19) is indeed satisfied
...
19) is satisfied we have to show that the integral exists
...
19) is satisfied, we have Aup Alow < ,
so that 0 Ä A
A < for any , meaning that A D A which we shall denote
simply as A
...
24)

On the other hand, for any choice of the internal points k we have the integral sum
˛ satisfying Alow Ä ˛ Ä Aup , which, when combined with inequality (4
...
However, Aup Alow < for any
division obeying  < ı due to condition (4
...
But, by virtue of condition (4
...
Q
...
D
...
x/
to be integratable
...
5
...
e
...
3) exists if f
...


Proof
...
25)

kD1

where mk and Mk are the minimum and maximum values of the function f
...
Because the function f
...
x0 / f
...

Let x0 be a point where f
...
Then, we can say that
ˇ 0
ˇf x

ˇ
f x00 ˇ D jmk

Mk j D Mk

mk D fk <

0

as long as xk < ı:

4
...
b

Alow D

n
X

a/, we can rewrite (4
...
b

a/ D ;

which corresponds to the sufficient condition for the integral to exist (from the
previous Theorem 4
...
Therefore, continuous functions do fully
satisfy that theorem
...
E
...

In fact it appears that piecewise continuous functions are also integrable
...
Moreover, the value of the integral does not change if the value of the
function f
...
4
...
3)
...
6 (Average Value Theorem)
...
x/ is continuous
between a and b, then
ˆ

b

f
...
b

a/ ;

(4
...


Proof
...
Let us now estimate the integral sum from below,
n
X

f
...
b

a/ D m ;

k D M
...
k / k Ä

kD1

where
obtain

D b

n
X

Mk D M

kD1

n
X
kD1

a is the length of the interval
...
x/dx Ä M

m Ä
a

H)

mÄ

a

f
...
27)

188

4 Integral

i
...
the value of the integral divided by the lengths of the interval , let us call
this ratio ; is bracketed by the minimum and maximum values of the function
on the interval: m Ä Ä M
...
18), there should exist a point within the interval at which the function
f
...
e
...
/, since continuous functions
take on all values continuously within the interval
...
26)
...
E
...

This theorem is called the average value theorem because it shows that f
...
x/dx
:
b a

f
...
28)

Theorem 4
...
It is usually necessary to be able
to put an upper and lower boundaries to an integral
...
x/dxˇ Ä
ˇ

ˆ

b

jf
...
x/dx Ä

jf
...
x/j dx:
a

(4
...
This follows immediately from the integral being an integral sum and the
inequality (1
...
k / xk ˇ Ä
jf
...
Q
...
D
...
3 Main Theorem of Integration: Indefinite Integrals
We mentioned above that it is possible to relate the definite integral to a derivative,
and this relationship forms a very powerful tool for calculating integrals which
bypasses calculation of the integral sums
...
Indeed, consider a continuous function f
...
x/ D
a

x

f
...
30)

4
...
It is always a good idea to follow this simple rule to avoid
mistakes
...
x/
...
According to the definition of the derivative,
F
;
x!0 x

F 0
...
x C x/

xCx

F
...
x0 /dx0

a

x

f
...
31)

a

Because of the additivity property (4
...
x0 /dx0 :

(4
...
4
...
e
...
32)
...
x/ is continuous,
and hence we can benefit from Theorem 4
...
4
...
32)

f
...
x C x/

x D f
...
33)

190

4 Integral

where is somewhere between x and x C x, i
...

number between 0 and 1
...
/ x
D lim
D lim f
...
x C #x/ D f
...
34)

F 0
...
x/ considered as a function of its upper limit
gives the integrand f
...
In this sense, the integration is inverse to differentiation
...
Indeed, what it
tells us is that in order to calculate an integral of f
...
x/ which derivative is exactly equal to f
...

Of course, if such a function is found, then any function F
...
x/ D f
...
This general
relationship between the integrand f
...
x/ whose derivative is
equal to f
...
x/dx D F
...
x/ D f
...
35)
Here F
...
x/ and C is an arbitrary
number (a constant)
...

Problem 4
...
Let F
...
x/
...
x/ can only differ by a
constant
...
Also note that the only function whose
derivative is equal to zero is a constant function
...
64)
...
Since
 ˛C1 Ã0
x
D x˛ for any ˛ ¤ 1;
˛C1
then we can immediately write this also as
ˆ
x˛C1
C C;
x˛ dx D
˛C1

˛¤

1:

(4
...
ln x/0 D x 1
...
ln jxj/0 D x 1
...
If, however, x < 0, then

...
ln
...
1/ D ;
x
x

4
...
e
...
Therefore, we arrive at the following indefinite integral:
ˆ

dx
D ln jxj C C:
x

(4
...
9
...
38)

ˆ
ex dx D ex C C;
ˆ

dx
D arctan x C C ;
1 C x2
ˆ

ˆ
p

dx
1

x2

(4
...
40)

(4
...
42)

Now we are finally at the position to derive the formula for calculating definite
integrals
...
We found above that its derivative, F 0
...
x/
...
x0 /dx0 D F
...
To find C, we
set x D a in the formula above and use the fact that the integral between identical
limits is equal to zero
...
a/
...
x0 /dx0 D F
...
a/:

a

It is now convenient to replace x as the upper limit with b, and we arrive at the main
formula of the integral calculus (the so-called Newton–Leibniz formula):
ˆ

b

f
...
b/
a

F
...
x/jb :
a

(4
...
x/ between the end points of the integration interval,
a and b; we shall be using it frequently
...
x/ whose derivative is equal to the integrand f
...
x/ at b and a provides us with the final result
...
x/ (ignoring
any constant C) and calculate the difference of it at b and a
...
x/ D x2 and
f2
...
In the case of f1
...
36): F1
...
We can ignore the constant to obtain:
ˆ

1

0

ˇ1 Â 3 Ã
x3 ˇ
x
x dx D ˇ D
ˇ
3 0
3 xD1
2

Â

x3
3

Ã

13
3

D
xD0

03
1
D ;
3
3

which is the same result as we got in Sect
...
1 using the method of the integral sum
...
x/ can be considered as a linear combination of f1
...
x/ D 1
(a constant function)
...
1

0/ D

2
3

1D

1
:
3

Problem 4
...
Verify the following results:
ˆ

1
1

2ex C x3 dx D 2eC2e 1 I

ˆ

a

x2

3

b dx D

0

a7
7

3a5 b
Ca3 b2 ab3 :
5

Formula (4
...
It may fail for
functions which are piece-wise continuous
...
2) which is discontinuous at x D 0,
as shown in Fig
...
1(b)
...
Using formally Eq
...
43), we then would have an
integral between 1 and C1 equal to zero:
ˆ

1
1

sgn
...
x/j1 1 D xjxD1


...
The obtained result makes perfect sense as
the sign function is an odd function, and integration of an odd function between
symmetric limits should give zero, see (4
...
Moreover, we can also split the integral

4
...
x/dx D

1

0
1

ˆ

...
C1/ dx D
...
43)
even for discontinuous functions
...
As our second example, let
us consider a function
f1
...
x/ ;

f
...
The indefinite integral of f1
...
x/ D ex C C,
while it is F2
...
x/
...
43):
ˆ

1
1

f
...
x/dx D
D1

0
1

ˆ
2ex dx C
2e

1

0

1

ex dx D 2ex j0 1 C ex j1 D 2 1
0

e

1

C
...
So, a general conclusion from these examples must be that if a function
makes a finite jump (has a discontinuity) at some point c which is within the
integration region, a < c < b, then one has to split the interval into regions where
the function is continuous (a Ä x < c and c Ä x Ä b), and then integrate separately
within each such region as we did above; applying the Newton–Leibniz formula in
those cases may give incorrect results
...
3 ln jxj C C2 /
3
D

x2 dx C 3

ˆ

dx
x

ˆ
2

sin xdx


...

In many cases already simple algebraic manipulations of the integrand may result
in a combination of elementary functions which can be integrated:
Ã
ˆ Â
ˆ
ˆ
2
dx
D x C 2 ln jxj C C;
1C
dx D dx C 2
x
x
Ã
ˆ
ˆ
ˆ
ˆ Â
e xC1
1
dx D
1 C x dx D dx C ex dx D x C ex C C:
e x
e

ˆ

xC2
dx D
x

Problem 4
...
Prove the following identities:
ˆ

ˆ

2

cot xdx D x

cot x C C ;

tan2 xdx D

x C tan x C C:

[Hint: express cotangent and tangent functions via sine and cosine functions
...
2x C 1/3 dx D 2x4 C 4x3 C 3x2 C x C C ;

ˆ
sin
...
x C a/ C C:

[Hint: use the binomial formula in the first and Eqs
...
46) and (2
...
]
ˆ
ˆ
1
1
x
x
sin2 dx D
...
x C sin x/ C C:
2
2
2
2
[Hint: use Eq
...
44)
...
Some of the most
frequent ones will be considered in the next section
...
4 Indefinite Integrals: Main Techniques

195

4
...
This requires finding a function
which, when differentiated, gives the integrand itself
...
That may be useful since by looking at this table “in reverse” it
is possible to find solutions of many indefinite integrals (and it is a good idea for
the reader to make his/her own table of various indefinite integrals while reading the
rest of this chapter)
...
We need
to develop various techniques for solving integrals
...
It is of course sufficient to discuss indefinite integrals, and this is what we
shall be doing in the first instance
...
4
...

Consider an integral:
ˆ
f
...
x/ C C;
where F 0
...
x/
...
t/ be a function of a variable t, so that F
...
x
...
Using the chain rule, we can write:
d
dF dx
F
...
t// D
D F 0
...
t/ D f
...
t/ D f
...
t// x0
...
t/ D
f
...
t// x0
...
x
...
t/
...
x
...
t/dt D F
...
t// C C D F
...
x/dx;

i
...
it is exactly the same integral we are seeking to solve
...
t/ and of

196

4 Integral

dx with x0
...
We note that the differential for x
...
t/dt and it is indeed denoted dx
...
1
Before we consider various cases in a more systematic way, let us first look at
some simple examples
...
The simplest choice is to introduce a new variable
2
t D x2 which turns the rather complex exponential function ex into a simpler one,
0
et
...

0
Alternatively, we can solve t D x2 for x
...
Here we
2
2
2
explicitly made use of the fact that dx is the differential of x
...
xdx/ D

1
e dt D
2
2
t1

It is easy to check that, indeed, 1 ex
2
integrand of I
...
t C 5/ =2 and hence x2 1 D t2 =4 C
5t=2C21=4
...
2x
16

C

5
...


4
...
x/) and
the connection of the differentials dt and dx
...
t/ (of course, only if needed and possible) together with other relevant
algebraic manipulations (if not too cumbersome) may also be written there
...
We shall be using this
method frequently
...
12
...
ax C b/ dx D
cos
...
ax C b/nC1
C C;

...
n C 1/

ˆ

n

ˆ

p

dx
a2
ˆ
ˆ

x2

D arcsin

cos
...
ax C b/ C C;
a

x
C C;
a
ˆ

1
dx
D ln jax C bj C C;
ax C b
a

tan xdx D

ln jcos xj C C;

cot xdx D ln jsin xj C C;
ˆ

p
1
x3 1 C x2 dx D
1 C x2
15
ˆ
ˆ

3=2

1
1
1
sin dx D cos C C ;
2
x
x
x
dx

...
x C 3/
3

3

ˆ

3x2

2 ;
ˆ

C C:

x2

x
1
dx
D arctan C C;
2
Ca
a
a

1
1
cos dx D
2
x
x

sin

1
C C;
x

198

4 Integral

Problem 4
...
Let the function f
...
e
...
x C X/ D f
...
Prove that for any a
ˆ

ˆ

aCX

f
...
x/dx:

0

a

(4
...
x/ D sin
...


4
...
2 Integration by Parts
This method is based on the following observation: if u
...
x/ are two
functions, then, as we know well,
...
Therefore, the function F
...
x/v
...
x/ D u
...
x/ C u0
...
x/:
ˆ

b

a

uv 0 C u0 v dx D uvjb ;
a

or, since the integral of a sum is a sum of integrals:
ˆ

b

a

ˆ

uv 0 dx D uvjb
a

b

vu0 dx:

(4
...
46)

a

where we introduced the differentials du D u0 dx and dv D v 0 dx of the functions u
...
x/, respectively
...
For indefinite
integrals the analogue expressions are the following:
ˆ

ˆ
udv D uv

ˆ
vdu

or

0

uv dx D uv

ˆ

vu0 dx:

(4
...


4
...
x/ is differentiated and, at the same time, instead of the function v 0 , we have v
...

This interchange may help in calculating the original integral in the left-hand side
...
x/, and various
“divisions” or “factorisations” of it into u
...
x/ may be possible; it is not
known which one would succeed, so sometimes different divisions are needed to try
and then see which one would work
...
In some cases more than one
integration by parts is required, one after another
...

´
As an example, let us calculate the indefinite integral I D xex dx
...
If first we try taking u D ex and dv D xdx, then
u0 D ex and v D x2 =2 (remember, the constant C is introduced at the last step and
hence is now ignored), so that we get
ˆ

ˆ
xex dx D

ex
xdx
ex
„ƒ‚… „ƒ‚… D „ƒ‚…
u

ˆ

u

dv

x2
2
„ƒ‚…
v

1
x2
ex dx D x2 ex
2 „ƒ‚…
2
„ƒ‚…
v

1
2

du

ˆ

x2 ex dx:

We see that, instead of getting rid of the x in the integrand, we raised its power to two
and got x2 instead
...
However, our unsuccessful
result may give us a hint as to how to proceed: it shows (if we look from right to left)
´
´
that the integral x2 ex dx can in fact be made related to the integral xex dx with the
lower power of x:
ˆ

2 x

2 x

x e dx D x e

ˆ
2

xex dx;

(4
...
x/, so that upon differentiation in the right-hand side we would
get u0 D 2x, i
...
one power less
...
It is easy to see that
...
xex /0
...
e
...

If we now need to calculate the integral with x2 ex , then two integrations by parts
would be needed: one to reduce the power of the x from two to one as in (4
...
14
...
2x C 5/3 e
ˆ
ˆ

2x

dx

x3 ex dx

x4 ex dx

4x3 C 36x2 C 111x C 118 e

Answer:

Answer:

Answer:

x4

x3

3x2 C 6x

4x3 C 12x2

2x

CC I

6 ex C C I
24x C 24 ex C C :

Problem 4
...
Using integration by parts, derive a recurrence relation In D
´
xn ex nIn 1 for the integral In D xn ex dx (where n D 0; 1; 2; : : :)
...
1/k

nŠ

...


4
...
16
...
x/e˛x dx,
Pn
...
ax C b/ dx and
´
Pn
...
ax C b/ dx, where Pn
...
Note that
more general arguments to the exponential, sine and cosine functions can indeed be
used as these arguments can always be replaced by t using the appropriate change
of variables; the polynomial will remain a polynomial of the same degree (but with
different coefficients)
...
It can also be used
for integrating functions which are products of a polynomial and ln x or any of the
inverse trigonometric functions as illustrated below:
ˆ

ˇ
ˇ
ˇ
ˇ
1
ˇ u D arctan x ; du D 1Cx2 dx ˇ
arctan xdx D ˇ
ˇ D x arctan x
ˇ
ˇ
dv D dx ; v D x
D x arctan x

ˆ

xdx
1 C x2

1
ln x2 C 1 C C;
2

where in the last integral we have made a change of variable t D 1 C x2 , dt D 2xdx
...
17
...
1
4

2 ln x/ C C :

The last example which we shall consider is of the following two integrals:
ˆ
IS D

ˇ
ˇ
ˇ
ˇ
u D ex ; du D ex dx
ˇ
ˇD
e sin xdx D ˇ
dv D sin xdx ; v D cos x ˇ
x

D

ˆ
e cos x C
x

ex cos xdx

ex cos x C IC

and
ˆ
IC D

ˇ
ˇ
ˇ u D ex ; du D ex dx ˇ
ˇ
ˇ D ex sin x
e cos xdx D ˇ
dv D cos xdx ; v D sin x ˇ
x

D ex sin x

ˆ
ex sin xdx

IS :

It is seen that a single integration by parts relates one integral to the other
...
cos x C sin x/

IS ;

4
...
cos x C sin x/ :
2

As a by-product, we also have
IC D ex sin x

IS D

1 x
e
...
x/ with several of exponential or trigonometric
functions is dealt with in more or less the same way:
ˇ
ˇ
ˆ
ˆ
ˇ
ˇ
u D x ; du D dx
x
ˇ D xIS 1 ex
...

The integral we have arrived at is a difference of the familiar IS and IC integrals,
which allows us to finally write:
Â
Ã
ˆ
1
1
1
xex sin xdx D xIS

...
1
2

x/ cos x C x sin x C C:

Problem 4
...
Calculate
Ä
ˆ
1
xex cos xdx Answer: ex f
...
x
2
Ä
1
Answer: ex
...
x
1/ f
...
x C 1/ sin xg C C I
1/ sin x C
...
19
...
1/m mŠ

...
n C 1/ In;m 1
...
20
...
1

x/n xm dx D

nŠmŠ
:

...
n C m C 1/
Jn 1;m by writing
...
1=x 1/ x
...
21
...
x/f 0
...
x/2 C
C
...
df =dx/ dx D f 0
...
Indeed,
ˆ

f
...
x/dx D

ˆ
fdf D

f
...


4
...
3 Integration of Rational Functions
In the case of rational functions f
...
x/=Qm
...
x/ and
Qm
...
As we saw in Sect
...
3
...
x/ D

1

...
x/ D

Ax C B

...
Before we consider an example of more complex rational
functions, let us first make sure that we know how to integrate the above two
functions
...
x/ (for k D 1) is trivial and leads to a logarithm
...
The integration of g1
...
4 Indefinite Integrals: Main Techniques

205

conjugate roots (see Sects
...
5 and 2
...
2)
...
x/ is easily integrated
...
x 1/ C 1
ˆ
ˆ
dt
tdt
C3
:
D2
t2 C 1
t2 C 1

2x C 1
dx D
2
x
2x C 2

ˆ

2x C 1

The first integral is calculated using the substitution y D t2 C 1, while the second
integral leads to arctan t:
ˆ
x2

2x C 1
dx D ln 1 C t2 C 3 arctan t C C
2x C 2
D ln x2

Problem 4
...
Calculate
Ä
ˆ
x 1
dx Answer:
x2 C 4x C 6
ˆ
x2

3x C 2
dx
6x C 10

2x C 2 C 3 arctan
...
3

x/

3
ln x2
2

6x C 10 C C :

Calculation of the integral gk
...
Firstly, we build
the complete square as in the previous case of k D 1 and make the corresponding
change of variables
...
t2 C a2 /k

...
t2

C

a 2 /k

D

1
2
...


2

206

4 Integral

The second integral, rk , is calculated with the following trick leading to a recurrence
relation whereby rk is expressed via the same type of integrals with lower values
of k:
ˆ

dt

rk D
1
D 2
a


...
t2

D

1
a2

ˆ
ˆ

dt

C

t 2 C a2

a 2 /k 1

t2 dt


...
t2

C

a 2 /k

D

rk 1
a2

1
Jk ;
a2

where the integral Jk we take by parts:
ˇ
ˇ
ˇ
ˇ
u D t ; du D dt
ˇ
ˇ
Jk D
Dˇ
ˇ D thk
k
ˇ dv D t t2 C a2
dt ; v D hk ˇ

...
1 k/
2
...
1 k/
2
...
1 k/

t
2
...
This method allows expressing recursively
the given rk to rk 1 , then the latter to rk 2 , etc
...
1=a/ arctan
...
This way it is possible to
express any of the gk integrals for any integer k via elementary functions
...
x2

2x C 2/2

ˆ
dx D

h
ˆ

D

2x C 1

...
t2 C 1/2

ˇ
ˇ
ˇt D x 1ˇ
ˇ
ˇ
i2 dx D ˇ
dt D dx ˇ

dt D 2h2 C 3r2 :

Here
ˆ
h2 D

ˇ
ˇ
ˆ
ˇ y D t2 C 1 ˇ
dy
ˇ
ˇD 1
Dˇ
D
2 C 1/2
dy D 2tdt ˇ
2
y2

...
t2 C 1/

4
...
t2 C 1/2

D

t2 C 1

t2


...
t2 C 1/2

D arctan t J2 ;

where the integral J2 we calculate by parts:
ˆ
J2 D
D

ˇ
ˇ
ˇ
ˇ
u D t ; du D dt
ˇ
ˇ
Dˇ
ˇ
2
2 C 1/2
ˇ dv D t t2 C 1
dt ; v D h2
...
t
ˆ
1
t
1
1
t
dt
t
C
D
C J1 D
C arctan t:
2 C 1/
2C1
2 C 1/
2 C 1/
2
...
t
2
2
...
x2

2x C 2/

2

dx D

3
3x 5
C arctan
...
x2 2x C 2/
2

1/ C C:

Problem 4
...
Calculate
ˆ

x

"

1


...
x2 C 4x C 6/2
ˆ

3x C 2

...
3
2

x/

#

CC I

24 7x
CC :
2
...
The calculation consists of first
decomposing the rational function into elementary fractions as discussed at the
beginning of this section and especially in Sect
...
3
...

As an example, we calculate the integral
ˆ
JD

3x2 C x

...
x2

1
C 1/

dx:

We start by decomposing the fraction
...
(2
...
x 1/
ˆ
ˆ
1
xdx
3
dx
2
2
...
x

1
arctan x C C:
2

ln x2 C 1

1/

Problem 4
...
Calculate the following integrals of rational functions:
ˆ

Ä

x2 x C 5
dx
1/
...
2x

C
ˆ

x3 C x2 C 1
2


...
x2

C 4/

dx

ˆ

ˆ

dx
1 x3

dx
1 C x4

x2 dx
1 C x4

ˆ

Ä
Answer:

19
ln j2x
50

Ä
Answer:
C

ˆ

11
3
C
ln j2 C xj
5
...
x

1/

C

41
x
arctan
50
2

3
19
ln x2 C 4 C
ln jx
25
25

1j C C I

1
2x C 1
1 1 C x C x2
p arctan p
CC I
C ln
6

...
4 Indefinite Integrals: Main Techniques

209

4
...
4 Integration of Trigonometric Functions
Integrals containing a rational function R
...
The trick is
based on replacing the sine and cosine functions by rational expressions using the
substitution
t D tan

x
;
2
D

so that dt D
1 2
t C 1 dx
2

dx
2 cos2
H)

x
2

D

x
x
sin2 2 C cos2 2
dx
x
2 cos2 2

dx D

2dt
;
1 C t2

(4
...
(2
...
50)

It is clear that any rational function of the sine and cosine functions will be
transformed by means of this substitution into some rational function of t
...

As an example we shall calculate the following integral:
ˆ

ˇ
ˇ ˆ
x
ˇ
ˇ
1 C 2t= 1 C t2
2dt
1 C sin x
t D tan 2
ˇ
dx D ˇ
ˇ dx D 2dt= 1 C t2 ˇ D
1 C cos x
1 C
...
1 C t2 / 1 C t2
Ã
ˆ
ˆ Â
2t
1 C t2 C 2t
D
dt D
dt
1C
1 C t2
1 C t2
ˇ
ˇ
ˆ
ˆ
ˇ
ˇ y D 1 C t2 ˇ
ˇ
dy
2tdt
ˇDtC
D t C ln jyj D t C ln ˇ1 C t2 ˇ
DtC
Dˇ
ˇ dy D 2tdt ˇ
1 C t2
y
ˇ
ˇ
ˇ
ˇ
ˇ 1 ˇ
x
x
xˇ
xˇ
x
ˇ
ˇ
ˇ
ˇ
2 ln ˇcos ˇ C C:
D tan C ln ˇ1 C tan2 ˇ D tan C ln ˇ 2 x ˇ D tan
ˇ cos ˇ
2
2
2
2
2
2

We have introduced the constant C only at the very last step
...
25
...
sin x C cos x/ C ln
...
1 C sin x/2
3 cos 2 C sin 2
ˆ

ˆ

i
h
x
dx
Answer: tan C C
1 C cos x
2
Ä
ˆ

...
1 C cos x/
3
...
2x/
CC :
4 cos
...

For instance, if the rational function R
...
e
...
sin x; cos x/ D R
...
2 C sin x/CC:
ˇ dt D cos xdx ˇ D
2 C sin x
2Ct

4
...
sin x; cos x/ D cos x=
...
If, instead, the rational function is odd with respect to the sine
function, R
...
sin x; cos x/, then the substitution t D cos x is
found to be useful:
ˇ ˆ
ˇ
ˆ
ˇ t D cos x ˇ
1 t2
sin3 x
ˇD
ˇ
dx D ˇ

...
26
...
(2
...
61) may be found useful, especially if the trigonometric functions
of different arguments are present
...
52), we can
easily take the following integral:
ˆ

ˆ
sin
...
9x/ dx D
D

1
Œcos
...
2x/
4

cos
...
20x/ C C:
40

Sometimes, by reducing the power of the sine or cosine functions via sine
and/or cosine functions of a different argument, integrals containing powers of the
trigonometric functions can be calculated:
ˆ

2

ˆ

cos xdx D
ˆ

sin2 xdx D

ˆ

x
1
1
Œ1 C cos
...
2x/ C C;
2
2
4
1
Œ1
2

cos
...
2x/ C C;
4

where we made use of Eq
...
44)
...
(2
...
3x/
dx D
sin x

ˆ
ˆ

D

3 sin x cos2 x
sin x
3 cos2 x

sin3 x

dx

sin2 x dx D x C sin
...
This
can always be done if the arguments are of the form nx with n being an integer
...
27
...
6x/ cos
...
5x/
dx
sin x

Answer:

Ä
Answer:

1
cos x
2

x C sin
...
11x/ C C I
22
1
sin
...
4
...
ex / of the exponential function can also be transformed, using
the new variable t D ex , into an integral of the rational function R1
...
t/=t, and
hence can always be integrated
...
1 2t/

The last integral is calculated, e
...
, by decomposing the rational function into
elementary functions A=t and B=
...
1

2ex /2 C C:

2 ln j1

2tj D ln ex

2 ln j1

2ex j

4
...
28
...
x/dx ŒAnswer: cosh
...
x/dx ŒAnswer: sinh
...
cosh x/ C C D

tanh
...
x/dx
ˆ
ˆ

tanh2
...
x/dx Answer: ln
...
4
...
e
...
However, there exists a class of irrational functions which, upon
an appropriate substitution, can be transformed into an integral with respect to
a rational function, and hence are integrable
...
x/1=m D R x; m y
...
x/ (only the latter argument of course makes the whole function irrational)
...
x/ D
...
gx C f / is a simple rational function
of x with some constant coefficients a, b, g and f
...
e
...
x/ could be in any rational power n=m (with
integer n)
...
t/ D

ftm b
;
gtm C a

(4
...
Indeed, in this case,
s

ˆ
R x;

m

!
ˆ
ax C b
dx D R
...
t/dt;
gx C f

(4
...
t/ is a rational
function of t and R is rational with respect to both of its arguments
...
x; z/ D
...
z C 1/,
z D y1=2 (i
...
m D 2) and y D x C 1:
ˇ
ˇ
ˇ ˆ
ˇ 2
ˆ p
ˇy D t C 1ˇ
ˇt D x C 1ˇ
xC1 1
t 1
ˇ
ˇ
ˇD
ˇ
2tdt D ˇ
p
dx D ˇ
dy D dt ˇ
2tdt D dx ˇ
tC1
xC1C1
Ã
ˆ Â
ˆ
2
2

...
y 2/ dy D 2
y 3C
dy D y2 6y C 4 ln jyj
D
y
y
Á
Á
p
p
D x 4 1 C x C 1 C 4 ln 1 C x C 1 C C;
p
since y D 1 C t D 1 C x C 1
...
x 1/
...
t
1/
!
ˇ
ˇ
Ã
ˆ Â 2
ˆ
ˇ1 C tˇ
6t
1
t
dt
ˇ
ˇ
D ln ˇ
D t
dt D 2
3t2
t2 1
1 tˇ

...
x 1/
...
x 1/
...
x 1/
...
x 1/
...
4 Indefinite Integrals: Main Techniques

215

where the constant ln 3 has been dropped as we have to add an arbitrary constant C
(introduced at the last step) anyway
...
29
...
t 1/2
3

dx

q
3

...
x

1/2

where t3 D

ˆ Â

x 1
xC1

Ã2=3
dx

Ä
Answer:

x 1
I
xC1
2t2
t3

1

2t C 1
4
C p arctan p
3
3

x 1
2 1 C t C t2
ln
:
C C ; where t3 D
3
xC1

...
e
...
53)

Euler studied these types of integrals and proposed general methods for their
calculation based on special substitutions which bear Euler’s name
...
Three possibilities can be considered, although in practice only two
would be needed
...
54)

can be used (with either sign)
...
30
...
Demonstrate by the direct
calculation that the following identities are valid:
(continued)

216

4 Integral

Problem 4
...
55)

After the t integral is calculated, one has to replace it with an appropriate
p expression via x, which follows immediately from Eq
...
54) as t D
p
ax2 C bx C c
ax
...
54) using
the minus sign for definiteness:
ˇp
ˇ
ˆ p
ˇ
2t C t2 C 2 ˇ
2t C t2 C 2
2 C 2x C 2dx D ˇ x2 C 2x C 2 D
; dx D 2
x
dtˇ
ˇ
2
...
1 C t/2 ˇ
ˇ
ˇ
ÃÂ
Ã
ˆ
ˆ Â
2
ˇy D 1 C tˇ
2t C t2 C 2
2t C t2 C 2
1
2t C t2 C 2
ˇ
dt D
dt D ˇ
D
ˇ dy D dt ˇ
2
...
1 C t/2

...


Problem 4
...
Using this method, calculate the following integrals:
ˆ

dx
p
2 C 2x C 2
x
ˆ

xdx
p
x2 C 2x C 2

h
Answer:

ˇ
ˇ
i
p
ˇ
ˇ
ln ˇ1 C x C x2 C 2x C 2ˇ C C I

h
Answer:

p
x2 C 2x C 2

ˇ
ˇ
i
p
ˇ
ˇ
ln ˇ1 C x C x2 C 2x C 2ˇ C C :

4
...
56)

Problem 4
...
Prove that in this case the required transformations are
given by:
xD

b

p
2 ct
;
t2 a

dx D 2

p
bt
ax2 C bx C c D
p

and t D

ax2 C bx C c

bt ˙

p 2 p
ct ˙ ca


...
57)

p Á
c =x
...
x

x1 /
...
x

x1 /

(4
...


Problem 4
...
Prove that in this case the required transformations are as
follows:
xD

t 2 x1
t2

and t D

p
2a
...
x1
; dx D
dt and ax2 C bx C c D
2
2
a
t2

...
x

x2 / t
;
a
(4
...


p
p
Note that if x1 D x2 (repeated roots), then ax2 C bx C c D a jx x1 j (a
positive value of the square root is to be assumed) and the square root disappears, i
...

the integrand is a rational function without any additional transformations
...
x
p
D a
...
x

x1 /
...
x

x2 /

x
x

x2
x2

x1
;
x2

i
...
the substitution of the type (4
...
x x1 / =
...

Finally, it is easy to see that Euler’s second substitution is in fact redundant
...
If, however,
a < 0, then the parabola corresponding to the square polynomial is either located
entirely below the x axis (or just touches it at a single point) or its top is above it
...
In the latter case there must be two real roots x1 and x2 , in which
case the third substitution becomes legitimate
...

The consideration above leads us to a conclusion that all integrals with the
Á
p
function R x; ax2 C bx C c can be transformed into a rational form and hence
integrated in elementary functions
...
For instance, transforming to the complete square and then
performing a straightforward change of the variable may suffice as illustrated in
the example we have already considered above using a different method:
ˆ

ˇ
ˇ ˆ
ˇt D x C 1ˇ
dt
ˇ
ˇD
D
Dˇ
p
q
p
2 C 2x C 2
dt D dx ˇ
x
1 C t2

...
31
...
34
...
2x 1/ dx
p
x2 C 2x C 2

h
Answer:

p
2 x2 C 2x C 2

3

ˇ
ˇ
i
p
ˇ
ˇ
ln ˇ1 C x C x2 C 2x C 2ˇ C C :

4
...
Of
course, these particular integrals can always be taken using the Euler’s substitutions,
but simpler trigonometric substitutions may result in less cumbersome algebra
...
g
...
x=a/2 ˇ D a2
1
ˇ
ˇ

sin2 t cos tdt

Ä

1
a2
t
C sin
...
60)

Note that it is legitimate here to choose the positive value of the cosine function
during intermediate manipulations since a Ä x Ä a and hence one can choose
=2 Ä t Ä =2 when cos t 0
...

Problem 4
...
Calculate the following integral:
ˆ p

x2

a2 dx

Ä
Answer:

xp 2
x
2

a2

p
a2 ˇ
ˇ
ln ˇx C x2
2

ˇ
ˇ
a2 ˇ C C :

Problem 4
...
Calculate the integral:
ˆ p
x2 C a2 dx

Ä
Answer:

ˇ
p
xp 2
a2 ˇ
ˇ
ˇ
ln ˇx C x2 C a2 ˇ C C :
x C a2 C
2
2
(4
...
37
...
x/ C C:

(4
...
5 More on Calculation of Definite Integrals
As we have already discussed in Sect
...
3, it is easy to calculate a definite integral if
the corresponding indefinite integral is known, and many methods to help with this
task have been reviewed in the previous Sect
...
4
...


4
...
1 Change of Variables and Integration by Parts
in Definite Integrals
We learned two important techniques of calculating indefinite integrals in Sect
...
4,
a change in the integration variable and integration by parts, and of course these can
be used for calculating definite integrals as well
...

´b
Change of Variables
...
x/dx
...
t/ is used, then an integral with
respect to the variable t is obtained, and, if we are lucky, the t-integral is calculated,
the t is replaced back (by solving the equation x D x
...
Two comments are in order here as far as definite integrals are concerned
...
The benefit of this is that one can avoid
calculating t via x and hence replacing t back with the x; instead, the boundary values
of t can be immediately used after the t-integral is calculated to get the required
result (which is a number)
...

Secondly, one has to make sure that the substitution x D x
...
t/ is also continuous
...
The following examples illustrate this subtle point
...
5 More on Calculation of Definite Integrals

221

Of course, this integral is zero because the integrand is an odd function and the
limits of integration form a symmetric interval around x D 0, see also Eq
...
8), i
...

the above obtained result makes a lot of sense
...
The same result as before
has been obtained; good! However, let us now consider the integral of x2 instead:
ˇa
ˆ a
x3 ˇ
2
2
I2 D
x dx D ˇ D a3 ;
ˇ
3 a
3
a
while if we used the substitution t D x2 , the result would have been zero as the two
limits will be the same again! This paradoxical result originates from the fact that the
p
substitution we used defines a multivalued function t D ˙ x, and this contradicts
one of the necessary conditions needed for the substitution to work
...
The same is true for I1 ; the correct result
we obtained for I1 was accidental
...

Another caution needs to be given concerning the requirement that both the
function x
...
t/ are to be continuous within the integration
interval
...
When x D
we have t D 0;
however, when x D , the same result t D 0 is obtained
...
Therefore, formula (4
...
4
...
3 for more details)
...
The subtle points here are that in the first integral
the upper limit corresponds to x D 2
0 (as x < 2 ) which results in t D
tan
...

The above examples indicate that care is needed when applying a substitution;
the latter should satisfy a number of conditions, and must satisfy them along the
whole integration interval as otherwise the result may be wrong!
Integration by Parts
...
4
...
2
...
g
...
1/ sin
...
cos x/j1 1 D 0;

which is to be expected as the function x cos x is odd and the interval is symmetric
...

Consider, for instance, the following integral:
ˆ
In D

=2

cosn xdx;

(4
...
n 1/ cosn 2 x
...
n 1/
cosn 2 x sin2 xdx
„
ƒ‚
…
0
D0

ˆ

D
...
n

1/ In

=2

cosn

0
2


...
5 More on Calculation of Definite Integrals

223

which gives an equation for In , namely:
1

n

In D

n

In 2 :

´ =2
Since I0 D 0 dx D =2, we obtain, for instance: I2 D
...
4 1/ I2 =4 D 3 =16, etc
...
3

0

=2

cos xdx D sin xj0

1/ I1 =3 D 2=3, I5 D
...


Problem 4
...
Prove that for even n D 2p,
I2p


...
p 1/Š
4

Â

2p

1
p

Ã
;

while for odd values of n D 2p C 1,
I2pC1 D

4p
...
2p C 1/Š

Problem 4
...
Show that the same recurrence relation is obtained for the
integral
ˆ
Jn D

=2

sinn xdx;

0

and that for both even and odd values of n the same results are obtained as
for In
...
5
...
/ and b
...
x; / itself may depend on some parameter, let us call it ,
ˆ
I
...
/

a
...
xI /dx;

(4
...
We know from Sect
...
3 that if only the upper limit
depends on as b D , one obtains I 0
...
/, i
...
the value of the integrand
at x D
...
(4
...

We start by writing the corresponding definition of the derivative:
I
...
/

I 0
...
xI C  /dx
aCa

ˆ

f
...
xI C  /dx C

D
aCa

ˆ

ˆ

b

f
...
xI C  /dx
a

bCb

C

b

f
...
xI C  /
a

aCa

f
...
xI C  /dx
a

ˆ

bCb

C

f
...
x/ and b D b0
...
According to our discussion in Sect
...
3
(see specifically Eq
...
33)), we can write for the second and the third integrals:
ˆ

aCa

f
...
a1 I C  /a D f
...
a1 I /

a0
...
xI C  /dx D f
...
b1 I / C f 0
...
/ ;

a

ˆ

bCb

b

where a < a1 < a C a, b < b1 < b C b and in the second passage in both lines
we have made use of the Lagrange formula (3
...
a1 I C  / D f
...
a1 I /
0

f
...
b1 I / C f
...
5 More on Calculation of Definite Integrals

225

with < < C  and < < C 
...
xI / with respect to its second variable,
...
xI C  / f
...
a1 I / C f 0
...
/


a
C f
...
b1 I /

b0
...
/
ˆ b
df
...
bI / b0
...
aI / a0
...
xI /dx D
(4
...
/
d
a
It is seen that this result indeed generalises the one we obtained earlier in Sect
...
3:
if a and f do not depend on and b D , formula (4
...

Sometimes, by introducing a parameter artificially, the integral of interest may be
related to another; if the expression for the latter integral is known, this trick allows
us to calculate the former
...
!x/ dx:
0

Of course, this can be calculated by doing three integrations by parts; instead, we
shall consider a function
ˆ =2
Á
1
;
J
...
!x/ dx D sin !
!
2
0
and notice that upon differentiation of the cosine function cos
...
!/ three
times with respect to !, we get x3 sin
...
Therefore, we can write:
Ä
Á
d3
d3 1
ID
sin !
J
...
2/ h
 Ã
...
1/
1
1
sin !
C3
D
sin !
!
2
!
2
 Ã
...
2/
Ái
...
5
...
x; / of two variables
...
(4
...


226

4 Integral

D

6
!4
1
!

Á
6
3
cos !
C 2
2
!3 2
2
!
Á3
Á
cos !
;
2
2

sin !

Á

C

Á2
2

sin !

Á
2

which is the required expression (which can still be simplified further if desired)
...
47)
...
40
...
t

/ v
...
t/
particle velocity, and K
...
˛t/, the so-called friction kernel which
we assume decays exponentially with time
...
The integral here means that the particle motion
depends on all its previous velocities, i
...
the memory of its motion is essential
to correctly describe its dynamics
...
g
...
Now, denote the integral term as a function y
...
5
...
x/ does not have singularities anywhere
within the integration interval a Ä x Ä b, including the boundary points themselves
...

We shall start by considering the upper limit being C1
...
b/ D

b

f
...
5 More on Calculation of Definite Integrals

227

which we assume exists for any finite b
...
b/, with respect to its upper limit b
...
x/dx D lim

b!1

a

f
...
66)

a

Let us state the definition of its convergence
...
x/dx is said to converge to a number F, if for any
such A > a, that
ˇ
ˇ
ˇF
ˇ

ˆ

A

a

ˇ
ˇ
f
...
x/dx be close (within ) to F
...
x/dxˇ < ;
ˇ

´1
stating that the integral A f
...

Similarly, one defines the improper integrals with the bottom limit being 1,
ˆ

ˆ

b
1

b

f
...
x/dx;

(4
...
x/dx D lim

b!1

ˆ

Ã
f
...
68)

a

Problem 4
...
Give the definition of the improper integral with the bottom
limit being 1
...
(4
...
In each case the integrals may either converge (the
limits exist) or diverge (are infinite)
...
x/ is a non-negative function,
then any of the above improper integrals corresponds to the area under the curve of
the function within the corresponding intervals a Ä x < C1, 1 < x Ä b or
1 < x < C1, respectively
...


228

4 Integral

When calculating the improper integrals one can still use the Newton–Leibnitz
formula (4
...
Indeed, if F 0
...
x/, then, for instance,
ˆ

ˆ

1

f
...
x/dx D lim ŒF
...
a/

b!1

a

D lim F
...
a/ D F
...
a/;

(4
...
x/dx D lim

ˆ

f
...
a/ D F
...
b/

a! 1

b!1

lim
...
b/

i
F
...
1/:

(4
...
As our first
example let us calculate the integral which has a finite upper limit:
ˆ

b

e x dx D

I
...
Therefore,
ˆ 1
e x dx D lim 1 e b D 1:
b!1

0

As the limit exists (i
...
it is finite), it is to be taken as the value of the integral
...
71)
x
1
where is positive
...
In this case
ˆ
I

>1

D lim

b!1

D

1
1

1

b

ˇb
dx
x C1 ˇ
ˇ D 1
D lim
b!1
x
C 1 ˇ1
1
D

1
1

> 1 (strictly larger than
Â
lim

b!1

Ã

1
b

1

1

;

since the power
1 of b in 1=b 1 is positive and hence this term tends to zero in
the limit
...
ln b
1
b!1
b!1
x

ln 1/ D lim ln b D 1;
b!1

4
...
e
...
Finally, if 0 <
ˆ
I

<1

b

D lim

b!1

1

< 1, then

ˇb
dx
x C1 ˇ
ˇ D 1
D lim
b!1
x
C 1 ˇ1
1

lim b1

b!1

1 D 1;

and the integral diverges again
...
71) only converges
for strictly > 1
...
cos x/jb D lim
...

The above examples tell us that an improper integral can be calculated in exactly
the same way as the corresponding proper ones, and the value of the integral is still
given as the difference of the function F
...
The
only difference is in working out the value(s) of F
...


Problem 4
...
Check the values of the following improper integrals using
their appropriate definitions:
ˆ

1

1
e sin xdx D I
2
ˆ 0
ˆ
x
e dx D 1 I

ˆ

1

x

0

1

0
1

1=

dx
D I
1 C x2
2

1
1
sin dx D 2:
x2
x

Problem 4
...
Using integration by parts, derive a recurrence relation for the
following integral5 :
ˆ
€
...
n C 1/ D nŠ
...


230

4 Integral

It is also possible to define an integral of a function which has a singularity6
either at the bottom and/or top limits, and/or between the limits
...
x/
is singular at the top limit b, i
...
f
...
x/dx D lim

!0

a

f
...
72)

a

Here the function is finite for a Ä x Ä b
for any > 0, and hence the integral
before the limit is well defined
...

Similarly, if the singularity of f
...
a/ D ˙1,
then we consider
ˆ

ˆ

b

b

f
...
x/dx:

(4
...
x/dx D lim

1 !0

a

ˆ

1

b

f
...
x/dx:
cC

(4
...

It is essential here that both limits are taken independently; this is indicated by using
7
1 ¤ 2 in the two integrals
...
x/ at a single isolated point x D c will not change the
value of the integral
...
(4
...
x/ at x D c does not appear in this
case at all; it could in fact be different from either of the two limits limx!c˙0 f
...

As an example of a singularity inside the integration interval, consider the
function f
...
This function is singular at x D 0, so that
whether it is integratable around this point needs to be specifically investigated
...
4
...
4
...


