Search for notes by fellow students, in your own course and all over the country.
Browse our notes for titles which look like what you need, you can preview any of the notes via a sample of the contents. After you're happy these are the notes you're after simply pop them into your shopping cart.
Document Preview
Extracts from the notes are below, to see the PDF you'll receive please use the links above
Introductory Calculus Notes
Ambar N
...
Sengupta 11/6/2011
Contents
Preface
...
10
1 Sets: Language and Notation
1
...
1
...
1
...
1
...
5 Integers and Rationals
...
6 Cartesian Products
...
7 Mappings and Functions
...
8 Sequences
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
1 The Real Line
...
2 The Extended Real Line
...
1 Upper Bounds and Lower Bounds
...
2 Sup and Inf: Completeness
...
3 More on Sup and Inf
...
1 Intervals
...
2 Neighborhoods
...
3 Types of points for a set
...
4 Interior, Exterior, and Boundary of a Set
...
5 Open Sets and Topology
...
6 Closed Sets
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
Sengupta 11/6/2011
4
...
8
Open Sets and Closed Sets
...
40
5 Magnitude and Distance
5
...
5
...
5
...
5
...
1 Limits, Sup and Inf
...
2 Limits for 1/x
...
3 A function with no limits
6
...
6
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
41
41
42
43
43
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
45
46
49
51
52
54
7 Limits: Properties
7
...
7
...
7
...
7
...
7
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
8 Trigonometric Functions
8
...
8
...
3 Reciprocals of sin, cos, and tan
...
4 Identities
...
5 Inequalities
...
6 Limits for sin and cos
...
7 Limits with sin(1/x)
...
8 Graphs of trigonometric functions
...
9 Postcript on trigonometric functions
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
9 Continuity
85
9
...
85
9
...
85
DRAFT Calculus Notes 11/17/2011
9
...
4
9
...
6
10 The
10
...
2
10
...
4
10
...
Two examples using Q
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
Intermediate Value Theorem
Inequalities from limits
...
Intermediate Value Theorem: a second formulation
Intermediate Value Theorem: an application
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
91
92
93
94
95
96
11 Inverse Functions
99
11
...
99
11
...
102
11
...
103
12 Maxima and Minima
107
12
...
107
12
...
110
12
...
111
13 Tangents, Slopes and Derivatives
13
...
13
...
13
...
13
...
13
...
13
...
13
...
13
...
√
13
...
13
...
13
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
114
...
117
...
120
...
121
...
123
...
125
14 Derivatives of Trigonometric Functions
127
14
...
127
14
...
129
6
Ambar N
...
3 Derivative of tan is sec2
...
1 Differentiability implies continuity
...
1 Using the sum rule
...
2 Using the product rule
...
3 Using the quotient rule
...
1 Initiating examples
...
2 The chain rule
...
1 Sums
...
2 Products
...
3 Quotients
...
141
...
143
19 Proving the Chain Rule
145
19
...
145
19
...
146
20 Using Derivatives for Extrema
20
...
20
...
20
...
20
...
Exercises on Maxima and Minima
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
151
152
153
155
160
164
21 Local Extrema and Derivatives
167
21
...
167
Review Exercises
...
1 Rolle’s Theorem
...
2 Mean Value Theorem
...
3 Rolle’s theorem on R∗
...
171
...
174
DRAFT Calculus Notes 11/17/2011
23 The
23
...
2
23
...
175
Negative derivative and decreasing nature
...
179
24 Differentiating Inverse Functions
181
24
...
182
25 Analyzing local extrema with higher derivatives
185
25
...
185
25
...
188
26 Exp and Log
26
...
26
...
26
...
26
...
26
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
1 Convex and concave functions
...
2 Convexity and slope
...
3 Checking convexity/concavity
...
4 Inequalities from convexity/concavity
...
5 Convexity and derivatives
...
6 Supporting Lines
...
7 Convex combinations
...
1 Examples
...
2 Proving l’Hospital’s rule
...