:

4
...
e
...
In the case of D 1, we have
ˆ
I

1

D

D1

0

dx
D lim
!0
x

ˆ

1

dx
D lim
...
e
...
In this case
ˆ
I

<1

D
0

1

dx
D lim
!0
x

ˆ

1

dx
1
D lim
!0 1
x

1

1

D

D 1
...
So,
the above integral only converges for 0 < < 1
...
44
...
Answers:
2
...
]

=2,

,

Problem 4
...
Consider a function f
...
If the
integral of f
...
x/
is continuous in any of the singular points (F 0
...
x/), show that formally
the Newton–Leibnitz formula (4
...
e
...
x/dx D F
...
a/:

(4
...
]

8

It is said that it diverges logarithmically
...
46
...
x/ is continuous at all singularity points as otherwise
the limits from the left and right at the singularity points would be different and
not cancel out
...
Show using
the definition of the improper integral of a function with singularity at x D 0
that
ˆ

1

x
1

1=3

dx D 0:

On the other hand, using the Newton–Leibnitz formula (4
...
1/2=3 D
...
x/ D x2=3 is continuous at x D 0
...
x/ may not always be calculated using the Newton–Leibnitz
formula as this requires knowing the indefinite integral, i
...
the function F
...
x/ D f
...
Still, it is frequently useful to know if the given improper integral
converges even though its analytical calculation is difficult or even impossible
...
Below we shall consider several useful criteria which can be
used to test the convergence of improper integrals
...
8 (Comparison Test)
...
x/
and g
...
x/ Ä g
...
Both functions may also have a finite number of singularities within the
´b
´b
interval
...
x/dx converges, then the integral a f
...


Proof
...
We define two functions of the upper limit:
ˆ
F
...
x/dx
a

and G
...
x/dx:
a

4
...
x/ and g
...
X/ and G
...
Indeed, for instance by taking any positive X > 0 we can write:
ˆ

XCX

F
...
x/dx D

X

ˆ

a

ˆ

XCX

D F
...
x/dx C

f
...
x/dx D F
...
X/;

X

where the integral F is obviously positive since f
...

Now since f
...
x/, then F
...
X/ as integrals are defined as the integral
sums
...
x/ converges
...
X/ when X ! C1 exists; moreover, since G
...
e
...
C1/ < M with some finite M
...
X/ is never larger than G
...
g
...
Hence, the function F
...
19 in Sect
...
4
...
e
...
C1/ can only be finite, i
...
the integral of
f
...
This proves the first part of the theorem that from convergence
of the integral of g
...
x/
...
x/ diverges
...
x/ converges, the integral
of f
...
However,
this contradicts with the fact that the integral of f
...
Therefore,
our assumption concerning the convergence of the integral of g
...
Q
...
D
...
Note that the theorem cannot establish convergence of the integral of f
...
x/ diverges; the former
integral may either diverge or converge and this requires further consideration
...
x/ D
e x sin2 x containing the infinite upper limit
...
x/ D e x such that for any x we have f
...
x/ since
sin2 x Ä 1
...

a e

234

4 Integral

Problem 4
...
Investigate the convergence of the integral
ˆ 1
x ˛
dx
1 C x2
1
for various values of a positive parameter ˛
...
Answer: ˛ > 1
...
48
...
]
When investigating convergence, it is useful to approximate the integrand around
the singularity point to see if the integral with the limits just around that singularity
converges
...
x/ D x ˛ = 1 C x2 and f2
...
In the first case we have to
investigate the convergence at x ! 1
...
This function is well defined at x ! 1 (and tends
to zero there) only if ˛ > 1
...

For small x we can replace sin x with x, which results in the integrand proportional
to x 1=2C2 D x3=2 , which is integrable around x D 0
...
8 requires the two functions to be necessarily
non-negative
...
x/ changes its sign, then this test cannot be
applied
...
9
...
x/dx (where either the
function f
...
This integral converges if the integral a jf
...
x/ converges
...
Indeed, since the integral of jf
...
x/j dx < M with
some positive M
...
(4
...
4
...
x/dxˇ Ä
jf
...
e
...
E
...


a

´b
a

f
...


4
...
x/j dx
´b
follows convergence of a f
...
This condition of convergence is stronger than an ordinary convergence
´b
´b
of the integral a f
...
x/j dx may not converge even if the
´b
integral a f
...

´b
Problem 4
...
Consider an improper integral a f
...
x/ bounded from below and above, jg
...
Prove
´b
that the improper integral a f
...
x/dx also converges absolutely
...
50
...
x/ is limited from below and
1 g
...
[Answer: converges for > 1
...
51
...
x/e dx
...
]
Problem 4
...
Investigate the convergence of the following integrals:
ˆ
0

1

ˆ

sin x
dx and
1 C x2

1

1

[Answers: converges; converges for

cos x
dx with
x

> 0:

> 1
...
5
...
x/ which lies within the interval a < x < b and has there a
singularity at an internal point c (here we assume that a and b are either finite or
infinite)
...
(4
...
e
...
If this improper integrals
diverges, it might in some cases be useful to constrain the limiting procedure to
having 1 D 2 D and hence perform a single limit ! C0
...
x/dx or
f
...
x/dx C
f
...
76)
P
a

a

!0

a

cC

This value may be assigned to some of the improper integrals which otherwise
diverge
...

Consider, for instance, the following improper integral (a < c < b):
ˆ b
dx
:
ID
x c
a

236

4 Integral

In the sense of the formal definition of Sect
...
5
...
1/
a

c
2

c
C
...
1/

At the same time, its principal value is well defined:
Ĉ
I D lim

!0

a

c

ˆ

dx
x

c

C

b

cC

Ä
D lim ln

dx
x

!0

c

c

a

C ln

b

c

D ln

b
c

c
;
a

as the two terms with ln cancel out exactly
...
53
...
x c/ n dx with n > 1
being a positive integer and c lying between a and b
...
When does this integral exist? [Answer: n must be odd;
i

...
a c/1 n =
...
]
It can also be shown that a more general integral
ˆ
ID
a

b

f
...
x/ is continuous everywhere between a and b, is well defined
in the principal value sense
...
x/
x

f
...
c/
c

ˆ
a

b

dx
x

c

:

Here the first integral is not improper anymore since the integrand is a continuous
function at x D c with the limiting value of f 0
...
3
...

The regularisation we performed above, when discussing the improper integrals
with the integrand having a singularity, is not the only possible
...
68) with both infinite boundaries
...


4
...
x/dx D lim
f
...
77)
a!1

1

a

As an example, consider the following integral:
ˆ

1

ID
1

ˆ

xdx
D lim
2C1
a! 1
x

D lim

lim

a! 1

b!C1

lim

b!C1

1
ln b2 C 1
2

D lim ln b2 C 1
b!1

a

b

xdx
C1

x2

ln a2 C 1

lim ln a2 C 1 :

a! 1

This expression is ill-defined as it contains a difference of two infinities
...

´1
Problem 4
...
Investigate whether improper integrals
1 sin xdx and
´1
cos xdx exist in the principal value sense
...
]
1
Problem 4
...
Show that
ˆ
P

E
E


...
x

b/

D

1
a

ˇ
ˇ
ˇ
...
E C b/ ˇ
ˇ
ln ˇ
b ˇ
...
E b/ ˇ

irrespective of whether the points a and/or b are inside or outside the
integration interval
...
6 Applications of Definite Integrals
Here we shall consider some applications of definite integrals in geometry and
physics
...
g
...


238

4 Integral

The underlying idea in all these applications is the same; therefore, it is essential
to get the general principle, and then it would be straightforward to apply it in any
other situation: we have to sum up small quantities of something to get the whole
of it, e
...
small lengths along a curved line to calculate the length of the whole
line, small volumes within some big volume bounded by some curved surface or
the work done over a small distance summed over the whole distance
...
The problem here is that the formula
A D Fqall would only make sense if F was constant; if this is not the case and
F
...

So, the idea is to discretise the values of q and construct the Riemann sum similar to
the one in Eq
...
1) which we discussed when doing the calculation of an area under
the curve
...
Most easily, these points can be chosen
equidistant, i
...
qi D qini C iq, where i D 0; 1; 2; : : : ; n and q D
...

Then, the quantity of interest corresponding to q is approximately calculated as
Ai ' F
...
e
...
q/ can be approximated as a constant F
...
The final quantity
is obtained by summing up all the contributions,
A'

n
X

F
...
q/
...
6
...
x/ within
the interval a Ä x Ä b as shown in Fig
...
4(a)
...
To do this, we first divide the line into n small segments by points
Ai , where i D 0; 1; 2; : : : ; n, and then connect the points by straight lines (shown
as dashed)
...
xi /2 C
...
4
...
Since the change yi can be related to
the derivative of the function y D f
...
e
...
xi /xi (recall our
discussion in Sect
...
2, Eq
...
9), and Sect
...
8), we can write
q
q
li '
...
f 0
...
f 0
...
6 Applications of Definite Integrals

239

Fig
...
4 (a) A curved line between points
...
a// and
...
b// is divided by points Ai (i D
0; : : : ; n) into n curved segments that are then replaced by straight lines (shown by the dashed
lines)
...
Its length is approximated by the length li of the
straight line connecting the points Ai and AiC1

Summing up these lengths along the whole curve and taking the limit of n ! 1 (or
maxi fxi g ! 0), we arrive at the useful exact formula for calculating the length of
a curved line:
s
 Ã2
ˆ bq
ˆ b
dy
1C
dx D
1 C
...
x//2 dx:
(4
...
To make sure that this
calculation always yields a positive result, one has to integrate from the smallest
to the largest values of the x
...
The equation of
a circle centred at the origin is x2 C y2 D R2
...
This will have to be multiplied by 4 to give the whole circle
p
circumference
...
y0 /2 dx D R

ˆ

R

0

dx
p
DR
R 2 x2

ˆ
0

1

dt
p
D R arcsin tj1 D R ;
0
2
1 t2

which yields for the whole length a well-known expression of l D 2 R
...
Although what we have just done can formally be
considered as yet another substitution, we have actually used the polar coordinates
(see Sect
...
13
...

This consideration brings us to the following point: a line can also be considered
by means of the polar coordinates in which case the angle is used as the driving
coordinate along the line
...
1
...
3) r D r
...
/ cos and y D r
...
/d D r0
...
/d D r0
...
79)

p
r 0
...
80)

and therefore
ˆ max
ˆ
lD
dl D
min

max

ˆ
p
dx2 C dy2 D

min

max

min

Check this simple result using Eq
...
79)
...
The equation of the circle is simply r D R
(i
...
r does not depend on ), so that r0 D 0 and
ˆ

2

lD
0

p

0 C R2 d D 2 R:

Problem 4
...
Show that the length of the spiral in Fig
...
20(d) given in the
polar coordinates as r
...

Problem 4
...
Show that the length of the single revolution of the
Archimedean spiral, Fig
...
20(c), given in polar coordinates as r D a (with
changing from 0 to 2 ), is equal to
lD

ah p
2
1C4
2

2

C ln 2 C

p
1C4

2

Ái

:

Problem 4
...
Show that the length of the snail curve specified via r D
a
...
1
...

More generally, if the curve on the x y plane is specified parametrically as
x D x
...
t/, then
ˆ tmax p
lD
x0
...
t/2 dt;
(4
...
6 Applications of Definite Integrals

241

since dx D x0
...
t/dt
...
t/, y D y
...
t/
...
dx2 C dy2 / C dz2 D x0
...
t/2 C z0
...
82)

(see Fig
...
5) and therefore the length of the line
ˆ
lD

tmax

p
x0
...
t/2 C z0
...
83)

tmin

Problem 4
...
Show that the length of the three revolutions of the spiral,
specified via x
...
/ D b and shown in
y
...
4
...

Problem 4
...
Derive a formula for the maximum length of a wire of radius r
that can be coiled around a cylinder of radius R and length L in a single layer
...
4
...


Fig
...
5 For the derivation of
the diagonal dl of a
three-dimensional cuboid
formed by three sides
(increments along the three
Cartesian axes) dx, dy and dz

Fig
...
6 (a) Three
revolutions of the spiral
...
4
...
x/ and
y D y2
...
6
...
x/ is a positive function, then the area under it (i
...
the area
bounded by the curve above and the x axis below) and between the vertical lines
´b
x D a and x D b is given by the definite integral a y
...
This result is easily
generalised for an area bounded by two curves y D y1
...
x/ as (see
Fig
...
7):
ˆ
AD

b

Œy2
...
x/ dx:

(4
...
61
...
y/ and x D x2
...

These formulae, however, are not very convenient if the figure has a more
complex shape
...
e
...
t/ and y D y
...
4
...
To derive the corresponding formula in this case, we have to carefully
consider regions where for the same value of x there are several possible values of
y, i
...
the shaded regions in Fig
...
8(a)
...
The green shaded region in case A in Fig
...
8(b) is
to be fully included, and its area is dA D y
...
t C dt/ x
...
t/x0
...
The to-be-excluded area in case B can formally be written using the same
expression, but dA in this case will be negative (since x
...
t/ and will then
be subtracted)
...
6 Applications of Definite Integrals

243

Fig
...
8 For the calculation of the area of a figure whose bounding curve is specified by parametric
equations x D x
...
t/ with the parameter t
...
(b) Several particular cases (see text)
...
It is clear now that if we sum up all these areas, the correct expression for
the whole area enclosed by the bounding curve is obtained:
ˇˆ
ˇ
ADˇ
ˇ

tb

ta

ˇ
ˇ
y
...
t/dtˇ ;
ˇ
0

(4
...
The regions that need to be excluded are excluded in this
case automatically (explain, why)
...

As an example, let us calculate the area enclosed by a circle of radius R centred
at the origin
...
/ D R cos
and y
...
R sin /
...
2 / d D R2
...
62
...
/ D a cos and y
...
Note that when
a D b, the ellipse turns into a circle of radius a, and the correct expression for
its area, a2 , is indeed recovered
...
/ (see Sect
...
13
...


244

4 Integral

Fig
...
9 For the derivation of
formula (4
...
/ in
polar coordinates

In this case one can use Eq
...
85) with the polar angle as the parameter (see
Problem 4
...
However, a simpler method exists
...
4
...
The area of the sector is then calculated assuming that the figure OAB
can be treated approximately as a triangle with the height h ' r
...
/ tan
...
/ of the triangle (recall
that for small angles, tan ˛ ' ˛, see Eq
...
37))
...
/ =2
...
e
...
/d :

(4
...
/ D R is a constant, the angle changes between
0 and 2 , so that the same expression for the circle area is obtained as above, A D
1 2
2
2R 2 D R
...
63
...
(4
...

Let
...
4
...
/ cos and y D r
...
Calculate the area of the triangle OAB by
!
!
considering the absolute value of the vector product of the vectors OA and OB,
and then expand the obtained expression in terms of  recovering A D
r2
...

Problem 4
...
Derive Eq
...
86) from Eq
...
85)
...
/ D r
...
/ D r
...
/); also make use of the fact that the angle changes by an integer amount
of when traversing around the whole figure
...
6 Applications of Definite Integrals

245

Problem 4
...
Show that the area enclosed by the three-leaf rose shown in
Fig
...
20(a) is A D a2 =4
...
6
...
6 after developing concepts of a function of several variables and
of double and triple integrals, we shall learn how to calculate volumes of general
three-dimensional bodies
...
Here we shall consider some
of the most common cases
...
z/ of the cross section at each
value of z is known, see Fig
...
10(a)
...
z/dx, so
that the whole volume is given by:
ˆ z2
VD
A
...
87)
z1

where z1 and z2 and the smallest and the largest values of the z which are necessary
to specify the object under discussion
...
If we choose the z axis along the symmetry axis of the
cylinder, then the cross section at any point z along the cylinder height will be the a
circle of the same radius R with the area A
...
It does not depend on the z,
leading to the volume
ˆ h
Vcylinder D
R2 dz D R2 h:
(4
...
4
...
z/
...
(b) A sphere of radius R
...
4
...
A cross section of the sphere perpendicular to the z axis is a circle
of radius r D R sin Â, where the angle  is related to z via z D R cos Â, i
...
r D
q
R 1
...
z/ D r2 D
we obtain for the volume:
Ã
Â
ˆ R
4 3
R3
R :
D
2R3 2
R2 z2 dz D
Vsphere D
3
3
R

R2

z2 , and

(4
...
66
...
4
...
90)

Note that the equation of the ellipsoid is given by
x Á2
y Á2
z Á2
C
C
D 1:
a
b
c

(4
...
90) makes perfect sense: when the ellipsoid is fully
symmetric, i
...
when a D b D c, it becomes a sphere, and the volume in this
case goes directly into the expression (4
...

Problem 4
...
Show that the volume of a cone shown in Fig
...
11(a) is
Vcone D

1 2
1 3 2
R hD
h tan ˛:
3
3

(4
...
4
...
(b) A solid obtained by a full revolution around the
z axis of the curve z D f
...
6 Applications of Definite Integrals

247

Fig
...
12 For the calculation
of the volume of a triangular
pyramid (tetrahedron)
A1 B1 C1 D

Problem 4
...
Show that the volume of a tetrahedron A1 B1 C1 D (also called a
triangular pyramid), shown in Fig
...
12, is given by the following formula:
Vtetrahedron D

1
S4 c sin ˛;
3

(4
...
e
...
[Hint: Let the base A1 B1 C1 of the
pyramid be in the x y plane
...
Demonstrate, using the similarity of
the triangles 4A1 B1 C1 and 4A2 B2 C2 , that the area of the intermediate triangle
4A2 B2 C2 , expressed as a function of the parameter t, is given by S4
...
c t/2 , where h D A1 F1 is the height of the base drawn towards the
side a D B1 C1
...
]
Problem 4
...
Next consider the triangular pyramid of the previous problem
!
!
!
as created by three vectors: A1 B1 , A1 C1 and A1 D
...
93) can also be written as
Vtetrahedron D

1ˇ ! h !
ˇ
ˇ A1 D A1 B1
6

1 ˇh ! ! !iˇ
!iÁˇ
ˇ
ˇ
ˇ
A1 C1 ˇ D ˇ A1 B1 ; A1 C1 ; A1 D ˇ ;
6
(4
...
e
...
Since the triple
product corresponds to the volume of the parallelepiped formed by the three
vectors, see Eq
...
41), we see that the volume of the tetrahedron is one sixth of
the volume of the corresponding parallelepiped
...
x/ around the z axis, the so-called solids
of revolution, see Fig
...
11(b)
...
(4
...
z/, is obtained by solving the equation
z D f
...
e
...
z/
...
z/:
(4
...
4
...

Therefore, in this case x
...
tan ˛/ z and the integration above yields
exactly the same result as in Eq
...
92)
...
70
...

2
Problem 4
...
Show that the volume of a solid of revolution obtained by
rotating around the z axis the curve z D ax4 with 0 Ä x Ä R is V D
...


4
...
4 A Surface of Revolution
A general consideration of the surface area of three-dimensional bodies requires
developing a number of special mathematical instruments, and we shall do it in
Chap
...
These will allow us to calculate the area of the outer surface of arbitrary
bodies
...

Consider an object created by rotating around the z axis a curve z D f
...
4
...
Above we have considered its volume
...
Looking at the surface layer created between the cross sections
at z and z C dz, we can calculate its area as a product of 2 r
...
z/ D x
...
z/,
as above) and the layer width, dl
...
x/ may
have a curvature
...
4
...
1 (and,
specifically, see Fig
...
4(b)) when considering the length of a curve
...
dx/2 C
...
z/2 dz:
Therefore, the surface area of the solid of revolution enclosed between z1 and z2 is
obtained by summing up all these contributions, i
...
by integrating:
ˆ z2
q
2 x
...
x0
...
96)
AD
z1

4
...
The equation of a sphere of radius
R and centred at the origin, see Fig
...
10(b), is expressed by x2 C y2 C z2 D R2
...
The upper hemisphere of the sphere can be considered as a revolution
p
of the curve z D R2 x2 around the z axis
...
z/ D R2 z2
...
z/ D
z= R2 z2 and we can write:
s
ˆ R p
z2
Ssphere D 2
2
R2 z2 1 C 2
dz
R
z2
0
s
ˆ Rp
ˆ R
R2
D4
R2 z2
dz D 4 R
dz D 4 R2 : (4
...
72
...
4
...
98)

Problem 4
...
Show that the area of the side surface of a solid of revolution
obtained by rotating around the z axis the parabola z D ax2 with 0 Ä x Ä R is
h
i
A D 2
...

Problem 4
...
A torus of radius c with the tube of radius R, see Fig
...
13, can
be formed by rotating the circle of radius R (shown in the figure as the cross
section of the torus tube in the x z plane) around the z axis
...
The equation of the
p
circle
...
The plus sign corresponds to
the outer part of the surface obtained by rotating the right half of the circle
with c Ä x Ä c C R, while the minus sign gives the inner surface of the torus
due to revolution of the left side of the circle, c R Ä x Ä c, around the z axis
...
z/ is multivalued, the two surfaces have to be considered
separately
...
c ˙ 2R/, where the
plus sign corresponds to the outer and the minus to the inner surfaces
...
99)

250

4 Integral

Fig
...
13 A torus of radius c
with the tube of radius R can
be obtained by rotating a
coloured circle of radius R
displaced from the origin by
the distance c along the x axis

Problem 4
...
Consider a solid of revolution created by the line z D f
...
Let
the line be given parametrically as z D z
...
t/
...
t/
...
t//2 C
...
t//2 dt;
(4
...
Note that this formula is in a sense more general than (4
...
It could be any line in space, not necessarily the Cartesian axes, if the
appropriate parametric equations are provided
...
76
...
4
...
The convenience of this representation is shown
when using Eq
...
100) instead of (4
...
Show that the same
result (4
...


4
...
5 Simple Applications in Physics
Of course, the language of physics is mathematics and hence integrals (comprising
an essential part of the language) play an extremely important role in it
...


4
...
Consider a particle performing a one-dimensional movement along
the x axis
...
t1 t2 /
...
To generalise this result for the case
of variable velocity, we can use the concept of integration: (i) divide up the time
into small intervals ti , (ii) assuming that the velocity vi within each interval is
constant, calculate the distance xi ' vi ti the particle passed over each such
small time interval (4xi calculated that way is approximate for finite ti ), and (iii)
then sum these distances up to get the whole distance passed over the whole time
...
t/dt:

(4
...
t/ D v0 C at, which
linearly increases (a > 0) or decreases (a < 0) with time from the value of v0 (at
t D 0)
...
v0 C a / d D v0 t C at2 :
sD
2
0
Here the acceleration a D F=m is assumed to be constant, which happens when the
force F acting on the particle of mass m is constant; e
...
during a vertical motion
under the Earth’s gravity or because of the friction during the sliding motion on a
surface
...
x/x
...
102)
x1

Problem 4
...
A body of mass m is taken from the surface of the Earth to
infinity
...
The gravitational force acting
between two bodies which are a distance x apart is F D GmM=x2
...

What is the minimum possible velocity the rocket needs to develop in order to
overcome the Earth’s gravity?

252

4 Integral

If two functions are equal for all values of x within some interval, f1
...
x/,
´b
´b
then obviously their integrals are equal as well, a f1
...
x/dx, where the
points a and b lie within the interval
...
x/ D f2
...

Problem 4
...
Consider a particle of mass m moving along the x axis under
the influence of the force F
...
The Newtonian equation of motion is p D F
...
Integrate both
P
sides of this equation with respect to x between the initial, x0 , and final, x1 ,
points along a particle trajectory to show that
p2
1
2m

p2
0
D
2m

ˆ

x1

F
...
[Hint: express p via the derivative of p over x first
...
For linear objects such as a rod of length L oriented along
the x axis and positioned between coordinates x D 0 and x D L, the linear density
is defined as the derivative
...
Here m
...
Correspondingly, m '
...
The total mass is calculated by summing up these
contributions along the whole length of the rod and taking the limit x ! 0, i
...
by
the integral
ˆ
mD

x2


...
103)

x1

If we are interested in calculating the centre of mass of the rod, this can also be
related to the appropriate definite integral
...
For a continuous rod, we
iD1
split it into small sections of length x and mass m '
...
6 Applications of Definite Integrals

253

Fig
...
14 To the calculation of the centre of mass of: (a) a thin curved line specified by the
equation y D f
...
Each of these sections can be treated as a point mass
for sufficiently short sections
...
x/dx:

(4
...
x/, see Fig
...
14(a)
...
xxm ; ycm /, which are determined
by considering small lengths
q
dl D


...
dy/2 D

q

1 C
...
x//2 dx

along the line and then summing up all the contributions:
xcm

1
D
m

ycm D

1
m

ˆ

x2

x1

ˆ

x2

q
x
...
f 0
...
x/
...
f 0
...
105)

x1

where m is the mass of the line and
...
x; f
...


254

4 Integral

Problem 4
...
Show that if the line is specified parametrically in the threedimensional space as x D x
...
t/ and z D z
...
cm/
form, as r D r
...
cm/ D r˛
D
...
cm/
r˛ D

1
m

ˆ

t2

q
r˛
...
t/
...
t//2 C
...
t//2 C
...
t//2 dt;

t1

where ˛ D x; y; z designates the three Cartesian components of the vectors and

...

Problem 4
...
Show that the coordinates of the centre of mass of an arch of
a unit linear density, radius R and angle 2˛, shown in Fig
...
14(b), is given by

...

Problem 4
...
Show that the coordinates of the centre of mass of a uniform
sector of radius R and angle 2˛, shown in Fig
...
14(c), is given by
...
82
...
Show
hR4 ! 2 =4
...
r/ D !r
...
83
...
If is the water density, show that
the total force exerted on any side surface of the disk is F D h gR2 , where g
is the Earth’s gravitational constant
...
84
...
Show that in this case
F D g hR2

Â
2

C arcsin

h
R

Ã

p
C h 2 R2

h2 C

2 2
R
3

h2

3=2

:

However, the mass and the centre of mass of more complex two- and even
three-dimensional objects are defined via their corresponding surface and volume
densities
...


4
...
An equation of the ideal gas reads PV D nRT, where P is pressure,
V volume, T temperature, n the number of moles and R is the gas constant
...
V2 V1 /
...
This happens in the isobaric process when the V changes
due to a change in temperature, but the pressure remains the same
...
Again, the solution to this problem can easily be found using the
integration: divide the whole process into small volume changes V during which
the pressure can approximately be considered constant, calculate the work W '
PV done during this small volume change, and then sum up all these changes and
take the limit V ! 0, which results in the integral:
ˆ
Wgas D

V2

PdV;

(4
...
V/
...
85
...
86
...
The number density of the gas would be greatest near the
Earth’s surface and then it will get smaller and smaller as the distance z from
the surface is increased:
Â

...
Show that the weight P D mgas g of the air in a vertical tube
above the Earth’s surface of cross section area A and height H, where mgas is
the total mass of air in the tube, is P D A
...
Rewrite this equation as
P=A C PH D P0 and then interpret this result
...
4
...
t/ specified parametrically, and (b) a vertical uniformly charged straight line
with the field calculated distance x away from it

Electrostatics and Magnetism
...
t/ in a three-dimensional space
...
t/ D dq=dl may vary along the line
...
x0
...
y0
...
z0
...
4
...
The electrostatic field dE at point M, whose position
is given by the vector rM , is equal to dE D dqg=g3, where g D rM r is the
vector connecting dl and the point M
...
e
...
rM / D
line

ˆ
D

t1

t0

g
dq D
g3
rM
jrM

ˆ

g
dl
3
line g
q
r
...
t/
...
t//2 C
...
t//2 C
...
t//2 dt:
r
...
107)

We have ignored here the irrelevant pre-factor 1=4 0
...
4
...

Let the element dl D dz of the line with the charge dq D dl D dz be distance
z from the point on the line which is closest to the point M
...
It is clear that the field components along the z axis from the
two line elements dl symmetrically placed at both sides of the connecting line at
z and z would cancel each other
...
6 Applications of Definite Integrals

257

component of the field along the x axis
...
x; 0; z/ and the
required component of the field Ex is:
ˆ

1

Ex D

1

ˆ
D x
ˆ
D

x

ˇ
ˇ
ˇ
ˇ z D x tan ˛
ˇ
ˇ
dz D ˇ
2 ˇ
dz D xd˛= cos ˛

...
x2 C x2 tan2 ˛/
ˆ
cos3 ˛
d˛ D
cos2 ˛
x

xd˛
cos2 ˛
=2
=2

cos ˛d˛ D

2
;
x

i
...
the field decays with the distance x from the charged line as 1=x
...
t/ in a three-dimensional space, see Fig
...
16(a)
...
rM / D
line

J Œn g
dl D J
g3

ˆ

t1
t0

Œn
...
t/
g
...
x0
...
y0
...
z0
...
108)

Fig
...
16 For the calculation of the magnetic field B at point M due to the wire in which the
current J is running: (a) a general case of a line r D r
...
87
...
4
...

Problem 4
...
Consider a single coil of wire of radius R placed within the
x y plane as shown in Fig
...
17(a)
...
Show
that the magnetic field B at a point M along the symmetry axis of the circle (i
...

along the z axis) at a distance z from it is
B D
...
R2 C z2 /3=2

:

[Hint: the vector n D e is tangential to the circle as shown in Fig
...
17(b);
also, g D z r, where r D Rer (er being a unit vector along r), and z D zk is
the vector along the z axis
...
zk Rer / is needed
...
Physics presents numerous
examples of such quantities, but in order to consider them we need to build up
our mathematical muscle by considering, in the first instance, functions of many
variables and multiple integration
...
4
...
The close-up of a piece of
the wire within the x y plane is shown in (b)
...
7 Summary

259

the so-called differential equations which connect derivatives of functions of interest
and the functions themselves into a single (or a set of) equation(s)
...
You may appreciate
now that differentiation and integration, together with infinite series and differential
equations (still yet to be considered), represent the basic language of the calculus
used in physics
...
7 Summary
As it should have become clear from the previous chapter, any “well-behaved”
function written in an analytical form can be differentiated, however complex
...

There are many methods to integrate specific classes of functions, some of which
we considered in this chapter:
• the change of variable, and there are a number of different ones suited for
particular types (classes) of functions;
• integration by parts, several of them may be needed;
• sometimes integration by parts establishes an equation for the integral being
calculated;
• some integrals can be calculated by introducing parameters in the integrand;
if the differentiation with respect to the parameter yields a known integral,
then the required integral is obtained by differentiating it; not one but many
differentiations are sometimes required
...
For instance, indefinite integrals can be
expanded into infinite series; some definite integrals can be calculated using more
sophisticated methods such as integral transforms or complex calculus
...

x
x
Many integrals have been studied and solved over last centuries, and several
comprehensive tables of integrals exist which list these results
...
S
...
M
...
It is a good idea
to consult this and other books when needed
...

Finally, it is a good idea to remember that definite integrals of almost any complexity
can be calculated numerically
...
x/ of a
single variable
...
In practical
applications functions of more than one variable are frequently encountered
...
x; y; z/, and may also be a function of time, t, i
...
it depends on four variables
...
1 Therefore, one has to generalise our definitions of the
limits and derivatives for function of many variables
...
In the next chapter
the notion of integration will be extended to this case as well
...
1 Specification of Functions of Many Variables
If a unique number z corresponds to a pair x and y of variables, defined within their
respective intervals, then it is said that a function of two variables is defined
...
x; y/ or z D f
...
This is a direct way in
2
which a function can be defined, e
...
by a formula like z D x2 C y
...
x; y; z/ D 0, a solution z
of which for every pair of x and y provides the required correspondence (if proven
unique, i
...
that there is one and only one value of z for every pair of the variables x
and y)
...
x; y/ can also be specified parametrically as

1

In fact, in the latter case we have three functions, not one, for either of the fields, as each
component of the field is an independent function of all these variables
...
Kantorovich, Mathematics for Natural Scientists, Undergraduate Lecture
Notes in Physics, DOI 10
...
u; v/, y D y
...
Functions of more than two variables are defined in a similar way
...
x/, it can be represented
as a graph of y vs
...
e
...
Correspondingly, functions z D f
...
x; y; z/
...

Before making a general statement on functions of arbitrary number of variables,
let us consider here some examples of functions of two variables, which is the
simplest case
...
x

x0 /2 C b
...
z

z0 /2 D 1:

The equation defines not one but two functions z D f
...
x

x 0 /2

b
...
Depending on the
coefficients a, b and c, different types of surfaces are obtained
...
x0 ; y0 ; z0 /
defines the central point of the surfaces and can be chosen to coincide with the origin
of the coordinate system, i
...
when x0 D y0 D z0 D 0; their non-zero values simply
correspond to a shift of the surfaces by the vector r0 D
...
Hence, we shall
assume in the following that x0 D y0 D z0 D 0, and consider a simpler equation
ax2 C by2 C cz2 D 1:

(5
...
1
...
There are two surfaces here describing the upper
p
(z 0) and the lower (z Ä 0) hemispheres via z D ˙ R2 x2 y2
...
1
...
1 Specification of Functions of Many Variables

263

Fig
...
1 Some of the second order surfaces: (a) ellipsoid; (b) one-pole (one sheet) hyperboloid;
(c) two-pole (two sheets) hyperboloid; (d) saddle-like elliptic paraboloid

x Á2
C
˛

 Ã2  Ã2
y
z
C
D1
ˇ

(5
...
5
...
To convince ourselves of this, let us consider
cross sections of this shape with various planes perpendicular to the Cartesian axes
...
(5
...
Therefore, we conclude that ˛ Ä x Ä ˛, ˇ Ä y Ä ˇ and
Ä z Ä
...
In the cross section by
either of these planes we obtain a line with the equation
...
y=ˇ/2 D d 2 ,
where d2 D 1
...
This line is an ellipse
x Á2
y Á2
C
D1
A
B
with the semi-axes A D ˛d and B D ˇd
...
e
...

Therefore, at z D
the ellipse is simply a point, but then, as the plane z D g
moves down closer to the x y plane, we observe the ellipse in the cross section
whose size gradually increases as we approach the x y plane
...


264

5 Functions of Many Variables: Differentiation

5
...
3 One-Pole (One Sheet) Hyperboloid
If now a D 1=˛ 2 > 0, b D 1=ˇ 2 > 0, but c D 1= 2 < 0, then we obtain the
equation
 Ã2  Ã2
x Á2
y
z
C
D 1;
(5
...
5
...


5
...
4 Two-Pole (Two Sheet) Hyperboloid
When choosing 1 in the right-hand side of Eq
...
3), a very different figure is
obtained called two-pole (or two sheet) hyperboloid which is shown in Fig
...
1(c)
...
4)
˛
ˇ
Problem 5
...
Using the method of sections with planes perpendicular to the
coordinate axes, investigate the shape of the figure given by Eq
...
3)
...
2
...
(5
...
Are there
any limitations for the values of z?

5
...
5 Hyperbolic Paraboloid
This figure is described by the equation
x Á2
˛

 Ã2
y
D z:
ˇ

(5
...
Considering first crossing planes x D g (which are
perpendicular to the x axis), we obtain the lines z D y2 =ˇ 2 C
...
Similarly, cutting with planes
y D g (which are perpendicular to the y axis) results in parabolas z D x2 =˛ 2
...
Finally, planes z D g > 0 (parallel to the x y plane) result in

5
...
x=A/2
...
When z D g < 0,
then hyperbolas
...
y=B/2 D 1 are obtained which are rotated by 90ı with
respect to the first set
...
5
...
This point is usually called the saddle point or, in physics, the transition
state (TS)
...
2 Each
minimum corresponds to a stable configuration of the system
...
e
...
The energy
barrier  along the path would determine the rate with which such a transition
can happen as the rate is approximately proportional to exp
...
Sometimes there is
only a single so-called “reaction” coordinate connecting the two minima along
the minimum energy path, and going along this direction requires overcoming the
energy barrier, i
...
along this direction one has to move over a maximum
...
In other words, the
crossing point is nothing but a saddle on the complicated potential energy surface E
considered as a function of its all coordinates
...
x; y; z/ of an adatom on the surface can be considered
merely as a function of its two Cartesian coordinates, x and y, determining its lateral
position, and of its height z above the surface as schematically shown in Fig
...
2
...
0; 0; z0 / and

...
a=2; 0; zTS / which is exactly
half way through; it corresponds to the bridge position of the adatom
...
In practice, however, a single Cartesian coordinate rarely
serves as the reaction coordinate for the transition and is instead specified in a more
complicated way
...


266

5 Functions of Many Variables: Differentiation

Fig
...
2 Transition of an adatom between two hollow sites on a simple cubic lattice
...
(a) Three-dimensional view in which the initial and
final (dimmed) positions of the adatom are shown
...
The x coordinate serves as the
reaction coordinate here

5
...
x/ of a single variable
at a point x0 (Sect
...
4
...
1)
...

One definition was based on building a sequence of points fxn ; n D 1; 2; 3; : : :g D
fx1 ; x2 ; x3 ; : : : ; xn ; : : :g which converges to the point x0 ; this sequence generates the
corresponding functional sequence of values ff
...
x/, and the
limit of the function as x ! x0 is the limit of this functional sequence
...
Another definition was based on the “
ı” language: we called A the limit
of f
...
/, such
that for any x within the distance ı from x0 (i
...
for any x satisfying jx x0 j < ı)
the number A differs from f
...
e
...
x/ Aj < is implied
...

Both these definitions are immediately generalised to the case of a functions of
more than one variable if we note that a function f
...
x1 ; : : : ; xN /,
i
...
as f
...
x/ when
x ! x0 , with x0 being some other vector
...
x1 ; : : : ; xN /; correspondingly,
the distance jx x0 j between the points x and x0 in the one-dimensional space is
replaced with the distance (1
...
e
...
x; x0 / D
jx x0 j
...
x/
when x ! x0 sounds exactly the same
...
2 Limit and Continuity of a Function of Several Variables

267

Fig
...
3 This illustration shows that for a function f
...
The limit would exist if and only if any path
towards x0 , as depicted in Fig
...
3, generates the corresponding functional sequence
converging to the same limit A
...
Consider as our first example the limit of
the function of two variables f
...
x C y/ at the point
...
0; 0/
...

By changing the constant k different paths towards the centre of the coordinate
system are taken; however, a well-defined limit exists for any value of k
...
1 C k/
1 C k x!0
lim

Consider now, using the same set of paths, the function f
...
For
this function the limit does depend on the value of the constant k:
lim

x!0;y!0 x2

k
xy
kx2
D
D lim 2
:
2
2/
x!0 x
...
0; 0/ does not
exist
...
3
...
x; y/ D arctan
...
x; y/ points: (a)
...
0; 1/; (c)
...
0; 0/
...
[Answers: (a) =4; (b) 0; (c,d)
do not exist
...
2
...
2 are immediately generalised
to the case of the functions of many variables, e
...
if the limit exists it is unique and
that one can manipulate limits algebraically
...
2
...
3): namely, the function
f
...
e
...
x/ D f
...
6)

x!x0

Points where the function f
...

For continuous functions the change y of the function y D f
...
e
...
x C x/

x!0

x!0

f
...
7)

Many properties of continuous functions of several variables are also analogous to
those of functions of a single variable
...
x/ is continuous
within some region in the N-dimensional space (where x is defined) and takes there
two values A and B, then it takes in this region all values between A and B, i
...
for
any C lying between A and B there is a point xC such that f
...
In particular,
if A < 0 and B > 0, then there exists at least one point x0 where the function
f
...

Problem 5
...
Consider the function z D f
...
Show that the change
of the function z corresponding to the change x and y of its variables can
be written in the form:
z D y2 x C 2xyy C ˛x C ˇy;
where both ˛ and ˇ tend to zero as
...
Note that the choice of ˛
and ˇ here is not unique
...
3 Partial Derivatives: Differentiability
When in Sect
...
1 we considered a function of a single variable, y D f
...
This definition cannot be directly
generalised to the case of functions of many variables as x in this case is an Ndimensional vector
...
3 Partial Derivatives: Differentiability

269

At the same time, we can easily turn a function of many variables into a function
of just one variable by “freezing” all other variables
...
x1 ; x2 ; : : : ; xN / D f
...
x1 C x1 ; x2 ; : : : ; xN /

f
...
This derivative is
called partial derivative of the function y with respect to its first variable x1 , and
is denoted @y=@x1
...
x/ due to its
single variable x1
...
Correspondingly, dividing the partial change
of the function i y by the change of the corresponding variable, xi , and taking the
limit, we arrive at the partial derivative
ˇ
@y
i y
dy ˇ
ˇ
D lim
D
xi !0 xi
@xi
dx ˇx1 ;:::;xi

1 ;xiC1 ;:::;xN

with respect to the variable xi
...

x
Problem 5
...
Calculate all partial derivatives of the functions:
Ä
f
...
x; y/ D ex

2 Cy2

Answer:

@f
@f
D cos x sin y;
D sin x cos y I
@x
@y

Ä
@f
@f
2
2
2
2
D 2xex Cy ;
D 2yex Cy I
Answer:
@x
@y

1
f
...
6
...
r/ D U
...
e
...
A satellite of mass M which is orbiting the Earth of
mass M0 experiences the potential field U
...
Determine the force acting on the satellite
...
5
...

In Sect
...
2 we obtained a general expression for a change y of the function
y D f
...

There exists a generalisation of this result for a function of many variables
...
x; y/; a more general case is straightforward
...
x; y/ is differentiable at
...
(3
...
8)

where ˛ and ˇ tend to zero as both x and y tend to zero, while A and B depend
only on the point
...
This formula states that the first
two terms in the right-hand side are of the first order with respect to the changes of
the variables, x and y, while the last two terms are at least of the second order
(cf
...
4)
...
7
...
8)
...

It is obvious from this definition that if the function is differentiable, i
...
if its
change can be recast in the form (5
...

I
...
the function z D f
...
It is also clear that both partial derivatives
exist at the point
...
Indeed,
ˇ
Ä
y
y
z ˇ
@z
ˇ
D lim
C˛Cˇ
D lim A C B
ˇ
x!0 x yD0
x!0
@x
x
x
D lim
...
3 Partial Derivatives: Differentiability

271

and similarly
ˇ
Ä
@z
x
x
z ˇ
ˇ
D lim
Cˇ
D lim A CBC˛
@y y!0 y ˇxD0 y!0 y
y

xD0

DBC lim ˇjxD0 D B;
y!0

where, while calculating the partial derivative, we kept the other variable constant,
i
...
considered its change to be equal to zero before taking the limit
...
8) can be written in
a more detailed form as
z D

@z
@z
x C y C ˛x C ˇy:
@x
@y

(5
...

In a general case of a function y D f
...
x1 ; : : : ; xN / the
previous formula is straightforwardly generalised as follows:
y D y
...
x/ D

N
X Â @y
iD1

@xi

Ã
xi C ˛i xi ;

(5
...

Importantly, existence of the partial derivatives does not yet guarantee the
differentiability of the function, i
...
the reverse statement may not be true
...
0; 0/, however its
partial derivatives do not exist there: indeed, when, for instance, we are interested
p
in the @z=@x, then at this point y D 0 and hence z D x2 D jxj, which does
not have the derivative at x D 0 (since the derivatives from the left and right are
different)
...
Also, similarly to the case of functions of a single variable,
differentiability is a stronger condition than continuity: continuity of a function at a
point is not yet sufficient for it to be differentiable there
...
1 (Sufficient Condition of Differentiability)
...
x; y/ to be differentiable at the point
...

Proof
...
x C x; y C y/ f
...
xCx; yCy/ f
...
x; yCy/ f
...
11)

272

5 Functions of Many Variables: Differentiation

Here the first term is the partial change x z of the function z with its second variable
kept constant at the value of y C y, while the second term gives the partial change
y z with the first variable kept constant at x
...
64) for
the corresponding partial changes:
ˇ
(5
...
x C #x; y C y/ x; y zˇx D fy0
...
Note that derivatives
above are, in fact, the partial derivatives, and we have used simplified notations
for them: fx0 or fy0
...
) As noted above, these kinds of simplified notations are frequently met in the
literature and we shall be using them frequently as well
...
x; y C y/ D fy0
...
xC#x; yCy/ D fx0
...
Substituting
the above results into Eqs
...
12) and (5
...
9),
proving the differentiability of the function z
...
E
...

Hence, if the partial derivatives exist and are continuous, the function of two
variables is differentiable
...
8
...

Above we have introduced first order partial derivatives
...

Taking the partial derivative with respect to the variable x of the x-partial derivative
yields the second order partial derivative:
 Ã
@ @z
@2 z
00
D 2 D fxx :
@x @x
@x
However, if we take a partial derivative with respect to y of the partial derivative
@z=@x, or vise versa, two mixed second order partial derivatives are obtained:
 Ã
 Ã
@ @z
@2 z
@2 z
@ @z
00
00
D fyx and
D fxy :
D
D
@y @x
@y@x
@x @y
@x@y
If this process is continued, various partial derivatives can be obtained
...
3 Partial Derivatives: Differentiability

273

As an example, consider z
...
x
z0 D ˛e˛
...
x

z00 D
yx
z00 D
xx

@ 0
z D˛ 2 e˛
...
We have:
@ 0
z D ˛ 2 e˛
...
x
@y x

@ 0
z D ˛ 2 e˛
...
x
@y xy

y/

I z000 D
yyx

y/

y/

y/

I

I

I z000 D
xyy

@ 00
z D ˛ 3 e˛
...
x
@y yx

y/

;

y/

I

etc
...
9
...
x; y/ D x3 y2 C x2 y3
...
10
...
x; t/ D p
e
4 Dt

x2 =4Dt

;

(5
...
Check by direct differentiation that this
function satisfies the following (the so-called partial differential) equation:
1 @w
@2 w
D 2:
D @t
@x

(5
...

It follows from the example and the problem, that the partial derivatives can be
taken in any order, the result does not depend on it
...

Theorem 5
...
Consider a function z D f
...
If
00
00
its mixed second order partial derivatives fxy and fyx exist and are continuous
in some vicinity of a point
...
15)

274

5 Functions of Many Variables: Differentiation

Proof
...
x C x; y/

f
...
y/;

which we can formally consider as a function
...
The mixed
00
derivative fyx will then require calculating the partial difference of the function
...

I
...
the total change needed for the mixed derivative will be
yx f D y D
...
x C x; y/


...
x C x; y C y/

f
...
x; y/ :

(5
...

Similarly, let us construct the full change xy f of the function f required for the
00
calculation of the mixed derivative fxy
...
x; y C y/

f
...
x/;

and then the additional change with respect to the other variable:
xy f D x ' D '
...
x; y C y/

'
...
x C x; y C y/

f
...
x; y/ :

(5
...
One can see now that the two expressions (5
...
17) are exactly
the same, xy f D yx f (although with terms arranged in a different order), which
already suggests that the mixed derivatives must be the same
...
64) for each of the
changes of the function:

...
x C Â1 x; y/
yx f D y D

0
y
...
x C Â1 x; y C Â2 y/

and, similarly,
'
...
x; y C #1 y/

and

0
00
xy f D x ' D 'x
...
x C #2 x; y C #1 y/;

where numbers #1 , #2 , Â1 and Â2 are each between 0 and 1
...


5
...
Tangent Plane

275

Since both changes of the function f
...
x C #2 x; y C #1 y/ D fyx
...
x C Â1 x; y C Â2 y/ D fyx
...
x C #2 x; y C #1 y/ D fxy
...
x; y/ D fyx
...
E
...


Problem 5
...
Generalise this theorem for a function of N > 2 variables
...
]
Problem 5
...
Generalise this theorem for mixed derivatives of any order
...
4 A Surface Normal
...
x; y/ at point M0
...
5
...
To this
end, we shall consider lines lying within the surface
...
The line DC is obtained
by changing x and keeping y D y0 , its equation is z D z
...
Similarly, the line
AB with equation z D z
...
Both
lines cross exactly at the point M0
...
5
...
x; y/, at point M0
...
x; y/k
...
x0 ; y/) along the AB curve (x D x0 , y
!
is variable), see Fig
...
4(b)
...

As the point M approaches the point M0 , the angle must approach the value of
=2
...
0; y; z/, where y D y y0 and z D z
...
x0 ; y0 /:
cos

!
z
ny C nz y
M M n
ny y C nz z
ˇ 0!ˇ
p
D
D
r
D
Á2 :
ˇ
ˇ
jnj y2 C z2
ˇM0 M ˇ jnj
z
jnj 1 C y

In the limit when the point M approaches M0 (or, which is equivalent, y ! 0), the
angle ! =2, i
...
cos ! 0
...
Similarly, considering a point M somewhere
on the DC line, when y D y0 and x is allowed to change, we obtain an equation
@z
nx C nz @x D 0
...
x; y/
which is not parallel to the z axis anywhere in the vicinity of the given point), we
obtain for the normal:
Â
Ã
Ã
Â
@z @z
@z
@z
n D nx ; ny ; nz D
nz ; nz ; nz D nz
; ; 1
@x
@y
@x @y
Â
Ã
@z @z
H) n D
; ; 1 ;
(5
...

Once we have an expression for the normal, we can also write an equation for the
tangent plane touching our surface at point M0 as shown in Fig
...
4(a)
...
(1
...
x

x0 / C ny
...
z

z0 / D 0;

where the components of the normal are given by Eq
...
18)
...
x

x0 /

@z
@x

Ã

Â
C
...
19)

M0

Note that derivatives above are calculated at point M0 and hence are constants
...
5 Exact Differentials

277

As an example of the application of this result, consider the cone specified by the
p
equation z D x2 C y2
...
(5
...
x=z; y=z; 1/
...
13
...
e
...
x; y; z/
...
14
...
3) at
the point
...

Problem 5
...
The same for the two-pole hyperboloid (5
...

Problem 5
...
The same for the saddle (5
...

Problem 5
...
Derive the result (5
...
First,
consider the vectors3 x D @r=@x and y D @r=@y, where r D xi C yj C
z
...
These vectors, see Fig
...
4(a), are tangential to the curves DC and AB,
respectively, at the point M0
...
Up to a scale factor and (possibly) a sign this
is exactly the same result as in Eq
...
18)
...
x; y/
...
y; z/ and y D y
...
An arbitrary surface
requires a somewhat more general approach; this will be done later on in Sect
...
4
...


5
...
x; y/ due to change of its variables
dx and dy, see (5
...
Sect
...
2)
...
20)

In the general case of a function y D f
...
x1 ; : : : ; xN / the
differential is
dy D

N
X @f
dxi :
@xi
iD1

(5
...
@x=@ ; @y=@ ; @z=@ /
...
x; y/dx C B
...
x; y/
...
x; y/ whose differential (5
...
x; y/dx C B
...
Of course, it is not at all obvious that for
any functions A and B this would be necessarily true; if this is not true, the form
is called inexact differential
...

Theorem 5
...
The form A
...
x; y/dy is the exact differential dz of
some function z D z
...
22)

Proof
...
Indeed, if there was a
function z such that its differential dz is given precisely by formula (5
...
x; y/ D @z=@x and B
...
Let us now differentiate
A with respect to y and B with respect to x:
@A
@
D
@y
@y

Â

@z
@x

Ã
D

@2 z
@y@x

and

@B
@
D
@x
@x

Â

@z
@y

Ã
D

@2 z
:
@x@y

We have arrived at two mixed derivatives where differentiation is performed in the
opposite order
...
x; y/ satisfies the required conditions
of Theorem 5
...
e
...
22)
...
22) is satisfied
...
x; y/ such that its differential dz has
exactly the given form Adx C Bdy
...
x; y/ such that
A
...
23)

This can always be done
...
22), this is also equal to @B=@x,
which means that
 Ã
Â
Ã
@B
@ @g
@
@g
@g
D
Ch
...
24)
H)
B
D 0 H) B D
@x
@x @y
@x
@y
@y

5
...
y/ must be some function of y only
...
22)
...
y/ such that p0
...
y/
...
y/ can be always found, at least in principle, as it is an indefinite
integral of h
...

Now we are ready to provide a working guess for the function z
...
x; y/ D g
...
y/
...
Indeed,
@g
@z
D
DA
@x
@x

and

@z
@g
@g
D
C p0
...
y/ D B;
@y
@y
@y

as required
...
E
...

The second part of the theorem suggests also a constructive method of finding
the function z D z
...
22) is
satisfied
...
Here A
...
x; y/ D 2x3 y
...
x; y/
...
To find this function, we first consider the condition
@z=@x D A D 1 C 3x2 y2
...
Thus,
ˆ
z
...
x; y/dx D x C x3 y2 C C
...
y/ since its derivative with respect to
x is obviously zero
...
e
...
y/ D 2x3 y C C0
...
y/ D 0, so C
...
This results in the final expression for the function sought
for: z D x C x3 y2 C C
...

Although in the above example C0
...
y/ may still be some function of y
...
18
...
x; y/, and then find this
function
...
]
Problem 5
...
The same for the form 6xy C 5y2 dx C 3x2 C 10xy dy
...
]
Problem 5
...
Show that the criterion (5
...
x/dxi

iD1

to be the exact differential of some function y D f
...
N

for any i ¤ j:
1/=2 such equations to be satisfied
...
6 Derivatives of Composite Functions
Consider a differentiable function y D f
...
x1 ; : : : ; xN /
...
e
...
t/,
and each of these functions can be differentiated with respect to t
...
t/ a function of a single variable t and it is reasonable to ask a question
of how to find its derivative dy=dt
...
t C t/ y
...
Once t is changed by t, each of the functions xi
...
t C t/ xi
...
Because the function y D f
...
(5
...
t/ D lim
C ˛i
C lim ˛i lim
t!0 t
t!0
t!0 t
@xi
t
@xi t!0
iD1
iD1
0

5
...
t/ are continuous
...
t/ D

N

(5
...
t/ and then sum up all the contributions
...
Indeed,
in this case y is a composition y D y
...
xi
...
t/ D
y0
...
t/ according to the chain rule (3
...
Therefore, formula (5
...
x/ individually (i
...
assuming that other arguments
are constants, and hence using the partial derivative symbol for the y0
...
It is clear from this discussion that our
result (5
...
(3
...
e
...

As an example, consider z D x2 y2 C sin x sin y with x D t2 and y D t3
...
x; y/ with their respective expressions via t at the very
beginning and then differentiate with respect to t:
z
...

As a more physical example, let us consider a gas of atoms
...
Statistical properties of such a gas can be described by
the distribution function f
...
We define a sixdimensional phase space formed by the particles coordinates r D
...


282

5 Functions of Many Variables: Differentiation

Then f
...
e
...

The total change of the number of atoms in that six-dimensional box in time can
be due to a possible intrinsic time dependence of the distribution function itself,
and because the atoms move in and out of the box
...
r; p; t/ with respect
to time, considering all seven its arguments f
...

Therefore, f can actually be written as f
...
r
...
t/; t/
is a 7 dimensional vector
...
t/ D dr=dt D p=m
P
and p
...
Therefore, the total derivative of this composite function, according
to Eq
...
25), is:
Â

df
dt

Ã

3

D
tot

3

3

3

X @f p˛ X @f
@f
@f dt X @f dr˛ X @f dp˛
C
C
D
C
C
F˛ ;
@t dt ˛D1 @r˛ dt
@p˛ dt
@t ˛D1 @r˛ m ˛D1 @p˛
˛D1

where ˛ D 1; 2; 3 designates the three Cartesian components of the vectors
...
26)
D
dt tot
@t
@r m
@p
Here the second and the third terms in the right-hand side are scalar products of
vectors: @f =@r is the vector
...
4 The change of f in time is due to collisions between
atoms
...

Problem 5
...
Calculate z0
...
[Answer: 2ze 2t 1 C 8e 2t
...
22
...
r; t/
is given as a function of both particle momentum p and its position r, i
...

H
...
r; t/
...
[Hint:
P
P
the Newton’s equations of motion read p D F D @U=@r, where p D dp=dt
...
5
...


5

Due to Sir William Rowan Hamilton
...
6 Derivatives of Composite Functions

283

The derivative (5
...
Sometimes this formula is written in a symbolic form via operators:
N
N
X dxi @y
X dxi @
dy
D
D
dt
dt @xi
dt @xi
iD1
iD1

!
O
y D Dy:

(5
...
t/
C x0
...
t/
N

@
@xN

(5
...
,
multiplied by t-derivatives of the variables themselves
...

Now we can consider a more general case of the composite functions
...
x; y/, when the
variables are both functions not of one (as we have done before) but two variables:
x D x
...
u; v/
...
Consider first the partial
derivative of z D z
...
u; v/ ; y
...
e
...
In that case z may be thought of as effectively depending only on a single
variable u, and hence our general result (5
...
29)

where we have again used simplified notations for the partial derivatives, e
...
y0 D
u
@y=@u, etc
...
30)

This result is easily generalised to a more general case of functions of more than
two variables each of which is also a function of arbitrary number of variables
...
23
...
x/, where x D x
...

v
Problem 5
...
Similarly for z D z
...
u; v/
...
25
...
u; v; x/, where x D x
...

Problem 5
...
Consider z D z
...
u; v/ and y D y
...
Prove
the following expressions for the second order derivatives of z with respect to u
and v:
i
h
2
2
;
(5
...
32)
(5
...
g
...

xy
Problem 5
...
Polar coordinates on the plane
...
Show that differential operators with respect to the
Cartesian coordinates and polar coordinates are related as follows:
@
@
@
D cos ' C sin '
@r
@x
@y

and

@
@
@
D r sin ' C r cos ' :
@'
@x
@y

(5
...
x; y/ with x D x
...
r; '/ as given
by the equations for the polar coordinates, and then use Eqs
...
29) and (5
...
]
Then apply these relationships to z D x2 y2 and calculate its partial derivatives
with respect to r and '
...

Finally, let us consider implicit functions and their derivatives
...
x; y/, which
is specified by means of a function of three variables f
...
x; y; z
...
35)

If we could solve this equation for the function z D z
...
However, this may not be the case (i
...

the equation above cannot be solved analytically) and hence a general method is
required
...
6 Derivatives of Composite Functions

285

fact that its third argument z is also a function of these two) is zero
...
36)
@x ˇtot
@x
@z @x
@x
@f =@z
Note that the partial derivative @f =@xjtot in the left-hand side corresponds to the total
derivative of f with respect to the variable x, including its dependence on x via its
third variable z
...
e
...
Similarly one obtains
@z
D
@y

@f =@y
:
@f =@z

(5
...
28
...
g
...
x; y; z; w/ D 0, where w D w
...

Problem 5
...
Consider a function w D w
...
x; y; z
...
As we
already know, the first order derivative is given by:
@w
@h
@h @z
D
C
:
@x
@x
@z @x

(5
...
39)

and similarly for the derivatives with respect to y
...
40)

As an example, consider the function z D z
...
In this case f D x2 C y2 C z2 1, and we
obtain:
2

@z
D
@x

@f =@x
D
@f =@z

2x
D
2z

x
z

and

@z
D
@y

y
:
z

286

5 Functions of Many Variables: Differentiation

This result can be compared with that obtainedp the direct calculation: solving
via
the algebraic equation for z, we obtain z D C 1 x2 y2 , so that the required
partial derivatives can be immediately calculated
...

Problem 5
...
Consider equations x D r cos ' and y D r sin ', defining the
polar coordinates
...
Solve
the obtained algebraic equations with respect to the derivatives of r and ', and
hence show that:
@r
D sin ';
@y

@r
D cos ';
@x

@'
D
@x

sin '
r

and

@'
cos '
D
:
@y
r

(5
...
(5
...

Indeed,
@2 f
@
D
2
@r
@r

Â

@f
@r

Ã
D

Â
Ã
@f
@
@f
cos ' C sin '
:
@r
@x
@y

Since we differentiate with respect to r, the other independent variable ' is to be
kept constant, and hence the cosine and sine functions need to be left alone:
@2 f
@
D cos '
2
@r
@r

Â

@f
@x

Ã

@
C sin '
@r

Â

Ã
@f
:
@y

The functions @f =@x and @f =@y are some functions of x and y, and hence we can
again use Eq
...
34), yielding
Â
Ã
Ã
Â
@2 f
@ @f
@
@ @f
@
C sin ' cos ' C sin '
D cos ' cos ' C sin '
@r2
@x
@y @x
@x
@y @y
D cos2 '

@2 f
@2 f
@2 f
@2 f
C sin ' cos '
C sin2 ' 2 ;
C cos ' sin '
2
@x
@y@x
@x@y
@y

and, since the mixed derivatives do not depend on the order in which the partial
derivatives are taken, we obtain:
@2 f
@2 f
@2 f
@2 f
C sin2 ' 2 ;
D cos2 ' 2 C 2 cos ' sin '
@r2
@x
@y@x
@y

5
...
42)
This is of course the same as repeating twice the operator @=@r, Eq
...
34), i
...


...

Problem 5
...
Show that
@2
D
@' 2

à Â
Ã
Â
@
@ 2
@
@
C
r sin ' C r cos '
r cos ' C sin '
;
@x
@y
@x
@y

@2
D
@r@'

sin '

(5
...
44)

Problem 5
...
Calculate all three second derivatives of the function z
...
(5
...
44)
...

Problem 5
...
Using the above formulae, demonstrate by a direct calculation
that
@2 f
1 @2 f
1 @f
@2 f
@2 f
C 2 2 D 2 C 2:
C
@r2
r @r
r @'
@x
@y

(5
...
x; y/
...
There are much more
powerful methods to change variables in this kind of differential expressions
...


288

5 Functions of Many Variables: Differentiation

Problem 5
...
Consider the spherical coordinates
x D r sin  cos ;

y D r sin  sin

and z D r cos Â:

Show by differentiating these equations with respect to x, y and z that
@Â
1
@
1 sin
@r
D cos ;
D cos  cos ;
D
;
@x
@x
r
@x
r sin Â
@Â
1
@
1 cos
@r
D sin  sin ;
D cos  sin ;
D
;
@y
@y
r
@y
r sin Â
@r
D cos Â;
@z

@Â
D
@z

1
sin Â;
r

@
D 0:
@z

Problem 5
...
In quantum mechanics Cartesian components of the operator
of the angular momentum of an electron are defined as follows:
Ã
Â
O x D i„ y @ z @ ;
L
@z @y

Ã
Â
@
@
x
;
i„ z
@x @z

O
Ly D

O
Lz D

Â
@
i„ x
@y

Ã
@
y
:
@x

Show that in the spherical coordinates the operators are (these operators act
on functions depending only on the angles so that the r dependence can be
ignored):
Â
O
Lx D i„ sin

cos @
@
C
@Â
tan  @

Ã

O
Ly D i„

;

Â
cos

sin @
@
C
@Â
tan  @

Ã
;

@
O
Lz s D i„ :
@
Finally, demonstrate that the square of the operator of the total angular
momentum is:
Â
Ã
Ä
@
1 @2
1 @
O
Ox Oy Oz
L2 D L2 C L2 C L2 D „2
sin Â
C
:
sin  @Â
@Â
sin2 Â @ 2
Consider now, quite generally, three quantities x, y and z which are related by a
single equation f
...
This means that any one of them may be considered
as a function of the other two
...
x; y/ and x D
x
...
46)

5
...
47)

y

Here the variable written as a subscript to the partial derivatives is considered
constant while differentiating
...
46) in the second term of dx
above:
 Ã
 à " Ã
 à #
@x
@z
@x
@z
dx D
dy C
dx C
dy
@y z
@z y
@x y
@y x
"Â Ã
 à  Ã#
 à  Ã
@x
@x
@x
@z
@z
dx C
C
dy:
D
@z y @x y
@y z
@z y @y x
Examine the coefficient to dx
...
(3
...
To have the left-hand side equal to
the right one, we then need the expression in the square brackets to be zero which
gives us an interesting identity between various partial derivatives in this case:
 Ã
 à  Ã
@x
@z
@x
C
D 0:
(5
...
36
...
(5
...
49)
@z y @y x @x z

Next, let us examine relationships between derivatives of four functions x, y, z
and t, from which only two (any two) are independent, i
...
any other two can be
considered as functions of them
...
47):
 Ã
 à " Ã
 à #  Ã
 Ã
@x
@t
@t
@x
@t
@t
dx C
dy D
dy C
dz C
dy
dt D
@x y
@y x
@x y
@y z
@z y
@y x
"Â Ã Â Ã
 à  Ã
 Ã#
@t
@t
@x
@x
@t
dy C
D
C
dz:
@x y @y z
@y x
@x y @z y

290

5 Functions of Many Variables: Differentiation

In the last term y is kept constant and hence t can be considered a function of x D
x
...
@t=@z/y :
"Â
dt D

@t
@x

à Â
y

@x
@y

Ã

Â
C

z

@t
@y

Ã#

Â
dy C

x

@t
@z

Ã
dz:
y

Now, we compare this result with the one in which t D t
...
50)

z

Numerous applications of the derived identities (5
...
50) can be found in
thermodynamics, and several typical examples of these the reader can find in the
next section
...
7 Applications in Thermodynamics
The formalism we have developed is quite useful in thermodynamics
...
Once a particular pair of variables is
chosen, any other variable appears a function of these two
...
The thermodynamic potential A D A
...
51)
@X Y
@Y X
Above, for convenience, as was done in the previous section and is customarily done
in thermodynamics, we have written partial derivatives with a subscript indicating
which particular variable is kept constant
...
Note that above dA is the exact differential with respect to
the corresponding variables X and Y
...
7 Applications in Thermodynamics

291

We shall now consider four most useful thermodynamic potentials indicating in
each case the appropriate pair of variables corresponding to them: (i) the Helmholtz
free energy
FDU

TS

[variables T; V];

(5
...
53)

(iii) the enthalpy
H D U C PV

[variables S; P]I

(5
...

Note that entropy can also be considered as a thermodynamic potential
...
Note that the heat Q is not a thermodynamic potential and its change
Q D TdS is not an exact differential (that is why instead of dQ we write Q), but
dS D Q=T is
...

Subtracting8 d
...
TS/ D d
...
TS/ D
...
TdS C SdT/ D

SdT

PdV:

(5
...
56)

T

Note that the last equation is in fact an equation of state of the system which
relates P, T and V together, making any one of them a function of the other two
...
Indeed,

The factor 1=T, which turns an inexact differential into an exact one, is called the integrating
factor
...

7

8

We are using here the product rule written directly for the differential of the function TS
...
57)
This last formula is called the Maxwell relation
...
37
...
58)
Problem 5
...
Similarly for the Gibbs free energy:
Â

dG D VdP SdT;

Â

Ã

@G
VD
@P

;

SD

T

@G
@T

Â

Ã
;
P

@V
@T

Â

Ã
D
P

@S
@P

Ã
:
T

(5
...
39
...
60)
Problem 5
...
Prove the following relationships:
Â
UDF T

@F
@T

Ã

Â
I

V

@U
@V

Ã

Â
D

PCT

T

@P
@T

Ã

Ä
I

V

@
@T

 Ã
F
T

D
V

U
I
T2
(5
...
62)

At the next step we introduce various thermodynamic observable quantities
which can be measured experimentally
...
63)
@T P
@T P
@T P

5
...
(5
...
60)); (ii) thermal expansion and thermal pressure
coefficients
 Ã
 Ã
1 @V
1 @P
˛V D
and ˛P D
I
(5
...
65)

S

Simple relationships exist between all these quantities which can be established
using the mathematical devices we developed at the end of the previous section, see
Eqs
...
48) and (5
...
We start by considering V as a function of P and T, i
...
V D
V
...
(5
...
V; T; P/ H)
...

We have:
 à  Ã
 Ã
 Ã
@V
@T

...
@P=@T/V
@P T
where we inverted the derivative
...

Upon employing the definitions for ˛V , T and ˛P given above, this immediately
yields:
V˛V
P˛P

V

T

D0

H)

˛V D P˛P

T:

Another useful relationship is obtained by first applying (5
...
S; P; V/ H)
...
@S=@P/V
I

...
66)

similarly, using the correspondence
...
y; z; x/, we get
Â

@T
@V

à Â
P

@V
@P

Ã

Â
C

T

@T
@P

Ã

Â
D 0 H)

V

@V
@P

Ã
D
T


...
@T=@V/P


...
@P=@T/V
(5
...

Therefore,
S
T

D


...
@S=@P/V
...
@S=@T/V
cV
D
D
D :

...
@S=@V/P
...
@S=@T/P
cP

294

5 Functions of Many Variables: Differentiation

So far we have used formula (5
...
Let us now illustrate an application of Eq
...
50) which
connects four functions
...
S; T; P; V/ H)
...
57) in the last passage
...
41
...
50) with the correspondence
...
t; y; z; x/; then employ Eq
...
48) for
...
x; y; z/ and the last formula
in Eq
...
59)
...
8 Directional Derivative and the Gradient of a Scalar Field
Here we shall generalise further our definition of the partial derivative
...
In fact, it is also possible
to define the partial derivative along an arbitrary direction in space
...

Suppose there is a scalar function U
...
x; y; z/ (it is frequently said a
“scalar field”) that is continuous in some region V in a three-dimensional space
(generalisation to spaces of arbitrary dimensions is straightforward)
...
M/ U
...
Now, suppose the line L is defined
parametrically in a natural way as x D x
...
l/ and z D z
...
5
...

This means that the change of U along the curve can be written, in the limit of the

5
...
5
...
r/ along the
curve L

point M approaching M0 , as
U
U
...
l C l//
@U
D lim
D lim
l!0 l
l!0
@l
l

U
...
l//

D

@U dx
@U dy
@U dz
C
C
@x dl
@y dl
@z dl

by the rule of differentiation of a composite function
...
5
...
Therefore,
@U
@U
@U
@U
@U
@U
@U
D
lx C
ly C
lz D
cos ˛ C
cos ˇ C
cos ;
@l
@x
@y
@z
@x
@y
@z

(5
...
lx ; ly ; lz / is the unit vector tangent to the curve L at point M0 , and ˛, ˇ
and are the angles it forms with the three Cartesian axes
...
42
...
e
...

The derivative considered above is called directional derivative of the scalar field
U along the curve L
...
For
some directions the directional derivative at the same point M0 can be smaller, for
some larger
...

If we introduce the gradient of the scalar field U as the vector
g D grad U
...
69)

296

5 Functions of Many Variables: Differentiation

then it follows from Eq
...
68) that the directional derivative can be written as a
scalar product
@U
D
...
70)

where # is the angle between g and l
...
Thus, the gradient of a scalar field points in the direction of the
most rapid increase of the field, and its magnitude gives the rate of the increase
...
e
...

The gradient of U can also be written, quite formally, as a product of the so-called
del operator
rD

@
@
@
iC jC k
@x
@y
@z

(5
...
Indeed,
Ã
Â
@
@
@U
@U
@U
@
iC jC k U D
iC
jC
k;
rU D
@x
@y
@z
@x
@y
@z
which is exactly grad U
...

As an example, let us calculate the gradient of U
...
r/, where f
...
The calculation of the gradient requires calculating derivatives
of the function fp with respect to Cartesian components x, y and z; these are easily

...
e
...
e
...


Problem 5
...
The gravitational potential of a galaxy decays with the distance r from it as '
...
Calculate the
force F D mr'
...
What is the direction of the force? State whether the interaction
between the two objects corresponds to repulsion or attraction
...
]

5
...
44
...
x;p z/ D xyz at the
y;
point
...
1; 1; 1/
...
]
Problem 5
...
Consider the scalar field
...
(a) Evaluate
the derivative of along the direction e D
...
x; y/; (b)
find the direction of the most rapid increase of at the point
...
y C 6x/ = 5; (b)
...
16; 48; 0/
...
46
...
1; 1; 1/
...
1; 2; 1/
...
2; 2; 1/ and 5= 6
...
r/ in p dimensions
...
r/ would indicate the
direction of the most rapid decrease of f from that point
...
Calculate the
gradient there, g1 D gradf
...
Near the minimum, the gradient will be very small (exactly equal to zero
at the minimum, see Sect
...
10), the steps ır would become smaller and smaller, so
that after some finite number of steps a point very close to the exact minimum will
be reached
...
Note that the steps 1 , 2 ; etc
...

For instance, this method may be used to calculate the mechanical equilibrium of
a collection of atoms (e
...
in a molecule or a solid)
...
r/ is related to the force F acting on a particle
moving in it via F D grad U D @U=@r D rU (all these notations are often
used)
...
Moving atoms in the direction of forces leads
to a lower energy
...
This idea is at the heart of all modern
methods of material modelling
...

Consider a surface in 3D that is specified by the equation f
...
g
...

Theorem 5
...
The normal vector to the surface f
...
x; y; z/
...
Because of the equation f
...
Let us assume that the coordinates x and y are independent, i
...

z D z
...
Then, consider the gradient of f :
rf D

@f
@f
@f
i C j C k:
@x
@y
@z

To calculate the necessary derivatives, we first differentiate both sides of the
equation f
...
x; y; z
...
x; y/ dependence explicitly:
 Ã
 Ã
@f
@f @z
@f @z
@f
@f
@f
C
D 0 and
C
D0
D
D
@x tot
@x
@z @x
@y tot
@y
@z @y
(zeros in the right-hand sides are due to the zero right-hand side of f
...

Hence, the partial derivatives entering the gradient above are:
@f
D
@x

@f @z
@z @x

and

@f
D
@y

@f @z
;
@z @y

@z
i
@x

Ã
@z
jCk :
@y

and hence the gradient becomes:
@f
rf D
@z

Â

Comparing this result with Eq
...
18), we immediately see that rf is indeed directed
along the normal n to the surface z D z
...
x; y; z/ D 0)
...
E
...

As an example of this application of the gradient, we calculate the normal to
the surface x2 C y2 D z at the point A
...
Here f
...
A/ D 2
...


Problem 5
...
Using the definition of the gradient, prove its following properties:
r
...
UV/ D UrV C VrU;

where U and V are scalar fields
...
V/ D

dU
rV;
dV
(5
...
9 Taylor’s Theorem for Functions of Many Variables

299

5
...
3
...
For simplicity, we shall
only consider the case of a function z D f
...

What we would like to do is to write an expansion for f
...
x0 ; y0 / in terms of x D x x0 and y D y y0 which are
variations of the function arguments around that point
...
3
...
60) for the functions of a single variable which we already
know
...
t/ D f
...
73)

of a single variable 0 Ä t Ä 1
...
x; y/
is obtained by calculating F
...
e
...
t/, and then taking away
f
...
Sine F
...
62), i
...
the Taylor’s formula around the point t D 0, in
terms of t D t:
F
...
0/ C F 0
...
0/t2
C
2

C

F
...
0/tn
F
...
/tnC1
C
;
nŠ

...
74)

where the last term is the remainder RnC1
...
By taking
t D 1 the function F
...
x0 C x; y0 C y/ D f
...
x; y/
around the point
...

Obviously, F
...
x0 ; y0 /
...
t/, we notice that F
...
x; y/ with the arguments being linear
functions of t, i
...
x
...
t/ D y0 C ty with @x=@t D x and
@y=@t D y
...
25) for the derivative of a composite
function of two variables (N D 2), we have:
Â
Ã
@F @y
@F
@F
@
@
@F @x
F
...
t/:
@x @t
@y @t
@x
@y
@x
@y
0

(5
...
t/
...
Indeed, the second
derivative can be calculated by differentiating the just obtained first derivative and
noticing that both functions @F=@x and @F=@y are again composite functions of t:

300

5 Functions of Many Variables: Differentiation

Ã
 Ã
 Ã
@F
d @F
d @F
@F
x C
y D x
C y
@x
@y
dt @x
dt @y
Ã
 2
Ã
 2
@2 F
@F
@2 F
@ F
y C y
x C 2 y
x C
D x
@x2
@y@x
@x@y
@y
Ã
Â
2
2
@ 2
@2 F
@
2 @ F
2 @ F
C
...
t/;
D
...
t/ D
dt
00

Â

where again we have written the result of the differentiation using an operator; in
fact, this is the same operator as in our result (5
...
9

Problem 5
...
Use induction to prove that generally
Ã
Â
@ n
@
F
...
t/ D x C y
F
...
76)

for any n D 1; 2; 3; : : :
...
n/
ˇ
F
...
t/ˇ
D x C y
f
...
77)
while the remainder term at t D 1 yields
1
RnC1
...
n C 1/Š

ˇ
Â
Ã
ˇ
@ nC1
@
ˇ
x C y
F
...
x0 C #x; y0 C #y/ :

...
78)
Correspondingly, the Taylor expansion, after taking t D 1 in Eq
...
74), reads:
f
...
x0 ; y0 / C F 0
...
0/
C
2

C

F
...
0/
C RnC1 ;
nŠ

(5
...

9
Recall, that a similar trick with operators we have already used in Sect
...
6 when illustrating the
derivation of the Leibnitz formula (3
...


5
...
64) for the case of a function of two variables
...
(5
...
79)
for n D 0, we get:
Ã
Â
@
@
f
...
x; y/ D f
...
x0 ; y0 / C fx0 x0 ; y0 x C fy0 x0 ; y0 y;

(5
...
Here we differentiate f
...
Another important formula which we shall need below corresponds to
the n D 1 case:
f
...
x0 ; y0 / C fx0
...
x0 ; y0 / y
i
1 h 00 0 0
00
00
fxx x ; y
...
y/2 ;
C
2

(5
...

Problem 5
...
Derive the n D 2 Taylor expansion of f
...

The formulae we have obtained here can in fact be generalised for functions of
any number of variables; this, however, requires developing more algebraic results;
in particular, a generalisation of the Binomial expansion to forms
...
1
...

Then, the method developed above can be basically repeated almost without change
...


5
...
We shall only consider here in detail the case
of functions of two variables, z D f
...

Similarly to functions of a single variable, we define the function z D f
...
x0 ; y0 /, if there exists a positive radius R such that
for any point M
...
e
...
M; M0 / D
...
y y0 /2 < R), the value of the function is f
...
x0 ; y0 /
...


302

5 Functions of Many Variables: Differentiation

Fig
...
6 For the prove of the
necessary condition of a
maximum of a function
z D f
...

Here the function has a
maximum at the point M0 and
the lines AB and DC along
the surface z D f
...
10
...

Consider Fig
...
6 where the function z D f
...
If
we fix the first variable x of the function at x0 and allow only the other variable to
change, we would draw the line DC on the surface associated with our function;
this line will correspond to the function g
...
x0 ; y/ which is a function of only
a single variable y
...
y0 / D f
...
y/ has a maximum at the
point y0
...
y/ is equal to zero at y D y0
...
x0 ; y/ calculated at the point y D y0
...
x/ D f
...
5
...
x0 / D fx0
...

Therefore, we conclude that the necessary condition for a function to have a
maximum is that its partial derivatives are equal to zero at that point:
fx0
...
x0 ; y0 / D 0:

(5
...
We see that the conditions (5
...
3
...
Solving
two equations (5
...
x0 ; y0 / which might be a minimum
or maximum of the function
...
Note
that the stationary point(s) obtained may be either minimum, maximum or neither
of these (see below) since these conditions are only the necessary ones
...


5
...
x/, where x D
...

Indeed, if at the point x0 D x0 ; : : : ; x0 the function has a minimum (maximum),
1
N
then by fixing all variables but one, say xi , we arrive at a function of a single variable
g
...
83)

By repeating this process for every i D 1; : : : ; N we arrive at N such conditions
which give N equations; solving these equations yields the stationary point(s) x0
which might correspond to extrema of the function
...
84)

As an example, let us find the minimum distance d between two lines specified
by the vector equations r D r1 C t1 a1 and r D r2 C t2 a2 , where t1 and t2 are
the corresponding parameters
...
t1 ; t2 / D
...
x1 C t1 a1x

t2 a2x /2 C y1 C t1 a1y

x2

C
...
85)

with respect to the two parameters t1 and t2
...
r1 C t1 a1
@t1
@d2
D
@t2

2a2
...
Solving these equations with
1
2
respect to the two parameters (e
...
solve for t1 in the first equation and substitute
into the second), we obtain
t1 D

a2
...
a1 a2 /
...
a1 a2 /

2

;

t2 D

a2
...
a1 a2 /
...
a1 a2 /2

:

304

5 Functions of Many Variables: Differentiation

Problem 5
...
Substituting the found parameters into the distance square
in (5
...
(1
...


5
...
2 Characterising Stationary Points: Sufficient Conditions
Of course, similar to the case of a single variable function, we still have to establish
the sufficient conditions for the minimum and maximum
...
Analogue to this situation is the saddle in Fig
...
1(d)
...
(5
...
If we, however, consider a
similar function z D x2 C y2 , then at the same point it obviously has a minimum:
indeed, z0 D 2x and z0 D 2y give a single solution x D y D 0 at which z
...
x; y/ > 0, i
...
the function at any point in the
vicinity of the point
...
Hence, the necessary conditions (5
...
(5
...

Consider a function z D f
...
x0 ; y0 / D
0
fy
...
x0 ; y0 /
...
x; y/
f
...
x; y/ can then be written using the n D 1 case
Taylor expansion (5
...
x/2 C 2Fxy xy C Fyy
...
86)

where for convenience we introduced the following notations:
00
00
Fxx D fxx
...
x0 C #x; y0 C #y/ ;
00
Fyy D fyy
...
87)

We shall rearrange the terms in (5
...
10 Introduction to Finding an Extremum of a Function

305

#)
Â
Ã2 # " Â
Ã2
Fxy
Fxy
Fyy
Fxy
2
C

...
y/
Fxx
Fxx
Fxx
Fxx
"Â
Ã2
Â
Ã#
Fxx
y 2
Fxy
D
y CD
;
(5
...
Note that here Fxx , Fxy and Fyy (and hence D) still
depend on the particular point
...
To proceed we have to
assume that all three second derivatives of the function f
...
x0 ; y0 /
...
x0 ; y0 / ; Fxy Dfxy
...
x0 ; y0 /

0 0
and D0 D Fxx Fyy

0
Fxy

2

:

0
Assume first that Fxx > 0 and D0 > 0
...
x0 ; y0 / (i
...
small enough x and y) such that Fxx > 0 and
D > 0 for any
...
It is seen now from Eq
...
88) that z > 0
and hence the point
...
Conversely if Fxx < 0 but still
D > 0, then we have a maximum
...
Indeed, let us go along a particular direction
in the x y plane, given by y D kx with some fixed k
...
86) reads
z D

1

...
This means that the
parabola g
...
Then, depending on the sign of Fyy , the parabola either has a minimum
or maximum as a function of k, but either way, it is positive for some values of
k and negative for some others
...
x/2 ), while going along
other directions in which the parabola is negative we have z

...
Hence, in some directions the function z
...
x0 ; y0 /, as in the other it is going down, i
...
the function has neither
minimum nor maximum at this point
...
x; y/ has a minimum or
a maximum at a point
...
89)

at the point
...
90)

The function does not have an extremum if D < 0
...
1
...
15 of Volume II
...
Setting its first derivatives z0 D
x
10x 3y and z0 D 3x C 12y to zero, we obtain a possible point for a minimum
y
or a maximum as x D y D 0
...
To characterise this
point, we calculate the second derivatives, z00 D 10, z00 D 3 and z00 D 12, and
xx
xy
yy
the determinant D D 10 12
...
Since D > 0 and z00 > 0, we have
xx
a minimum at the point
...
It is easy to see that this result makes perfect sense
...
0; 0/ point
and hence has a minimum there
...
To illustrate this point, let us consider two simple functions:
z D x4 C y4 and h D x3 C y3
...
0; 0/, where the function is
equal to zero and (please, check!) all second order derivatives are equal to zero as
00
well
...
However, as is easily seen, the function
z > 0 away from the point
...
x; y/ changes sign when crossing zero (e
...
take y D 0 and move x from
negative to positive values) and hence has neither minimum nor maximum
...
10 Introduction to Finding an Extremum of a Function

307

Problem 5
...
Show that the function z D 2x2 C 2xy C 3y2
minimum at the point
...


Problem 5
...
Show that the function z D 27x2
have an extremum at the point
...


y has a

5y2 C 6xy C 12x

9x2

Problem 5
...
Show that the function z D
maximum at the point
...


2x

5 has a

25y2 C 30xy

Problem 5
...
Show that the function w D x2 C y2 C
...
1=3; 1=3/
...
55
...
This way N pairs of data points
...
Assuming a linear dependence of y over x, i
...
y D axCb, the “best”
fit to the measured data is obtained by minimising the error function
D

N
X


...
91)

iD1

Demonstrate by direct calculation that the “best” constants a and b are
determined by solving the equations
Sxx a C Sx b D Sxy

and Sx a C Nb D Sy ;

where
Sx D

N
X
iD1

xi ;

Sy D

N
X

yi ;

Sxx D

iD1

N
X

x2
i

and Sxy D

iD1

N
X

xi yi :

iD1

Then show that the error
...
[Hint: note that
1 XX
xi
2 iD1 jD1
N

NSxx
and is hence always positive
...
10
...
e
...
x; y; z/ is sought on a surface specified by the equation
g
...

Sometimes, this problem can simply be solved by relating one of the variables to
the others via the equation of the condition, e
...
z D z
...
g
...
x; y/ D h
...
x; y//
...
From this condition, z D 1 x y, and
hence we have to find an extremum of the function of two variables, w D x2 C y2 C

...
We know how to do this, see Problem 5
...
Hence, at the point

...
x; y; z/ experiences a minimum
...
56
...
x; y; z/ D x2
y2 z2 C xy C yz C xz C 3
...
[Answer: a
maximum at
...
]
Problem 5
...
Find a conditional extremum of the function h
...
x C y C z/ on the cone z D x2 C y2
...
]
The method above is only useful if the constraint equation(s) can be solved
analytically with respect to some of the variables so that these could be eliminated
from the function explicitly
...
To this end,
we consider a problem of finding extrema of a function h D h
...
x; y; z/ D 0
...
x; y/ and
hence defines a function of two variables w
...
x; y; z
...
Therefore, one
can use formula (5
...
Next, formulae (5
...
40) can be employed for
calculating the corresponding second order derivatives and hence characterising the
stationary point(s)
...
x; y/
...
Differentiating both sides of the
condition equation g
...
92)

5
...
93)

which, of course, are exactly the same as Eqs
...
36) and (5
...
5
...
The
second order derivatives can then be calculated by differentiating the condition
equation twice and using Eqs
...
38)–(5
...
94)

(5
...
96)

The obtained equations fully solve the problem
...
(5
...
Hence, we can write the required equations for
the stationary point(s) as follows:
h0
h0
h0
y
x
D 0 D z:
g0
gy
g0
x
z

(5
...
x; y; z/ D 0
giving one or more stationary points (or none)
...
Here g D x C y C z 1, and hence g0 D g0 D g0 D 1 with all second
x
y
z
order derivatives of g equal to zero
...
The stationary point is obtained by calculating
h0 D 2x, h0 D 2y and h0 D 2z, yielding 2x=1 D 2y=1 D 2z=1, i
...
x D y D z,
x
y
z
which, together with the constraint equation, gives the same point as we obtained

310

5 Functions of Many Variables: Differentiation

above using the direct calculation:
...
Now, we need to calculate the
second order derivatives to characterise the stationary point
...
x; y/ are equal to zero, we obtain from Eqs
...
39) and (5
...
1/ C 2
...
1/
...

Problem 5
...
Find a conditional extremum of the function h
...
[Answer: a maximum at

...
]
Problem 5
...
Find a conditional extremum of the function h
...
x C y C z/ on the cone z D x2 Cy2 using the above method
...
57
...
10
...
(5
...
Note that sometimes it is more convenient to consider three equations
instead of the two by introducing the fourth unknown equal to either of the
fractions in (5
...
98)

The three equations above together with the condition equation are sufficient to find
the required four unknowns: x, y, z and
...
x; y; z/ D h
...
x; y; z/:

(5
...
The
three equations (5
...
g
...
10 Introduction to Finding an Extremum of a Function

311

frequently used in practical calculations since it is extremely convenient: it treats all
the variables on an equal footing and also allows the introduction of more than one
condition easily
...