225
...
228
...
1 From areas to integrals
...
2 The Riemann integral
...
3 Refining partitions
...
4 Estimating approximation error
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
233
233
235
237
239
8
Ambar N
...
5 Continuous functions are integrable
...
6 A function for which the integral does not exist
...
7 Basic properties of the integral
...
1
30
...
3
30
...
5
Fundamental Theorem of Calculus
Fundamental theorem of calculus
...
Using the fundamental theorem
...
Revisiting the exponential function
...
1 Riemann sums for 1 dx
x2
31
...
31
...
31
...
31
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
245
...
246
...
252
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
257
257
260
261
264
266
32 Integration Techniques
32
...
32
...
32
...
4 Using trigonometric substitutions
...
5 Integration by parts
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
269
...
276
...
282
...
290
33 Paths and Length
33
...
33
...
33
...
33
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
324
DRAFT Calculus Notes 11/17/2011
9
Preface
These notes are being written for an introductory honors calculus class, Math
1551, at LSU in the Fall of 2011
...
(In
fact if I had to choose a subtitle for these notes, it would be ‘An Anticalculus-text Book’
...
We also avoid several stranger
aspects of the universe of calculus texts, such as counterintuitive notions of
what counts as ‘local maximum’ or obsessing over ‘convex up/down’, and
stay with practice that is consistent with the way mathematicians actually
work
...
We prove several results in sharper
formulations than seen in calculus texts
...
We follow a consistent strategy of using suprema
and infima, which form a running theme through the historical development
of the real line and calculus
...
For various corrections and comments I thank Justin Katz
...
Sengupta 11/6/2011
Introduction
There are two fundamental notions that led to the development of calculus
historically: (i) the measurement of areas of curved regions, and (ii) the study
of tangents to curves
...
Lengths of line segments are measured by comparing to a chosen ‘unit’
of length
...
This works very well for a rectangle
...
This leads naturally to
area of a rectangle = product of the lengths of its sides
...
This strategy fails when we think of a curved region, such as a disk
...
(Whether such slicing and reassembly of
regions is really possible, and in what sense, is a truly difficult problem
...
Consider non-overlapping squares lying
inside a disk
...
On the other hand, if we cover the disk with squares, which spill over to the
outside of the disk, and add up the areas of these squares we obtain an over
estimate of the area of the disk:
area of disk ≤ areas of polygonal regions covering disk
...
DRAFT Calculus Notes 11/17/2011
11
This idea, of pinning down a value by realizing it as being squeezed in
between overestimates and underestimates is an enormously powerful idea,
running all through the foundations of calculus
...
Returning to the disk, it turns out that there is indeed such a unique
value lying between the underestimates and the overestimates
...
Indeed, the ratio of the area of the disk to that of the square on the radius
is the fundamental constant denoted
π
...
Next, drawing a square
with width given by the diameter of the circle gives an overestimate for the
area: 22
...
Archimedes, working with a 96-sided polygon obtained the estimates:
10
1
3 <π<3
...
The value of π expressed as a decimal is:
π = 3
...
where the dots mean that the decimal continues endlessly
...
This expression, though
quite concrete, is frustrating in that there is no direct specificaton of, say,
what the 25-th decimal entry is
...
Sengupta 11/6/2011
or the enormously efficient but mysterious Ramanujan formula
π = √ P∞
2 2 n=0
9801
(4n)!(1103+26390n)
(n!)4 (396)4n
...
The number π, originating in geometry, appears in a vast array of contexts
in physics, chemistry, engineering and statistics
...
The great power of integral calculus is illustrated by the fact that it
turns the genius of Archimedes’ method into an utterly routine calculation
that a student of calculus can do in moments
...
It is, however, an especially abstruse area in mathematics
...
1
...
These objects are
called the elements of the set
...
This is a typical way of displaying
a set: list its elements, separated by commas, within the braces { and }
...
For instance,
{x : x is a prime number less than 10}
is the same as the set
{2, 3, 5, 7}
...