Therefore, it is essential that we consider this method for a general case
...
We would like
to determine all its stationary points subject to p (where 1 Ä p Ä N 1) conditions
gk x1 ; : : : ; xnCp D 0, where k D 1; : : : ; p
...
(5
...
100)

The total differential of every condition must also be zero because gk D 0:
dgk D

nCp
X @gk
iD1

@xi

dxi D 0;

k D 1; : : : ; p:

(5
...
102)

where k (with k D 1; : : : ; p) are arbitrary numbers and we introduced an auxiliary
function
ˆ D w
...
x1 ; : : : ; xN / :

(5
...
e
...
Then, we request that the contribution to the differential
dˆ in (5
...
e
...
104)

312

5 Functions of Many Variables: Differentiation

Since the first n variables x1 ; : : : ; xn are independent and the above expression
should be zero for any dxi , we arrive at the n equations:
@ˆ
D 0;
@xi

i D 1; : : : ; n:

(5
...
(5
...
105) that both conditions can be written
simply as
@ˆ
D 0;
@xi

i D 1; : : : ; n C p;

(5
...
e
...
103) instead of the original function w
...
In other words,
all variables can be considered as independent and all constraints (conditions)
combined into the auxiliary function (5
...
Note that there are exactly n C 2p D
N Cp equations for N variables fxi g and p Lagrange multipliers: N equations (5
...

As an example of using Lagrange multipliers, let us find the minimum distance
between a point M
...
If P
...
M; P/2 D
...
y

yM /2 C
...
e
...
As we only have a single condition, there will be one
Lagrange multiplier , and we consider the auxiliary function
ˆ D
...
y

yM /2 C
...
Ax C By C Cz

D/ ;

for which the stationary point is determined by three equations:
@ˆ
D2
...
z
@z

zM /

@ˆ
D 2
...
107)

and the equation of the plane
...
AxM C ByM C CzM
A2 C B2 C C 2

D/

:

(5
...
10 Introduction to Finding an Extremum of a Function

313

Using now Eqs
...
107) we can obtain the coordinates of the point P on the plane
which is nearest to the point M:
x D xM C

1
A;
2

y D yM C

1
B;
2

z D zM C

1
C:
2

(5
...
(1
...


Chapter 6

Functions of Many Variables: Integration

6
...
4
...
Here we shall generalise the idea of integration for
functions of more than one variable
...


6
...
1 Definition and Intuitive Approach
Consider a thin 2D plate lying in the x y plane, Fig
...
1(a), the area density
of which is
...
We would like to calculate the total mass of the plate
...
x; y/ D
...
If the
plate was uniform, then its total mass could be obtained simply by multiplying
its surface density by its area
...
6
...
Within
each such segment the density can be approximately treated as a constant yielding
its mass equal to Mkj Akj , where the point Mkj k ; j is chosen somewhere
inside the
...
The total mass of the whole plate is then equal to the sum over all
segments:
X
k;j

Mkj Akj D

X

k; j

xk yj :

k;j

© Springer Science+Business Media, LLC 2016
L
...
1007/978-1-4939-2785-2_6

315

316

6 Functions of Many Variables: Integration

Fig
...
1 2D plate lying in the x

y plane

Fig
...
2 For the calculation
of the volume of a cylinder
oriented along the z axis

Then, we consider all segments becoming smaller and smaller (so that their number
increases), and then in the limit of infinitesimally small segments we arrive at the
double integral
ˆ ˆ

...
1)

k;j

which gives the exact mass of the plate
...

Another example where a similar concept is employed is encountered when we
would like to calculate the volume of a cylindrical body bounded by the x y plane
at the bottom and some surface z D z
...
6
...
Again,
we divide the bottom surface into small regions of the area Akj , choose a point
Mkj
...
1 Double Integrals

317

that in the limit of infinitesimally small regions tends to the double integral
ˆ ˆ
VD
z
...

Since the double integral is defined as a limit of an integral sum, it has the same
properties as a usual (one-dimensional) definite integral
...
Theory of the Darboux sum developed in
Sect
...
2 can be directly applied here as well almost without change
...
x; y/ to be integrable in region A
is its continuity, although a wider class of functions can also be integrated
...
6 in Sect
...
2 is valid for double integrals, i
...

ˆ ˆ
f
...
; / A;
(6
...
; / belonging to the region A, i
...
the point M lies
somewhere inside A
...
1
...
x; y/dxdy can be calculated simply by considering one
definite integral after another, and this, as we shall argue presently, may be done
in any order in the cases of proper integrals (for improper integrals see Sect
...
1
...

Indeed, consider a plane region A bounded from the bottom and the top by
functions y1
...
x/, respectively, as depicted in Fig
...
3(a)
...
6
...
4
...


318

6 Functions of Many Variables: Integration

whole region stretches from x D a to x D b
...
1) is considered as a
limit of an integral sum, and obviously, the sum would only make sense if it does
not depend on the order in which we add contributions from each region
...
However, as
we shall show below, a certain order in summing up the regions will allow us to
devise a simple method of calculating the double integral via usual definite integrals
calculated one after another
...
Let mkj
and Mkj be the minimum and the maximum values of the function f
...
kj/, i
...
mkj Ä f
...
kj/
...
kj/
region (i
...
between yj 1 and yj ) we arrive at the inequality:
ˆ

yj

mkj yj Ä

yj

f
...
e
...
2 Since this is the case for any j, we can sum these inequalities for all j values,
giving:
X

ˆ
mkj yj Ä

j

H)

X

y2
...
k /

f
...
k / Ä

j

X

X

Mkj yj

j

Mkj yj ;

(6
...
k / and y2
...

The integral in the middle of the inequality is some function g
...

Since the inequality is valid for any choice of the value of k (as long as it lies
within the interval between xk 1 and xk ), it can be replaced simply by x lying within
the interval:
X
X
mkj yj Ä g
...


6
...
x/ dx Ä @
Mkj yj A xk :
xk

j

1

j

Since this is valid for any k, we can sum up these inequalities over all values of k,
which yields:
ˆ b
XX
XX
mkj xk yj Ä
g
...
Now we
take the limit of all regions Akj D xk yj tending to zero; both Darboux sums
converge to the value of the double integral, since according to our assumption
the double integral exists
...
x; y/dxdy D
g
...
x/

y1
...
x; y/dy dx D

y2
...
x/

f
...
4)

Here we recalled that the g
...
x; y/ over dy taken between y1
...
x/
...
x/, and the largest, y2
...
Two ways of writing the double integral shown on the right
in Eq
...
4) demonstrate this explicitly in two different ways (both are frequently
used)
...

The method we just described basically suggests that in order to calculate the
double integral, we may first integrate over a (hatched) vertical stripe in region A
shown in Fig
...
3(a), which corresponds to the limit of the corresponding integral
sum over j for the fixed k :
ˆ y2
...
k ; y/dy D xk g
...
k / D xk
y1
...
k / D
g
...

The arrows in Fig
...
3(b) serve to indicate symbolically that the first integration is
performed in the direction of the arrows (i
...
along y), and then the contributions
from all such integrations are summed up by means of taking the x integral
...
The obtained formula provides us with a practical recipe
for calculating double integrals: first, we fix the value of x and calculate the y integral
between y1
...
x/ (these give the span of y for this particular value of x); then,
we perform the x integration between a and b
...
x/ and y2
...
k / and y2
...
k / and y2
...
g
...
The consideration is easily
generalised to more complex integration regions when this is not possible: in those
cases the region should be divided up into smaller regions each satisfying the
required condition
...

Example 6
...
I Calculate the volume of the hemisphere of radius R centred at zero
with z 0, see Fig
...
4(a)
...
For the given x; y, the height of the hemisphere is z
...
Thus, the
volume of the hemisphere is obtained from the double integral
ˆ ˆ
VD

ˆ

ˆ

R

z
...
For the given x, the variable y spans
to
p
between
R2 x2 and C R2 x2 , see Fig
...
4(b), since at the base of the

p
Fig
...
4 (a) A hemisphere bounded by the surfaces z D 0 and z D
R2 x2 y2 used in
Example 6
...
(b) The base of the hemisphere showing the span of y ( the vertical thin dashed line)
for the given value of x
...
1 Double Integrals

321

hemisphere we have a circle x2 C y2 D R2
...
e
...
6
...
e
...
J
Above, we performed the y integration first, that followed by the x integration
...
Indeed, assuming that
the same region in Fig
...
3(a) can also be described by the boundary lines x D x1
...
y/ bounding the whole region from the left and right, respectively, see
Fig
...
3(c), we can start by summing up contributions along the x axis first, i
...
from
all little regions that have the same value of j lying somewhere between the division
points yj 1 and yj , and then sum up all these contributions as shown schematically
in Fig
...
3(c)
...
x; y/dxdy D
r
...
y/

x1
...
x; y/dx dy D

ˆ

d

x2
...
y/

c

f
...
5)

where the whole span of y is between c and d
...
y/
...
In some cases useful general
relationships can be obtained by using this “symmetry” property
...
g
...
y/ and x2
...
x/ and y2
...

Example 6
...
ICalculate the area bounded by curves shown in Fig
...
5
...
(4
...


´p
a2

t2 dt D

1
2

hp
t a2

t2 C a2 arcsin

t
a

i

C C, see

322

6 Functions of Many Variables: Integration

Fig
...
5 The integration
region in Example 6
...
First of all, we note that it is a triangle, and hence its area should be
equal to one (the horizontal base is equal to 2 and the vertical height to 1)
...
We first try the method of
Fig
...
3(b)
...
1 C x/ dx C

ˆ

1

0

dy

0
1

0


...
Now, if using the other method of Fig
...
3(c), we only need to
calculate a single double integral as the whole integration region can be dealt with
at once
...
Hence, we obtain:
ˆ

ˆ

1

1 y

dy
0

y 1

ˆ
dx D
0

1

ˆ
Œ1

y


...
1

y/ dy D 1:

The result is the same, although the calculation may seem easier
...
1
...
Use both methods, when either the y or
the x-integrations are performed first
...
]
Problem 6
...
Calculate the double integral:
ˆ 1 ˆ 1 x
ye
dx:
dy
x
0
y
[Hint: Change the order of the integration so that the y integral goes first, paying
special attention to the limits, and sketch the way the integration region is
“scanned” in both cases as in Fig
...
3(b, c)
...
]

6
...
1
...
We have argued that
one can do the integration in any order for these integrals: either over x first (the fast
variable) and over y second (the slow), or other way round
...
e
...
4
...
3
...

For that let us consider the following improper integral with respect to t
containing the function f
...
x/ D

1

f
...
6)

a

This integral can be treated as an improper integral depending on the parameter x
(cf
...
4
...
2)
...
e
...
We are going to investigate such improper integrals here
...

We gave already a definition of the convergence of improper integrals in Sect
...
5
...

Let us apply this definition to our case: the integral (6
...
x/
for the given value of x, if for any > 0 there always exists A > a, such that
ˇ
ˇ
ˇF
...
x; t/dtˇ <
ˇ

or

ˇˆ
ˇ
ˇ
ˇ

A

1

ˇ
ˇ
f
...
7)

This condition states that the integral converges to F
...
However, in some cases it won’t
...
6) is said
to converge to F
...

There is a very simple test one can apply in order to check whether the
convergence is uniform or not, due to Weierstrass
...
1 (Weierstrass Test)
...
x; t/ which is continuous
with respect to both of its variables can be estimated from above as jf
...
t/ for all values of x and t, and the function M
...
6) converges uniformly with respect to x
...
Indeed, what we need to prove is that the residue of the integral appearing
in Eq
...
7) can be estimated by some chosen using the bottom limit A which is
independent of x
...
x; t/dtˇ Ä
ˇ

ˆ

1

1

M
...
x; tj dt Ä

A

A

It is seen that for the given one can always choose such value of A that the integral
of M
...
The value of A chosen in this way does not
depend on x, and hence, the integral converges uniformly
...
E
...

´1
This simple test can be illustrated on the integral 0 e xt dt
...
x; t/ D
e xt and can be estimated from above for any x > 1 as f
...
Therefore,
M
...
The integral therefore converges uniformly within this interval of x
...
3
...
a/
ˆ

...
e/

1

0

1

ˆ

dt

...
b/

sin
...
˛ > 0/ I
x2 C t 2


...
xt/
dt
...
f /

ˆ

1

0

sin2
...
xt/ dt
...
> 1/ :

...
]
Problem 6
...
Show that the integral
ˆ
E1
...
E1
...
[Hint: make a substitution u D t=x and then estimate
the function e xu =u for u > 1 by e u
...
1 Double Integrals

325

Next we shall prove a theorem which states under which conditions the function
F
...
6) is continuous
...
2
...
x; t/ is continuous with respect both of its variables, and
the integral (6
...
x/
...
Indeed, the function F
...
x/ D
F
...
This means that for any > 0 one can find a positive ı such that for any
x in the proximity to x0 , i
...
jx x0 j < ı, one has jF
...
x0 /j <
...
We then have:
jF
...
x; t/dt
f
...
x0 /j D ˇ
ˇ
ˇ
a
a
ˇˆ A
ˇ
ˆ 1
ˆ 1
ˇ
ˇ
Dˇ
Œf
...
x0 ; t/ dt C
f
...
x0 ; t/ dtˇ
ˇ
ˇ
a
A
A
ˇˆ 1
ˇ ˇˆ 1
ˇ
ˆ A
ˇ
ˇ ˇ
ˇ
Ä
f
...
x0 ; t/ dtˇ :
jf
...
x0 ; t/j dt C ˇ
ˇ
ˇ ˇ
ˇ
a

A

A

Since f
...
x; t/ f
...
This fact allows estimating the first integral in the righthand side above as less than or equal to 1
...
The other two integrals can be
both estimated using some 2 > 0 due to uniform convergence of the improper
integral in question
...
x/

F
...
A

a/ C 2

2

D ;

as required
...
Once 1 is known, then the
value of ı is determined from the continuity of the function f
...
Q
...
D
...


326

6 Functions of Many Variables: Integration

Theorem 6
...
If f
...
6) F
...
x/dx D

x2

dx f
...
Here
ˆ

x2

ˆ
F
...
x; t/ :

dx

x1

a

x1

We need to prove that the difference
ˆ
LA D

ˆ

x2

ˆ

A

F
...
x; t/

x1

a

tends to zero as A ! 1
...
x; t/

x1

a

dt f
...
x; t/

dx

a

x1

x2

dt
ˆ

A

dx

ˆ

A

dt f
...
x; t/ :

x1

The first and the last double integrals are the proper ones, and since a proper double
integral does not depend on the order of integration, they cancel out
...
x; t/ :

dx
x1

(6
...
x; t/ˇ Ä
ˇ

x2

x1

ˇˆ 1
ˇ
ˇ
ˇ
ˇ
dx ˇ
dt f
...
x2

x1 / D ;

1

since the integral over t of f
...
Therefore, by choosing
arbitrary > 0, we calculate 1 D =
...
This means that for any > 0 one
can always find such A > a that jLA j <
...
8) converges to zero in the A ! 1 limit, as required
...
E
...


6
...
5
...
4)
ˆ
0

ˆ

1

E1
...
x/ D

1

t 1 e t dt:

x

[Hint: change the order of the integration
...
Moreover, they
are also valid for the improper integrals due to singularity of the integrand f
...


6
...
4 Change of Variables: Jacobian
We know from Chap
...
Remarkably,
it is also possible to perform a change of variables in a double integral, but this time
we have to change two variables at once
...

Let us assume that the calculation of a double integral can be simplified by
introducing new variables u and v via some so-called connection equations:
x D x
...
u; v/

(6
...
u; v/ coordinate
system
...
6
...

In order to change the integration variables
...
u; v/, we have to find
the relationship between the areas dxdy and dudv in the two systems
...
6
...
u; v/ system may be much
simpler than in the
...
6
...
u; v/ system, Fig
...
7(a)
...
u; v/, B
...
u C du; v C dv/
and D
...
These points
will transform into points A0 , B0 , C0 and D0 , respectively, in the
...
9)
...
A/ D
...
u; v/ and y D y
...
10)

Indeed, e
...
the point B0 is the image of the point B in the
...
u; v C dv/ D x
...
Therefore, in the vector form we obtain the
first relation in (6
...
All other points are obtained accordingly, and in the vector
notations their coordinates are given by Eq
...
10)
...
x; y/ system, which can approximately be considered as a
parallelogram, we can calculate the vectors corresponding to its sides:
!
A0 B0 D r B0

r A0 D

@r
dv D
@v

Â

@x
@y
dv; dv; 0
@v
@v

Ã

and
Â
Ã
@x
@y
@r
du D
du; du; 0 ;
@u
@u
@u
Â
Ã
!
!
@x
@y
@r
du D
du; du; 0 D A0 D0
B0 C 0 D
@u
@u
@u
!
A 0 D0 D r D0

r A0 D

6
...
x; y/ system, which is the image of the corresponding
shaded rectangular region in the
...
6
...
@x=@v/ dv
...
@x=@v/ dv
...
@x=@u/ du
...
@x=@u/ du
...
@x=@v/ dv
...
@x=@u/ du
...
u; v/j dudv;
@u @v ˇ

where
J
...
x; y/
@x @y
D
@
...
11)

@
...
The notation @
...
Thus,
the areas in the two systems, which correspond to each other by means of the
transformation (6
...
x; y/ ˇ
ˇ dudv;
ˇ
dA D ˇ
(6
...
u; v/ ˇ

and the change of the variables in the double integral is thus performed using the
following rule:

†

ˇ
ˇ
ˇ @
...
x
...
u; v// ˇ
@
...
x; y/dxdy D

(6
...
x; y/ system transforms into the region  in the
...
Note that the absolute value of the Jacobian must be used in the formula
since the area must be positive
...
6
...
We shall transform the integral to the polar coordinates
...


330

6 Functions of Many Variables: Integration

Fig
...
8 A quarter of a circle
in the
...
r; / system of the polar
coordinates (a)

which the integration region D (a quarter of a circle) turns into a rectangle with sides
R and =2, shown in Fig
...
8(a)
...
Hence, the span of is the same for any r, and hence
in the
...
The old coordinates are
related to the new ones via x D r cos , y D r sin , so that the Jacobian can be
calculated as
ˇ
ˇ ˇ
ˇ
ˇ @x=@r @y=@r ˇ ˇ cos
@
...
u; v/
and the double integral becomes:
ˆ ˆ
ID


...
r sin /
...
The integrand is a product of
two functions, the limits are constants, so that it does not matter which integral is
calculated first:
! Â Ã
ˇ =2
ˆ R
à ˆ =2
R4 sin2 ˇ
R4
3
ˇ
:
ID
r dr
cos sin d
D
D
ˇ
4
2 0
8
0
0
Let us remember: for the polar coordinate system the Jacobian is equal to r and the
area element in this system
dA D rdrd :

(6
...
6
...
The area dA of the hatched region ABCD (which can
approximately be considered rectangular) is a product of the sides AB and AD, where
AB D dr and AD D r sin
...
14)
...
3
...
6
...

Solution
...
The Jacobian is J D r, and the volume becomes:

6
...
6
...
e
...
1 in the previous section using
direct integration over x and y
...
6
...
[Answer:
2 =˛
...
7
...

Problem 6
...
Show using Cartesian coordinates, that the double integral
´´
0 half circle of radius R centred at the
S ydxdy D 0, where S is the x
origin
...


332

6 Functions of Many Variables: Integration

Fig
...
10 For the calculation
of the electrostatic potential
from a disk in Problem 6
...
9
...
Do the calculation using both methods as well
...
10
...
6
...
The disk is uniformly charged with the
surface charge density
...
][Answer:
...
]
Problem 6
...
A thin disc of radius R placed in the x y plane with its centre at
p
the origin has the surface mass density
...

Assuming that the density decays sufficiently quickly with increasing r, i
...
that
˛
1=R, calculate the total disc mass using the polar coordinates
...
]
In physics applications, the following, the so-called Gaussian integral, is
frequently used:
ˆ

1

e

˛x2

r
dx D

˛

1

:

(6
...
5 Consider a product of two Gaussian integrals,
in which we shall use x as the integration variable in the first of them and y in the
second:
ˆ 1
ˆ 1
2
2
G2 D
e ˛x dx
e ˛y dy:
1

5

1

The corresponding indefinite integral cannot be represented by a combination of elementary
analytical functions
...
2 Volume (Triple) Integrals

333

Of course, it does not matter which letters are used as integration variables, but using
these makes the argument more transparent: the product of these two integrals can
be also considered as a double integral across the whole x y plane, in which case
we can use the polar coordinates in an attempt of calculating it:
G2 D

ˆ ˆ

e
ˆ

D2
so that G D

p

C1

˛
...


Problem 6
...
Prove that
ˆ 1
h
exp a
...
x

r
i
y/2 dx D

1

Ä
aCb

exp

ab

...
16)

Problem 6
...
Prove that the diffusion probability for the Brownian motion of
a particle in a solution, introduced by Eq
...
13), satisfies the equation
ˆ
w
...
17)

This equation shows that the probability to find the particle at the point
x at time t, if it was at x0 at time t0 , can be represented as a sum of
all possible trajectories via an intermediate point x0 at some intermediate
time t0 , where t0 < t0 < t
...


6
...
2
...
Similarly, triple integrals appear in problems related to volumes of
three-dimensional objects
...
k ; j ; Ál /xk yj zl ;

k;j;l

where k , j and Ál are some points within the intervals xk , yj and zl along
the Cartesian axes
...

For proper integrals it does not matter which integration (over x, y or z) goes first,
second and third, this is dictated entirely by convenience of the calculation (the
integrand and the shape of the integration region)
...
r/ D f
...
x; y; z/dxdydz;

(6
...
x/

y1
...
x;y/

z1
...
x; y; z/dz dy dx D

dx
a

ˆ

y2
...
x;y/

dy
y1
...
x;y/

f
...
19)

where we first fix x and y and calculate the integral over z between the corresponding
minimum and maximum z values z1 and z2 for the given x and y; after that, the y
integration is performed for the fixed x value; the x integration is performed last
...
4
...
6
...

Solution
...
When fixing z (see Fig
...
11), we arrive at
a circular cross section of the hemisphere by the plane parallel top x y plane at
the
this z value
...
So, the
p
span of x (for the given z) will be p
between ˙ R2 z2 ; while the span of y for the
given (fixed) x and z is between ˙ R2 z2 x2
...
(4
...


6
...
6
...
1/
2

z

2

R

2

R2
2

z

z2 arcsin p

x
R2

z2 arcsin
...
Note that this calculation performed in Cartesian
coordinates was rather cumbersome
...
2
...
Suppose, we would like to introduce new variables
...
x; y; z/ using connection equations:
x D x
...
u; v; g/ and z D z
...
20)

The integration region V in the
...
u; v; g/ system
...
u; v; g/ system that is obtained by advancing by du,
dv and dg along the corresponding coordinate axes u, v and g from the point
A
...
x; y; z/ in the
...
6
...


336

6 Functions of Many Variables: Integration

Fig
...
12 A small cuboid in
the
...
x; y; z/ system the cuboid ABCDEFGH transforms into the parallelepiped
A0 B0 C0 D0 E0 F 0 G0 H 0
...
x; y; z/ are:
!
@r
A 0 D0 D
dv
@v

!
@r
du ;
A0 B0 D
@u

!
@r
and A0 E0 D
dg:
@g

Then the volume of the parallelepiped is given by the absolute value of the mixed
product of these three vectors (see Eq
...
41)):
ˇÂ
Ä
ˇ @r
@r
ˇ
dV D ˇ
du
dv
@u
@v

ˇ
ˇ
Ãˇ ˇ
...
@y=@u/ du
...
@x=@v/ dv
...
@z=@v/ dv ˇ
ˇ
ˇ ˇ
@g
ˇ
...
@y=@g/ dg
...
u; v; g/j dudvdg;

(6
...
x; y; z/
J
...
u; v; g/
ˇ @x=@g @y=@g @z=@g ˇ

(6
...
As you can see, for three-dimensions it is defined
in a very similar way to the two-dimensional case of Eq
...
11)
...
x; y; z/ ˇ
ˇ dudvdg:
ˇ
f
...
u; v; g// ˇ
@
...
23)

As an example, let us again calculate the volume of the hemisphere in Fig
...
4
...
r; Â; / here,
x D r sin  cos

;

y D r sin  sin

and z D r cos Â;

6
...
The Jacobian of the
transformation
ˇ
ˇ
ˇ @x=@r @y=@r @z=@r ˇ
ˇ
ˇ
@
...
r; Â; /
ˇ @x=@ @y=@ @z=@ ˇ
ˇ
ˇ
ˇ sin  cos
sin  sin
cos  ˇ
ˇ
ˇ
D ˇ r cos  cos r cos  sin
r sin  ˇ D r2 sin Â:
ˇ
ˇ
ˇ
ˇ r sin  sin r sin  cos
0
It is a useful exercise to use the definition of the 3 3 determinant (Sect
...
6) and
verify the above result for the Jacobian
...
cos Â/0 D
R :
3
3

As you see, all three integrals are completely independent yielding the same result
as in Examples 6
...
3 and 6
...

It is useful to remember: the Jacobian for the spherical coordinate system is
r2 sin Â, and the volume element
dV D r2 sin ÂdrdÂd :

(6
...
14
...
x; y; z/ about
the z axis is given by
ˆ ˆ ˆ
Iz D
x2 C y2
...
x; y; z/dxdydz

is the body mass
...
6
...
[Hint: use the cylinder coordinates

...
]
[Answer: M D R2 h, Iz D 1 MR2
...
15
...
r; ; z/
...
]
3
5
Problem 6
...
A uniform sphere of radius R and density is rotated around
an axis passing through its centre with the angular velocity !
...
4=15/ R5 ! 2
...
3 Line Integrals
6
...
1 Line Integrals for Scalar Fields
The consideration in this section generalises the one we made in Sect
...
6
...

Consider a scalar function (a scalar field) f
...
x; y; z/ that is specified
in the three-dimensional space
...
Then, consider a curved line AB that
is specified parametrically as
x D x
...
u/ ;

z D z
...
25)

where u is a parameter (e
...
time t in which case the line may be a particle trajectory
in the 3D space)
...
Then, let us divide the line into
n sections by points M1 , M2 , : : :, Mn 1 ; the beginning and the end of the line are
marked by points M0 D A and Mn D B, respectively, see Fig
...
13
...
Finally, consider the sum
n
X

f
...
If the limit
exists, then it does not depend on the way the curved line is divided up and how the
points Ni are chosen within each segment, and the sum converges to what is called
a line integral between points A and B
...
x; y; z/dl D

ID
AB

f
...
26)

6
...
6
...

Points N1 ; N2 ; : : : ; Nn are
chosen somewhere inside
each such segment

where dl is the length of the differential section of the line related to the changes
dx, dy and dz of the Cartesian coordinates due to the increment du of the parameter
(e
...
dx D
...
u/du and so on), and f
...
x; y; z/ on the line at the point where the length l is reached (e
...
measured from
the initial point A); obviously, the value of l is directly related to the value of u
...

If the line is closed, then the following notation is frequently used:
˛
f
...
82) (see also Fig
...
5):
dl D

p

q
dx2 C dy2 C dz2 D


...
u//2 C
...
u//2 C
...
u//2 du;

so that a working expression for the integral (6
...
x
...
u/; z
...
x0
...
y0
...
z0
...
27)

a

Note that the above defined line integral possesses all properties of definite integrals
...
x; y; z/dl D

AB

f
...
28)

340

6 Functions of Many Variables: Integration

To make sure that this is the case in practice, you must follow the rule: if you
integrate from smaller to larger values of the u, then the change du > 0 and hence dl
will automatically be positive
...

Let us now consider some examples
...
The
circle can be specified on the x y plane as x D R cos and y D R sin with
changing from zero to 2
...
Note that since this is a planar problem, z0
...
Then, the length
is obtained as a line integral with f
...
x0
...
y0
...

In our second example we shall determine the mass of a non-uniform wire which
has a shape of a spiral and is given in polar coordinates as r D e , with the angle
changing from 0 to 1, see Fig
...
20(d)
...
/ D ˛ , where ˛
´
is a parameter
...
Introducing polar
coordinates, we arrive at the following definite integral:
ˆ
lD˛

1
0

where x
...
/ cos D e
ˆ 1 q

...
x0
...
y0
...
/ D e

sin , so that

sin /2 C
...

In our third example, we will calculate the line integral
˛
I D
...
6
...
We split the integral into three
contributions:

6
...
6
...
x C 2y/ dl D

ˆ


...
x C 2y/ dl C

OB

BA


...
x/2 dx;
dx
while y should be used for AO as the parameter, with the line increment being
v
u
 Ã2 ! p
u
p
dx
2 C dy2 D tdy2 1 C
D 1 C x0
...
x/ D 0:
ˆ

ˆ

...
x C 2 0/ 1 C 0

0

Along the line BA we have: y D x C 1, y0
...
x C 2y/ dl D
BA


...
2

1

0

ˇ1
x2 ˇ
1
xdx D ˇ D :
ˇ
2 0
2

1, so that


...
1

p
x/ dx D 2 2

1

x//

p
1 C
...
Finally, along
the line AO we have: x D 0 and x0
...
x C 2y/ dl D
AO

ˆ

...
0 C 2y/ dy D

0

1

2ydy D 1:

342

6 Functions of Many Variables: Integration

Again, we integrated from the smaller (y D 0) to the larger (y D 1) value of the y
3
here
...

2
Problem 6
...
Find the length of a two-dimensional spiral specified parametrically in polar coordinates
...
[Answer: 2 a 1 C ˛ 2 =4 =˛
...
18
...
[Answer: 3
...
19
...
x; y/ D
...
1; 1/
...
61) may be of use
...
]
Problem 6
...
Calculate the mass of ten revolutions of the unfolding helix
wire given parametrically by x D e cos , pD e sin and z D e and with
y
the linear mass density D ˛
...
2
1/ e20 C 1
...
3
...
However, one may also consider a vector
function F
...
x; y; z/ specified in the 3D space, the so-called vector field
...
If
we would like to calculate the work done by the force on the particle, we may split
the trajectory into little directional straight segments l that would run along the
actual trajectory (see again Fig
...
13), calculate the work A D F
...
l/ l:

Then, we can make the segments smaller and smaller, and as the limit of each of the
segments length tends to zero, we shall arrive at the line integral
ˆ
ID

ˆ
F
...
l/dx C Fy
...
l/dz;

(6
...
5
...
Note that the
increments dx, dy and dz are not independent, but correspond to the change along

6
...
This line integral is called a line integral for vector fields or of the second
kind
...
In the first
method the line L is specified parametrically as x D x
...
u/ and z D z
...
u/du, and similarly
for dy and dz, so that
ˆ

ˆ
Fx
...
l/dy C Fz
...
u/x0
...
u/y0
...
u/z0
...
30)

where e
...
Fx
...
x
...
u/; z
...
u/ and Fz
...
Of course, the integral can also be split into three independent
integrals (since the whole integral is a limit of an integral sum, and the sum can be
split into three), each of them to be calculated independently, e
...

ˆ

ˆ
Fx
...
x
...
u/; z
...
u/du D
L

Fx
...
u/du;

L

and similarly for the other two integrals
...

In the second (Cartesian) method, projections of the line L on the Cartesian
planes x y, y z and x z are used
...
l/dx C Fy
...
l/dz D Fx
...
l/dy C Fz
...
31)
L

L

L

L

Here, each of the integrals in the right-hand side is completely independent and
can be calculated separately
...
g
...
e
...
Basically, this method is similar to the first
one; the main difference is that u D x is used for the calculation of the dx integral,
u D y for the dy, and u D z for the dz integrals, respectively, i
...
different parameters
are employed for each part of the line integral
...
6
...
1 by Eq
...
26)
...
l/ dl D
F
...
32)
AB

BA

344

6 Functions of Many Variables: Integration

since the directional differential length dl changes its sign in this case
...
By noting that dl D mdl, where m is the
unit vector directed tangentially to the line L at each point and along the direction
of integration, and dl is the differential length, we can write:
ˆ

ˆ
F
...
l/ m dl;

(6
...
l/ D
F
...

Consider several examples
...
5
...
4
...

from zero to 2
...
One revolution corresponds to a change of
dx D

R sin d ; dy D R cos d

and dz D bd ;

so that
ˆ

2

JD
0

D

2R

ˆ
ΠxR sin
2

ˆ
0

0

C zb d D

yR cos

sin d
...
6
...
6
...

Solution
...
Along OB dy D 0
and
ˆ 1
JOB D
2xydx D 0 since y D 0:
0

Along BA y D 1 x, so, by considering x as the parameter u, we have dy D
y0
...
3 Line Integrals

ˆ
JBA D

1

0

345

2xydx C x2
...
1

x/

x2 dx D

 2
x
2
2

3

Ãˇ0
x3 ˇ
ˇ D 0:
3 ˇ1

Note that, opposite to the example in the previous subsection, the direction of
integration here does matter,7 and hence along the path BA the integral was
calculated from x D 1 to x D 0 corresponding to the change of x when moving
from point B to A
...
J
Note that, as we shall see later on, this result is not accidental as the integrand is
the total differential of the function F
...
Indeed,
dF D

@F
@F
dx C
dy D 2xydx C x2 dy :
@x
@y

´
Problem 6
...
Calculate the line integral L F dl along the straight line
connecting points
...
1; 1; 1/ for the vector field F D
...

[Answer: 3=2
...
22
...
0; x; 0/ in
moving a particle between two points A
...
0; 1; 0/ along a circle of
unit radius and centred at the origin
...
]
´
Problem 6
...
Evaluate the line integral
x2 dx C 3xydy along the sides of
the triangle ABC in the anticlockwise direction
...
0; 0/, B
...
2; 2/
...
]
´
Problem 6
...
Evaluate F dl along the parabola y D 2x2 between points
A
...
1; 2/ if F D xyi y2 j
...
]
Problem 6
...
A particle moves in the anticlockwise direction along a square,
see Fig
...
15, by the force F D xyi C x2 y2 j
...
[Answer: 0
...
6
...
6
...
26
...
yi C xj/ = x2 C y2
...
]
¸
Problem 6
...
Evaluate the integral L F dl taken in the anticlockwise
direction along the circle of radius R centred at
...
[Answer:
aR2
...
28
...
y; x; z/ in moving
a particle from
...
1; 0; / along the helical trajectory specified via
x D cos t, y D sin t, z D t
...
1 C =2/
...
3
...
6
...
This may happen if, for instance, a function has singularities at some
points which must be removed
...


6
...
6
...
4 (Green’s Theorem)
...
x; y/dx C Q
...
34)

where the double integral in the left-hand side is taken over the region S,
while the line integral in the right-hand side is taken along its boundary L in
the anticlockwise direction
...
Consider a simply connected region S bounded from the bottom and the top
by two functions y D y1
...
x/, Fig
...
17(a), and two functions P
...
x; y/ defined there
...
Then, the double integral over the region
of @P=@y is
ˆ ˆ
S

ˆ b
@P
...
x; y2
...
x/
a
ˆ b
ˆ b
P
...
x// dx
P
...
x// dx
D

@P
dxdy D
@y

ˆ

ˆ

b

y2
...
x; y1
...
x; y2
...
x; y1
...
x; y/dx D

P
...
x; y/dx;
ABCDA

where ABCDA is the boundary of the region S
...


348

6 Functions of Many Variables: Integration

On the other hand, this special region can be considered as bounded by functions
x D x1
...
y/ on the left and the right of the region S instead, as shown
in Fig
...
17(b)
...
x; y/
dx D
dy
ŒQ
...
y/; y/
@x
c
x1
...
x2
...
x1
...
y/

c

ˆ
D

Q
...
y/; y/ dy

c

ˆ

d

c

Q
...
y/; y/ dy C
c

ˆ

ˆ

Q
...
y/; y/ dy
d

Q
...
x; y/dy D
DAB

Q
...
Now
we subtract one expression from the other to get the required Eq
...
34)
...
6
...
This result is, however, valid
for any region, since it is always possible to split the arbitrarily complicated region
into subregions each of the same type as considered above, i
...
that vertical and
horizontal lines cross the boundary of each part at two points at most
...
To illustrate this idea of generalising the region shape,
consider the region shown in Fig
...
17(c): the region S is split into three regions
S1 , S2 and S3 by a vertical line
...
Thus, for each of the
regions the above Green’s formula has been proven to be valid:
ˆ ˆ Â
S1

@Q
@x

S2

@Q
@x

S3

@Q
@x

ˆ ˆ Â
ˆ ˆ Â

Ã
˛
@P
dxdy D
P
...
x; y/dy;
@y
ABFA
Ã
˛
@P
dxdy D
P
...
x; y/dy;
@y
BCDB
Ã
˛
@P
dxdy D
P
...
x; y/dy:
@y
BDEFB

If we now sum up these equations, we will have the double integral over the
whole region S in the left-hand side, and the line integral over all boundaries in
the right-hand side
...
Therefore, the contributions from the added (vertical) line, which is run

6
...
Q
...
E
...
6
...
The
idea of the proof is actually simple: we first consider an extension of the functions
P
...
x; y/ inside the holes (any will do), then apply Green’s theorem to the
whole region including the holes and subtract from it Green’s formula for each of
the holes
...
x; y/ D 2xy and Q
...
6
...
We know from Example 6
...
Consider now the left-hand side of Green’s formula:
ˆ ˆ Â

@Q
@x

Ã
ˆ ˆ
@P
dxdy D

...
e
...

In our next example, we shall verify the Green’s formula for P
...
x; y/ D 4xy and the region which is a semicircle of radius R lying in the y 0
region as shown in Fig
...
18
...
4y 2y/ dxdy D
2ydxdy
@y
ˆ ˆ
ˆ R
ˆ
D

...
cos /0 D
:
3
3

On the other hand, consider the line integral
˛
x2 C y2 dx C 4xydy

Fig
...
18 Semicircle region

0

sin d

350

6 Functions of Many Variables: Integration

along the boundary of the region
...
The first integral is most easily
calculated using the polar coordinates x D R cos and y D R sin (using as
a parameter representing the semicircle):
ˆ
R2 x0
...
/ d
semicircle

ˆ

D
ˇ
ˇ
Dˇ
ˇ

R3 sin

0

C 4R3 cos2

ˇ
ˆ
ˇ
t D cos
ˇ D R3
dt D sin d ˇ
1

d D R3

sin

ˆ
0

1

4t2

4 cos2

1
...


Problem 6
...
Consider a planar figure S with the boundary line L
...
35)

L

[Hint: in each case find appropriate functions P
...
x; y/
...
30
...
/ and contains the origin inside it
...
/d ;

(6
...

Derive this expression using both the first and the last formulae for the area
given in (6
...

Some problems related to the application of Eq
...
36) can be found in Sect
...
6
...


6
...
3
...
6 that the line integral over a specific closed path was zero
...
One may wonder when line integrals over a closed loop are equal to zero for
any choice of the contour L
...
5
...
Then,
the following four statements are equivalent to each other:
1
...
x; y/dx C Q
...
37)

L

´
2
...
x; y/dx C Q
...
dU
...
x; y/dx C Q
...
the following condition is satisfied:
@P
@Q
D
:
@y
@x

(6
...
We shall prove this theorem by using the scheme: from 1 follows 2 (i
...

1!2), then 2!3, 3!4 and then, finally, 4!1 again
...
We assume that Eq
...
37) is satisfied for any closed loop L
...
Indeed, take a closed loop ABCDA as
shown in Fig
...
19(a)
...
x; y/dx C Q
...
6
...
(b) To the proof of
the step 2!3 of Theorem 6
...
x; y/dx C Q
...
x; y/dx C Q
...
x; y/dx C Q
...
x; y/dx C Q
...
x; y/dx C Q
...
x; y/dx C Q
...
e
...
Since the initial
closed loop was chosen arbitrarily, then the integral will be the same along any path
from A to C
...
Now we are given that the line integral does not depend on the particular
path, and we have to prove that dU D P
...
x; y/dy is the exact differential
...
x0 ; y0 /
...
x; y/, we can calculate
ˆ
U
...
x;y/

A
...
x1 ; y1 /dx1 C Q
...
39)

Note that we have used integration variables x1 and y1 here to distinguish them from
the coordinates of the point B
...
x; y/
...
6
...
x C x; y/ U
...
x;y/
ˆ C
...
x0 ;y0 /
A
...
The first integral above
in the right-hand side is taken along the path from A to C, while the second one goes
from A to B
...
Then, the part
of the path A ! B cancels out, and we have
1
U
D
x
x
D

1
x

ˆ

C
...
x1 ; y1 /dx1 C Q
...
x;y/

ˆ

1
x

ˆ

C
...
x1 ; y/dx1

B
...
x1 ; y/dx1 D P
...
Here we
have used the (average value) Theorem 4
...
4
...
Note
also that above the dy1 part of the line integral disappeared since along the chosen

6
...
In the limit of x ! 0 the point ! x,
hence we have that @U=@x D P
...

Similarly, one can show that @U=@y D Q
...
Therefore, we find that
the differential of the function U
...

3!4
...
5
...
(5
...
Since dU is an exact differential, then
Â

@
@y

@U
@x

Ã

@
@P
must be equal to
D
@y
@x

Â

@U
@y

Ã
D

@Q
;
@x

as required
...
Since the region S is simply connected, we can use the Green’s
formula (6
...
x; y/dx C P
...
However, due to condition
4, Eq
...
38), the double integral in the right-hand side above is zero for any choice
of L, i
...
the closed loop integral for any L is zero, i
...
the statement 1 is true
...
Q
...
D
...
6
...
3)
...
A classical example is the vector field
AD

x2

y
x
iC 2
j:
2
Cy
x C y2

It is seen that
@Ax
y2 x2
@Ay
D
;
D
2
@y
@x

...
5 above
...
sin / C
1
1

ˆ

...
x; y/ and Ay
...
The
theorem is not valid since the contour (the unit radius circle) has the origin inside it
...

We have already discussed in Sect
...
5 a method of restoring the function U
...
x; y/dx C Q
...
Formula (6
...
Moreover, as the integral
can be taken along an arbitrary path connecting points
...
x; y/, we can
find the simplest route to make the calculation the easiest
...
x0 ; y0 / !
...
x; y/ as shown in Fig
...
20
...
x; y/ D

y

P
...
x; y1 /dy1 :

(6
...
x; y/ is defined up to an arbitrary constant since the choice of the initial
point
...
The information about the initial point should combine
into a single term to serve as such a constant
...
x; y/ from its differential dU D 2xydx C x2 dy as an
example
...
38) is satisfied:
@P
@
@Q
@ 2
D

...
Then, using Eq
...
40), we obtain
Fig
...
20 For the calculation
of the exact differential

6
...
x; y/ D

y

2x1 y0 dx1 C
x0

2

Â

x dy1 D 2y0
y0

x2
2

x2
0
2

Ã

C x2
...
e
...
x; y/ D x2 y C C is a solution
...

Problem 6
...
Verify that x2 y3 dx C x3 y2 dy is the ¸exact differential of some
function U
...

Problem 6
...
Find the function U
...
[Answer: U D x3 y3 =3 C C
...
33
...
x; y/ D x3 y3 =3CC and compare
it with the one given in the previous problem
...
34
...
x;y/

U
...
x0 ;y0 /

sin x1 sin y1 dx1 C cos x1 cos y1 dy1

does not depend on the particular path that connects two points A
...
x; y/
...

Problem 6
...
Verify the Green’s formula for P
...
x; y/ D
2x2 y and the region of a square of side 2 shown in Fig
...
15
...
4 Surface Integrals
So far we have discussed double integrals that are calculated on planar regions
...
This is the subject of this section
...


6
...
1 Surfaces
Generally, a line in space can be specified parametrically as x D x
...
u/ and
z D z
...
A surface S can also be specified in the same
way, but using two parameters:

356

6 Functions of Many Variables: Integration

x D x
...
u; v/ and z D z
...
41)

so that any point lying on the surface is represented by a vector
r
...
u; v/i C y
...
u; v/k:

(6
...
u; v/ !
...
e
...
x; y; z/ for each set
...
u; v/
give the same point
...