The notation
x∈y
13
14
Ambar N
...
Thus,
3 ∈ {2, 3, 5, 8}
...
Sets x and y are said to be equal, written
x=y
if every element of x is an element of y and every element of y is an elements
of x; in other words, sets are equal if they have the same elements
...
If we
happen to repeat elements in one display or if we display the elements in a
different order it does not change the set
...
1
...
(1
...
Notice that the set
{∅}
is not empty: it contains one element, that being ∅
...
Thus:
{∅} =
6 ∅
(1
...
DRAFT Calculus Notes 11/17/2011
15
In fact, {∅} has a name
...
(1
...
This, of course, is 2:
def
2 = {∅, 1}
...
4)
In this way we obtain the numbers
0 = ∅, 1, 2, 3,
...
Here we have identified 0 and the empty set, but conceptually one thinks of
0 as ‘the number of elements’ in the empty set rather than ∅
...
The negative numbers
−1, −2, −3,
...
In fact, virtually every object in mathematics is a set
...
It is also good to
keep in mind that our notion of numbers, both for counting and for ordering
(such as listing items as first, second, third, etc
...
1
...
16
Ambar N
...
Note that every set is a subset of itself:
x ⊂ x
...
Sometimes one gets confused between a ∈ b and x ⊂ y
...
For example,
3 ∈ {1, 3, 5}
but 3 is not a subset of {1, 3, 5}
...
But this is unusual
...
Here is a starter instance of this:
Proposition 1
...
1 The empty set is a subset of every set:
∅⊂x
for all sets x
...
We argue by contradiction
...
This would mean that ∅ contains some element that is not in x
...
Thus we have a contradiction, and so ∅ is in fact
a subset of x
...
4
17
Union, Intersections, Complements
The union of sets x and y is the set obtained by pooling together their
elements into one set
...
For example,
{1, 3, 5} ∪ {3, 5, 6, 7} = {1, 3, 5, 6, 7}
...
For example,
{1} ∪ {1, 2} ∪ {1, 2, 3} ∪
...
The intersection of sets x and y is the set containing the elements that
are both in x and in y, and is denoted
x ∩ y
...
The intersection can be empty of course:
{2, 4, 5} ∩ {1, 3, 7} = ∅
...
One can take intersections of more than two sets as well:
{1, 5, 3, 6} ∩ {2, 3, 4} ∩ {3, 8, 9} = {3}
...
Then the complement of any given subset A ⊂ X is the set of elements of X not in A:
Ac = {p ∈ X : p ∈
/ A}
...
5
Integers and Rationals
The numbers 0, 1, 2, 3,
...
(1
...
Sengupta 11/6/2011
Taking ratios of integers yields the rational numbers
...
For example, −34/15 is
rational
...
(1
...
Moreover, there is an order
relation on Q, telling us which of two given rationals is bigger
...
√This is a much larger
set than Q, as it contains irrational numbers such as 2 and π, in addition
to all the rational numbers
...
6
Cartesian Products
The displays {a, b} and {b, a} describe the same set, but sometimes we need
to express a appearing first followed by b
...
There are a number of ways to construct a set out of a and b, reflecting the
distinction between them; a simple (though certainly not obvious) strategy
is to define
(a, b) = {{a}, {a, b}}
...
7)
We can check that, with this definition,
(a, b) = (c, d) if and only if a = c and b = d
...
Next, from sets A and B we can construct the set of all ordered pairs (a, b),
drawing the first entry a from A, and the second entry b from B:
A × B = {(a, b) : a ∈ A,
b ∈ B}
...
8)
DRAFT Calculus Notes 11/17/2011
19
This is called the Cartesian product of A with B
...
The Cartesian product of a set A with itself is denoted A2 :
A2 = A × A
...
9)
Thus the plane, coordinatized by real numbers, can be modeled mathematically as
R2 = {(x, y) : x ∈ R, y ∈ R}
...
10)
1
...