For instance, a spherical surface of radius R can be specified by equations:
x D R sin  cos ; y D R sin  sin ; z D R cos Â;
where 0 Ä Â Ä

and 0 Ä

<2 :

Here  and are spherical angles defined as in Fig
...
19(c)
...
For instance, if we restrict  to the interval
0 Ä Â Ä =2 only, then we shall obtain the upper hemisphere, but to specify the
whole sphere  has to be between 0 and
...

We say that r
...
41) and
their first derivatives
@x
@y
@z
@r
@x
@y
@z
@r
D
i C j C k and
D
iC
jC
k
@u
@u
@u
@u
@v
@v
@v
@v

(6
...

In order to understand what this means, let us calculate a normal to the surface
at a point M0
...
6
...

The line CM0 D D r
...
u0 ; v/ obtained by setting
u D u0 and allowing only v to change; they both meet at the point M0
...
6
...
u; v/
...
4 Surface Integrals

u

D

357

@r
...
u0 ; v/
D x0 ; y0 ; z0
v v v
@v

uDu0

are tangent to the two lines and lie in the tangent plane to our surface at point M0
(see Sect
...
4)
...
44)
N
...
u/ y
...
u/ ˇ :
ˇ x0
...
v/ z0
...
u/ z0
...
u/ x
...
u/ y
...
u0 ; v0 / D ˇ 0
ˇ y
...
v/ ˇ
ˇ z0
...
v/ ˇ
ˇ x0
...
v/ ˇ
D

@
...
x; y/
@
...
u; v/i C Jy
...
u; v/k:
@
...
u; v/
@
...
45)

It is seen that the normal is expressed via the Jacobians that form components of it
in Eq
...
45)
...
g
...
y; z/
:
@
...
Equivalently, one can use formula (6
...

Also note that the normal N may not (and most likely will never) be of unit length,
although one can always construct the unit normal n D N= jNj from it if required
...
The latter is ensured by the condition
that the derivatives (6
...

Let us now see how this technique can be used to obtain the normal of the side
surface of a cylinder of radius R of infinite length running along the z axis
...
e
...
; z/ with 0 Ä < 2 and 1 < z < 1
...
45):
Jx D

ˇ
ˇ R cos
@
...
; z/

ˇ
0ˇ
ˇ D R cos
1ˇ

and Jz D

; Jy D

ˇ
ˇ0
@
...
; z/

ˇ
ˇ R sin
@
...
; z/

ˇ
R sin ˇ
ˇ D R sin ;
ˇ
0

ˇ
R cos ˇ
ˇ D 0;
0 ˇ

(6
...
R cos / i C
...
For instance, for D =2 we have r? D Rj and for
D we obtain r? D Ri
...
x; y/
...
(6
...
e
...
u; v/;
which gives for the Jacobians:
ˇ 0ˇ
ˇ0 z ˇ
@
...
u; v/
v
ˇ 0 ˇ
ˇz 1ˇ
@
...
u; v/
v

Jx D

@z
D
@u

@z
;
@x

@z
D
@v

@z
;
@y

ˇ
ˇ1
@
...
u; v/

ˇ
0ˇ
ˇ D 1;
1ˇ

where we replaced back u; v by x; y, so that
N
...
47)

Recall that exactly the same expression for the surface normal was obtained by
us earlier in Eq
...
18), the result which was derived using a rather different
consideration
...
36
...
x; y/ D
3
...
Using x and y as the coordinates u and v specifying the surface,
calculate the appropriate Jacobians and obtain an expression for the surface
outward normal as a function of x and y
...
]

6
...
37
...
u; / via
the connection relations: x D u2 1, y D usin and z D ucos
...
[Answer: N D u; 2u2 sin ; 2u2 cos
...
g
...
The surfaces consisting of such smooth pieces are called piecewise smooth
surfaces
...
Indeed, if we take a membrane,
i
...
a surface that caps a closed line L, see Fig
...
22(a), then two normals n1 and
n2 D n1 can be defined at each point
...
The same is true for the other (lower
side) normal n2
...
A classic example
of a non-orientable (one-sided) surface is the “Möbius8 strip” (or “leaf”) shown in
Fig
...
23
...


Fig
...
22 (a) The surface S caps a closed loop line L that serves as its boundary
...
The traverse direction along the line L corresponding to the
upper side (the normal vector n1 ) is also shown: at each point along the path the surface is on
the left which corresponds to the right-hand rule shown in (b)

8

The strip was also discovered independently by Johann Benedict Listing
...
6
...
6
...
u; v/ system into the
...
4
...
e
...
u; v/, using two parameters u and v
...
The driving idea to solve this problem is as follows
...
u; v/ coordinate system, in which the values of u and v form a plane
surface region †, and draw horizontal and vertical lines u D ui and v D vj there to
form a grid as shown in Fig
...
24(a)
...
u; v/ grid lines will form coordinate lines
ri
...
ui ; v/ and rj
...
u; vj / on the actual surface S in the
...
6
...

Consider now a small surface element uv in the
...
ui ; vj /, the hatched area in Fig
...
24(a)
...
6
...
For small
u and v the actual surface element can be replaced by the corresponding surface
element in the tangent surface built by the tangent vectors

6
...
u; vj /
u and
@u

v v

D

@r
...
6
...
ui ; vj / uv
q
2
2
2
D Jx
...
ui ; vj / C Jz
...
(6
...
This way the
whole surface S will be represented by small planar tiles smoothly going around it
...
ui ; vj / C Jy
...
ui ; vj /uv;

i;j

which, in the limit of little sections tending to zero, gives the required exact formula
for the surface area as a double integral over all allowed values of u and v:
AD

ˆ ˆ
ˆ ˆ q
2
2
2
Jx
...
u; v/ C Jz
...
u; v/j dudv;
†

†

(6
...

Let us illustrate the formula we have just derived by calculating the surface area
of a hemisphere shown in Fig
...
4(a)
...
(6
...
The required Jacobians in

ˇ
ˇ
ˇ R cos  sin
@
...
Â; /
ˇ
ˇ
ˇ R sin  R cos  cos ˇ
@
...
Â; /
ˇ
ˇ
ˇ R cos  cos R cos  sin ˇ
@
...
Â; /
Jx D

D R2 sin  cos  sin2

C cos2

D R2 sin  cos Â:

362

6 Functions of Many Variables: Integration

The normal is then given by9
N
...
R sin Â/ r;
(6
...
Â; /j D
...
Therefore, the surface of
the hemisphere, according to Eq
...
48), is
ˆ 2
ˆ ˆ
ˆ ˆ
ˆ =2
2
2
sin ÂdÂ
d
AD
dA D
R sin  dÂd D R
0

2

D 2 R
...

Remember this expression for the differential surface area in the spherical
coordinates,
dA D R2 sin ÂdÂd :

(6
...
In
spherical coordinates (and assuming a constant radius R) we have x D R sin  cos ,
y D R sin  sin and z D R cos Â
...
6
...

It can approximately be considered rectangular with the sides AD D rd and CD D
FD d D R sin Âd , so that the area becomes dA D R2 sin ÂdÂd , which is exactly
the same expression as obtained above using Jacobians
...
6
...
4 Surface Integrals

363

Fig
...
26 (a) Elliptic cone bound from above by the plane z D R; (b) paraboloid of the height h2 ;
(c) triangle

Problem 6
...
Show that if the surface is specified by the equation z D z
...
48) becomes:
ˆ ˆ

s

AD
†

Â
1C

@z
@x

Ã2

Â
C

@z
@y

where † is the projection of the surface S on the x

Ã2
dxdy;

(6
...


As an illustration of application of Eq
...
51), we shall calculate the surface area
p
of the elliptic cone z D x2 C y2 > 0 cut by the horizontal plane z D R, see
Fig
...
26(a)
...
Therefore, the surface area
ˆ ˆ p
p ˆ ˆ
p
2dxdy D 2
dxdy D 2 R2 ;
AD

v plane) is a

circle

where we have used R2 for the area of a circle of radius R (the “spread” of x; y)
...
39
...
6
...
Write the equation for the surface in a parametric form x D x
...
u; v/ and z D z
...
Obtain the expression for the outer normal to the surface N
...
][Answer: N D

...
]
Problem 6
...
Repeat the previous problem by considering the polar coordinates
...
Notice that the length of N will change, but not the
direction
...
41
...
6
...
[Answer:
h
i
3=2
4h2 C 1
1 =4
...
42
...
Note that the same result is obtained although the
length of the vector N changed with this choice of the parameters describing
the surface
...
43
...
4
...
c C R cos / cos  ; y D
...
52)

where 0 Ä Â < 2 and 0 Ä
< 2
...
Â; / and (b) the entire surface area
...
c C
R cos /
...
]

6
...
3 Surface Integrals for Scalar Fields
If we are given a scalar field f
...
Mij /Aij
ij

with Aij being the surface area of the ij-th surface element
...
r/dA:
(6
...
4 Surface Integrals

365

Here points Mij are on the surface
...
Therefore, the surface integral (6
...
If the surface is specified parametrically via Eq
...
42), the
surface integral is calculated similarly to Eq
...
48) as follows:
ˆ ˆ

ˆ ˆ
f
...
r
...
u; v/ C Jy
...
u; v/dudv

ˆ ˆ
D

†

f
...
u; v// jN
...
54)

In particular, if the surface is defined via equation z D z
...
r/dA D
S

†

Â

f
...
x; y// 1 C

@z
@x

Ã2

Â
C

@z
@y

Ã2
dxdy:

(6
...
We use spherical coordinates to specify the surface, i
...
using the
angles  and in place of the two parameters u and v
...
(6
...
In the first octant
0 Ä Ä =2 and 0 Ä Â Ä =2, so that the integral becomes:
ˆ ˆ
JD
D R3

Â


...
2Â/
4

Ã
0

=2


...
44
...
Show by calculating the
electrostatic potential
ˆ ˆ

...
e
...
x; y; z/ there)
...
6
...
Â; / when integrating over the surface of the sphere
...
6
...
44

Problem 6
...
Show that the potential of the sphere from the previous problem
at the distance z outside the sphere (i
...
for z > R) is given by D Q=z, where
Q D 4 R2 is the total charge of the sphere
...
46
...
r/ D
S


...
r/ D ˛ x2 C y2
...
Show that outside the sphere and at the distance D
(where D > R) from its centre

...
r/ dA D 8 ˛R =3,
and hence the potential far away from the sphere ' Q=D, as it would be for
a point charge Q localised in the sphere centre
...
4
...
6
...
2, one
can also consider surface integrals for vector fields
...
4 Surface Integrals

367

Fig
...
28 Flow of a liquid
through the surface A
during time dt

Let F
...
M/ D N
...
M/j at each point M belonging to S
...
F n/ dA D

Fn dA D

F dA:

(6
...
If the flux integral is taken over a closed surface S, the
–
notation for the surface integral is frequently used instead
...
6
...
7), let us consider a flow of a liquid
...
6
...
57)

where vn D v n is the projection of the velocity vector on the normal n to the
surface
...
If, however, the surface was not planar and
the velocity distribution not uniform, then the same result would have been obtained
only for a small surface area A which can approximately be assumed planar, and
the velocity distribution across it uniform
...
Hence, the flux integral
of the velocity field through a surface gives a rate of flow of the liquid through that
surface area, i
...
the volume of the liquid flown through it per unit time
...
If the surface S is specified parametrically as x D
x
...
u; v/ and z D z
...
(6
...

At the same time, the unit normal to the surface n D N=jNj, hence we can write:
ˆ ˆ

ˆ ˆ
F dA D


...
jNj dudv/ D
...
x
...
u; v/; z
...
u; v/ dudv
ˆ ˆ

D

Fx Jx C Fy Jy C Fz Jz dudv:

(6
...
yz; xz; xy/ through the outer surface of the cylinder x2 C y2 D R2 , 0 Ä z Ä h
...
Thus, in this case u D
and v D z, so that, see Eq
...
46),
Jx D R cos ; Jy D R sin
and we obtain
ˆ ˆ
Flux D
d dz
...
47
...
2; 0; 0/,
...
0; 0; 2/ as shown in
Fig
...
26(c), where F D
...

[Hint: the general equation for the plane is axCbyCcz D 1, where the constants
a, b and c can be found from the known points on the plane
...
]

6
...
56) for vector
fields in which it is expressed via a sum of double integrals taken over Cartesian
planes x y, y z and x z
...
Consider
the vector field F D Fx ; Fy ; Fz , then
ˆ ˆ

ˆ ˆ
Fx nx C Fy ny C Fz nz dA

F dA D
ˆ ˆ
D

ˆ ˆ
Fx nx dA C

ˆ ˆ
Fy ny dA C

Fz nz dA;

(6
...
Note that in each of them
the integration is performed over the actual surface, i
...
x, y and z are related to
each other by the equation of the surface
...
(6
...
47) (for the two ways the surface can be specified), we have:
Jx
...
n; i/ D
jNj
2
2
2
Jx
...
u; v/ C Jz
...
@z=@x/2 C
...
n; j/ D

or q

nz D cos

1 C
...
@z=@y/2

b

or q

(6
...
u; v/
Dq
jNj
2
2
2
Jx
...
u; v/ C Jz
...
n; k/ D

;

;

(6
...
u; v/
Dq
jNj
2
2
2
Jx
...
u; v/ C Jz
...
@z=@x/2 C
...
62)

These are components of the normal vector n along the x, y and z axes, and ˛ D
c
c

...
n; j/ and D
...
Then, nz dA D dA cos projects the surface area dA on
the x y plane, Fig
...
29, and thus can simply be replaced by ˙dxdy
...
Similarly,
nx dA D ˙dydz and ny dA D ˙dzdx, so that the integrals (6
...
6
...
6
...
Its projections on the
three Cartesian planes are
also shown in (b) by light
brown (Sxy ), green (Sxz ) and
red (Syz ) colours

ˆ ˆ

ˆ ˆ
F dA D ˙

S

ˆ ˆ
Fx dydz ˙

ˆ ˆ
Fy dzdx ˙

Syz

Fz dxdy;

Sxz

(6
...

This expression gives a practical way of calculating the flux integrals by
essentially breaking them down into three double integrals with respect to the three
projections on coordinate surfaces, and then calculating each one of them via
ˆ ˆ
ˆ ˆ
Fx
...
x
...
x; y; z/dzdx D
ˆ ˆ

Sxz

Fy
...
x; z/; z/ dzdx;
ˆ ˆ

Fz
...
x; y; z
...
x2 z; xy2 ; z/ through
the outer surface of z D x2 C y2 , 0 Ä z Ä 1, x 0 and y 0, see Fig
...
30
...
Note the minus sign before the dxdy integral
...
4 Surface Integrals

371

while in the case of Iz it is directed downwards (in the direction of z), so that we
have to take the negative sign for the dxdy integral
...
We obtain:
ˆ ˆ

x2 zdydz D

Ix D
ˆ
D

Syz
1

ˆ

dy

y2

0

ˆ ˆ
ˆ

Sxz
1

ˆ

dx
0

1

x2

dz z

y2 zdydz

z
y2 z D

dz z

xy2 dxdz D

Iy D
D

1

ˆ ˆ
ˆ

Â

1

dy
0

ˆ ˆ

y2
2

Ã
D

4
;
21

D

1
;
12

x2 dxdz

x z
x2 x D

y6
1
C
3
6

ˆ

Â

1

dx
0

x
2

x3 C

x5
2

Ã

and, finally, the last integral Iz is most easily calculated using polar coordinates

...


Problem 6
...
Calculate the flux of the magnetic field B D
...
[Answer: R2
...
4
...
6
...
3
...


372

6 Functions of Many Variables: Integration

Theorem 6
...
Let S be a simply connected non-planar
piecewise smooth orientable (two-sided, see Fig
...
22(a)) surface with the
closed boundary line L
...
r/, Fy
...
r/ are all
continuous together with their first derivatives everywhere in S including its
boundary, then
˛

ˆ ˆ
F dl D

curl F dA;

L

(6
...
r/ D Fx
...
r/j C Fz
...
65)
curl F
...
Note
that the surface boundary line L is traversed in a direction that corresponds
to the chosen side of the surface to be always on the left (according to the
right-hand rule, see Fig
...
22(b))
...
64) can also be written
differently as
˛
Fx dx C Fy dy C Fz dz
L

ˆ ˆ Â

D
S

@Fz
@y

@Fy
@z

Ã

Â
dydz C

@Fx
@z

@Fz
@x

Ã

Â
dxdz C

@Fy
@x

@Fx
@y

Ã
dxdy;
(6
...
6
...
4
...
Consider first a surface S of a rather special shape which is specified by
the equation z D z
...
6
...

The surface boundary L will become the boundary Lxy in the projection Sxy
...
e
...
x/,
which we assume does exist and which, together with the equation for the surface
z D z
...
Then, consider the x-part of the
line integral (see the text around Eq
...
31)):
˛
JP D

˛
Fx
...
x; y
...
x; y
...
67)

6
...
6
...
Using Green’s formula (6
...
68)

Note that the derivative
...
x; y/ also depends on y
...
(6
...
62)
...
n; j/ D q

b

@z=@y
1 C
...
@z=@y/2

nz D cos
...
@z=@x/2 C
...
n; j/

b;

cos
...
(6
...
n; j/
dxdy:
@z cos
...
69)

374

6 Functions of Many Variables: Integration

On the other hand, consider the following surface integral over the surface S:
ˆ ˆ Â
S

D

Ã
@Fx
@Fx
c
cos
...
n; j/ dA
@y
@z
!
ˆ ˆ
c
@Fx @Fx cos
...
n; k/dA
@y
@z cos
...
n; j/
dxdy;
@y
@z cos
...
6
...
n; k/dA D Cdxdy
...
69) has been obtained, i
...
we have just proven that
ˆ ˆ Â

˛
JP D

Fx
...
n; j/
@z

Ã
@Fx
cos
...
70)

A similar consideration for the same surface S but specified via y D y
...
r/ gives
ˆ ˆ Â

˛
JR D

Fz
...
n; i/
@y

Ã
@Fz
c dA;
cos
...
71)

and for the surface S given via x D x
...
r/, one obtains instead:
ˆ ˆ Â

˛
JQ D

Fy
...
n; k/
@x

Ã
@Fy
ci/ dA:
cos
...
72)

Any piecewise smooth simply connected surface can always be represented as a
combination of elementary surfaces considered above
...
Let us add these three identities (6
...
72)
together
...
n; j/
cos
...
n; i/
cos
...
n; k/
cos
...
4 Surface Integrals

375

Ã
Â
Ã
@Fy
ci/ C @Fx @Fz cos
...
n;
D
@z
@z
@x
S
Â
Ã
@Fy @Fx
C
cos
...
curl F n/ dA D
curl F dA:
ˆ ˆ ÄÂ

@Fz
@y

S

b

S

If a more complex surface is considered, it can always be divided up into finite
fragments for each of which the above identity is valid separately
...
However, the internal
boundaries will be contributing twice, each time traversed in the opposite directions,
and hence there will be no contribution from them since the line integral changes
sign if taken in the opposite direction (cf
...
6
...
3 and especially Fig
...
17(c))
...
Q
...
D
...
g
...
6
...
In this case the line integral L is
considered along all boundaries, including the inner ones, and the traverse directions
should be chosen in accord with the right-hand rule as illustrated in the figure
...
73)
curl F
...
6
...
The traverse
directions, chosen with
respect to the right-hand rule
for the given direction of the
normal n (outward)
correspond to the surface
being always on the left

376

6 Functions of Many Variables: Integration

This determinant notation is to be understood as follows: the determinant has to be
opened along the first row with the operators in the second row to appear on the
left from the components of the vector field F occupying the third row
...
For instance, the x component of the curl of F is obtained as
ˇ
ˇ
ˇ @=@y @=@z ˇ
@
ˇ

...
65)
...

As an example, we shall verify Stokes’s theorem by considering the flux of the
p
vector field F D yi C xj C k through the hemisphere z D R2 x2 y2
0
centred at the origin
...
64)
...
Using polar coordinates, x D R cos and y D R sin with the angle as
a parameter, we obtain
˛

˛

ˆ
Fx dx C Fy dy D

F dl D
L

L

DR

2

ˆ
0

2

2
0

Œ
...
R sin / C
...
R cos / d

d D 2 R2 :

On the other hand, let us now consider the surface integral in the right-hand side of
Eq
...
64)
...
1

Ã
Â
@
...
y/
k
@y


...
R sin Â/ r, see Eq
...
49), so that,
following Eq
...
58), and using spherical coordinates,
ˆ ˆ

ˆ ˆ

ˆ ˆ

curl F dA D
S


...
k r/ dÂd

ˆ ˆ
D 2R
D 2R

ˆ ˆ

...
sin Â/
...
4 Surface Integrals

377

The same result is obtained also using Eq
...
59):
ˆ ˆ

ˆ ˆ
curl F dA D

ˆ ˆ
2
...

Therefore, if a vector field can be written as V D curl F, then the surface integral
ˆ ˆ

ˆ ˆ

˛


...
curl F n/ dA D

S

S

F dl
L

will not depend on the actual surface, only on its bounding curve L
...
xyz; x; z/ and the surface S being a hemisphere x2 C y2 C z2 D R2 bounded below
by the z D 0 plane
...
This is most easily calculated in polar
coordinates (x D R cos , y D R sin and z D 0):
˛

ˆ

2

D
0

circle

D

R

2

ˆ
Œxyz
...
R cos / d D
R

2

Â

0

2

ΠR cos
...
2 /
D
2
4
0

R2 :

Problem 6
...
Verify Stokes’s theorem for F D
...
6
...

Problem 6
...
Verify Stokes’s theorem for F D
...
˙1; ˙1; 0/
...
]
Problem 6
...
Consider a cylinder x2 C y2 D R2 of height h
...
Verify Stokes’s theorem for the surface of the cylinder if F D
...

[Answer: either of the integrals is zero
...
52
...
r/r D 0, where f
...


378

6 Functions of Many Variables: Integration

Problem 6
...
Calculate the flux of the vector field F D
...
e
...
6
...
[Answer:
=2
...
54
...
x; y; 2/ and the outward direction for each
face of the cube
...
]
Problem 6
...
Evaluate the surface (flux) integral over the outward surface of
the unit sphere centred at the origin with F D r
...
]
Problem 6
...
Calculate curl F, where F D
...
[Answer: 0
...
57
...
1; 1; 1/, where F D
...
[Answer:
...
]
¸
Problem 6
...
Use Stokes’s theorem to evaluate the line integral L F dl,
where F D
...
[Hint: cap the circle with the hemisphere
...
]
´´
Problem 6
...
Calculate, using Stokes’s theorem, the surface integral
SF
dA, where F D curl B with B D
...
Will the integral change if a different
surface is chosen which caps the same bounding curve? [Answer: 5 R2 ; no
...
60
...
e
...
´Calculate
´
F Dcurl B and show by the direct calculation of the surface integral, S F dA,
that it is equal to the same value as in the previous problem
...
61
...
y2 ; z2 ; x2 / and S is the surface
of the circular ellipsoid described by z D 1 x2 y2 with z 0, see Fig
...
33
...
]
¸
Problem 6
...
Use Stokes’s theorem to evaluate the line integral L F dl over
the unit circle x2 Cy2 D 1 in the z D 5 plane, where F D
...

[Answer: 3
...
6
...
4 Surface Integrals

379

6
...
6 Three-Dimensional Case: Exact Differentials
As in the planar case, Sect
...
3
...
7
...
r/ D
...
Then, the following four statements are
equivalent to each other:
1
...
74)

L

´
2
...
dU
...
x; y; z/dx C Fy
...
x; y; z/dz is the exact
differential, and
4
...
75)

or, which is equivalent (see Eq
...
65)), curl F D 0
...
This theorem is proven similarly to the planar case of Sect
...
3
...

• 1!2 : the proof is identical to the planar case;
• 2!3: essentially the same as in the planar case, the function U
...
x; y; z/ D

B
...
x0 ;y0 ;z0 /

Fx dx C Fy dy C Fz dz;

(6
...
x0 ; y0 ; z0 / is a fixed point;
• 3!4: the same as in the planar case;
• 4!1: this is slightly different (although the idea is the same): we take a closed
contour L and “dress” on it a surface S so that´L would serve as its boundary
...
However, the curl of F is
S

380

6 Functions of Many Variables: Integration

¸
equal to zero due to our assumption (property 4)
...
Since
the choice of L was arbitrary, this result is valid for any L in D
...
E
...

Note that Eq
...
76) is indispensable for finding exact differentials, similarly to the
two-dimensional case
...
This is most easily done by taking the integration path

...
x; y0 ; z0 / !
...
x; y; z/ which contains straight lines going
parallel to the Cartesian axis; it would split the line integral into three ordinary
definite integrals:
ˆ

ˆ

x

U
...
x1 ; y0 ; z0 /dx1 C

z

Fy
...
x; y; z1 /dz1 :
z0

(6
...
x0 ; y0 ; z0 /
...
40) from the
two-dimensional case
...
First of all, we check
that this expression is the exact differential
...

Therefore, if F D
...
x
ˇ
ˇ
ˇ yz zx xy ˇ

x/

j
...
z

z/ D 0;

which means that we can apply the results of the theorem
...
(6
...
x; y; z/ which would give rise to the differential dU D yzdx C
zxdy C xydz:
ˆ

ˆ

x

UD

y0 z0 dx1 C
x0

ˆ

y

y0

D y0 z0
...
y

xydz1
z0

y0 / C xy
...


6
...
4
...
Similarly, there is also a very general result relating the
integral over some volume V and its bounding surface S which bears the names
of Ostrogradsky and Gauss
...
8 (Divergence Theorem)
...
Fx ; Fy ; Fz / be a vector
field which is continuous (together with its first derivatives) in some threedimensional region D
...
78)

S

where S is the bounding surface to D, with the outward normal n, and the
scalar
div F D

@Fx
@Fy
@Fz
C
C
@x
@y
@z

(6
...


Proof
...
x; y/ and z D z2
...
6
...

Consider then the volume integral
Fig
...
34 A cylindrical space
region bounded from below
and above by surfaces
z D z1
...
x; y/,
respectively
...
x;y/

dxdy
z1
...
x; y; z2
...
x; y; z1
...

On the other hand, the surface integral over the whole surface of the cylinder
S D S1 C S2 C S3 (see the figure) for the vector field G D
...
x; y; z1
...
x; y; z2
...
e
...
Therefore,
—
ˆ ˆ
G dA D
ŒFz
...
x; y// Fz
...
x; y// dxdy;
S

Sxy

which is the same as the volume integral above
...
80)

where the vector G has been shown explicitly via its components
...

This is because the volume integrals for these regions will just sum up (the lefthand side), while the surface integrals will be taken over all surfaces
...

Similarly, if the region D has a cylindrical shape with the axis along the x axis
and the caps are specified by the equations x D x1
...
y; z/, then we
obtain

6
...
81)

Finally, for cylindrical regions along the y axis with the two caps specified by y D
y1
...
x; z/, we obtain
ˆ ˆ ˆ
D

@Fy
dxdydz D
@y

—

0

1
0
@ Fy A dA:
S
0

(6
...
e
...
Therefore, summing
up all three expressions (6
...
82), we arrive at the final formula (6
...
Q
...
D
...
63
...
xy; yz; zx/ and the
volume region of a cube with vertices at the points
...
[Answer:
each integral is zero
...
64
...
xy2 ; x2 y; y/ and S is a
circular cylinder x2 Cy2 D 1 oriented with its axis along the z axis, 1 Ä z Ä 1
...
]
Problem–
6
...
Use the divergence theorem to calculate the surface flux
integral S F dA over the entire surface of the cylinder x2 C y2 D R2 ,
h=2 Ä z Ä h=2, where F D
...
[Answer: hR2 6R2 C h2 =4
...
We shall discuss only two
of them
...
P; 0; 0/, F2 D
...
0; 0; P/
...
83)

We shall make use of this form of the Gauss’s theorem later on in Sect
...
7
...
Above
we have introduced a vector field
gradP D

@P
@P
@P
iC
jC
k;
@x
@y
@z

which is the gradient of the scalar field P
...
5
...


Problem 6
...
Use a similar method to prove another integral identity:
—

ˆ ˆ ˆ

...
a grad/ b ;

(6
...
a grad bx / i C a grad by j C
...
r/
...
]

6
...
Here we shall consider some of them
...
6
...


6
...
1 Continuity Equation
As our first example, we shall derive the so-called continuity equation
@
C div
...
85)

6
...
r/
...

Consider an arbitrary volume V of the fluid with the bounding surface S
...
v n/ dAdt D v dAdt:

(6
...
v n/ dA D v dA, where
n is the outer normal to the surface area dA, see Eq
...
57)
...
The total change (loss)
of mass during time dt due to the flow of the fluid through the whole surface S that
bounds the volume V is then the surface integral
—
dM D dt

v dA:

(6
...
e
...
e
...
t C dt/

...
t C dt/ M
...
The two expressions
should be equal to each other due to mass conservation:
—

ˆ ˆ ˆ
v dA D
V

S

@
dxdydz:
@t

Let us now transform the surface integral using the Gauss theorem of Eq
...
78):
ˆ ˆ ˆ

ˆ ˆ ˆ
div
...
v/ C
@t
V

386

6 Functions of Many Variables: Integration

Since the 3D region V is arbitrary, this integral can only be equal to zero if the
integrand is zero, i
...
we arrive at the continuity equation (6
...

Note that the divergence term in the continuity equation (6
...
v/ D

@ vy
@
...
vz /
C
C
@x
@y
@z

@vy
@
@
@
@vz
@vx
C vx C
C vy C
C vz
@x
@x
@y
@y
@z
@z
à Â
Ã
Â
@vy
@vz
@
@
@
@vx
C
C
vx C vy C vz
C
D
@x
@y
@z
@x
@y
@z
D

D div v C v grad ;
so that the continuity equation may also be written as follows:
@
C div v C v grad D 0:
@t

(6
...
For instance, due
to conservation of charge, the flux of charge in or out of a certain volume should be
reflected by the change of the charge density inside the volume, which is precisely
what the continuity equation (6
...
Indeed, consider a flow of particles
each of charge q through an imaginable tube of the cross section dA (infinitesimally
small)
...
v dA/ dt; this result is absolutely identical to the one above, Eq
...
86), which
we have derived for the mass of the fluid
...
(6
...

However, in this particular case it is convenient to rewrite this equation in a slightly
different form using the current density j
...
89)

S

The current corresponds to the amount of charge dQ flown per unit time, i
...

ˆ ˆ
dQ D Idt D dt

j dA:
S

Comparing this with expression (6
...
e
...
Hence, the continuity equation can be rewritten as follows:
@
C div j D 0:
@t

(6
...
5 Application of Integral Theorems in Physics: Part I

387

Fig
...
35 An object is placed
inside a uniform liquid
...
5
...
This is in
essence the law established by Archimedes
...

Indeed, the value of the force acting on a small area dA of the object surface that
is located at the depth z (see Fig
...
35) is dF D gzdA, where g is the Earth gravity
constant
...
The
components of the total force acting on the object from the liquid are obtained by
summing up all contributions from all elements of the object surface:
ˆ ˆ

ˆ ˆ

Fx D

ˆ ˆ

gznx dA ; Fy D

gzny dA and Fz D

S

gznz dA:

S

S

Now, consider the first integral
...
gz; 0; 0/ as a vector field,
then the force Fx can be written as the surface integral and hence we can use the
divergence theorem (6
...

Similarly, the force in the y direction is also zero, Fy D 0, so only the vertical force
remains
...
0; 0; gz/, then we can write:
—
Fz D

—
Gz nz dA D

S


...
6
...

This is a very powerful result since it is proven for an arbitrary shape of the object
immersed in the liquid!

Problem 6
...
An engineer is designing an air balloon with a specific ratio
D R=d < 1 of the radius R of its horizontal cross section in the middle and
of its half height, d
...
Show that the Cartesian coordinates of the points on the
balloon surface can be expressed in terms of the z and the polar angle as
follows:
xD

p
d2

z2 cos

p
d2

; yD

z2 sin

; zDz :

Then show that the corresponding Jacobians for the surface normal in terms of
the coordinates
...
If m is the maximum mass of a basket intended
to be used in the flights, is the surface density of the material to be used for
the balloon, and is the air density, give advice to the engineer concerning the
necessary conditions that the parameters and d should satisfy in order for
the balloon to be able to take off
...
6 Vector Calculus
6
...
1 Divergence of a Vector Field
We have already introduced the divergence of a vector field: if there is a vector field
F D
...
(6
...
However, this specific
definition is only useful in Cartesian coordinates
...
There are called curvilinear
coordinate systems (we shall discuss them in detail in Book II, Chap
...
So, if one

6
...

This general definition can be formulated via a flux through a closed surface
...
78), the surface integral is expressed via a volume
integral:
flux
D
volume

´´´
V

div FdV
;
V

where dV D dxdydz is a differential volume element
...
(6
...
Therefore,
ˆ ˆ ˆ
div F
...
p/V

H)

flux
D div F
...
Now, take the limit of the
volume V around the point M tending to zero
...
Therefore, the
ratio of the flux to the volume in the limit results exactly in div F
...
Thus, we have
just proven that
1
lim
V!0 V

—
F dA D div F
...
91)

S

This definition of the divergence can be used to derive a working expression for it
in general curvilinear coordinates (Volume II, Chap
...

Divergence can also be formally written via the del operator (5
...
92)

which indeed is the divergence (6
...

As an example of calculating the divergence, consider the vector field F D
r2
e
...
1; 1; 1/
...
2xyz/ C e

r2

r2

Á
@
e
xz C
@z


...
2xyz/ D

6xyze

r2

;

and at the point A we have: r F
...

The divergence obeys some important properties
...
F1 C F2 / D r F1 C r F2 :
Other properties are given in the Problems below
...
68
...
UF/ D U
...
rU/ ;

(6
...
Using this result,
calculate r
...
Compare your result with direct
calculation of the divergence of the vector field G D xy3 ; 0; x2 yz
...
]
Problem 6
...
One can also apply the divergence to the gradient of some
scalar field U
...
grad U/ D U D

@2
@2
@2
C 2C 2
@x2
@y
@z

Ã
UD

@2 U
@2 U
@2 U
C 2 C 2:
@x2
@y
@z
(6
...
70
...
grad / D  C r
where

and

r ;

are scalar fields
...
94) has a special name, Laplacian, and various notations
for it are frequently used in the literature
...
94), the notation r 2 D r r

6
...
6
...
6
...
r/ by formula (6
...
Similarly to the divergence, a more general definition to the curl can
also be given, and this is important as a tool to derive alternative expressions in other
coordinate systems
...
6
...
Enclose M by a closed line L lying
in the surface and consider the line integral of F along it divided by the area A of the
¸
closed loop, L F dl=A
...
64), we can write:
1
A

˛
F dl D
L

1
A

ˆ ˆ
curl F dA D
S

1
A

ˆ ˆ

...
95)

S

Above, S is the surface on † for which the line L serves as the boundary
...
According to the average value theorem,
1
A

ˆ ˆ

...
curl F
...
P/ n;
A

where the point P lies somewhere inside surface S
...
M/ at the point M and the
normal vector there:
˛
1
lim
F dl D curl F n:
(6
...
It can be used for the derivation of the
curl in a general curvilinear coordinate system (Chap
...

To understand better what does the curl mean, let us consider a fluid with the
velocity distribution v
...
An average velocity along some circular contour L can
be defined as
˛
1
vav D
v dl;
2 R L
where R is the circle radius
...
This result can now be compared with
Eqs
...
95) and (6
...
We see that curlv
...
The maximum intensity is achieved along the direction of the curl
of v
...
71) as a vector
product of r and F:
curl F D r

F;

(6
...
(6
...

As an example, consider the curl of the vector field F
...
We have:
ˇ
ˇ
ˇ i
j
k ˇ Â
ˇ
ˇ
@z
curl F D ˇ @=@x @=@y @=@z ˇ D
ˇ
ˇ
@y
ˇ x
y
z ˇ

Ã
@y
i
@z

Â

@z
@x

Ã
Â
@x
@y
jC
@z
@x

Ã
@x
k D 0;
@y

which is consistent with the intuitive understanding of the curl, as the field r is radial
(it corresponds to rays coming directly out of the centre of the coordinate system)
and hence does not have rotations
...
r/ D w

ˇ
ˇ
ˇ i j kˇ
ˇ
ˇ
r D ˇ wx wy wz ˇ D wy z
ˇ
ˇ
ˇ x y z ˇ

wz y i C
...
6 Vector Calculus

393

Fig
...
37 The vector field v
...
This may correspond, e
...
, to a velocity distribution of a
fluid in a vortex as shown in Fig
...
37
...
73) for
the curl, we write:
ˇ
ˇ
ˇ
ˇ
i
j
k
ˇ
ˇ
ˇ @=@x
curl v D ˇ
@=@y
@=@z ˇ
ˇ
ˇw z w y w x w z w y w xˇ
y
z
z
x
x
y
Ä
@
@
D
wx y wy x

...
wz x wx z/
wy z wz y k
C
@x
@y
D 2wx i C 2wy j C 2wz k D 2w:
This result is in complete agreement with what our intuitive understanding of the
curl: it shows the direction (along the vector w) and the strength (the velocity is
proportional to the magnitude w D jwj of w) of the rotation of the fluid around the
direction given by the vector w
...
g
...
F1 C F2 / D curl F1 C curl F2 :

(6
...
99)

394

6 Functions of Many Variables: Integration

This identity can be checked directly using an explicit expression (6
...
79) of the div:
r
...
71
...
UF/ D Ucurl F

F

grad U;

(6
...

Problem 6
...
Similarly, prove the identities:
curl grad U D 0I
curl curl F D grad div F

(6
...
102)

where F D Fx i C Fy j C Fz k is the vector Laplacian
...
6
...
Here, we shall consider the so-called conservative and solenoidal vector fields
since these possess certain essential properties
...
r/ is called conservative,
if one can introduce a scalar potential
...
The
field F
...
r/ such that F D
curl G D r G
...
This theorem which we shall discuss
in more detail below plays an extremely important role in many applications in
physics
...
6
...
1 Conservative Fields
If the field F is conservative with the potential
...
e
...
r/ D
...
Thus, the
conservative fields are determined by a single scalar function
...


6
...
r/ D qr=r3 , where
p
r D jrj D
x2 C y2 C z2 , is conservative
...
We
have @ =@x D qx=r3 , etc
...
xi C yj C zk/ D
r D E
...
So, the electrostatic potential in electrostatics (no moving charges)
is basically, from the mathematical point of view, a potential of the conservative
electric field (see Sect
...
7
...

The conservative field has a number of important properties that are provided by
the following

Theorem 6
...
Let the field F
...

Then the following three conditions are equivalent:
1
...
e
...
r/
such that F D r ;
2
...
e
...
the line integral L F dl D 0 for any closed contour L
...
We shall proof the theorem using the familiar scheme: 1 ! 2 ! 3 ! 1
...
(6
...

2 ! 3: take a closed contour L, then, using the Stokes’s theorem (6
...
¸
´B
3 ! 1: since L F dl D 0 for any contour L, then the line integral A F dl does
not depend on the particular integration path A
...
x; y; z/, but only on
the initial and final points A and B (see Sect
...
4
...
Therefore, the function
ˆ

...
x;y;z/

F dl

(6
...
x0 ;y0 ;z0 /

is the exact differential, and thus @ =@x D Fx , @ =@y D Fy and @ =@z D Fz , i
...

F D r , as required
...
E
...


396

6 Functions of Many Variables: Integration

Equation (6
...
r/ for a
conservative field from the field itself
...

It is essential that the region D is simply connected
...
r/ D
where

D

Á
yO C xO
i
j
2

;

p
x2 C y2 is a distance from the wire
...
However, the line integral
¸
2
2
L H dl taken along the circlex C y D 1 (e
...
for z D 0) is not equal to zero
...
ydx C xdy/ D

C cos
...
sin /

d D2

¤ 0:

The obtained non-zero result for the line integral is because the region is not simply
connected as the field H is infinite at any point along the wire x D y D 0
...

Of course, any closed loop line integral which does not have the z axis inside the
loop will be equal to zero as required by Theorem 6
...

Problem 6
...
Consider the vector field F D
...
Prove that it is
conservative
...
Check that indeed F D r
...
74
...
yz; xz; xy/
...
75
...
y; x; z/
...
]
D xy

z2 =2 C C
...
6 Vector Calculus

397

Problem 6
...
Given the vector force field F D
...
Explicitly calculate the line integral along the closed path
x2 C z2 D 1 and y D 0 using polar coordinates
...
x; y; z/ D
x2 C y2 z C C
...
x; y; z/
...
77
...
y C z; x C z; x C y/, showing that in this case U D xz C yz C xy C C
...
6
...
2 Solenoidal Fields
As was already mentioned, if a vector field A can be represented as a curl of another
field, B, called its vector potential, i
...
A D curl B D r B, then the field A is
called solenoidal
...
(6
...
104)

This condition serves as another definition of the solenoidal field
...
It is also a
sufficient one as the reverse statement is also valid: if divA D 0, then there exists a
vector potential B, such that A D curl B
...
x0 ; y0 ; z0 / and M
...
Bx ; By ; Bz /
defined in the following way:
ˆ
Bx D

z

ˆ
Ay
...
x1 ; y; z0 /dx1

x0

z

Ax
...
105)

Consider the curl of this vector field
...
curl B/x D
Az
...
x; y; z1 /dz1 :
@y
@z
@z
@z
x0
z0
The first term does not contribute since it does not depend on z; by differentiating the
second integral over the upper limit we get the integrand, see Sect
...
3
...
curl B/x D

@
@z

ˆ

z

Ax
...
x; y; z/:
z0

398

6 Functions of Many Variables: Integration

Similarly,

...
x; y; z1 /dz1 D Ay
...
curl B/z D

ˆ x
@Bx
@
D
Az
...
x; y; z1 /dz1 ;
@y
z0

@By
@x

ˆ

z

Ã
Ax
...
x; y; z0 / in the same way as for the
above two components, but in the two integrals taken over z1 the differentiation is to
be performed with respect to the variable which happens to be only in the integrand
...
4
...
2, the operators @=@x and @=@y can then be inserted inside the
integrals since the integration limits do not depend on the variables with respect to
which the differentiation is performed
...
curl B/z D Az
...
x; y; z0 /

ˆ z
@Ay
...
x; y; z1 /
dz1
dz1
@x
@y
z0
z0
ˆ zÄ
@Ay
...
x; y; z1 /
C
dz1 :
@x
@y
z0
z

Since we know that divA D @Ax =@x C @Ax =@y C @Ax =@z D 0, the expression in the
square brackets above can be replaced by @Az
...
curl B/z D Az
...
x; y; z1 /
dz1
@z1

D Az
...
x; y; z/

Az
...
x; y; z/:

Thus, we have shown that indeed curlB D A
...
Of course, the vector potential (6
...

Thus, the two conditions for A to be a solenoidal field, namely that A is curl of
some B and that divA D 0, are completely equivalent
...
x; y; z/ with arbitrary scalar field U
...
(6
...

Thus, the vector potential is defined only up to the gradient of an arbitrary scalar
field
...
6 Vector Calculus

399

xz cosh y; z sinh y; x2 y is

Problem 6
...
Show that the vector field F D
solenoidal
...
79
...
X; Y; Z/
...
r/ D q

r
jr

R
Rj3

q
at point r, where jr Rj D
...
y Y/2 C
...
Show that the
field is solenoidal anywhere outside the position of the charge q, i
...
that
div E
...
106)

Of course, the above statement is immediately generalised for the electrostatic
field of arbitrary number of point charges
...
10
...


Proof
...
Then, the flux through that surface
—

ˆ ˆ ˆ
F dA D

divF dxdydz D 0;
V

S

since the field is solenoidal, divF D 0
...
(6
...