This relation is not read as simply an equality of two quantities y and x3 +
x2 + 1, but rather as a procedure for computing one quantity from the value
of another:
given the value x = 2 we compute y = 23 + 22 + 1 = 13
...
Of course, the letters x and y are in themselves of no
significance; the same function is specified by
s = t4 + t2 + 1
...
specifies a ‘function of m’, where m runs over the positive integers
...
20
Ambar N
...
Note that to keep things unambiguous, f (t) should mean exactly one unique
value, and not multiple values
...
We can now turn to a formal definition that reflects this notion
...
If
(a, b) ∈ Gr(f ) we denote b by f (a):
b = f (a) means (a, b) ∈ Gr(f )
...
The term function is used normally, instead of mapping, in calculus
...
For
example, instead of saying the function, with domain and codomain the set
of real numbers and having graph
{(a, a2 ) : a ∈ R},
we often simply write
the function y = x2 ,
it being clear, unless otherwise spelled out, that x runs over all real numbers
...
1: Graph of y = x
y
y = x3 − x2 − 2x + 1
x
Figure 1
...
75) = −4
...
The graph of y = x is displayed visually as a straight line:
The graph of y = x3 − x2 − 2x + 1 is displayed visually as
Often in calculus, the codomain is clear from context (usually, the set
R of all real numbers) and we identify a function with its graph, ignoring
specification of the codomain
...
For example, for the function
y = x2
for all real numbers x,
the range is the set of all non-negative real numbers
[0, ∞)
...
Sengupta 11/6/2011
For the function 1prime we considered before, the range is the set {0, 1}:
Range(1prime ) = {0, 1}
...
This can be viewed as its graph:
{(x, y) ∈ R : x ∈ R, y ∈ R},
which is the circle of unit radius, centered at the origin (0, 0)
...
3: Graph of the circle x2 + y 2 = 1
...
8
23
Sequences
A function whose domain is the set
N = {1, 2, 3,
...
For example,
f : N → R : n 7→
1
n
is a sequence
...
In the preceding example we may describe it as
the sequence fn = 1/n,
or, more formally, as
the sequence (fn )n≥1 where fn = 1/n,
or, most simply, as
the sequence 1/n
...
The way to think about the sequence is to think of it as a list of its
values:
1
1
1
1
,
,
,
,
...
(There are infinitely many prime numbers and so this does indeed specify a
sequence
...
For example, the factorials
fn = n!
for all n ∈ {0, 1, 2,
...
24
Ambar N
...
Using these makes it possible to write many theorems
in a simpler way, without having a list of qualifiers of which situations need
to be excluded
...
1
The Real Line
The numbers 0, 1, 2,
...
From these it
is possible to construct negative numbers −1, −2,
...
Geometry leads us beyond the rationals and forces us to bring in other
numbers
...
P
corresponds to the
real number OP/OU
U
R
O
Figure 2
...
Sengupta 11/6/2011
Then for any point P on the line l we can think of the geometical concept
of the ratio
OP/OU,
where we take this to be negative if P is on the opposite side of O from
U
...
Thus they form a system of numbers called the real numbers
...
Similarly, we have points P for which OP/OU
is a rational such as −4/7
...
For example, consider a right angled triangle that has two sides of length
OU
...
It is a fact that 2 is not a rational number, in that there is
no rational number whose square is 2
...
The irrationals are also dense in the real line: between
any two distinct reals lies an irrational
...
2
The Extended Real Line
The extended real line is obtained by a largest element ∞, and a smallest
element −∞, to the real line R:
R∗ = R ∪ {−∞, ∞}
(2
...
We extend the order relation to R
by declaring that
−∞
(2
...
We define addition on R∗ as follows:
x + ∞ = ∞ = ∞ + x for all x ∈ R∗ with x > −∞
y + (−∞) = −∞ = (−∞) + y for all y ∈ R∗ with y < ∞
...
3)
(2
...