Next, consider a more general region D with the closed surface S, which contains
a hole inside as schematically shown in Fig
...
38
...
The plane will cross S at the line L,
and the surface S will also be broken down into two surfaces S1 (left) and S2 (right)
that both cap the line L on both sides
...
6
...
The plane P cuts it
into two regions D1 and D2 ,
and crosses the surface S at
the line L

Fig
...
39 Lines of flow of a
liquid in a tube

can be represented as a sum of two flux integrals: over the surface S1 and over S2
...
Therefore,
using the Stokes’s theorem, Eq
...
64), we have for each of the surfaces:
ˆ ˆ

ˆ ˆ
F dA D

S1

˛
curl B dA D

S1

B dl;
L1

where L1 is the boundary of S1 along the line L in the direction as indicated in
Fig
...
38
...
L2 also goes along the line L but with
the traverse direction opposite to that of L1 , Fig
...
38
...
Above, we have only discussed the case with a single “hole”; obviously,
the case of many “holes” can be considered along the same lines
...
E
...

Theorem 6
...
Consider a flow of a fluid
described by a vector field of velocity v that is solenoidal
...
6
...
It means that every line of flow does not cross S0 and

6
...
Now, consider two surfaces S1
and S2 on both sides of the tube with normals n1 and n2 directed outward as shown
...
However, the flux through S0 is also zero since at each point on the
surface, v is perpendicular to n0
...
n2 /dA D 0
S1

or

S2

S1

ˆ ˆ

S2

ˆ ˆ
v n1 dA D

S1

v
...
Thus,
the flux through any surface cutting the tube is conserved along the tube
...


6
...
3
...
r/ defined in some simply connected region D which has
non-zero divergence and curl, i
...
divA
...
r/ ¤ 0 and curl A
...
r/ ¤ 0,
where f
...
r/ are some rather general scalar and vector functions, respectively
...
11
...
r/ defined in a simply connected region D
can always be represented as a sum of a conservative, Ac
...
r/, fields, i
...

A
...
r/ C As
...
r/ D 0 and divAs
...
107)

Proof
...
r/:

Since the field is conservative (due to our choice, curl Ac D 0), there exists a scalar
potential U such that Ac D grad U
...
r/:

(6
...
e
...


402

6 Functions of Many Variables: Integration

Next, we consider the difference A Ac D As , and let us calculate its div and curl:
divAs D divA divAc D f f D 0
We see that the difference, A

and curl As D curl A curl Ac D curl A D g:

Ac is, in fact, solenoidal
...
E
...


We have assumed above that the functions g
...
r/ are not equal identically
to zero in region D
...
e
...
r/ would be conservative and solenoidal at the
same time
...
However, since A is also solenoidal, divA D div grad U D U D 0, which
means that the scalar potential U satisfies the so-called Laplace equation U D 0
...

Interestingly, the vector field satisfying curl A D 0 and divA D 0 is determined up
to the gradient of a harmonic potential: A0 D A C rU
...
The other condition,
curl A0

curl A D curl grad U D 0;

is satisfied for any U since curl of grad is always equal to zero, Eq
...
101)
...
r/

and curl A D g
...
109)

The method is based on a fact that a general solution of the Laplace equation  D
can be written via a volume integral:

...
r0 / dr0
:
jr r0 j

(6
...
r/ is defined
...
We will not prove Eq
...
110) here
...

5
...
2 there
...
6
...
1
...
110) of the Laplace equation  D can be justified, see specifically
Eq
...
119)
...


6
...
r/ such that Ac D grad U
...
r/ (note that
divAs D 0 by construction), the potential U satisfies the Laplace equation,
divAc D div grad U D U D f
...
r/ D

1
4

ˆ ˆ ˆ
V

f
...
111)

This formula solves for the Ac part, as U totally defines Ac
...
The latter
satisfies the equations:
divAs D 0 and curl As D curl A D g
...
112)

since curl Ac D 0
...
Moreover, the choice of G is not unique, as
G0 D G C grad ‰ is also perfectly acceptable for any scalar field ‰
...
On the other hand,
divG0 D divG C div grad ‰ D divG C ‰:
The field ‰ is arbitrary; we shall select it in such a way that ‰ D divG
...
This means that
we can always choose the vector potential G0 being solenoidal, i
...
with divG0 D 0
...
(6
...
It is seen that G0 satisfies
the vector Laplace equation, i
...
for each of its components we write a formula
analogous to Eq
...
111); in the vector form we have then:
ˆ ˆ ˆ
g
...
r/ D
:
(6
...
111) and (6
...
r/ D grad U
...
r/ :

(6
...
7 Application of Integral Theorems in Physics: Part II
In this section more applications in physics of the calculus we developed in this
chapter will be briefly outlined
...
7
...
We shall denote the electric and magnetic fields in vacuum as
E and H here
...
A point charge q creates
around itself an electric field E
...
A probe charge q0 placed in this field
experiences a force F D q0 E, where E is the electrostatic field due to q
...
80
...
Show that the flux integral of E due to the charge q through the
surface of the sphere is
—
E dA D 4 q:
Ssph

Note that the flux does not depend on the sphere radius
...
]
In fact, it is easy to see that the result is valid for any surface S, not necessarily the
sphere
...
6
...
3
...
Therefore, if we surround the charge with
a sphere Ssph which lies completely inside the surface S as shown in Fig
...
40, then
the flux through the sphere can be easily seen to be the same as the flux through S:
Fig
...
40 Point charge q is
surrounded by a sphere Ssph
which in turn is surrounded
by an arbitrary surface S

6
...
This is because inside the region of space V enclosed by
the two surfaces Ssph and S the field E is solenoidal, and hence the flux is zero since
—

—
E dA D

E dA
S

—

Ssph

ˆ ˆ ˆ
E dA D
V

SCSph

divE dV D 0;

where the closed surface S C Sph encloses the region V
...
But the
fields due to every individual charge sum up (superposition principle)
...
must satisfy the following equation:
—
X
E dA D 4
qi :
(6
...
Therefore,
—
ˆ ˆ ˆ
E dA D 4
dV:
V

S

The derived equation can in principle allow one to find the field E due to the
charge distribution
...
Such an equation can be obtained applying
the Gauss (divergence) theorem to the flux integral in the left-hand side:
—
ˆ ˆ ˆ
ˆ ˆ ˆ
E dA D
divE dV H)

...
Therefore, it must be that
divE D 4

:

This fundamental equation of electrostatics is the Maxwell’s first equation
...
116)

406

6 Functions of Many Variables: Integration

The electrostatic field is also a conservative field for which one can choose11
a potential , such that E D grad (see Sect
...
6
...
1; note that there only
a single charge was considered; generalisation to many charges is, however,
straightforward)
...
9 of Sect
...
6
...
1,
curl E D 0:

(6
...
The corresponding differential equation for
the potential
...
(6
...
118)

which is the familiar Laplace equation
...
r/ D q= ˇr rq ˇ
...
Then, we know that the result for many charges is obtained simply by
summing up contributions from all charges
...
r0 / dV 0 in each, where the volume is positioned near the point r0 with the charge
density being
...
Then the potential at point r due to this infinitesimal charge
will be
d D

dQ

...
r/ D
V


...
119)

This formula can be considered as a general solution of the Laplace equation of
Eq
...
118)
...


6
...
6
...

(b) A surface S is chosen so
that the line L is its boundary

Problem 6
...
Consider a set of point charges fqi I i D 1; : : : ; ng positioned at
fri I i D 1; : : : ; ng, and prove by direct calculation that everywhere, including
the positions of the charges themselves, curl E D 0
...
]
Moving charges create the magnetic field H, and the force acting on a probe
charge q0 moving with the velocity v in the magnetic field is given by the Lorenz
formula: F D
...
It was established
experimentally by Ampère that a closed loop line integral of H around a wire with
the current I is proportional to the current, see Fig
...
41(a):
˛
H dl D
L

4
I:
c

(6
...
g
...


Problem 6
...
Show using Eq
...
120) that the magnitude of the field H
around an infinite wire of the current I at a distance r from the wire is
H D 2I=cr
...
120) for any flow of the current and postulated
that the formula is valid for any closed-loop contour L surrounding the current I
...
(6
...
6
...
64) for the line integral, we obtain:
˛

ˆ ˆ
H dl D

L

ˆ ˆ
curl H dA

S

H)

curl H dA D
S

4
c

ˆ ˆ
j dA;
S

408

6 Functions of Many Variables: Integration

and hence
ˆ ˆ Â
curl H
S

4
j
c

Ã
dA D 0

H)

curl H D

4
j;
c

(6
...
This
is the next Maxwell’s equation corresponding to stationary conditions (no time
dependence)
...


Problem 6
...
Prove from this that the magnetic field is solenoidal, i
...

div H D 0:

(6
...
] This is yet another Maxwell’s equation
...
84
...
(6
...
This corresponds to
the continuity equation (6
...
[Hint: take the
divergence of both sides
...
116), (6
...
121) and (6
...

As we have seen in the last problem, they do not contradict the continuity equation
...
r; t/ and current j
...
This results in the
fields H
...
r; t/ depending both on r and t
...
6
...
This observation can be formulated in the following way: the rate of change
of the magnetic flux through S is equal to the electromotive force, i
...

1@
c @t

˛

ˆ ˆ
H dA D
S

E dl:

(6
...
85
...
124)

6
...
117) for the case of non-stationary sources (charge
and/or current densities)
...
(6
...
He realised that a displacement current need to be added related to
the rate of change of the electric flux through the surface S:
˛
H dl D
L

1@
4
IC
c
c @t

ˆ ˆ
E dA:

(6
...
86
...
126)

Problem 6
...
Show that Eq
...
124) does not contradict Eq
...
122)
...
]
Problem 6
...
Show that Maxwell equations do not contradict the full continuity equation (6
...
[Hint: apply divergence to both sides of Eq
...
126)
...
116), (6
...
124) and (6
...
Introduction of the displacement current (the
last term in the right-hand side of (6
...

Firstly, without that term the continuity equation would not be satisfied
...
Discussing now this
latter point, let us consider Maxwell’s equations in vacuum where there are no
charges and currents, i
...
D 0 and j D 0
...
(6
...
124),
c
...
(6
...
(6
...
This finally gives:

410

6 Functions of Many Variables: Integration

1 @2 E
D E:
c2 @t2

(6
...


Problem 6
...
Starting from Eq
...
124), derive similarly the wave equation
for the magnetic field:
1 @2 H
D H:
c2 @t2

(6
...
127) and (6
...

Another interesting point worth mentioning is the symmetry of the equations
with respect to the two fields, E and H (compare Eq
...
116) with (6
...
124) with (6
...
The absence of their
magnetic analogues in Eqs
...
122) and (6
...

In practice, instead of the six quantities (three components of E and three of H)
it is more convenient to work with only four fields: one scalar and one vector
...
Since H is solenoidal, Eq
...
122), there exists
a vector potential A
...
Replacing H in Eq
...
124), we
obtain:
Â
Ã
1 @A
curl
C E D 0;
c @t
i
...
the field inside the brackets is conservative
...
r; t/ (the minus sign is chosen due to historical reasons)
such that
1 @A
C E D grad
c @t

H)

ED

grad

Problem 6
...
Show now that replacing A and
A0 DA C grad U

and

0

D

1 @A
:
c @t

(6
...
130)

respectively, does not change the fields E and H
...


6
...
7
...
This could be the case
of solute molecules dissolved in a solvent (a solution), or some interstitial foreign
atoms in a crystal lattice
...
g
...
The flow of particles per unit
time through a surface area dA D ndA with the normal n is proportional to the
directional derivative of the concentration C
...
That is, the total mass passed through dA is
dm D D

@c
dAdt D D
...
rC dA/ dt;
@n

(6
...
(5
...
Above, D is the diffusion constant
...
The total mass passed through a closed
surface S is then
—
dM D dt
DrC dA;
S

and we used the minus sign to indicate that the diffusion out of the volume V
enclosed by the surface S results in the reduction of mass (i
...
the direction into
the volume is considered as positive)
...
t/ D
V CdV in the volume:
dM D

ŒM
...
t/ D

@M
dt D dt
@t

ˆ ˆ ˆ
V

@C
dV:
@t

The two expressions for the dM must be equal to each other
...
D grad C/ D

@C
:
@t

(6
...
If the
diffusion coefficient is constant throughout the entire system, then it can be taken
out of the divergence, and we obtain in this case

412

6 Functions of Many Variables: Integration

C D

1 @C
;
D @t

(6
...

Problem 6
...
The reader may have noticed that our derivation of the diffusion equation was very similar to that of the continuity equation in Sect
...
5
...

Derive Eq
...
132) directly from the continuity equation (6
...
90),
extracting the needed flux j of mass from the Fick’s law, Eq
...
131)
...
92
...
g
...
Check that the function
1
e
C
...
134)

is a solution of the corresponding diffusion equation
...
This solution
corresponds to the spread of particles distribution with time when initially (at
t D 0) the particles were all at x D 0
...
This time,
heat Q is transferred from regions of higher temperature T
...
We start by stating an experimental fact (Fourier’s law) about
the heat conduction: the amount of heat passing through a surface dA per unit time
is proportional to the gradient of temperature there:
dQ
@T
D Ä dA;
dt
@n
where Ä is the thermal conductivity, @T=@n D n grad T is the directional derivative
showing the change of temperature in the direction perpendicular to the normal n to
the surface dA
...
If the energy flow goes across the boundary surface S outside in the
direction along its normal n (directed outwards), the temperature inside the volume
is reduced and hence obviously @T=@n < 0
...
Ä grad T/ dV;

dt
V

6
...
This is how we define dQ, and for this the minus sign is necessary; the
divergence theorem was used in the last passage
...

On the other hand, the heat dQ is the one lost by the volume over time dt due to
decrease of its temperature
...
dT/ dV D

CV

@T
dVdt
@t

of heat (again, this quantity is positive as the time derivative of the temperature is
negative)
...
The
total loss in the volume is then
ˆ ˆ ˆ
@T
dV:
dQ D dt
CV
@t
V
The two expressions for dQ must be equal, which results in the heat transport
equation:
CV

@T
D div
...
135)

This heat transport equation, given the appropriate boundary and initial conditions,
should provide us with the temperature distribution in the system T
...

For a homogeneous system, the constant Ä does not depend on the spatial variables
and can be taken out of the divergence
...
136)

where D D Ä=
...
Comparing the two equations (6
...
135), we see that these are practically identical
...


6
...
3 Hydrodynamic Equations of Ideal Liquid (Gas)
Let us consider an ideal liquid (it could also be a gas), i
...
we assume that the forces
applied to its any finite volume V with the boundary surface S, can be expressed via
pressure P due to the external (with respect to the chosen volume) part of the liquid
...
e
...
In other words,
the total force acting on the volume V is
—

ˆ ˆ ˆ
PndA D

gradP dV;

(6
...
83)
...
e
...
F

gradP/ dV:

V

According to Newton’s second law, this force results in the acceleration of the liquid
volume given by
ˆ ˆ ˆ
V

dv
dV;
dt

where v
...

Therefore, one can write:
ˆ ˆ ˆ
ˆ ˆ ˆ
dv
dv
dV D
D F gradP:

...
138)
Note that here the acceleration in the left-hand side is given by the total derivative of
the velocity
...
Therefore,
similarly to our consideration of the derivative of the distribution function of a gas
in Sect
...
6 (see Eq
...
26)), we can write:
dv
dt

Â
7 !

dv
dt

Ã
D
tot

@v @v
@v @v @x @v @y @v @z
@v
@v
C
C
C
D
C vx C vy C vz :
@t @x @t @y @t @z @t
@t @x
@y
@z

The coordinate derivatives term above is normally written in the following short
form as a dot product:
@v
@v
@v
vx C vy C vz D v grad v;
@x
@y
@z

6
...
Finally, we arrive at the following equations:
@v
C v grad v D F
@t

1

gradP:

(6
...
85) or (6
...
Given the equation of states for the liquid,
P D f
...
e
...


Chapter 7

Infinite Numerical and Functional Series

In Sect
...
9 we considered summation of finite series
...
Here an is a
general (we say “the n-th”) term of the series which is constructed via a some kind
of formula depending on an integer n D 1; 2; 3; : : :
...
2
...
3
...
Our interest here is in
understanding whether or not one can define a sum of such an infinite series,
SD

1
X

an ;

(7
...
e
...
It is also
possible that each term of the series, an
...
x/ of the sum
...
e
...
x/ of the sum itself
...
Kantorovich, Mathematics for Natural Scientists, Undergraduate Lecture
Notes in Physics, DOI 10
...
1 Infinite Numerical Series
We say that the series (7
...
2)

nD1

has a well-defined limit
...
form a numerical sequence
which we considered in Sect
...
2
...
1) converges
if the corresponding sequence of its partial sums, SN , has a well-defined limit at
N ! 1
...
2
...
3, and can serve as a simple example
...
The series (1
...
n C 1/
NC1
nD1
As an example of a diverging series, let us consider the so-called harmonic series
1C

1
1
1
C C C
2
3
4

D

1
X1
nD1

n

:

(7
...
On the other hand, if we assume that the series converges to S, then
limN!1
...
This proves that the
harmonic series actually diverges
...
1
...
(1
...
Answer: 3=4
...
1 Infinite Numerical Series

419

Problem 7
...
Prove that a combination of a geometric and arithmetic progressions, see Eq
...
69), has a well-defined limit for jqj < 1 and any r:
1
X
iD0


...
1

q/2

:

There are several simple theorems which establish essential properties of converging infinite numerical series
...
1
...
1) converges to S, one can remove any first m
terms in the series, and the rest of the series,
1
X

an ;

nDmC1

would still converge
...
2
...
1) of an converges to S, then the series of
bn D can converges to cS, where c is (generally complex) number
...
3
...
˛an C ˇbn / also converges to ˛Sa C ˇSb
...
3
...
[Hint: consider the limits of the
appropriate partial sums
...
4 (Necessary Condition for Convergence)
...

Proof
...
Q
...
D
...
Taking the limit

This is not a sufficient condition, only a necessary one
...


420

7 Infinite Numerical and Functional Series

However, we know that the series diverges
...


7
...
1 Series with Positive Terms
It is convenient to start a more detailed analysis from a particular case of series
which have all their terms positive, an > 0
...
5 (Necessary and Sufficient Condition for Convergence)
...


Proof
...
e
...
Indeed, because the sequence contains exclusively
positive terms, the partial sums form an increasing sequence 0 < S1 Ä S2 Ä S3 Ä

...
Therefore, once SN
has a limit, the numerical sequence converges to that limit
...
Indeed, since the
sequence of partial sums fSN g converges, it has a well-defined limit S
...
2
...
2)
...
E
...

There are several sufficiency criteria to check whether the series converges or not
...
6 (Sufficient Criterion for Convergence)
...
Then, if the second series n bn converges, then the first one,
n an , does as well; if the first series diverges, so does the second
...
Let AN and BN be the corresponding partial sums for the two series,
respectively; since for all n we have an Ä bn , then obviously AN Ä BN for any N
...
Hence, BN Ä B
...
e
...
5) also

7
...
Now we turn to the second statement where the first series diverges
...
Then, according to the first part of the
theorem, the first series should converge as well, which contradicts our assumption;
therefore, if the first series diverges, then so does the second
...
E
...

p
p
As an example, consider the series with an D 1= n
...
In fact, the series n n ˛ diverges for any 0 < ˛ Ä 1
...
4
...
5
...
]

n

n

converges
...
[Hint: use the

Theorem 7
...
If
D lim

n!1

anC1
< 1;
an

the series converges; if
> 1, it diverges; if
D 1, then this test in
inconclusive (the series may either converge or diverge, more investigations
are required)
...
Let us first consider the case of < 1
...
What is essential for us here
j < , i
...

is the second part: anC1 =an < C
...
e
...
Therefore, aNC1 < aN , aNC2 < aNC1 < 2 aN , etc
...
Therefore, the series without the first N terms converges, and so does
the whole series (Theorem 7
...

1

This test is due to Jean-Baptiste le Rond d’Alembert and normally bears his name
...
We can again write that
< anC1 =an < C ,
or
< anC1 =an
...

Therefore, anC1 =an > 1 or anC1 > an
...
4), i
...
the
original series indeed diverges
...
E
...

Nothing can be said about the case of D 1, the corresponding series may either
diverge or converge, more powerful criteria mustp used to establish convergence
...
Using the ratio test, we have:
p
r
anC1
1= n C 1
n
p
D lim
D 1:
D lim
D lim
n!1 an
n!1
n!1
1= n
nC1
Therefore, the ratio test is not powerful enough to establish convergence or
divergence of this particular series
...
6
...
Therefore, the case of D 1 is also
inconclusive in this case
...
7
...

Problem 7
...
Prove using the ratio test that the series with an D nn =nŠ
diverges
...
24)
...
8 (The Root Test2 )
...


N

1,

(7
...
In the first case, an Ä qn with q < 1 for any n
N
...


7
...
Therefore, the series with an and n
N converges as well because
of Theorem 7
...
As a finite number of terms before the aN term do not effect the
convergence (Theorem 7
...
In the second case, an > 1
and hence an does not tend to zero as n ! 1 yielding divergence of the series
(Theorem 7
...
Q
...
D
...
e
...
However, it may not
be straightforward to apply the root test, and hence it is used less frequently than
p
the ratio test
...

Problem 7
...
Prove that the series with an D e ˛n sin2
...
Try both the ratio and root tests
...


Theorem 7
...
Consider a series with
positive non-increasing terms, i
...

a2

a1

a3

an

> 0:

anC1

Next let us define a function y D f
...
e
...
n/ D an for any
n D 1; 2; 3 : : :
...
Then the convergence of the series can be
concluded depending on whether the integral
ˆ 1
f
...
5)
AD
1

converges or not
...
So, consider the partial sum of our series3 :
SN

3

1

D f
...
2/ C f
...
N

1/:

Note that some ideas of the proof are similar to those we encountered when considering Darboux
sums in Sect
...
2 (specifically, see Fig
...
2)
...
7
...
, while in (b) the integral is
larger than the sum of functions at points 2, 3, 4, etc
...
7
...

We also see from the figure that SN 1 will definitely be larger than the area AN under
the curve of the function f
...
e
...
On the other hand, let us now consider the sum
SN

f
...
2/ C f
...
4/ C

C f
...
7
...
e
...
1/ D SN a1 < AN
...
x/dx < SN

1

D SN

aN ;

(7
...
If the integral (7
...
x/ > 0, the partial integral is an increasing
function, i
...
ANC1 > AN , and AN should be bounded from above4) and hence SN
a1 < AN < A, i
...
SN < A C a1
...
5, converges
...
5) is equal to infinity (diverges)
...
6) it then follows that SN > AN C aN
...
Therefore, SN as well can be made arbitrarily big, i
...
it does
not have a limit
...
Q
...
D
...

This series was mentioned above in Problem 7
...
Here f
...
4
...
3, that improper integrals are understood as limits
...
1 Infinite Numerical Series

425

converges or diverges
...
4
...
3: the integral
converges only for ˛ > 1 and diverges otherwise
...
In particular, the harmonic series (7
...
7)

converges (˛ D 2)
...
10
...
7)
...
]
Problem 7
...
Using the integral test, prove that the series with an D

...

Problem 7
...
Using the integral test, prove that the series with an D ne
converges
...
1
...
3) diverges
...
8)

(7
...
converges, let us consider the function
y D 1=x for 1 Ä x Ä n
...
xk / D 1=k
...
Indeed, the
exact area under the curve is obviously
ˆ
Sn D

ˆ

n

y
...
7
...
Fig
...
1)

By choosing rectangles which go above the curve, see Fig
...
2, and summing up
their areas (note that the base of each rectangular is equal exactly to one), we shall
construct an approximation,
C
Sn D 1 C

1
1
C C
2
3

C

1
n

1

;

which is larger than the actual area by
C
vn D Sn

Sn D 1 C

1
1
C C
2
3

C

1
n

1

ln n D un

1
:
n

Obviously, vn is positive and forms an increasing sequence as positive extra areas
are always added due to new rectangulars when n is increased
...
Indeed, by choosing
the lower rectangulars, we can construct an approximation to the area
Sn D

1
1
C C
2
3

C

1
< Sn D ln n
n

H)

vn < 1

1
< 1:
n

Therefore, according to Theorem 7
...
Its numerical value is given above
...
1
...
Before we touch upon
general series where the sign of terms may change in a general way, it is instructive
to consider another special case of numerical series in which the sign of terms
alternates, i
...


7
...
10)

i
...
an D
...

For this type of series there exists a very simple sufficient convergence test which
we formulate as the following theorem:

Theorem 7
...
For the series (7
...
e
...

Proof
...
Then, consider a partial sum with an
even number of terms:
S2N D a1 C a2 C

C a2N D
...
a3

ja4 j/ C

C
...
11)

Since terms in the series do not increase in their absolute value, each difference
within the round parentheses is positive, and hence the partial sums S2N form an
increasing numerical sequence
...
ja2 j

a3 /


...
ja2 j

a3 / C
...
ja2N 2 j
C
...
Hence, S2N is
bounded from above by its first element, and, therefore, the numerical sequence
of S2N has a limit
...
Since both partial sums have their limits
and these coincide, this must be the limit of the whole numerical series
...
E
...

It also follows from the above proof that the overall sign of the sum of the series
S D limN!1 SN is entirely determined by the sign of its first term, a1
...
(7
...
Since the sequences S2N and S2NC1 converge to the same limit (if that limit

428

7 Infinite Numerical and Functional Series

exists), this proves the statement that was made
...


Theorem 7
...
Consider again the alternating series from the previous
theorem
...


Proof
...
The sum of all dropped terms is
S>N D aNC1 C aNC2 C

...
jaNC2 j

jaNC3 j/

< jaNC1 j D aNC1 ;

where we have assumed that the first dropped term, aNC1 , is positive
...
Since the first term,
aNC1 , is positive, then S>N > 0 as well, i
...
we obtain 0 < S>N < aNC1
...
jaNC4 j

D

jaNC1 j C
...
e
...
e
...

Combining both cases, we can write that jS>N j < jaNC1 j, as required
...
E
...

As an example, we shall consider the series

1

1
1
C
2
3

1
C
4

D

1
X
...
12)

This series looks like the harmonic series (7
...

We recall that the harmonic series diverges
...
12) actually
converges as it fully satisfies the conditions of Theorem 7
...
n C 1/ and

...
As will be shown in Sect
...
3
...
12)
corresponds to the Taylor’s expansion of ln
...
7
...
2)
...
1 Infinite Numerical Series

429

7
...
4 General Series: Absolute and Conditional Convergence
Here we shall consider a general infinite series in which signs of its terms neither
are the same nor they are alternating:
a1 C a2 C a3 C

D

1
X

an :

(7
...
12 (Sufficient Criterion)
...
14)

nD1

constructed from absolute values of the terms of the original series, converges,
then so does the original series
...
Consider the first N terms of the series (7
...
14) with the corresponding partial sums being SN and AN
...
13) may have different
0
signs
...
Obviously, SN D SN SN
0
00
and AN D SN C SN
...
Similarly, negative values of the negative terms also form
00
a subsequence with the partial sum SN
...
14)
0
00
converges, i
...
AN has a limit, then both partial sums SN and SN must be bounded
0
00
from above (both contain positive terms): SN < AN < A and SN < AN < A
...
4 on subsequences)
...
Q
...
D
...
13
...
1/nC1
D
n˛
nD1

converges (absolutely) for any value of ˛ > 1
...
7
...
1 can now be applied to general series
with the only change that absolute values of the terms of the series are to be used
...
14) converges if
ˇ
ˇ
ˇ anC1 ˇ
janC1 j
ˇ<1
D lim ˇ
n!1 jan j
n!1 ˇ an ˇ
lim

and diverges if the limit of the ratio of absolute values is larger than one
...
However, convergence of the series (7
...
13), and this proves the statement made
...
14
...


P1
nD1

xn =n converges

Problem 7
...
Using the ratio test, show that the series with an
xn = n2 C n converges for jxj < 1 and diverges for jxj > 1
...
12 provides only a sufficient condition for convergence
...
However, the series may converge, but the corresponding
series of absolute values of its terms may diverge
...
For instance, the series (7
...
This distinction has a very important meaning based on the following

Theorem 7
...
If a series converges absolutely, terms in the series may be
permuted arbitrarily without affecting the value of the sum
...
Consider first a series n an of positive terms an > 0 which partial sum,
AN , converges to A, i
...
limN!1 AN D A
...
5 The new series created in this way, n bn , consists
of elements bn > 0
...

Consider the first N terms in the second series, its partial sum we shall denote BN
...

Since amongst the M terms there will most likely be some other terms as well, and

5

Obviously, if a finite number of terms is permuted, these terms can be removed from the series
without affecting its convergence
...
1 Infinite Numerical Series

431

all elements of the series are positive, one can write: BN Ä AM
...
e
...
e
...

Conversely, let us consider the first N 0 elements of the first series; the second
series would contain these elements within its first M 0 terms, and hence AN 0 Ä BM0
...
This yields A Ä B
...
So, we
conclude that if all terms are positive,6 then one can indeed permute terms in the
series, and the new series created in this way would converge to the same sum
...
e
...
an ˙ bn /,
since this would simply correspond to a permutation of elements of the combined
P
P
series n an ˙ n bn D A ˙ B
...
We permute an infinite number of its elements arbitrarily and obtain
P
a new series n bn
...
bn C jbn j/

and S2 D

X1
n

2


...
15)

Since either of the series is built from only positive (more precisely, non-negative)
elements, we can apply to them the statements made above, i
...
we can permute their
elements arbitrarily without affecting their sums, and manipulate them
...

Next, we shall permute the elements of the bn -series precisely in the opposite
way to the one we have used initially when deriving the bn -series from the an -series;
then, the order of terms would correspond to the original an series
...
an C jan j/

0
and S2 D

X1
n

2


...
(7
...
Since
0
0
the terms are non-negative, S1 D S1 and S2 D S2
...
E
...

A simple corollary from this theorem is that two absolutely converging series can
be algebraically summed or subtracted from each other term-by-term
...


432

7 Infinite Numerical and Functional Series

A˙B D

X

an ˙

n

X
n

bn D

X


...
Here A< includes all negative terms, while A> all positive; both sums
converge independently
...
e
...
16)
as stated by the following theorem:

Theorem 7
...
One can multiply terms of two absolutely converging series
P
P
A D
n an and B D
n bn term-by-term, and the resulting series would
converge to the product of the two sums, AB
...
Let us start by assuming that both series consist of positive elements only
...
g
...
(7
...
e
...
The new series
converges to C D AB
...
We can always find a number M such that all elements ak and bl
we find in CN are contained in the partial sums AM and BM
...
e
...
The last inequality
follows from the convergence of the two series, i
...
AN Ä A and BN Ä B
...

Conversely, let us consider now two partial sums AN 0 and BN 0 with the first N 0
elements of the two original series, and construct the product AN 0 BN 0 of all their
elements
...
However, it
would also contain other elements, which means that AN 0 BN 0 Ä CM0 Ä C
...
Combining the two inequalities we get AB D C
...
e
...

According to the above proof, the sum of its absolute values converges to jAj jBj D
jABj, which means that the original product series converges absolutely
...
Since the product series converges absolutely, according to

7
...
13, we are allowed to permute its terms arbitrarily without
affecting the sum, i
...
the following manipulation is legitimate:
X
n

cn D

X

ak

k

X

!
bl

D

l

X

ak
...
Q
...
D
Therefore, we have proved that if two series converge absolutely to the values
S1 and S2 , then: (i) a series obtained by summing (or subtracting) the corresponding
terms of the two series converges absolutely to S1 CS2 (or, correspondingly, S1 S2 );
(ii) a series obtained by multiplication of the two series converges absolutely to S1 S2 ;
(iii) the order of terms in a series can be changed arbitrarily without affecting its
sum
...

Interestingly, this is not the case for conditionally converging series
...

To illustrate the point of the dependence of the sum of a conditionally converging
series on the order of terms in it, consider the series (7
...
Let us reorder the terms
in it in such a way that after each positive term we shall put two negative ones picked
up in the correct order:
Â
à Â
Ã
1
1 1
1 1 1
C
D 1
C
4
2 4
3 6 8
Â
à Â
Ã
1
1
1
1
1
1
C
C
C
5 10 12
7 14 16
Â
à Â
à Â
à Â
Ã
1 1
1
1
1 1
1
1
D
C
C
C
C
2 4
6 8
10 12
14 16
Ã
Â
1
1 1
1 1
1 1
1
1
D
C
C
C
C
D S;
1
2
2
3 4
5 6
7 8
2

SD1

1
1
C
2
3

i
...
S can be made equal to the half of itself
...

The Coulomb potential of a lattice of point charges (the so-called Hartree
potential) is an example of a conditionally converging series
...
7
...
The Coulomb potential at some point x away from the charges is
VCoul
...
7
...
Each unit cell
(indicated) contains two oppositely charged atoms with charges Q > 0 and Q < 0
...
This series does not converge absolutely
...
However, the potential of positive and negative
charges does converge conditionally, although the result would depend on the order
of the terms in the sum above
...
2 Functional Series: General
Now we shall turn our attention to specific infinite series, called functional series,
in which the general form of the n-th term an is a function an
...
Correspondingly, as each term of the series is a
function of x, i
...
an
...
x/ D a0
...
x/ C a2
...
x/:

(7
...
The difference now is that we deal with not just a single
value of the x or even its finite set of values, but with x defined continuously in the
interval a < x < b
...
But the range
of questions is much wider than that in fact
...
e
...
x/
after integration or differentiation? Finally, we shall consider a special and very
important class of functional series called power series and show how functions can
be expanded into such series, the so-called Taylor series
...
2 Functional Series: General

435

our discussion of the Taylor theorem started in Sect
...
7
...
The treatment of this simple case
can be extended to functions of several variables almost without change
...
2
...
It is also possible that convergence within some interval of x is
uniform, i
...
it does not depend on x (cf
...
6
...
3)
...
Therefore, we shall only be considering this case in what
follows
...
2
...
3): it is said that the series n an converges to S if for any positive
one can always find a number N such that for any N > N the partial sum SN
would be different from S by no more than , i
...
jSN Sj <
...
e
...
x/
...
e
...
This is the idea of this essential new notion
...
x/ D

1
X
...
It is known from Theorem 7
...
11), i
...

ˇ
ˇ
ˇ
ˇ
1
1
1
ˇ
jS
...
x/j Ä ˇ
ˇx C N C 1ˇ Ä N C 1 < N ;
where we have made use of the fact that x > 0
...
x/ SN
...
N C 1/ <
1=N D
...
Therefore,
this series converges uniformly in the interval 0 < x < 1
...

Consider the first N terms of it
...
x/

SN
...
In this case, therefore, the series converges
non-uniformly
...
But before discussing this, we shall provide one simple
test for the uniform convergence
...
15 (Due to Weierstrass)
...
x/ of a functional series
for all values of x within some interval satisfy the inequality
jan
...

P
Proof
...
Consider now the residue of our functional series for an
arbitrary x within the given interval:
ˇ 1
ˇ
1
1
ˇ X
ˇ
X
X
ˇ
ˇ
an
...
x/j <
jSN
...
x/ SN
...
Note that here we chose N by the numerical
P
series n cn , and hence it is the same for all values of x from the interval
...
Q
...
D
...
2 Functional Series: General

437

In the so-called Fourier series an
...
The above theorem then establishes the uniform convergence of the two
series based on the convergence of the series composed of the pre-factors ˛n and ˇn ,
since
ˇ
ˇ
nx ˇ
nx ˇ
ˇ
ˇ
ˇ
ˇ
ˇ˛n cos
ˇ Ä j˛n j and ˇˇn sin
ˇ Ä jˇn j :
l
l
It also follows from Theorem 7
...
x/j, would also converge uniformly; moreover,
the given series n an
...


7
...
2 Properties: Continuity
Here we shall consider an important question about the continuity of the sum S
...
x/
...
16
...
x/ in the
interval a < x < b
...
x/ of the series are continuous functions
at a point x0 belonging to the interval, then the sum of the series S
...

Proof
...
x/ D S
...
x/

SN
...
Then,

S
...
x/
Ä jSN
...
x0 / C SN
...
x0 /j

SN
...
x/j C jSN
...
18)

Since the series converges uniformly in the interval of interest (which includes the
point x0 ), then one can always find such N for the given > 0 that jSN
...
Each of the functions an
...
Therefore,
for any =3 one should be able to find such ı > 0 so that from jx x0 j < ı follows
jSN
...
x0 /j < =3
...
18) into:
jS
...
x0 /j Ä jSN
...
x0 /j C jSN
...
x0 /j <

3

C

3

C

3

D ;

which proves the continuity of the sum of the series at point x0
...
E
...

It is worth noting that condition of uniform convergence of the series on the
interval a < x < b was essential for the proof of the theorem
...
x/ is equal to the sum of the limits taken term-by-term:
X
X
an
...
x/:
(7
...

P
Theorem 7
...
Suppose the series n an
...
x/ on the interval a < x < b, and the terms of the series, an
...
x/ D bn :

x!x0

Then limx!x0 S
...
(7
...


P
n

bn
...
We shall first prove one statement of the theorem, namely that the series
P
n bn converges
...
x/j Ä
cn (Weierstrass test)
...
x/ P the limit, we get jbn j Ä cn
...

Let B D BN C BN be the sum of the latter series with the corresponding partial
sum BN
...
x/

Bj D jSN
...
x/

BN j Ä jSN
...
x/jCjBN j :
(7
...
x/j < =3 for all x;
P
since the n bn series converges, then one can always find such N that jBN j <
=3
...
x/ all have a well-defined limit at x ! x0 ,
then for any one can always find such ı > 0 that from jx x0 j < ı follows
jan
...
Therefore,
ˇ
ˇ ˇ N
ˇ N
N
ˇ
ˇX
X ˇ ˇX
ˇ
ˇ ˇ
ˇ
an
...
x/ bn ˇ
jSN
...
x/ bn j <

nD1

N
X
nD1

3N

D :
3

7
...
(7
...
x/

Bj Ä jSN
...
x/j C jBN j <

3

C

3

C

3

D :

(7
...
Q
...
D
...
2
...

P
Theorem 7
...
Consider a uniformly converging series S
...
x/ on
the interval a < x < b
...
x/ dx D
c

Xˆ
n

d

an
...
22)

c

where c and d > c lie within the interval between a and b
...
Since the series converges, we can write:
S
...
x/ C SN
...
x/dx D
c

N
Xˆ
nD1

ˆ

d

an
...
x/dx:

(7
...
x/j < for all x
...
x/dxˇ Ä
dx D
...
x/j dx <
ˇ
ˇ
c

c

c

440

7 Infinite Numerical and Functional Series

This means that the integral of the residue SN
...
This proves that at this limit the last term in Eq
...
23)
tends to zero and hence the integral of the sum is equal to the sum of the integrals,
i
...
the integration and summation can be interchanged in the case of the uniformly
converging series
...
E
...

Now we shall prove a similar statement for differentiation
...
19
...
x/ D
n an
...
x/ have
continuous derivatives a0
...
If the series containing derivatives of the terms
Pn
of the original series, n a0
...
x/ of the original series has a well-defined derivative
S0
...
x/:
n

(7
...
The series n a0
...
x/, will be a
n
continuous function (Theorem 7
...
According to Theorem 7
...
x/ an
...
x0 / explicitly
...
x/ converges to S
...
x/

an
...
x/

X

an
...
x/

S
...
x/

S
...
25)

c

Let us now differentiate both sides of Eq
...
25) with respect to x (recall Eq
...
34)
or a more general Eq
...
65)), and we obtain: S1
...
x/; as required
...
E
...

So, uniformly converging functional series can be both integrated and differentiated term-by-term (in the latter case the series of derivatives must also converge
uniformly)
...
3 Power Series

441

Problem 7
...
In quantum statistical mechanics a single harmonic oscillator
of frequency ! has an infinite set of discrete energy levels n D „!
...
The probability for an oscillator to be in the n-th state is given by
1
exp
...
Show that the so-called partition function Z of the oscillator is
given by:
ZD

1
X

e


...
/ converges uniformly with respect to
...


7
...
x/ D ˛n xn with the coefficients ˛n forming an infinite numerical sequence
...
x/ D ˛0 C ˛1 x C ˛2 x C

D

1
X

˛n xn :

(7
...
A more general power series
can be considered around an arbitrary point x D a,
S
...
x

a/ C ˛2
...
x

a/n :

(7
...
26) by a simple shift and
hence this generalisation is not really necessary for proving general statements and
theorems
...
3
...


Theorem 7
...
If the power series (7
...
e
...

Proof
...
e
...
Consider now the
series constructed of absolute values of the terms of the original series taken at the
value of x satisfying the inequality jxj < jx0 j:
1
X
nD0

j˛n xn j D

ˇ ˇn
ˇ ˇn
1
1
X
X
ˇxˇ
ˇxˇ
M0
;
M0 ˇ ˇ D M0
qn D
j˛n xn j ˇ ˇ <
0 ˇ
ˇx ˇ
x0 ˇ
1 q
0
nD0
nD0
nD0

1
X

where q D jx=x0 j < 1 and the sum of its powers forms a converging geometrical
progression which sum we know well, Eq
...
7)
...
e
...
Q
...
D
...
21 (Due to Abel)
...
26) diverges at x0 , it
diverges for any x satisfying jxj > jx0 j, i
...
for x < jx0 j and x > jx0 j
...
17
...

Schematically the convergence and divergence intervals are shown in Fig
...
4
...
3 Power Series

443

Fig
...
4 (a) If a series converges at x0 , it also converges (and converges absolutely) within the
interval shaded; (b) if the series diverges at x0 , it also diverges for any x in the two shaded intervals

Theorem 7
...
If a series converges at some x0 ¤ 0 and also is known
to diverge at some x1 (obviously, jx1 j must be larger than jx0 j, otherwise it
would contradict the Abel’s theorems), then there exists such positive R that
the series converges for any jxj < R and diverges for any jxj > R (nothing
can be said about x D R though)
...
Since the series converges at some x0 , it must converge at any x within
the interval jx0 j < x < jx0 j according to Theorem 7
...
21)
...
7
...
Then, according to the
Abel’s theorem, it will also converge for any x satisfying jx2 j < x < jx2 j ,
i
...
in the interval which completely includes the previous convergence interval
jx0 j < x < jx0 j, see Fig
...
5(b)
...
Assuming that the series diverges at x3 , we have to conclude
from the Abel’s theorem that it will diverge anywhere when x > jx3 j and x < jx3 j,
i
...
the new divergence intervals would completely incorporate the previous ones,
and the boundaries between the diverging and converging intervals would move
towards each other, see Fig
...
5(c)
...

Since the intervals are exclusive of their boundary points, nothing can be said about
the points x D ˙R
...
E
...

The positive number R is called the radius of convergence
...
If the series
converges only at x D 0, then R D 0:
It is easy to see now that the radius of convergence can be calculated via
ˇ
ˇ ˛n
R D lim ˇ
n!1 ˇ ˛

nC1

ˇ
ˇ
ˇ:
ˇ

(7
...
7
...
22
...
(b) Assuming that it also converges
at some x2 which lies between jx0 j and jx1 j, would inevitably widen the convergence interval
according to the Abel’s theorem
...

Problem 7
...
Prove the above formula by applying the ratio test
...
19
...
29)

where Sup means the maximum value achieved after taking the limit
...
20
...

Problem 7
...
Show that R D 1 for the series with an D
...
2n C 1/Š
...
22
...
1/n x2n =
...

Problem 7
...
Show that R D 1 for the series with ˛n D 1=n
...
However, we studied both series before: at x D 1, the power
series becomes the harmonic series about which we already know that it diverges; at
x D 1, the alternating harmonic series appears for which we also know that it converges (conditionally)
...
1/n =n
...
3 Power Series

445

7
...
2 Uniform Convergence and Term-by-Term Differentiation
and Integration of Power Series
Power series present the simplest and very important example of the uniformly
converging functional series and hence can be integrated and differentiated termby-term
...


Theorem 7
...
If R is the radius of convergence of the power series (7
...
e
...

P
Proof
...
e
...
Also, for any jxj < r we have
jan
...
15), means that our
series indeed converges uniformly
...
E
...