For multiplication we set
x · ∞ = ∞ = ∞ · x for all x ∈ R∗ with x > 0
x · (−∞) = −∞ = (−∞)·x for all x ∈ R∗ with x < 0
...
5)
(2
...
However, playing it very carefully, it
turns out to be useful to set
∞·0=0·∞=0
(−∞) · 0 = 0 · (−∞) = 0
...
7)
This convention is quite useful when we study integration theory, but it
should not be used in other parts of calculus, such as the study of limits
...
For example, 0/0 is undefined not because somehow we have arbitrarily decided not to define it, but any definition for it would be a dead end, of no
use elsewhere in mathematics
...
8)
whenever either side of these equations exist (that is we don’t have ∞+(−∞)
appearing)
...
Sengupta 11/6/2011
Chapter 3
Suprema, Infima, Completeness
In this chapter we examine a fundamental property of the real line that
distinguishes it from the rationals and that makes much of calculus possible
...
3
...
A point p ∈ R∗ is an upper bound of S if it lies to the right of S:
x≤p
for all x ∈ S
...
Note that 9 is also an upper bound
...
If we had restricted
ourselves to working only with real numbers then R would not have an upper
bound
...
Here is a mind twister for the empty set: 3 is an upper bound for ∅
...
Suppose 3 were not an upper
bound of ∅
...
29
30
Ambar N
...
Thus, 3 must
indeed be an upper bound for ∅
...
Now we turn to the flip side of the concept of upper bound
...
Thus, for example, for the set
(8, 9] ∪ [11, ∞)
all p ≤ 8 are lower bounds
...
The value −∞ is a lower bound for every subset of R∗
...
3
...
The smallest upper bound of S, that is the least upper bound
called the supremum of S and denoted
of S, is
sup S
...
The largest lower bound, that is the greatest lower bound, of S is called
its infimum and denoted
inf S
...
DRAFT Calculus Notes 11/17/2011
31
If you think about the empty set, a strange thing happens
...
Similarly,
inf ∅ = ∞
...
(3
...
If we want to stay within just the set of real numbers, the statement is a
little bit more complicated so as to rule out all the infinities:
If A ⊂ R is not empty and has an upper bound then it has a supremum,
(3
...
The completeness of R is a crucial property
...
Many of the great useful
results of calculus would fail on Q just for this reason
...
But in another, more constructive ,approach, where R is constructed out of set theory, completeness is a
property that is proved as a fundamental theorem about R
...
3
More on Sup and Inf
Consider a non-empty set S ⊂ R∗
...
Then any lower
bound of S is ≤ p and every upper bound of S is ≥ p:
any lower bound of S ≤ p ≤ any upper bound
...
32
Ambar N
...
If S contains just one point then the inf and sup coincide: for example,
inf{3} = 3 = sup{3}
...
(3
...
Consider sets
B ⊂ A ⊂ R∗
...
Any upper bound of A is ≥ all elements
of A and hence is ≥ all elements of B
...
In particular,
the least upper bound of A is an upper bound of B
...
So, of course,
the least upper bound of B is ≤ sup A
...
(3
...
(‘Decreases’ is in a lose sense here, as it may
happen that sup A is equal to sup B
...
Picking a smaller set increases the infimum, with qualifiers as before
...
5)
Chapter 4
Neighborhoods, Open Sets and
Closed Sets
In this chapter we study some useful concepts for studying the concept of
nearness of points in R∗
...
1
Intervals
An interval in R∗ is, geometrically, just a segment in the extended real line
...
More
officially, an interval J is a non-empty subset of R∗ with the property that
for any two points of J all points between the two points also lie in J: if
s, t ∈ J, with s < t, and if s < p < t then p ∈ J
...
Consider any point p strictly between a and b
...
Since b > p,
the point p is not an upper bound, and so there is a point t ∈ J with p < t
...
Since s, t ∈ J it follows that p, being between s and t, is also in J
...
Thus we have the
33
34
Ambar N
...
1)
def
[a, b) = {x ∈ R∗ : a < x < b}
...
Thus a closed interval contains it
two endpoints, while an open interval contains neither endpoint
...
2
Neighborhoods
A neighborhood of a point p ∈ R is an interval of the form
(p − δ, p + δ)
where δ > 0 is any positive real number
...
2, 1
...
A typical neighborhood of 0 is of the form
(−, )
for any positive real number
...
For example,
(5, ∞]
is a neigborhood of ∞
...
An example is
[−∞, 4)
Observe that if U and V are neighborhoods of p then U ∩ V is also a
neighborhood of p
...
)
Observe also that if N is a neighborhood of a point p, and if q ∈ N then q
had a neighborhood lying entirely inside N
...
5, and we can form the neighborhood (2, 3) of 2
...
Here is a simple but fundamental observation:
Distinct points of R∗ have disjoint neighborhoods
...
2)
This is called the Hausdorff property of R∗
...
5, 5
...
3
Types of points for a set
Consider a set
S ⊂ R∗
...
36
Ambar N
...
The point ∞ is also an interior point
of E
...
For example, for the set E above, points −5, 7, and −∞ are exterior to E
...
Thus p is a boundary point of S if every neighborhood of p intersects
both S and S c
...
Next consider the set
{3} ∪ (5, ∞)
The boundary points are 3, 5, and ∞
...
Example For the set A = [−∞, 4) ∪ {5, 9} ∪ [6, 7), decide which of the following
are true and which false:
(i) −6 is an interior point (T)
(ii) 6 is an interior point (F)
(iii) 9 is a boundary point (T)
(iv) 5 is an interior point (F)
Ecxerise For the set B = [−∞, −5) ∪ {2, 5, 8} ∪ [4, 7), decide which of the following are true and which false:
(i) −6 is an interior point
(ii) −5 is an interior point
(iii) 5 is a boundary point
(iv) 4 is an interior point
(v) 7 is a boundary point
...
4
37
Interior, Exterior, and Boundary of a Set
The set of all interior points of a set S is denoted
S0
and is called the interior of S
...
The set of all points exterior to S is the exterior of S, and we shall denote
it
S ext
...
3)
Recall that a point p is on the boundary of S if every neighborhood of
the point intersects both S and S c
...
Thus
∂S = ∂S c
...
4)
The interior of the entire extended line R∗ is all of R∗
...
Moreover,
∂∅ = ∅
...
If you take any neighborhood U of any point in R∗ then U contains
both rational numbers and irrational numbers
...
Example For the set A = [−∞, 4) ∪ {5, 9} ∪ [6, 7),
(i) A0 = [−∞, 4) ∪ (6, 7)
38
Ambar N
...
But if we decide to work only inside R then the boundary of G is just {3}
...
5
Open Sets and Topology
We say that a set is open if it does not contain any of its boundary points
...
The set
(3, 4]
is not open, because it contains 4, which is a boundary point
...
The entire extended line R∗ is open, because it has no boundary points
...
Notice then that every point of an open set is an interior point
...
Thus for an open set S each point has a neighborhood contained entirely
inside S
...
Viewed in this way, it becomes clear that the union of open sets is an
open set
...
Exerise Check that the intersection of the sets (4, ∞) and (−3 5) and (2, 6) is
open
...
The set of all open subsets of R∗ is called the topology of R∗
...
6
Closed Sets
A set S is said to be closed if it contains all its boundary points
...
But
[4, 5)
is not closed because the boundary point 5 is not in this set
...
But, viewed as a subset of R it is closed
...
40
Ambar N
...
The empty set ∅ is also closed
...
4
...
If S is open then its boundary points are all outside S:
∂S ⊂ S c
...
Thus, for S to be open we must have
∂(S c ) ⊂ S c ,
which means that S c contains all its boundary points
...
Thus, if a set is open then its complement is closed
...
Thus,
Theorem 4
...
1 A subset of R∗ is open if and only if its complement is
closed
...