This in turn means that the power series represents a continuous function for any
r Ä x Ä r with the positive r < R (Theorem 7
...
Note that nothing can be said
about the boundary points x D ˙R where a special investigation is required
...
For instance, the series
x

x3
x2
C
2
3

D

1
X
...
30)

at x D 1 becomes the alternating harmonic series (7
...
7
...
3, converges (although not absolutely)
...
30)
converges uniformly for any 0 Ä x Ä 1, and hence its sum is a continuous function
for all these values of x
...
1 C x/ and hence at x D 1 converges to ln 2
...
x/ D n ˛n xn uniformly
converges for any x from the semi-open interval 0 Ä x < R, where the function S
...
x/ at x ! R 0 (from the left) and then arrive at the converging
series at x D R
...
x/ will be continuous in the closed interval
0 Ä x Ä R
...
18 and 7
...
7
...
3 it can
be integrated and differentiated term-by-term
...
19, the
series containing derivatives of the terms of the original series should also converge
uniformly
...
23, for this it is sufficient to demonstrate that
the series of derivatives has the same radius of convergence as the original power
series
...
n C 1/ anC1 x ˇ D lim ˇ 1 C 1 anC1 x ˇ D lim ˇ anC1 x ˇ :
lim ˇ
ˇ b ˇ n!1 ˇ
ˇ n!1 ˇ
ˇ n!1 ˇ a ˇ
n 1
n!1
nan x
n
an
n
n
P
But the original series of n an xn converges for any x between R and R and hence,
according to the same ratio test, the last limit is smaller than one:
ˇ
ˇ
ˇ
ˇ
ˇ anC1 xnC1 ˇ
ˇ
ˇ
ˇ D lim ˇ anC1 x ˇ < 1:
lim ˇ
ˇ a xn ˇ n!1 ˇ a ˇ
n!1
n

n

Hence, the ratio test establishes convergence of the series of derivatives within the
same interval of the original power series
...

It is seen from this discussion that the power series can be differentiated any number
of times; each series thus obtained would have the same radius of convergence
...
3
...
x/ into a power series (cf
...
3
...

First of all, we prove that if f
...
x/ D

1
X

˛n xn D ˛0 x0 C ˛1 x1 C ˛2 x2 C ˛3 x3 C

;

(7
...
Indeed, suppose there exists another expansion of the
same function,
f
...
32)

7
...
In fact, they can
be differentiated many times:

f
...
x/ D

1
X

n˛n xn

1

D

nD1

1
X
nD1

nŠ

...
33)

f
...
x/ D

1
X

1/˛n xn

n
...
n

nD2

2/Š

˛n xn

D 2 1˛2 x0 C 3 2˛3 x1 C 4 3˛4 x2 C

f
...
x/ D

1
X

n
...
n

2/˛n xn

3

D

nD3

1
X
nD3

0

1

D 3 2 1˛3 x C 4 3 2˛4 x C

;

nŠ

...
34)

˛n xn

3

;

(7
...
The k-th term has the form:
f
...
x/ D

1
X

n
...
n

k C 1/˛n xn

k

D

nDk

D kŠ˛k x0 C

1
X
nDk


...
n

k/Š

˛n xn

k

;

(7
...
32)
...
x/ starts from the n D k term
...
(7
...
32);
we immediately obtain that ˛0 D ˇ0
...
x/, we obtain ˛1 D ˇ1 ; continuing this process, we
progressively have ˛2 D ˇ2 , ˛3 D ˇ3 , etc
...
e
...

This proves the statement made above
...
to the function f
...
Indeed, putting x D 0 again in Eqs
...
31), (7
...
35) yields,
respectively:
f
...
1/
...
2/
...
3/
...
36) we generally have f
...
0/ D kŠ˛k , i
...
the expansion of f
...
x/ D f
...
1/
...
2/
...
3/
...
n/
...
37)

which is called Maclaurin series, and, of course, this is not accidental
...
62)
f
...
0/ C

f 00
...
0/ 2
f 0
...
n/
...
x/;
nŠ

(7
...
x/, where
f
...
/ nC1
f
...
#x/ nC1
x
x ;
D

...
n C 1/Š

RnC1
...
39)

is the remainder term with 0 < # < 1
...
(7
...
38), we immediately notice that the first n terms in both of them coincide
...
This close relationship between the two results allows
one to formulate the necessary and sufficient conditions at which the Maclaurin
series of the function f
...
x/ itself
...
24
...
37) to converge to f
...
39) of the Maclaurin formula would tend to zero as
n ! 1
...
We shall first prove sufficiency, i
...
we start by assuming that limn!1
RnC1
...
Then f
...
x/ C RnC1
...
x/ contains the first

...
n C 1/ first terms
...
x/ D 0 by our
assumption, then
lim RnC1
...
x/

n!1

n!1

H)

SnC1
...
x/

lim SnC1
...
x/ D f
...
x/ at each point x between R and R
...
We start from the assumption that the
series (7
...
x/, and hence limn!1 SnC1
...
x/ for R < x < R
...
3 Power Series

449

lim SnC1
...
x/

n!1

RnC1
...
x/

n!1

lim RnC1
...
x/

n!1

lim RnC1
...
Q
...
D
...
37) requires in each case investigating
the limiting behaviour of the remainder term (7
...
In this respect it is expedient
to notice that an important n-dependent part of a general term in the Taylor’s series
tends to zero as n ! 1, i
...

xn
D0
n!1 nŠ
lim

(7
...
This can be seen, e
...
, by applying the ratio test to the sequence
an
...
Since this series converges everywhere (the ratio test is a sufficient
condition), then necessarily (Theorem 7
...
Of
course, condition (7
...
x/ the
series (7
...
n/
...

As an example, let us consider the expansion of the function f
...
The
remainder term
"
#
Ä nC1
x
...
e /
RnC1
...
n C 1/Š

...
n C 1/Š
x!#x

as n ! 1 because of the necessary condition (7
...
Therefore, the expansion of
the exponential function has the form:
x2
C
e D1CxC
2Š
x

D

1
X xn
nD0

nŠ

:

(7
...


Problem 7
...
Show that
 Ãn
@
lim
e
x!0 @x

x2 =2

D

nŠ
2n=2
...
[Hint: expand first the
exponential function into the Maclaurin series
...
25
...
42)

By expanding the exponential function in the right-hand side and comparing
the coefficients at the same powers of the parameter u in both sides, show that:
M1 D K1 I

2
M2 D K2 C K1 I

3
M3 D K3 C 3K1 K2 C K1 :

Similarly one can consider expansion of other elementary functions into the
Maclaurin series and investigate their convergence
...
66)–(3
...
1 C x/ D x

sin x D x

cos x D 1

x3
x2
C
2
3

x3 x5 x7
C
C
3Š 5Š 7Š
x2 x4
C
2Š 4Š

C
...
1/nC1 x2n

...
1/n x2n
C

...
˛ 1/ 2
x C
2
1
X²Ã
D
xn ;
n


...
1/nC1

nD1

D

1
X
...
2n
nD1

1 2n 1

x
1/Š

1
X
...
2n/Š
nD0

˛
...
˛

2/
nŠ

xn
I
n
I

(7
...
44)

(7
...
˛

n C 1/

xn

(7
...
x/ D

1
X
nD0

 Ã
˛
are given by Eq
...
70); further,
n

...
2n C 1/
...
47)

7
...
x/ D

1
X

2


...
2n C 1/
...
x/ D

x2nC1 I

(7
...
1/n
x2nC1 :
2n C 1
nD0

(7
...
47)–(7
...

Problem 7
...
Consider an expression
...
1 C x/ˇ D
...

By expanding the left- and the right-hand sides using Eq
...
46) and then
comparing the coefficients in both sides to the same power of x, prove the
following identity:
n
X² àˇ Ã Â˛ C ˇ Ã
:
D
n
n k
k

(7
...
27
...
g
...
1 C x/ expansion is R D 1
...

Problem 7
...
Show that R D 1 for the sine and cosine functions
...
29
...
1 C x/˛
...
30
...

Problem 7
...
Show that R D 1 for the inverse sine and cosine functions
...
32
...
1 C x/ 1
...
1 C x/ by integrating the progression term-by-term
...
33
...
Obtain the
Maclaurin expansion of arctan x by integrating the progression term-by-term
...
34
...
Obtain
the Maclaurin expansion of arcsin x by integrating the progression term-byterm
...
35
...
x2

4x C 1/3

:

Perform the calculation using two methods: (1) use directly formula (7
...
In the case of the first function the
first terms of the expansion for cos x will also have to be used
...
36
...

Problem 7
...
Prove for all the cases above in Eqs
...
43)–(7
...
Explicit expressions for
the remainder term in each case are given at the end of Sect
...
7
...
38
...
(7
...
In particular,
one can obtain expansions into a power series of some integrals which cannot be
calculated analytically
...


Problem 7
...
Using the expansion of sin x=x, prove the following formula:
ˆ
0

x

sin t
dt D x
t

x3
x5
C
C
3Š 3
5Š 5

D

1
X
nD0


...
2n C 1/Š
...
51)

7
...
40
...
x/ D p

ˆ

x

e

t2

2
dt D p

0

Â
x

x3
x5
C
3 1Š
5 2Š

1
2 X
...
2n C 1/nŠ
nD0

Ã

(7
...
41
...
x/ D

2

cos t dt D

0

ˆ
S
...
1/n
x4nC1 ;

...
2n/Š

(7
...
1/n
x4nC3 :

...
2n C 1/Š

(7
...
n/ D 0;
2

y
relating the variable x, a function y
...
with each other
...
x/ that satisfies this
equation
...

ODEs can be characterised by:

• order—order n of the highest derivative y
...
g
...

A specific class of ODEs are linear ODEs if the unknown function and its derivatives
are there in the first power without any cross terms
...
Linear ODEs
are also very often met in practical applications in physics, biology, chemistry and
engineering
...
There are certain cases, however, for which the
necessary tricks have been developed and we shall consider some of them in this
chapter
...
in what follows)
...
g
...
x0 / at some point x0 , and/or the value of its derivative(s)
...

If different points x1 , x2 , etc
...
Kantorovich, Mathematics for Natural Scientists, Undergraduate Lecture
Notes in Physics, DOI 10
...
In this chapter we shall be only concerned with the former case; examples
of the latter are met, e
...
when solving partial differential equations of mathematical
physics which we do not consider here (see Book II, Chap
...
The solution in which
no arbitrary constant(s) is (are) present (and hence some specific information about
the unknown function was already applied) is called a particular solution
...
1 First Order First Degree Differential Equations
Consider an equation F
...
x/ is present
...
Then we have to study the ODE of the following simplest
general form:
dy
D f
...
1)

In this section several methods will be considered which enable one to find the
function(s) y
...


8
...
1 Separable Differential Equations
If f
...
x/=h
...
(8
...
x/
dy
D
dx
h
...
y/dy D g
...
2)

The next step needs some explanations
...
x/, like df D df dx D
dx
f 0
...
Here df D f
...
x/ corresponds to the change of f
...
This means that an expression
like g
...
(8
...
x/ D
´
g
...
x/ D g
...
y/dy there
can also be written as a differential dH
...
y/dy of some other function H
...
y/dy, with respect to the y
...
y/dy D g
...
y/ D dG
...
e
...
y/ D G
...
In
other words, we obtain the final result which formally solves the ODE in question:
ˆ

ˆ
h
...
x/dx C C;

8
...
It is seen that the above
solution can formally be obtained by integrating both sides of the equation for
differentials:
ˆ
ˆ
h
...
x/dx H)
h
...
x/dx;
and introducing a constant C after the integration
...
If the initial condition is known, y
...

As an example, consider equation
xy0

xy D y;

subject to the initial conditions y
...
Regrouping the terms, we can write:
x

dy
D y C xy
dx
H)

dy
D y
...
x C C1 / D eC1 ex
x

H)

y D Cxex :

Here we have come across a typical situation related to the arbitrary constant: after
integration, we introduced a constant C1 which upon rearrangement turned into an
expression eC1 in the formula for the solution y
...
Since C1 is a constant, so is
eC1
...
Applying
the initial condition y
...
Hence, the
particular integral reads y D xex 1
...
1
...
1/ D 0
...
]

458

8 Ordinary Differential Equations

Problem 8
...
Obtain the general solutions of
x2

dy
dx
D y2 :
dx
dy

What its particular solution would be if y
...
]

8
...
2 “Exact” Differential Equations
If the ODE can be written in the form
A
...
x; y/dy D 0;

(8
...
5
...
x; y/ of two variables x and y, i
...

dF D

@F
@F
dx C
dy D A
...
x; y/dy;
@x
@y

(8
...
x; y/ D C
that contains no derivatives
...
Of course,
if the equation F
...
x/, then the solution is obtained
explicitly
...

In order for the left-hand side of Eq
...
3) to be the exact differential, we should
obviously have:
@F
@F
D A
...
x; y/;
@x
@y

(8
...
5
...
6)

for the two functions A
...
x; y/ to satisfy
...
(8
...
g
...
5
...
Briefly, consider equation @F=@x D A
...
1 First Order First Degree Differential Equations

459

the variable y is fixed
...
x; y/dx C C1
...
x; y/ D

(8
...
y/ may still (and most probably will) be a function
of the other variable y
...
y/ is then obtained from the other
condition @F=@y D B
...
Once the function F
...
x; y/ D 0 according to the DE (8
...

We also note that the other method of solving for F
...
6
...
4
...

As an example of using the first method, consider the equation:
dy
D
dx

2xy
:
x2 C y2

First of all, we rewrite this ODE in the form of Eq
...
3):
2xydx C x2 C y2 dy D 0:
Next, we should check if the condition (8
...
x; y/ D 2xy; B
...
e
...
x; y/ D 2xy with respect to x to get
ˆ
FD

2xydx D 2y

x2
C C1
...
y/
2

(note that y is a constant in this integral and hence must be considered as a
parameter!)
...
y/ is obtained from @F=@y D B
...
y/ D x2 C
@y
@y
dy
dy
dy
that is easily integrated (since it is a separable equation) to give:
2

dC1 D y dy

ˆ
H)

C1
...
Thus, the function F
...
x; y/ D x2 y C

y3
C C:
3

Since the total differential of F
...
(8
...
Therefore, we obtain:
F
...
e
...
x/ is obtained by solving the following algebraic equation in y:
y3
D C2 ;
3

x2 y C

where C2 D C1 C is an arbitrary constant to be determined from initial conditions
or some other additional information available for y
...
Note that the solution
obtained can be checked by differentiating both sides of the equation above with
respect to x (remember to treat y as a function of x):
Â
Ã
3y2 dy
dy
dy
d
y3
2
D 2xy C x2 C y2
D 0;
x yC
D 2xy C x2 C
dx
3
dx
3 dx
dx
which is the original equation
...
3
...
[Answer: yx

y C sin
...
y/

cos x

2 sin y D C
...
1
...
3) does not satisfy the condition (8
...
x; y/, called an integrating factor, so that with the replacements
A
...
x; y/A
...
x; y/ ! I
...
x; y/, the necessary condition

8
...
I
...
x; y// D

...
x; y/A
...
Note that the original
ODE (8
...

As an example of using this method, let us try to solve the following equation:
dy
y
D :
dx
x
This is a separable equation for which the solution is easily obtained:
dy
dx
D
H)
y
x

ˆ

dy
D
y
ˇ y ˇ
ˇ ˇ
ln ˇ ˇ D 0
Cx

H)

ˆ

dx
H) ln jyj D ln jxj C ln jCj
x
H)

y D Cx:

However, it is instructive here to solve this equation with the method of the
integrating factor
...
6):
@A=@y D 1, but @B=@x D
1
...
x; y/ D x˛ yˇ
...
x; y/ D x˛ yˇC1 and
B
...
The idea is to find such ˛ and ˇ that the newly defined functions
A and B would satisfy the necessary condition:
@A
D
...
˛ C 1/ x˛ yˇ :

Comparing the two expressions, we get ˇ C 1 D ˛ 1 or ˇ D ˛ 2, a single
condition for two constants to be determined, ˛ and ˇ; for instance, one can take
˛ D 2 and then getting ˇ D 0, so that the required integrating factor becomes
I
...
Now the equation becomes the one containing the exact differential:
x 2 ydx

x 1 dy D 0

since

@
@
x 2y D
@y
@x

x

1

;

and hence can be integrated using the method of the previous subsection:
ˆ
F
...
y/ D yx
2C1

1

C C1
...
y/ D x

1

C

dC1
D x 1;
dy

so that dC1 =dy D 0 yielding C1
...
Therefore, F
...
We have obtained the same
general solution as above using this method
...
For
instance, the factor I
...
x; y/ and B
...


Problem 8
...
Find the integrating factor for the DE
x

dy
C y D 2x:
dx

[Hint: assume that the integrating factor is a function of x only and obtain a
separable ODE for it
...
x/ D
x 2
...
5
...
x; y/ to turn the ODE

...
[Hint: Try I
...
There is
only one possibility here
...
]
Problem 8
...
Solve the DE of the previous problem using the integrating
factor I
...
[Answer: 2xy C 3x2 y2 1 D 2Cy2
...
1
...
Specifically, consider ODEs
of the form of Eq
...
3) (not necessarily the exact differential equation!) in which
functions A
...
x; y/ are homogeneous functions of the same degree
...
x; y/ D

n

A
...
x; y/ D

n

B
...
8)

8
...
x/ )
z
...
x/=x
...
x; y/dx C B
...
x/ D z
...
x; y/ D A
...
1; z/ and B
...
x; zx/ D xn B
...
x; zx/ and B
...
Using these results,
we are now able to perform simple manipulations in the original ODE to obtain an
equation which is separable:
xn A
...
1; z/
...
1; z/dx C B
...
zdx C xdz/ D 0

H)


...
1; z/z C A
...
1; z/dz D 0

H)

dx
D
x

B
...
1; z/ C A
...
e
...

As an example, consider equation
dy
x2 y
D 3
:
dx
x C y3
In the unfolded form the equation becomes
x2 ydx C x3 C y3 dy D 0:
It is seen that both functions A
...
x; y/ D

x2 y and B
...
x/2
...
x; y/ D
...
y/ D

3

x2 y D
3

3

x Cy

3

3

A
...
x; y/;

and thus the substitution z D y=x should be appropriate: y D zx and dy=dx D

...
The obtained solution is a transcendental algebraic
equation with respect to y
...
Note that after applying the corresponding initial conditions to determine the
constant C1 , this equation can easily be solved numerically by plotting for each
value of x the functions of y which appear in the left- and right-hand sides and
looking for the point y where they intersect
...
7
...
y2

xy/dx D 0:

[Answer: y D x=
...
]
Problem 8
...
Show that the ODE
xy
dy
D 2
dx
x
y2
is of the homogeneous type and then solve it
...
x/ is determined from
the equation x2 =2y2 D ln jyj C C
...
1
...
x/y D q
...
9)

8
...
x/ and q
...
We shall show that this equation can
be solved for any functions p and q using the method of an integrating factor of
Sect
...
1
...

Indeed, our ODE, if rewritten in the required (unfolded) form (8
...
q
...
x/y/ dx C dy D 0;
obviously does not satisfy the condition (8
...
x/ and q
...
Consider
then an integrating factor I
...
x; y/
...
x/ C p
...
x; y/dy D 0;
and let us choose I
...
6) is satisfied:
@
@
ŒI
...
q
...
x/y/ D I
...
x; y/
@I
...
q
...
x/y/ C p
...
x; y/ D
:
@y
@x

It is seen that if the function I
...
x; y/ D I
...
x/I
...
x/
dI
H) p
...
x/dx D ln jIj ;

ˆ

ˆ
p
...
x/ D exp

p
...
10)

The integral here is an indefinite integral which is considered as a function of x
...
8
...
2
...
x; y/ D I
...
q
...
x/y/ and B
...
x/;
so that
ˆ
F
...
x; y/dy D I
...
x/;

dI
dC1
@F
...
x; y/ D I
...
q
...
x/y/ :
@x
dx
dx

466

8 Ordinary Differential Equations

However, the function I
...
x/I
...
x/q
...
x/ D
I
...
x/dx C C2 :
Thus, we have found the function completely:
ˆ
F
...
x/y C C1
...
x/y

I
...
x/dx C C2 :

Since it must be a constant anyway, we can write:
ˆ
I
...
x/q
...
x/y

ˆ
I
...
x/dx D C;

which can be solved for y to get (C is the final arbitrary constant):
y D I
...
x/q
...
11)

Problem 8
...
Check that if each integral in the above formula is understood
as a definite integral with the upper limit being the corresponding variable, the
bottom limits do not matter
...
0/ D 1
...
x/ D 1 and q
...
Therefore, the
function I
...
(8
...
x/ D e

´

dx

D ex ;

and thus the solution is obtained as
Ä
ˆ
y
...


8
...
10
...
x/ D

1
sin
...
sin x/ C sin x


...
x/ D 2x=3 C x3 =5 C C=x2
...
9) can also be obtained
using the method of variation of parameters which we shall find especially useful
in a more general case of linear second order ODE to be considered in Sect
...
2
...
2
...
x/y D 0 H)
D pdx H) y
...
x/dx D C I
...
12)

where C is an arbitrary constant
...
e
...
e
...
(8
...
12), but with the
constant C replaced by a function C
...
x/dx D C
...
x/ 1 :

y
...
x/ exp

We need to substitute this trial solution into the ODE (8
...
x/
...
x/I
...
x/I
...
x/ C p
...
x/I
...
x/
d
I
...
x/;

satisfies the equation
p
...
x/ 1 ;

which follows from Eq
...
10)
...
x0 / dx0 , i
...
with the x appearing in the upper limit,
we treated the integral as
and hence its derivative is p
...
After simplification, we obtain:
ˆ
C0
...
x/ 1 D q
...
x/ D q
...
x/ H) C
...
x/I
...
It is easy to see now that y
...
x/=I
...
(8
...


468

8 Ordinary Differential Equations

8
...
x/

dn y
dn 1 y
C an 1
...
x/

dy
C a0
...
x/:
dx

(8
...
x/ in the right-hand side depending only on x,
and is called inhomogeneous
...

A very general property of this equation is that it is linear
...
Indeed, if we know two solutions y1
...
x/
of this equation, then any of their linear combinations with arbitrary constants C1
and C2 , i
...

y
...
x/ C C2 y2
...
Generally, any n-th order ODE like (8
...
x/; : : : ; yn
...
x/ D C1 y1
...
x/ C

C Cn yn
...
x/

(8
...

These are determined using additional information about the solution, e
...
initial
conditions, i
...
known values of y
...
1/
...
n 1/
...
i/
...

Problem 8
...
Prove that the linear combination (8
...
13) is also its solution
...
12
...
x/y00 C a1
...
x/y D 0:

(8
...
x/ which brings this ODE into the form
...

2
[Answer: f
...
a1 =a2 / dx =a2
...
There
must be two linearly independent solutions of the homogenous linear second order
DE
...


8
...
1
...
x/ of a linear second order ODE (8
...
x/ is known
...
Since both y1 and y2 satisfy the ODE (8
...
16)

Consider now the determinant (called Wronskian)
ˇ
ˇ
ˇ y1 y2 ˇ
ˇ
W D ˇ 0 0 ˇ D y1 y0
2
y1 y2 ˇ

y2 y0 :
1

(8
...
(8
...
x/:
a2
...
x/
C a1
...
x/ D 0 ;
dx

which is separable and thus can be easily integrated:
dW
D
W

a1
...
x/

ˆ

dW
D
W

ˆ

a1
...
x/j D
a2
...
x/
dx;
a2
...
Hence, the Wronskian becomes
 ˆ
W
...


Ã
a1
...
x/

(8
...
x/ has been calculated, it is now possible to consider
Eq
...
17) as a first order ODE for y2
...
x/
2

y0
...
x/
1
y2
...
x/
y1
...
9) we considered before in Sect
...
1
...
(8
...
x/ D y0
...
x/ and q
...
x/=y1
...

1
This way of finding y2 , however, appears a bit tricky as one has to deal carefully
with the modulus of y1
...
A much simpler and straightforward way of obtaining
y2
...
x/ D u
...
x/ directly in the above equation:
u0 y1 C uy0
1

W
W
y0
1

...
x/ D

W
...
x/
1

which, after integration, gives immediately
ˆ
u
...
x/
dx:
y2
...
19)

This should be understood as an indefinite integral without an arbitrary constant:
this is because y2 D uy1 and a constant C in u
...
Hence,
ˆ
y2
...
x/y1
...
x/

W
...
x/
y2
...
x/
1

 ˆ

Ã
a1
...
x/
(8
...
Q
...
D
...
x/ D sin x
...
(8
...
x/ D 1 and, using Eq
...
20), we get:
ˆ
y2 D sin x

dx
D sin x
...
1

1

The minus sign does not matter as in the general solution y2 will be multiplied by an arbitrary
constant anyway
...
2 Linear Second Order Differential Equations

471

Problem 8
...
Consider the following linear second order ODE:
y00 C 4y0 C 4y D 0;
which first solution is y1 D e
xe 2x
...
14
...
21)


...
x/ C V
...
x/ D E
...
x/ is the particle wavefunction, V
...
Consider a discrete energy spectrum E for
which
...
Show that the energies E
in the discrete spectrum cannot be degenerate, i
...
there can only be a single
state
...
[Hint: prove by contradiction, assuming there are
two such solutions, 1 and 2 ; then, argue that 00 = for both should be the
same; then use integration by parts
...
2
...
However, it is not possible to obtain such a solution in the
cases of higher-order ODEs containing variable coefficients
...

We shall again limit ourselves with the ODEs of the second order; higher order
ODEs can be considered along the same lines
...
22)

where a0 , a1 and a2 are some constant coefficients
...
Indeed, substituting this trial function into the
ODE above, we have:
a2 p2 C a1 p C a0 epx D 0:

472

8 Ordinary Differential Equations

Since we want this equation to be satisfied for any x, we must have p to satisfy the
following characteristic equation:
a2 p2 C a1 p C a0 D 0;

(8
...

If there are two (different) solutions p1 and p2 , then two functions ep1 x and ep2 x
satisfy the same equation; note that the two functions are linearly independent in
this case
...
22)
...
To solve the equation
d2 y
dx2

8

dy
C 12y D 0;
dx

the trial solution epx is substituted into the equation to give p2 8p C12 D 0
...
Hence the general solution is
y
...
Therefore, a general solution is
y
...
x/ D C1 Œcos
...
2x/ C C2 Œcos
...
2x/

D A1 cos
...
2x/;
where A1 D C1 C C2 and A2 D i
...

Let us now solve the equation
d2 y
dx2

4

C2 / are two new arbitrary (and in general

dy
C 8y D 0:
dx

8
...
Therefore, a general solution is
y
...
2C2i/x C C2 e
...
2x/ C A2 sin
...
(8
...
e
...
p1 x/
...
1, the second solution can always be found from the
first via Eq
...
20) in the form of y2
...
x/y1
...
x/ is to be
determined from Eq
...
19)
...
18) in this case is
Â
W
...
23) are the same, we must have
a2 D 4a0 a2 and hence p1 D a1 =2a2 , i
...
W
...
2p1 x/
...
(8
...
x/ D

W
dx D
y2
1

ˆ

e2p1 x
dx D x;
e2p1 x

and hence the second solution is to be taken in the form: y2 D xy1 D x exp
...

Problem 8
...
Consider equation
y00

4y0 C 4y D 0;

which has repeated roots p D 2 in the characteristic equation p2 4p C 4 D 0,
i
...
y1
...
To obtain the second solution, try directly the substitution
y D u
...
Show that u00 D 0, and hence u
...

Finally, argue that it is sufficient to take y2
...

Problem 8
...
Consider the case of two distinct but very close roots of the
characteristic equation: p1 and p2 D p1 C ı
...

Problem 8
...
Consider the case of two distinct roots p1 ¤ p2 of the
characteristic equation
...
1 that from y1 D ep1 x
follows y2 D ep2 x
...
18
...
C1 x C C2 / e3x
...
2
...
13), i
...
with non-zero righthand side when the function f
...
Whatever the order of the ODE and the
form of the functions-coefficients an
...
x/, the construction of the general
solution of the equation is based on the following theorem:

Theorem 8
...
The general solution of a linear inhomogeneous equation is
given by
y
...
x/ C yp
...
24)

where
yc
...
x/ C

C Cn yn
...
x/ is the so-called particular solution
(integral) of the inhomogeneous equation
...
Let us start by introducing a shorthand notation for the left-hand side of
Eq
...
13) by defining an operator
L D an
...
x/ n
n
dx
dx

1
1

C

C a1
...
x/;
dx

(8
...
x/ standing on the right of it, produces the
following action:
Lg
...
x/

dn g
dn 1 g
C an 1
...
x/

dg
C a0
...
2 Linear Second Order Differential Equations

475

The operator L is linear, i
...

L
...
26)

where g1
...
x/ are two arbitrary functions
...
(8
...
x/:

(8
...
x/:

(8
...
x/

f
...
29)

i
...
y yp must be the general solution yc of the homogeneous equation Lyc D 0,
i
...
any solution has the form y D yc C yp
...
(8
...
Q
...
D
...
x/ which is the general solution of the homogeneous equation y00 Cy D 0
...
Trying a particular
solution in the form yp D Aex with some unknown constant A, we get after
substitution in the equation:
Aex C Aex D ex H) 2A D 1 H) A D

1
2

H)

yp D

1 x
e;
2

so that the general solution is
1
y D yc C yp D C1 cos x C C2 sin x C ex :
2
Note that if we take two specific solutions, say, 2 sin x C 1 ex and 3 cos x C 1 ex ,
2
2
obtained by choosing particular values for the arbitrary constants C1 and C2 , then the
difference 2 sin x C 3 cos x of the two solutions is in fact a complementary solution
of the homogeneous equation with the particular choice of the constants C1 D 3 and
C2 D 2, as expected
...
x/ itself
...


8
...
2
...
x/:
Several cases of f
...
x/ is a polynomial P
...
x/ D bn xn C C b1 x C b0 of the order n, then the
particular solution should be sought in the form of a polynomial of the same
order with unknown coefficients that must be determined by substituting it into
the equation and comparing terms with the like powers of x on both sides
...
x/ D A cos
...
!x/ is a linear combination of trigonometric
functions, try yp
...
!x/ C B1 sin
...
These unknown
coefficients are obtained by substitution into the ODE and comparing coefficients to sin
...
!x/ on both sides
...
x/, still both sine and cosine terms must be present in yp
...

• If f
...
x/ D Ae x as the
particular solution with some constant A to be determined by substitution into
the equation
...
x/ is given as a combination of all these functions, i
...

f
...
n/
...
!x/ C B sin
...

Also note that if f
...

Examples below should illustrate the idea of the method
...
1
...
2 Linear Second Order Differential Equations

477

Solution
...
Since the righthand side is the first order polynomial, the particular integral is sought as a first order
polynomial as well, yp D Ax C B
...
Ax C B/ D 1 C x H)

H)

A D 1=6
;
B D 11=36

so that the general solution
Â
Ã
1
11
y D C1 e C C2 e C
xC
:J
6
6
2x

3x

Example 8
...
I Find the particular solution of the equation
y00

5y0 C 6y D sin
...
We must find the particular solutions individually for each of the terms
in the right-hand side since they have different frequencies
...
4x/ we try the
function A sin
...
4x/, that gives upon substitution:
Π16A sin
...
4x/

5 Œ4A cos
...
4x/

C6 ŒA sin
...
4x/ D sin
...
4x/ C
cos
...
4x/ C
cos
...
J
Example 8
...
I Find the particular solution of the equation:
y00

5y0 C 6y D 5xex :

Solution
...
Ax C B/ ex
...
Ax C 2A C B/ ex
H)
H)
H)

5
...
Ax C B/ ex D 5xex


...
Ax C A C B/ C 6
...
A C B/ C 6B D 0
Ã
Â
15 x
5
xC
e J
yp D
2
4

A D 5=2
B D 15=4

There is a special case to be also considered when f
...
In these cases the method should be modified
...
4
...
If we try yp D Ae2x following the function type in the right-hand side,
then we get zero in the left-hand side since e2x , as can easily be checked, is a solution
of the homogeneous equation
...
However,
the required modification is actually very simple: if we multiply yp tried above by
x, it will work
...
4A C 4Ax/ e2x

5
...
J
The considered case corresponds to a single “resonance” as the exponent in the
right-hand side coincides with that of only one of the solutions of the homogeneous

8
...
If however both complementary exponents are the same, p1 D p2 , and the
same exponent happens to be present in the right-hand side, we have the case of
a double “resonance”
...

Example 8
...
I Find the particular solution of the equation:
y00

4y0 C 4y D xe2x :

Solution
...
Substituting yp D
...
Let us now try yp D x
...
2Ax C B/ C 4 Ax2 C Bx e2x
C 4 Ax2 C Bx e2x D xe2x

H)

4
...
2Ax C B/ C 4 Ax2 C Bx e2x

4
...
e
...
Now let us try yp D x2
...
3A C B/ x2 C 2
...
3A C 2B/ x2 C 2Bx C 4 Ax3 C Bx2 D x:
The terms containing x3 and x2 in the left-hand side cancel out; the terms with x
yield A D 1=6, while the terms with x0 result in B D 0, i
...
yp D x3 =6 e2x ,
which as easily checked by the direct substitution is the correct particular integral
of this ODE
...
19
...
x/ with the following right-hand side function f
...
3=5/ sin x

...
3=5/ e 3x
...
]
Problem 8
...
Obtain the general solution of the previous problem with
f
...
[Answer: y D C1 e 2x C C2 e x C x
...
]

480

8 Ordinary Differential Equations

Problem 8
...
Find the general solutions of the ODE y00
(a) f
...
x/ for

(b) f
...
5=2/ xex
...
22
...
C1 C C2 xC
x3 =6 e 3x C 2=27 x=9
...
23
...
3x/I

(b) y00 C 9y D 3 cos x C xex :

[Answer: (a) y D C1 cos
...
3x/
...
3x/; (b) y D C1
cos
...
3x/ C
...
5x 1/=50ex
...
24
...


8
...
2
...
Here we shall discuss a general case of the ODE with variable
coefficients,
a2
...
x/y0 C a0
...
x/;

(8
...
x/ provided the two solutions y1
...
x/ of the corresponding homogeneous equation are known
...
x/ D 0) and
hence obtain the corresponding complementary solution:
yc
...
x/ C C2 y2
...
31)

8
...
The idea of the method is to search for the
particular solution of the original equation in the form (cf
...
8
...
5)
yp
...
x/y1
...
x/y2
...
32)

which is basically obtained from (8
...
x/ and u2
...
Substituting the trial solution (8
...
(8
...
u1 y1 C u2 y2 / D f ;
1
1
2
2
or after some trivial manipulation:
u1 a2 y00 C a1 y0 C a0 y1 C u2 a2 y00 C a1 y0 C a0 y2
1
1
2
2
C a2 u00 y1 C u0 y0 C u00 y2 C u0 y0 C a2 u0 y0 C u0 y0 C a1 u0 y1 C u0 y2 D f :
1
1 1
2
2 2
1 1
2 2
1
2
The first two terms vanish since y1 and y2 satisfy the homogeneous equation by
construction
...
33)

Both functions u1
...
x/ are to be determined
...
This means that we are free to constrain the two functions
u1
...
x/ in some way, and this can be done to simplify Eq
...
33), so that it
can be solved
...
34)

because it would remove two terms from Eq
...
33) yielding simply
a 2 u 0 y0 C u 0 y0 D f :
1 1
2 2

(8
...
34) and (8
...
36)

482

8 Ordinary Differential Equations

where W D y1 y0 y0 y2 is the Wronskian of y1 and y2 , see Eq
...
17)
...
25 below)
...
Note that
when integrating these equations, arbitrary constants can be dropped due to the
specific form of Eq
...
32) in which the particular solution is sought: the terms to
be originated from those constants will simply be absorbed in the complementary
solution (8
...

Problem 8
...
Prove by contradiction that the Wronskian W D y1 y0
2
0 if the two functions y1
...
x/ are linearly independent
...
The
complementary solution is:
yc D C1 ex C C2 xex ;
since the characteristic equation gives repeated roots
...
ex C xex /
1

ex xex D e2x :

Then, Eq
...
36) can be written explicitly as
u0 D
1


...
x/ D

ex x 2 ex
Dx
e2x

2

dx
D
x

ln jxj ;

H) u2
...
ln jxj/ ex C

x

1

xex D

ex ln jxj

ex :

In fact, the second term, ex , will be absorbed by the complementary solution, so that
only the first term may be kept
...


ex ln jxj ;

8
...
3 Non-linear Second Order Differential Equations
Here we shall discuss a few simple tricks which in some cases are useful in solving
non-linear second order ODEs
...

Consider first an ODE of the form y00 D f
...
It does not contain y and y0 and is
solved by simply noticing that y00 D
...
Therefore, by introducing a new function
z
...
x/, the first order ODE z0 D f
...
x/ is
obtained which is integrated immediately to give
ˆ
f
...
x/ D

To obtain y
...
x/ which is another first order ODE,
solved similarly, finally yielding the solution
ˆ
y
...
x/dx C C2 D

ˆ

x

x1

dx1

dx2 f
...

As an example, consider Newton’s equation of motion for a particle moving
under an external force changing sinusoidally with time:
mR D A sin
...
e
...
0/ D
x
...
To solve the equation above, we introduce the velocity v D x, and then
P
P
the equation of motion would read simply v D
...
!t/
...
t/ D

ˆ

A
m

t

0

sin
...
!t/ :

Integrating the velocity again, we obtain the particle position as a function of time:
ˆ
x
...
/d D
m!

1
sin
...
y; y0 /,
i
...
the ODE lacks direct dependence on x
...
Indeed, we denote z D y0 and then notice that
y00 D

dz dy
dz
dz
d 0
y D
D
D z;
dx
dx
dy dx
dy

484

8 Ordinary Differential Equations

which yields
z

dz
D F
...
y/
...
y/ which may be integrated without difficulty via
ˆ

y0 D z
...
y/

H)

ˆ

ˆ
H)

dx

dy
D x C C2 :
z
...
y/ after the
first integration
...
t/ of a particle subject to an
external potential V
...
Let x0 and v0 be the initial (t D 0) position and velocity of
the particle, respectively
...
x/;
dx

mR D
x

where f
...
x/ is the external force due to the potential V
...
Using the
method described above, we introduce the velocity v D x and then write the
P
acceleration as a D x D v D v
...
x/
dx

H)

H)

mv

mv 2
2

ˆ

t

vdv D

m
0

2
mv0
D
2

ˆ
0

x

x

f x0 dx0

0

f x0 dx0 ;

which is nothing but the expression of conservation of energy: in the left-hand side
we have the change of the particle kinetic energy due to the work by the force given
in the right-hand side
...
0/
dx0

V
...
Solving
this equation with respect to the velocity, we find
s
v
...
x0 / dx0 ;

8
...
Therefore, integrating again, we obtain an equation for the
trajectory in a rather general form:
ˆ x
dx00
q
D t:
´
2 x00
2
x0
0 / dx0
v0 C m 0 f
...
y00 ; y0 ; y; x/ D 0 is a homogeneous function with respect
to its variables y, y0 and y00 (treated together, see an example below), then one can
introduce instead of y
...
x/ via
Ã
ˆ
z
...
37)
y
...
Indeed,
Ã
ˆ
0
z
...
x/ D yz and y00 D
...
x/ in the equation as demonstrated by
the following example
...
Using the substitution (8
...
y C zy/ D 0

H)

z2 C z0

z
...
x/ D C1 ex
...
x/ sought
for is found as
Ã
ˆ
C1 ex dx D exp
...
C1 ex / ;
y
...


Problem 8
...
Solve the following ODE using one of the methods considered
above:
(a) y00 D ex I

(b) y y00 C y0 D y0

2

:

[Answer: (a) y D C1 x C C2 C ex ; (b) y D C1 exp
...
]

486

8 Ordinary Differential Equations

8
...
g
...
x/ C p
...
x/ C q
...
x/ D 0;

(8
...
x/ and q
...
This is the so-called
canonical form of the ODE
...

We know that when the coefficients p
...
x/ are both constants, it is
possible to solve Eq
...
38) and express the solution in terms of elementary functions
(exponentials, sine and cosine)
...
x/ and q
...
38) in terms of elementary functions is
possible only in special cases; in most cases, the solution cannot be expressed in
terms of a finite number of elementary functions
...
2 This can be used either as an analytical tool or as a means of calculating the
solution numerically when the series is terminated at a finite number of terms
...
The point x D a is said to be an ordinary point of the differential
equation (8
...
x/ and q
...
e
...
k/
...
If the point x D a is not an ordinary point, then we
shall call it a singular point
...
x C 3/y00
...
x C 3/y0
...
x/ D 0;

(8
...
x/ D

1
x2

and q
...
x C 3/

(8
...
All other points in the interval 1 <
x < 1 are therefore ordinary points
...


2

In some cases the series may be finite
...
4 Series Solution of Linear ODEs

487

8
...
1 Series Solutions About an Ordinary Point
Suppose that the initial conditions y
...
a/ are given at the point x D a which
we assume is an ordinary point of Eq
...
38)
...
k/
...
k D 1; 2; 3 : : :/ in terms of y
...
a/,
by repeated differentiation of Eq
...
38) with respect to x
...
a/ follows
immediately from the equation itself
...
x/ via y
...
a/ and y00
...

Repeating this process, one can calculate any derivative y
...
a/ of y
...
Hence, we can formally construct a Taylor series about x D a
for our function y
...
x/ D

1
X

cr
...
r/
...
41)

which forms a solution of the equation; note that the initial conditions are satisfied
if c0 D y
...
a/
...
(8
...
a/ D C1 and c1 D y0
...

This solution would converge within an interval jx aj < R, where R is
the radius of convergence of the power series, Eq
...
41)
...
(8
...
This point follows from a more general discussion based on the
theory of functions of complex variables (see Sect
...
8, Book II)
...
However, this method is not convenient for actual applications
...
41)
into the differential equation (8
...
Indeed, p
...
x/ can be expanded into
the Taylor series around x D a (recall that these functions have no singularities at
x D a):
p
...
x

a/ C p2
...
x/ D q0 C q1
...
x

a/2 C

;

which, when substituted into the differential equation (8
...
2c2 C p0 c1 C q0 c0 / u0 C
...
4 3c4 C 3p0 c3 C 2p1 c2 C p2 c1 C q0 c2 C q1 c1 C q2 c0 / u2 C

D 0;

where u D x a
...
form a linearly independent set, this can only be possible
if all coefficients of the powers of u are equal to zero at the same time, which yields
an infinite set of algebraic equations for the unknown coefficients c2 , c3 , c4 , etc
...
The first equation gives c2 , the second equation gives c3 via c2 , the third
c4 via c2 and c3 , etc
...
So, given the values
of the initial coefficients c0 and c1 , it is possible to generate all the coefficients cr up
to any desired maximum value of the index r
...
x/ and q
...

Example 8
...
I Determine the series solution of the differential equation
y00

xy D 0;

(8
...
0/ D 1 and y0
...

Solution
...
e
...
x/ D

1
X

cr xr :

(8
...
x/, we find:
y0
...
x/ D

rD1

1
X

r
...
44)

rD2

Note that in y0
...
x/
...
43) and (8
...
42) gives
1
X
rD2

r
...
To collect terms with like powers of x, we need
to combine the two sums into one
...
4 Series Solution of Linear ODEs

489

term in the first sum (corresponding to r D 2), so that the two sums would contain
identical powers of x:
2c2 C

1
X

r
...
45)

rD0

Indeed, now the first sum contains terms x1 , x2 , etc
...
Once we have both sums starting from
the same powers of x, we can now make them look similar which would eventually
enable us to combine them together into a single sum
...
Making the replacement m C 1 D r 2 for the summation index in the
first sum of Eq
...
45) does the trick, and we obtain:
2c2 C

1
X


...
m C 2/cmC3 xmC1

mD0

1
X

cr xrC1 D 0:

rD0

Note that m D r 3 here, and hence it starts from the zero value corresponding to
r D 3 of the original first sum
...