Consider the open set (1, 5)
...
Exercise
...
Show that its complement is
open
...
8
Closed sets in R and in R∗
The set
[3, ∞)
is closed in R, but is not closed in R∗
...
Thus, when working with closed sets it
is important to bear in mind the distinction between being closed in R and
being closed in R∗
...
Chapter 5
Magnitude and Distance
5
...
1)
−x if x < 0
...
Note that
|0| = 0
and
|x| ≥ 0 for all x ∈ R∗
...
2)
and in fact, of course, x is equal to either |x| (if x ≥ 0) or −|x| (if x < 0)
...
41
42
Ambar N
...
In other words, |x| is x or −x, whichever is ≥ 0
...
5
...
3)
(5
...
The same works if both numbers are negative:
|(−3) + (−5)| = 8 = | − 3| + | − 5|
...
We can summarize this in the triangle inequality for magnitudes:
|a + b| ≤ |a| + |b|,
(5
...
Here is a proof of this: since |a + b| is the larger of a + b and −(a + b), we
just need to show that both of these are ≤ the sum |a| + |b|
...
Thus both a + b and −(a + b) are less or equal to |a| + |b|, and so the larger
of a + b and −(a + b) is ≤ |a| + |b|
...
5)
...
6)
for all a, b ∈ R∗
...
DRAFT Calculus Notes 11/17/2011
5
...
(5
...
8)
if d(a, b) = 0 then a = b
...
9)
and
There is a third, less obvious, property that is called the triangle inequality
that is of great use:
d(a, c) ≤ d(a, b) + d(b, c) for all a, b, c ∈ R
...
10)
This follows from the triangle inequality for mangitudes:
d(a, c) = |a − c| + |a − b + b − c| ≤ |a − b| + |b − c| = d(a, b) + d(a, c)
...
7) is completely natural and
intuitive but does not extend nicely to R∗
...
5
...
Clearly this neighborhood consists exactly
of those points x whose distance from p is less than δ
...
(5
...
Sengupta 11/6/2011
Chapter 6
Limits
The concept of limit is fundamental to calculus
...
For example, anyone would agree that x2 approaches 9 when x
approaches 3; but explaining exactly what this means is a subtle matter
...
But using out technology of
sup and inf makes it somewhat easy to come to grips with the exact meaning
of x2 → 4 asn x → 3
...
Clearly for such quantities, the point x itself must be in the domain of f
...
If U is
a neighborhood of p then part of U might not be inside S, and so when we
speak of the values of f on U we need to focus on f (x) for x ∈ U ∩ S
...
01,
...
Putting the
restrciction x ∈ U ∩ S makes it possible to talk about the value f (x)
...
1
Ambar N
...
Intuitively it is clear that as x approaches 2 the value g(x) approaches 23 = 8;
note that the actual value g(2), which is given to be 0, is irrelevant to this
...
x→2
We read this as “g(x) has the limit 8 as x approaches 2’
...
For this consider the
behavior of the values g(x) when x is restricted to some neighborhood, say
(1
...
5) of 2, again ignoring the actual value g(2):
{g(x) : x ∈ (1
...
5) and x 6= 2}
...
5, 2
...
5)3 = 15
...
5, 2
...
5)3 = 3
...
There is a simpler notation for these sups and infs:
sup
g(x) = 15
...
5,2
...
375
...
5,2
...
9, 2
...
For this we have
sup
g(x) = 9
...
9,2
...
9,2
...
859
DRAFT Calculus Notes 11/17/2011
47
As we have shrunk the neighborhood the supremum has decreases and the
infimum has increased
...
Indeed intuition, and in
this case easy verification, suggests that:
the limit limx→2 g(x) always lies between the sup and inf of the
values of g(x) on neighborhoods of 2 (always excluding the value
x = 2)
...
This provides us with an official definition of limit:
Definition 6
...
1 Let g be a function defined on a set S ⊂ R and let p be
any point in R∗
...