To stress the similarity even further, we can use in the first sum the same letter r for
the summation index instead of m (the summation index is a “dummy index” and
any symbol can be used for it!), in which case the two sums are readily combined:
2c2 C

1
X


...
r C 2/crC3 xrC1

rD0

H)

2c2 C

1
X

cr xrC1 D 0

rD0
1
X

Œ
...
r C 2/ crC3

cr  xrC1 D 0:

(8
...
47)

and, at the same time,

...
r C 2/ crC3 D cr

H)

crC3 D

cr
; r D 0; 1; 2; : : : :

...
r C 3/
(8
...
One can see that three families of coefficients are generated
...
; if we start from c1 , we generate c4 , c7 , etc; finally,
starting from c2 (which is zero), we generate c5 , c8 , etc
...

Since nothing can be said about the coefficients c0 and c1 , we have to accept that
these can be arbitrary
...
x/ D c0 x0 C c1 x1 C c2 x2 C

;

then two solutions are obtained, one starting from c0 , and another one from c1 :
y1
...
2 5/
...
2 5 8/
...
x/ D c1 x C c4 x4 C c7 x7 C
Ä
x7
x10
x4
D c1 x C
C
C
C
3 4

...
4 7/

...
4 7 10/

:

As expected, the coefficients c0 and c1 serve as arbitrary constants, so that
expressions in the square brackets can in fact be considered as linearly independent
solutions
y1
...
2 5/
...
2 5 8/
...
x/ D x C

x7
x10
x4
C
C
C
3 4

...
4 7/

...
4 7 10/

;
:

8
...
x/ D C1 y1
...
x/;
with arbitrary constants C1 and C2
...
0/ D 1 and y2
...
0/ D 0 and y0
...

1
2
The application of the initial conditions y
...
0/ D C2 D 0
gives the constants and hence the required particular solution (integral) is:
y
...
x/ D 1 C

x3
x6
x9
C
C
C
2 3

...
3 6/

...
3 6 9/

:

(8
...
49) can be investigated directly
using the ratioP with the help of the recurrence relation (8
...
Recall that an
test
infinite series 1 ak is convergent if limk!1 jakC1 =ak j < 1
...
49) the two adjacent terms in the series are c3k x3k and c3kC3 x3kC3 , which
gives
ˇ
ˇ
ˇ c3kC3 x3kC3 ˇ
c3kC3 3
1
ˇ
lim
lim ˇ
D0
jxj D jxj3 lim
ˇ c x3k ˇ D k!1 c
k!1
k!1
...
3k C 3/
3k
3k
(8
...
We see from this result that the series solution (8
...
J

8
...
2 Series Solutions About a Regular Singular Point
Let us now suppose that the differential equation (8
...
It can be shown (Sect
...
8, Book II) that one has to distinguish two cases: (i) if
the functions
...
x/ and
...
x/ are both “well-behaved” at x D a, then we
say that x D a is a regular singular point (RSP) of Eq
...
38)
...
38)
...
(8
...

The introduced classification of singular points we shall illustrate for the
differential equation (8
...
x/ and q
...
(8
...
It has two singular points: at x D 0 and x D 3
...
x

a/ p
...
x


...
x/ D


...
x

a/2 q
...
x


...
x/ D 3


...
For the singular point, a D 0, we similarly find that

...
x/ D

1
x

and
...
x/ D

3x
:

...
x 0/p
...

If x D a is an RSP of Eq
...
38), then the Taylor series solution (8
...
a/ and/or q
...
r/
...
However, it can be shown
that Eq
...
38) always has at least one solution of the Frobenius type (Sect
...
8,
Book II):
y
...
x

a/s

1
X

cr
...
x

a/rCs;

(8
...
The radius of
convergence R of the series (8
...
38)
...
51)
into the differential equation, collecting terms with the like powers of x and equating
the corresponding coefficients to zero
...
It is more transparent to
illustrate this method considering an example
...
7
...
52)

about x D 0
...
In this differential equation p
...
x/ D 1=2x, and hence x D 0 is a
singular point
...
Since the functions
xp
...
x/ D x=2 are both “well-behaved” at x D 0, it follows,
therefore, that x D 0 is an RSP
...
x/ D

1
X
rD0

cr xrCs

(8
...
4 Series Solution of Linear ODEs

493

with some yet to be determined number s
...
53) we find that
y0
...
rCs/xrCs

1

and y00
...
rCs/
...
54)

rD0

Note that, contrary to Example 8
...
The substitution of these results
in Eq
...
52) and combining the sums originating from y00 and y0 gives:
1
X

cr
...
2r C 2s

1/xrCs

1

C

rD0

c0 s
...
r C s/
...
55)

rD0

where the first (r D 0) term in the first sum has been separated out as the second
sum does not have a term with xs 1
...
(8
...
e
...
r C s/
...
r0 C s C 1/
...
r C s C 1/
...
The two
sums in (8
...
2s

1/ x

s 1

C

1
X

Œ
...
2r C 2s C 1/ crC1 C cr  xrCs D 0:

(8
...
This means that in each and every term in the
expansion above the coefficients to the powers of x must vanish, i
...
we should have
c0 s
...
r C s C 1/
...
r C s C 1/
...


(8
...
57)
...
More explicitly: c1 is proportional to c0 , c2 is proportional to c1
and hence eventually also to c0
...
Next, let us
consider the first equation, c0 s
...
It accepts as a solution c0 D 0
...
x/ D 0 of
the differential equation and is thus of no interest
...

Hence, as a0 ¤ 0, we arrive at an algebraic equation s
...
It is called indicial equation and gives two solutions for
s, namely: s D 1=2 and s D 0
...
57) and hence
build completely two linearly independent solutions
...
From (8
...
r C 1/
...
r C 3=2/
...
Therefore, for constructing a linearly
independent elementary solution, it is sufficient (and convenient) to set c0 D 1
...
x/ D

1
X

cr x

rC1=2

Â
p
D x 1

x2
C
1 3
1 2 3 5
x

rD0

Ã

x3
C
1 2 32 5 7

:
(8
...
r C 1/
...
x/ D 1

xC

x2
2 3

x3
C
2 32 5

:

(8
...
J
Problem 8
...
By using the recurrence relations for both values of s D 1=2
and s D 0, work out explicitly the first four terms in the expansions of y1
...
x/ and hence confirm expressions (8
...
59) given above
...
4 Series Solution of Linear ODEs

495

The next example shows that the Frobenius method can also be used even in the
case of expanding around an ordinary point
...
8
...
x/ C ! 2 y
...

Solution
...
However,
to illustrate the power of the Frobenius method, we shall solve the equation using
the generalised series expansion assuming some s:
yD

1
X

cr xrCs :

rD0

After substituting into the DE, we get:
1
X

cr
...
r C s

1/xrCs

2

C !2

rD0

1
X

cr xrCs D 0:

rD0

The following powers of x are contained in the first sum: xs 2 , xs 1 , xs , xsC1 , etc
...
, respectively
...
Anticipating that we
will have to combine the two sums together later on, we separate out the first two
“foreign” terms from the first sum:
c0 s
...
s C 1/sx

s 1

C

1
X

cr
...
r C s 1/x

rD2

2

Cc1 s
...
s 1/xs

rCs 2

2 ! r, so that the two sums can


...
r C s C 1/crC2 C ! 2 cr xrCs D 0:

rD0

(8
...
Then, from the first term we
conclude that s
...
Therefore, we need to consider two cases
...
From the last term in (8
...
r C 2/
...
It is easy to see that generally
c2n D
...
2n/Š

which can be proven, e
...
, by the method of mathematical induction
...
We then assume that it is also true for some value of n
...
nC1/ D c2nC2 D
D

!2
c2n

...
2n C 1/

!2
! 2n
! 2nC2

...
1/nC1
;

...
2n C 1/

...
2n C 2/Š

as required
...
x/ D
x ;

...
2n/Š
nD0

which as can easily be seen is the Taylor’s expansion of the cosine function cos
...

Let us now consider the case of s D 1
...
s 1/ D
0 in the first term in Eq
...
60); however, the second term there gives c1 D 0 as
s
...
Then the corresponding recurrence relation
crC2 D

!2
cr ; r D 0; 1; 2; : : :

...
r C 2/

leads to
y2
...
1/n
nD1

1
1
1 X
! 2n

...
!x/ ;

...
2n C 1/Š
! nD1

...
Note that since
c1 D 0 there will be no odd terms at all: c3 D c5 D
D 0
...
Concluding, the final
general solution of the harmonic oscillator equation is
y
...
!x/ C C2 cos
...
J

8
...
28
...

Consider now another possible proposition which is assuming that c1 ¤ 0
...

Example 8
...
I Solve the Legendre equation
1

x2 y00
...
x/ C l
...
x/ D 0;

(8
...
Here l > 0
...
First of all, we consider p
...
x/ to find all singular points of the DE
and characterise them:
p
...
1 C x/
...
x/ D

l
...
1 C x/
...
1 C x/p
...
1 C x/2 q
...
1 C x/

and

l
...
1

x/p
...
1

x/2 q
...
1

x/

l
...

Correspondingly, we seek the series expansion around x D 0 exactly in the
middle of the interval 1 < x < 1 since in this case the series is supposed to
converge for any x within this interval
...

Substituting the expansion
yD

1
X

cr xrCs

rD0

into the differential equation, we get:
1
X
rD0

cr
...
rCs 1/ 1

x2 xrCs

2

2

1
X
rD0

cr
...
lC1/

1
X

cr xrCs D 0:

rD0

(8
...
The first one,

The first term leads to two sums due to 1
1st D

1
X

cr
...
r C s

2

1/xrCs

D c0 s
...
s C 1/xs

1

rD0

C

1
X

1/xrCs 2 ;

cr
...
r C s

rD2

2 ! r0 ! r, is worked out into

after changing the summation index r
1st D c0 s
...
s C 1/xs

1

C

1
X

crC2
...
r C s C 1/xrCs ;

rD0

while the second sum is
1
X

2nd D

cr
...
r C s

1/xrCs :

rD0

We separated out the first two terms in the first sum above in order to be able
to combine the two sums together into a single sum
...
(8
...
s

1/xs

2

C c1 s
...
r C s C 2/
...
r C s/
...
r C s/cr C l
...
s

1/xs

2

C c1 s
...
r C s C 2/
...
r C s/
...
l C 1/ cr g xrCs D 0:

(8
...
s 1/ D 0, resulting
in two values of s
...
If s D 0, then c1 must be arbitrary
as well because of the second term in Eq
...
63) which is zero for any c1
...
r C 1/ l
...
r C 2/
...
64)

8
...
If we first start from c0 D 1, then we can generate
c2n coefficients with even indices (below we use k D l
...
l C 1/
D
2 1

k
;
2Š

c4 D

2 3 k
c2 D
4 3


...
x/ D 1

k 2
x
2Š


...
65)

containing even powers of x
...
x/ D x C

2

k
3Š

x3 C


...
2
5Š

k/

x5 C

:

(8
...
The recurrence relation in this case is:
crC2 D


...
r C 2/ k
cr :

...
r C 2/

Since the second term in Eq
...
63) should be zero, we have to set c1 D 0 since
s
...
Then, starting from c0 , we obtain coefficients c2n with
even indices and it can easily be checked that this way we arrive at y2
...
Thus, the second value of s does not lead to any new solutions
...
64):
ˇ
ˇ
Ä
ˇ crC2 xrC2 ˇ
r
ˇ
ˇ D r
...
r C 2/
...
r C 1/
...
x/ converges for 1 < x < 1, as expected
...
x/:
ˇ
ˇ
ˇ crC2 xrC2 ˇ
r
...
r C 1/
...

In general, either of the solutions diverges at x D ˙1
...
Indeed, consider first l being even
...
x/ only coefficients cr with even indices r are present
...
r C 1/ k
r
...
l C 1/
cr D
cr :

...
r C 2/

...
r C 2/

It is readily seen that when r D l (which is possible as both r and l are even integers)
we have clC2 D 0 because of the numerator in the equation above
...
will also be equal to zero, i
...
the solution y1
...
For, example,
if l D 2, then k D 2 3 D 6 and
c2 D

6
D
1 2

3;

c4 D

2 3 6
c2 D 0 ;
3 4

c6 D c8 D

D 0;

and thus y1
...
x/lD2 D 1 3x2
...
x/, will not terminate for any even l since in the
recurrence relation the numerator reads r
...
l C 1/ and is not equal to zero
for any odd r
...
x/ is of a polynomial form, while y1
...
The two polynomial solutions are directly proportional to the socalled Legendre polynomials
...
29
...
x/lD4 D 1

10x2 C

35 4
xI
3

Œy2
...
30
...
2/n
xn ;
nŠ
...
2/n
xn ;
nŠ
...
Using the ratio test, prove that either series converges for any x
...
4 Series Solution of Linear ODEs

501

Problem 8
...
Show that the series solutions of the DE
8x2 y00 C 10xy0


...
32
...
33
...
3x C 1/ y0 C y D 0
are
1
X
...
3x/2

...
34
...
]

8
...
3 Special Cases
The procedure described above always gives the general series solution of the
differential equation (8
...
If s2 D s1 C m, where m is a positive integer,

502

8 Ordinary Differential Equations

including zero, only one series solution y1
...
Here we shall consider this
case in some detail
...
x/ and q
...
x/ D
q
...
x

p
x

x0

2

x0 /

1

2

C

C p0 C p1
...
x

C q0 C q1
...
67)

x 0 /2 C

x0 / C q2
...
68)

These expressions take account of the fact that the singular point is regular and that
the limits
lim
...
x/ and

x!x0

lim
...
x/

both exist, as otherwise the criteria at the beginning of Sect
...
4
...
Correspondingly we shall consider the differential equation

...
x x0 / C p0
...
x x0 / C q0
...
38) upon multiplication on
...
x/ and q
...


(8
...
35
...
x/ D
...
x

x0 / C c2
...
(8
...
Then, collecting terms with like powers of x and setting the
coefficients to zero, derive the following equation for the number s:
f
...
s

1/ C p 1 s C q

2

D s
...
70)

Note that above we have defined a function f
...
Correspondingly, show that
the two roots of this quadratic equation, s1 and s2 , are related via:
s1 C s2 D 1

p 1:

(8
...
4 Series Solution of Linear ODEs

503

Problem 8
...
s C 1/ C p0 s C q

1

D 0;

c2 f
...
s C 1/ p0 C q 1  C p1 s C q0 D 0:
In fact, show that an equation for cn (n D 1; 2; 3; : : :) generally has the form:
cn f
...
s/ D 0 ;

n D 1; 2; 3; : : : ;

(8
...
s/ D ˛n s C ˇn is a linear polynomial in s with the coefficients ˛n and
ˇn which depend on the expansion coefficients of p
...
x/, as well as on
coefficients c1 ; : : : ; cn 1
...
(8
...
s C n/ ¤ 0 for any integer n D 1; 2; 3; : : :
assuming either s D s1 or s D s2
...
Indeed, assume that the smallest
root is s2 and the largest s1 D s2 C m, where m is a positive integer
...
70), we have f
...
s2 / D f
...
When considering
Eq
...
72) for the coefficients cn corresponding to the solution associated with s1 , we
need f
...
This condition is valid
since there could only be two roots of the quadratic equation (8
...
Consider now if it is possible to construct in the same way
a solution corresponding to s2 D s1 m
...
(8
...
s2 C n/ D f
...
It is seen
that for n D m the function f
...
s1 / D 0 and hence the corresponding
coefficient cn cannot be determined, i
...
the method fails
...
We can also include the case of m D 0 here as only one value of s is available
(as s1 D s2 ) and hence only one generalised series solution can be constructed
...
1 which states that the
second linearly independent solution can always be found using the formula
 ˆ
Ã
ˆ
1
y2
...
x/
exp
p
...
73)
y2
...
x/
...
67) for p
...
x/ D
...
x

x0 / D us1 P1
...
74)

504

8 Ordinary Differential Equations

for the first solution associated with s1 , to investigate the integral in Eq
...
73) and
hence find the general form of the second solution
...
(8
...
u/ is a Taylor expansion of some function of u
...
(8
...
We can write:
ˆ
p
...
u/;

where P2
...
Several other such functions will be introduced below, we shall
distinguish them with a different subscript
...
73) now reads:
 ˆ
exp

Ã
p
...
P2
...
u/;

so that the integrand in Eq
...
73) becomes:
1
exp
y2
1

 ˆ

Ã
u p 1 P3
...
u/du D 2s 2
u 1 P1
...
2s1 Cp

1/

0


...
u/

2u

2

C

:

But, because of Eq
...
71), we have
2s1 C p

1

D 2s1 C
...
(8
...
u/
0
1
p
...
u/du du D
m 1

C
D

 ˆ

ln u C u

C
m

m

h

ln u C

mC1 u

0

m

1

m

1

C

uC

1 =
...
u/:

;

8
...
x/ D us1 P1
...
u/ D

mu

s1

P1
...
u/ ;

and hence we can finally write returning back to x:
y2
...
x/ ln
...
x

x0 /s2 P6
...
75)

We see that the second solution should have, apart from the usual second term
with s2 , also an additional term containing the product of the first solution and the
logarithm function
...
x/
...
36
...
(i) First of all, show
that only one value of s D 0 exists
...
x/ D 1

x6
C
22 42 62

D

1
X
...
rŠ/2

:

(iii) Write the second solution in the form K0
...
x/ ln x C P
...
x/ already satisfies it, and hence derive the
following inhomogeneous equation for the function P
...
x/:

(iv) Finally, obtain a Taylor series expansion for P
...
x/ D

1
X
rD1

pr x2r D

x2
4

3x4
11x6
C
128 13824

with

pr D

pr 1
...
rŠ/2 r

where p0 can be set to zero
...
x/ contains only even powers of x
...
5 Examples in Physics
8
...
1 Harmonic Oscillator
In many physical problems it is necessary to find a general solution of the following
second order linear ODE
2
y00 C y0 C !0 y D Fd sin
...
76)

with respect to the function of time y
...

Depending on the physical problem, the actual meaning of y
...

A natural example of a physical problem where this kind of equation plays the
principal role is, for instance, a harmonic oscillator with friction subjected to a
sinusoidal external force; we shall be calling this a forced harmonic oscillator
...
g
...
4
...
4)
...
8
...
It contains a resistance R, an inductance L and a capacitance
C, and a voltage applied to the circuit is some sinusoidal function of time, e
...

V
...
!t/
...
t1 /dt1 D V
...
!t/ ;
t0

where i
...
Differentiating both sides of the equation
above, we get:
L

1
d2 i
di
C R C i D !V0 sin
...
76) with !0 D 1= LC, D
R=L and Fd D !V0 =L
...
The applied
voltage induces oscillations with a different frequency !
...

Fig
...
1 Electrical circuit
with a resistance R, an
inductance L and a
capacitance C connected to a
voltage source V
...
5 Examples in Physics

507

8
...
1
...
77)

Its solution,
y
...
!0 t/ C B cos
...
Indeed, after some
rearrangements,
Ä
p
B
A
A2 C B2 p
sin
...
!0 t/
A2 C B2
A2 C B2
p
p
D A2 C B2 Œcos
...
!0 t/ C sin
...
!0 t/ D A2 C B2 sin
...
t/ D

(8
...
Therefore, the solution of the homogeneous equation (8
...
t/ D D sin
...
79)

with two arbitrary constants: D is called the amplitude and
corresponds to, e
...
, a self-oscillation in a circuit
...
This solution

8
...
1
...
!t/ :

(8
...
In this case the harmonic oscillation (8
...

Problem 8
...
Using the trial function y D B1 sin
...
!t/ for
2
Eq
...
80), show that B1 D Fd = ! 2 !0 and B2 D 0, so that the general
solution becomes
y
...
!0 t C /

Fd
!2

2
!0

sin
...
81)

508

8 Ordinary Differential Equations

We see from the above problem that there are two harmonic motions: one due
to self-oscillation (angular frequency !0 ) and another due to applied excitation that
happens with the same frequency ! as the excitation signal itself; its amplitude B1
depends crucially on the difference between the squares of the two frequencies
...
e
...

If the two frequencies are very close, but are not exactly the same, say ! D !0 C
with small , then the motion overall represents beats
...
e
...
0/ D y0
...
Then, from Eq
...
81)
it follows that
D sin
...
/

Fd !
D 0:
2
! 2 !0

Note that D cannot be equal zero as in this case the second equation above would
not be satisfied
...
t/ D
D

Fd
Œ! sin
...
!0 C /2

2
!0

!0 sin
...
!0 C / sin
...
!0 t C t/

Fd
Fd
Œsin
...
!0 t C t/
Œ!0 sin
...
!0 t C t/ D
2
2!0
2!0
tÁ
tÁ
Fd
Fd
tÁ
sin
sin
cos !0 t C
'
cos
...
82)
!0
2
2
!0
2

'

To obtain the first expression in the last line we have used the trigonometric
identity (2
...
The signal y
...
(8
...
t/ cos
...
t/ sin
...
The effect of
this is demonstrated in Fig
...
2 and is called amplitude modulation
...
When the excitation frequency
is the same as the self-oscillation one, the trial function for the particular solution
has to be modified (called a “resonance”, which we talked about in Sect
...
2
...
1)
...
5 Examples in Physics

509

Fig
...
2 Amplitude modulation: Eq
...
82) plotted with

Fd =!0 D 1,

D 0:004 and !0 D 0:2

Problem 8
...
Using the trial function
yp
...
!0 t/ C B2 cos
...
(8
...
t/ D D sin
...
!0 t/ :
2!0

Fd =2!0 ,

(8
...
(8
...
The frequency !0 is sometimes called the resonance
angular frequency
...
(8
...
t=2/
!
when
D
2
2
t=2
2

! 0:

Note also that at resonance the second term in the solution (8
...
!t/ in Eq
...
80) and
cos
...
(8
...
Therefore, at resonance there is a shift of
=2 in phase
between the driving force and the oscillating signal following it
...
5
...
3 Harmonic Oscillator with Friction
Now let us consider a self-oscillating system with friction (e
...
the resistance R is
present in the circuit in Fig
...
1):
2
y00 C y0 C !0 y D 0:

(8
...
85)

and hence there are three cases to consider
...
e
...
=2/2 , and the
general solution is then
2

Under-Damping: If

y
...

D Ae

=2Ci!/t
t=2

C C2 e
...
!t C / ;

=2 i!/t

De

t=2

C1 ei!t C C2 e

i!t

(8
...
In this case
the motion is easily recognisable as an oscillation with frequency w (which may be
very close to !0 if
!0 ); however, its amplitude is exponentially decreasing as
shown in Fig
...
3(a) providing an exponentially decaying envelope for the sinusoidal
motion
...
8
...


8
...
e
...
t/ D C1 ep1 t C C2 ep2 t :
The oscillations are not observed, the motion comes to a halt very quickly
...
8
...

2
Critical Dumping: When 2 4!0 D 0, i
...
if the friction value is transitional
between the two regimes, the two roots p1;2 coincide and the solution has the form

y
...
C1 C C2 t/ e

t=2

;

and is similar in shape to the over-dumped case
...
5
...
4 Driven Harmonic Oscillator with Friction
Finally, we shall consider the most common case of the sinusoidally driven dumped
harmonic oscillator described by Eq
...
76)
...
t/ was
considered in the previous subsection and corresponds to either under-dumped,
over-dumped or critically dumped case
...
t/ dies away either slowly or quickly (depending on the value of
the friction )
...
The particular solution will survive until, of course, the
driving force ceases to be applied
...
39
...
!t/ C B cos
...
(8
...
!/2

and B D

Fd !
2 C
...
8
...
1
...
(8
...
t/ D D sin
...
87)

512

8 Ordinary Differential Equations

where the amplitude is
DD

p
A2 C B2 D q

and the phase

Fd
2 C
...
!/2

(8
...
89)

Thus, the signal remains sinusoidal with the frequency of the driving force; it is
called the steady-state solution since it continues to hold without change as long as
the driving force is acting
...
8
...
1
...
Therefore, we may anticipate that the amplitude of the
steady-state motion will depend critically on the interplay between the two frequencies ! and !0
...
8
...

Problem 8
...
More precisely, show that D is maximum when ! is equal to
the resonance frequency
r
!res D

2
!0

2

2

:

(8
...

The phase as a function of ! is shown in Fig
...
4(b)
...


Fig
...
4 Amplitude D (a) and the phase (b) of the steady-state solution (8
...
Here Q D !0

8
...
5
...
The
equation of motion for the drop is
m

dv
D mg
dt

v;

where
is a friction coefficient (due to air resistance), g the gravity constant
(acceleration in the gravity field of the Earth), v the particle vertical velocity
(positive if directed downwards)
...

This ODE can be solved by rearranging terms and integrating both sides:
ˆ
ˆ
ˆ
m
dv
dv
dv
D dt H)
D dt H)
D t C1
g mv
g mv
v gm
Â
Ã
gm
m
gm
ln v
D Ce t=m ;
D t C C1 H) v
H)
where C D e
the drop is

C1 =m

is another constant
...
t/ D Ce

t=m

C

gm

:

The particular solution is now obtained if we find the constant C from the additional
information given about the velocity, i
...
0/ D v0
...
0/ D v0 D C C

gm

H) C D v0

gm

:

If v0 D 0, then C D gm= , so that we finally obtain:
v
...
Then the velocity is increased;
however, after some time the acceleration slows down (due to air resistance) and the
velocity approaches a constant value of v1 D gm= , called terminal velocity
...
Note that
it is positive since it is directed downward
...
This
will be just another arbitrary constant!

514

8 Ordinary Differential Equations

Fig
...
5 For the derivation of the Tsiolkovsky equation
...
t/ of fuel,
while (b) after time t when the mass m of the fuel is burned off, the mass of the fuel left will
be m m
...
5
...
We consider an
idealised rocket of initial mass M0 D MR C m0 , where MR is the mass of the
rocket without fuel and m0 is the initial mass of all fuel (in fact, normally the fuel
determines the initial mass at the launch, i
...
m0
MR )
...
If v
...
t/ D MR C m
...
t/ is the mass of the fuel left at this time), then the equation of
motion of the rocket will be given by the second Newton’s law P=t D Fext ,
where P is the change of the whole system momentum over time t and Fext is an
external force
...
8
...
t C t/

P
...
M

m/
...
u

v/ m

Mv;

where u v stands for the actual velocity of the fuel in a fixed coordinate frame of
the Earth (note that u is the relative velocity of the escaping fuel with respect to the
rocket)
...

Balancing this change of the momentum with the external force of Earth gravity,
Mg, acting in the opposite direction to the (positive) direction of the rocket
movement, we obtain:

8
...
91)

which is the equation of motion sought for
...
Assuming,
for instance, that it is burned with the constant rate proportional to the initial
mass of the rocket, dM=dt D
M0 with some positive , we write dv=dt D

...
dM=dt/ D
M0
...
0/ D 0 and M
...
M0
M0

M
g
C

...
The maximum velocity is
achieved when the whole fuel is exhausted, in which case M D MR , giving
vmax D u ln

MR C m0
MR

g

m0
m0
' u ln
MR C m0
MR

g

;

where in the last passage we used the fact that most of the mass of the rocket is due
to the fuel
...
g
...


8
...
4 Distribution of Particles
Let us consider a random distribution of particles in a volume as shown in
Fig
...
6(a)
...
r/dr of finding a
closest particle to a given particle
...
Then, the probability to find a particle anywhere within that sphere is given

516

8 Ordinary Differential Equations

Fig
...
6 (a) To the derivation of the probability to find a closest neighbour around a chosen particle
in a volume: we draw a spherical shell of radius r and width dr around the particle
...
r/ for three values of the particles concentration n D 0:1; 0:3; 0:8

by summing up all such probabilities for all radii from zero to r:
ˆ

r

P
...
r/ D 1 P
...
Here dV D 4 r2 dr if the shell volume
and n is the particle concentration
...
r/dr
...
r/dr D Œ1

2

P
...
r/
D 1
r2

ˆ
0

r

p r0 dr0 4 n:
(8
...
r/ we now differentiate Eq
...
92) with
respect to r which yields:
Ä
d
d pÁ
D4 n
1
dr r2
dr

ˆ

r

p r0 dr0

H)

0

1 dp
r2 dr

2p
D
r3

4 np;

which can be rewritten as
dp
D
dr

Â

2
r

Ã
4 r2 n p:

(8
...
5 Examples in Physics

517

This ODE is separable and can be easily integrated:
Ã
ˆ Â
ˆ
2
dp
D
4 r2 n dr H) ln p D 2 ln r
p
r
Â
Ã
4 3
r n :
H) p D Cr2 exp
3

4 3
r nCC
3

The arbitrary constant C is determined by normalising the distribution as there will
definitely be a particle found somewhere around the given particle:
ˆ
0

1

p
...
r/ D 4 nr2 exp

Â

Ã
4 3
r n :
3
(8
...
8
...
It shows that initially p
...
e
...
r/) which is reduced with the increase of the concentration
...


Problem 8
...
Show that the two-dimensional analogue of formula (8
...
r/ D 2 nr exp

nr2 :

8
...
5 Residence Probability
Here we shall derive the residence kinetics, i
...
we shall consider a system, which
can change its current state (i
...
make a transition to another state) with a certain
rate R
...
The states are separated by an energy barrier, and
hence both are stable; the molecule may change its state, however, by jumping from
one state to another, but these events are not deterministic, i
...
they happen with
a certain probability depending on the transition rates between the two states and
the temperature
...

Suppose the lowest energy state of the adatom is above a surface atom; since there
are many surface atoms, the adatom may occupy many adsorption positions
...
The rate of such jumps depends on
the energy barrier and temperature
...
t/ for a system to remain in the current state
by the time t
...

If R is the rate (probability per unit time) for the system to move away (escape)
from the given state it is currently in, then the probability for this to happen over
time dt is Rdt
...

Therefore, the probability P
...
t/, and (ii) the system
remains in the same state over the consecutive time dt, the probability of this is
1 Rdt
...
t C dt/ D P
...
1
H)

H)

Rdt/
dP
D
dt

P
...
t/ D RP
...
t/ D C exp
...
Since at t D 0 our system
is in the current state with probability one, then C D 1
...
t/ D exp
...
Eventually the probability
to remain in the current state approaches zero, i
...
the system will definitely escape
at some point in time
...
42
...
t/
...
t/ D exp
0

Ã

t

R
...
5
...
g
...
The state A lies higher in energy than B (i
...
the state A is less
energetically favourable than B, so that there should be on average more of defects
B than A), but it is relatively stable
...
5 Examples in Physics

519

Fig
...
7 Double well
potential energy surface with
two minima: A and B
...
The exponential term gives a probability to overcome the barrier during
a single jump (attempt) that is increased with an increase of temperature T or a
decrease of the energy barrier (AB or BA), while gives the number of attempts
per unit time to jump
...
8
...

There will be certain concentrations nA and nB of defects in the two states at
equilibrium that should be established over some time
...
Indeed, the time dependence of the defects concentrations in
the two states should satisfy the following “equations of motion”:
dnA
D
dt

wAB nA C wBA nB and

dnB
D wAB nA
dt

wBA nB :

(8
...
Firstly, there is an incoming flux of defects wBA nB
jumping from the state B which is proportional to the existing concentration nB of
defects in state B; this term works to increase nA (the “gain” term)
...
The meaning of
the second equation is similar, but this time the balance is written for the defects in
the other state
...
g
...
n
dt

nA / :

(8
...
9):
dnA
C wnA D wBA n
dt

with

w D wAB C wBA :

Therefore, using the general method of Sect
...
1
...
t/ D exp

wdt D ewt

and
Â
nA
...
0/ D n (all defects were in the A state), then, applying this initial
condition, we obtain:
nD

wBA n
C C H) C D n
w

wBA n
;
w

and thus
nA
...
It shows that the concentration of defects in state A is reduced
over time approaching the equilibrium value (at t D 1) of nA
...

Consequently, the concentration of defects in state B
nB
...
t/ D Œn

nA
...
1/
...
(8
...


Index

A
Abel theorems on power series, 442
Abel’s formula, 469
absolute convergence, 429
absolute value estimate for integral, 188
adiabatic compressibility, 293
air resistance, 513
algebra, 14
alternating harmonic series, 428, 445
alternating series, 426
Ampère law, 407
amplitude modulation, 508
analysing functions, 160
analysing functions: steps, 169
Approximations of functions, 136
approximations of functions, 155
Archimedean spiral, 60
Archimedes law, 387
area of figure via line integral, 350
area of plane figure, 242
area of sphere, 361
area of surface, 360
arithmetic progression, 42
Arrhenius rate formula, 519
asymptotes of function, 167
average value theorem for integral, 187

B
beats, 508
Bessel equation, 505
binomial coefficients, 45, 141, 153
binomial formula, 44
biorthogonal basis, 34

Bolzano-Cauchy first theorem, 113
Bolzano-Cauchy second theorem, 114
Bose-Einstein statistics, 441
boundary conditions, 456
Brownian motion, 333
Brownian particle, 226

C
Cartesian coordinates, 10
center of mass, 252
changing variables in double integral, 327
changing variables in triple integral, 335
Chebyshev polynomials, 452
circle, 11, 22
classical nucleation theory, 173
collinear vectors, 26
combinatorics, 49
complementary solution of ODE, 474
complex numbers, 37
composition functions, 67
concave behaviour of function, 163
conditional convergence, 429
connection equations, 335
conservation of charge, 386
conservative field, 394, 406
conservative vector field, 394
continuity equation, 384, 408, 409, 412, 415
continuity of function of many variables, 268
continuity of functional series, 437
continuous function, 108
convex behaviour of function, 163
coplanar vectors, 26
cosine function, 18

© Springer Science+Business Media, LLC 2016
L
...
1007/978-1-4939-2785-2

521

522
cosine theorem, 20
critical dumping in harmonic oscillator, 511
critical points of function, 161
cross product of vectors, 28
crossing of line and plane, 65
crossing of two lines, 64
crossing of two planes, 66
cumulants, 450, 452
curl, 372, 391
curl via determinant, 375
curved lines, 58
cylinder coordinates, 337

D
Darboux sums, 182, 319
DE, 455
de Moivre’s formula, 98
decimal representation, 7
defects in crystal, 518
del operator, 296
derivative chain rule, 131
derivative geometric interpretation, 125
derivative of function, 125
derivative of inverse function, 132
derivative quotient rule, 130
Derivatives main theorems, 127
derivatives of composite functions, 280
Derivatives of higher orders, 140
determinant of matrix, 23
determinant of quadratic equation, 18
differential, 127
differential equations, 455
differentiation of functional series, 439
differentiation of power series, 445
diffusion constant, 411
diffusion equation, 273, 411
directed surface element, 367
directional derivative, 294
discontinuity of first kind, 110
discontinuity of second kind, 110
displacement current, 409
distance between lines, 65
distance between parallel planes, 66
distance between point and line, 62
distance between point and plane, 62
distribution of particles in volume, 515
divergence, 381, 388
divergence theorem, 381
division theorem for numerical sequences, 74
dot product of vectors, 27
double integral, 315
driven harmonic oscillator with friction, 511
driven oscillator without friction, 507

Index
E
eccentricity of ellipse, 22
electric field, 404
electromagnetism, 404
electrostatic field, 256
electrostatic potential, 365
electrostatics, 404
elementary geometry, 11
ellipse, 22
ellipsoid, 262
elliptic cone, 363
enthalpy, 291
entropy, 290
equation of line, 56
equation of motion, 513
equation of plane, 59
equation of state, 291
equilateral triangle, 13
error function, 453
Euler’s substitutions, 215
Euler-Mascheroni constant, 425
even function, 20, 178
exact differential, 277, 351, 379, 458
exact differential equations, 458
exhaust velocity, 514
expansion into solenoidal and conservative
fields, 401
exponential function, 90, 134
exponential functions, 112
extremum of function of many variables, 301

F
Faraday law, 408
ferromagnetic material, 172
Fick’s first law, 411
finite series, 41
fluid in vortex, 393
flux integral, 367
flux of liquid through surface area, 367
focal points of ellipse, 23
foci of ellipse, 23
force, 254
Fourier’s law, 412
Fresnel integrals, 453
friction coefficient, 513
Frobenius method, 492
function, 16
functional series, 434

G
Gamma function, 229
gas distribution function, 281

Index
Gauss theorem, 381
Gauss theorem variants, 383
Gaussian integral, 332
general power function, 84, 132
general power functions, 111
general solution of differential equation, 455
generalised binomial coefficients, 153, 450
geometric progression, 417
geometrical progression, 41
geometrical series, 78
Gibbs free energy, 291
gradient, 294
Green’s formula, 346

H
Hamiltonian, 282
harmonic motion, 507
harmonic oscillator, 441, 506
harmonic oscillator equation, 495
harmonic oscillator with friction, 510
harmonic series, 418
Hartree potential, 433
heat capacity, 292, 413
heat transport equation, 411
Heaviside function, 67
Helmholtz free energy, 291
Hermite polynomials, 452
homogeneous differential equations, 462
homogeneous equation, 467
homogeneous linear differential equations with
constant coefficients, 471
horizontal asymptote, 167
Hydrodynamic equation, 413
hyperbolic functions, 91, 112, 135
hyperbolic paraboloid, 264

I
ideal liquid, 413
ideal rocket equation, 514
improper integrals, 226
indefinite integral formula, 190
indefinite integrals, 188
indicial equation, 492, 494
induction, 4
inequalities, 15, 52
infinite geometrical progression, 78
infinite limits, 115
infinite numerical sequence, 71
infinite numerical series, 77
infinite series, 417
infinite series integral test, 423
infinite series ratio test, 421

523
infinite series root test, 422
inflection point of function, 165
inhomogeneous equation, 467
inhomogeneous linear differential equations,
474
initial conditions, 455
integrals depending on parameter, 223
integrating factor, 460, 465
integration by changing variables, 195, 220
integration by parts, 198, 222
integration of functional series, 439
integration of irrational functions, 213
integration of power series, 445
integration of rational functions, 204
integration of rational functions with ex , 212
integration of trigonometric functions, 209
inverse function, 69
inverse trigonometric functions, 98, 112, 135
irregular singular point, 491
isosceles triangle, 13
isotherm of gas, 171
isothermal compressibility, 293
ISP, 491
iterated integral, 317

J
Jacobian, 327, 335, 357, 361

K
kinetic Monte Carlo, 518
Kirchoff’s law, 506
Kronecker symbol, 34

L
L’Hôpital’s rule, 158
Lagrange formula, 152, 190, 301
Lagrange multipliers, 310
Laguerre polynomials, 452
Landau-Ginzburg theory, 172
Langevin equation, 226
Laplace equation, 406
Laplace operator, 412
Laplacian, 287, 390
least square method, 307
Legendre equation, 497
Legendre polynomials, 452, 500
Leibnitz convergence test, 427
Leibnitz formula, 226
Leibniz formula, 141, 154, 155
length of circle, 340
length of curved line, 238

524
length of parabola, 342
length of spiral, 342
Levi-Civita symbol, 30
limacons of Pascal, 60
limit of composition of functions, 107
limit of function, 100
limit of inverse function, 108
limit of numerical sequence, 72
limit theorems, 105
limits at infinity, 115
line integral, 338
line integral for scalar field, 338
line integral for vector field, 342
linear first order differential equations, 464
linear operator, 475
linear second order differential equations, 468
logarithmic function, 91, 112, 133
logic, 3
Lorenz force, 407

Index
number e, 86, 155
number , 11, 156
numbers, 5

O
oblique asymptote, 168
odd function, 20, 178
ODE, 455
one-pole hyperboloid, 264
one-side limit of function, 104
operator, 143, 474
ordinary differential equations, 455
ordinary point of differential equation, 486
orientable surface, 359
orthogonal vectors, 28
Ostrogradsky-Gauss theorem, 381
over-dumping in harmonic oscillator, 511

M
Maclaurin formula, 299, 448
Maclaurin series, 448
Maclaurin’s formula, 152
magnetic charges, 410
magnetic field, 257, 404
major axis of ellipse, 22
mass calculation, 252
mass conservation, 385
maximum of function, 161
Maxwell relations, 292
Maxwell’s equations, 404
Möbius strip, 359
mechanical equilibrium, 297
method of undetermined coefficients, 476
method of variation of parameters, 480
minimum of function, 161
minor axis of ellipse, 22
moment of inertia, 337
moments, 450, 452
monotonic function, 69
multinomial expansion, 52

P
parabola, 17
paraboloid, 364
parallel lines, 12
parallelogram, 12
partial derivative, 268
partial sum of numerical series, 77
particular integral of ODE, 474
particular solution of differential equation, 456
particular solution of ODE, 474
Pascal’s triangle, 44
periodic function, 19
points of discontinuity, 268
Poisson’s equation, 401
polar coordinates, 57, 329
polynomial, 16
polynomials, 79, 110
power series, 441
pressure, 255, 290
principal value Cauchy, 235
product theorem for numerical sequences, 74
proving by contradiction, 4
Pythagorean theorem, 18

N
Newton’s equation of motion, 483
Newton-Leibnitz formula, 228, 232
Newton-Leibniz formula, 191
non-linear second order differential equations,
483
normal to cylinder, 357
normal to sphere, 362
normal to surface, 275, 297, 357

R
radians, 11
radius of convergence, 443
random process, 450
rational function, 17
rational functions, 80
rational functions decomposition, 81
reciprocal basis, 34
rectangle, 21

Index
recurrence relation, 41, 154, 492, 494, 496, 498
regular singular point, 491
remainder term, 148
reminder term, 448
residence kinetics, 517
residence probability, 517
resonance, 478, 508, 512
resonance frequency, 509, 512
rhodonea curve, 60
Riemann definite integral, 177
Riemann integral sum, 176
right triangle, 13
right-hand rule, 359
roots of polynomials, 79
RSP, 491

S
saddle point, 161, 163, 265
scalar field, 294
scalar potential, 394
scalar product of vectors, 27
Schrödinger equation, 471
second order phase transition, 172
separable differential equations, 456
series solution of linear differential equations,
486
series with positive terms, 420
sign function, 68
similar triangles, 13
simply connected region, 346, 351
sine function, 18
sine theorem, 20
singular point of differential equation, 486
smooth surface, 356
snail curve, 60
solenoidal field, 397
solenoidal vector filed, 394
solids of revolution, 247
sphere, 262
spherical coordinates, 58, 336, 361
stationary points, 302
steady-state solution, 512
steepest descent, 297
Stokes’s theorem, 371
subsequence theorem for numerical sequences,
75
sufficient condition of differentiability, 271
surface, 355
surface charge density, 365
surface integral for scalar field, 364
surface integral for vector field, 366
surface integrals, 355
surface of cone, 249

525
surface of elliptic cone, 363
surface of revolution, 248
surface of sphere, 249
surface of torus, 249
surfaces of second order, 262

T
tangent plane, 275
Taylor expansion, 504
Taylor series, 446
Taylor’s formula, 146
Taylor’s theorem, 151, 299
temperature, 290
terminal velocity, 513
theorem on mixed derivatives, 273
theorems for numerical sequences, 73
thermal expansion coefficient, 293
thermal pressure coefficient, 293
thermodynamic potentials, 291
thermodynamics, 290
three dimensional space, 26
three-leaf rose, 60
torus, 249
total derivative, 414
total derivative of function, 283
transition probabilities, 519
transition rates, 519
transition state, 265
triangle, 12
trigonometric functions, 18, 93, 111, 134
triple integral, 333
Tsiolkovsky formula, 514
two-pole hyperboloid, 264
two-sided surface, 359

U
uncertainties when taking limit, 117
under-damping in harmonic oscillator, 510
uniform convergence, 323, 435
uniform convergence of improper integral, 323
uniform convergence of power series, 445
uniqueness theorem for numerical sequences,
76
unit base vectors, 28
unit step function, 67

V
van der Waals equation, 171
vector field, 342
vector potential, 394
vector product of vectors, 28

526
vector-columns, 36
vectors, 26
vectors-rows, 36
vertical asymptote, 167
volume of bodies, 245
volume of cone, 246
volume of cylinder, 245
volume of ellipsoid, 246
volume of solid of revolution, 248

Index
volume of sphere, 246, 320, 330, 334, 336
volume of tetrahedron, 247
W
wave equation, 410
Weierstrass convergence test, 436
work done, 255, 342
Wronskian, 469, 473, 482



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Title: Mathematics for Natural Scientists Fundamentals and Basics
Description: limit of a function is one of the fundamental notions in mathematics. Several cases need to be considered.

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