1)
x∈U ∩S,x6=p
The reason for using x ∈ U ∩ S is that g(x) is only be defined for x in the
set S
...
1) are over the empty set and so (6
...
Thus there is no possibility of the
limit existing if p is not a limit point of S
...
A note of caution: the above definition is not the standard one but is
equivalent to it
...
We look at two such simple examples now just to make sure the definition
produces values in agreement with common sense
...
Sengupta 11/6/2011
Take for a starter example, the constant function
K(x) = 5
for all x ∈ R
...
1
...
x→3
To check this consider any neighborhood of 3:
(3 − δ, 3 + δ),
where δ is any positive real number
...
x∈(3−δ,3+δ),x6=3
Thus the only value that lies between the sup and the inf is 5 itself, and
hence
lim K(x) = 5
...
We would like to make sure that Definition 6
...
1 does imply that limx→6 f (x)
is 6
...
Then
{f (x) : x ∈ (6 − δ, 6 + δ)} = {x : x ∈ (6 − δ, 6 + δ)},
which is just the interval (6 − δ, 6 + δ), but with the point 6 excluded
...
What value lies between these two no
matter what δ is ? Certainly it is 6:
inf
x∈(6−δ,6+δ),x6=6
f (x) < 6 <
sup
x∈(6−δ,6+δ),x6=6
f (x)
...
x→6
In fact, it is clear that
lim x = p,
x→p
for every p ∈ R∗
...
)
Before moving on the fancier explorations here is a warning on notation
...
Instead of limx→p f (x) we could just as well write limy→p f (y) or limw→p f (w):
lim f (x) = lim f (y) = lim f (w) = lim f (blah)
...
2
Limits for 1/x
Common sense shows that
1/x → 0,
We will first show that
as x → ∞
...
Since 1/x > 0 when x is positive, 0 is a lower bound for such 1/x and so
the greatest lower bound is ≥ 0:
1
≥ 0
...
Denote the inf by b:
1
...
Sengupta 11/6/2011
Suppose b > 0
...
So if we take some real number y larger than both 1/b
and t, say y = t + 1/b, then this x is in (t, ∞) and is also > 1/b and so
1
< b
...
x∈(t,∞) x
inf
This is impossible, since it is sayng b < b
...
x∈(t,∞) x
inf
This holds for all real t > 0
...
(It is a tiresome check to see that it holds also for
t ≤ 0, keeing in mind that x = 0 is excluded from the domain of 1/x
...
x→∞ x
lim
(6
...
(6
...
Taking 1/x only on the
domain of positive values of x we have
lim
1
= ∞,
x→0,x>0 x
lim
(6
...
x→0,x<0 x
lim
(6
...
x→0− x
lim
(6
...
You can check that in any
neghborhood of 0, excluding the value x = 0 itself, the sup of 1/x is ∞
whereas the inf of 1/x is −∞, and so there is no unique value between these
two extremes
...
3
A function with no limits
Recall the set Q of all rational numbers:
Q = {all rationals}
...
The same is true of the irrationals:
every open interval in R contains irrational points
...
7)
0 if x ∈
/ Q;
If you take any p ∈ R and any neigborhood U of p it is clear that
sup 1Q (x) = 1 and
x∈U,x6=p
inf
x∈U,x6=p
1Q (x) = 0
...
52
Ambar N
...
4
Limits of sequences
Recall that a sequence s is a function whose domain is the set N = {1, 2, 3,
...
The value of the function s
on the number n ∈ N is denoted by
sn
rather than s(n), and the function itself is generally written as
(sn )n≥1
rather than s
...
Thus
we can apply to the case of sequences, with S being N and p = ∞
...
n→∞
For example,
1
= 0
...
n→∞
(6
...
(6
...
, n},
has, and clearly this is more than n itself since each of the n elements itself
provides a subset
...
8) we have
1
= 0
...
10)
where SN is the N -th partial sum
SN = s1 + · · · + sN
...
n=0
We will not explore these ideas much further at this point
...
We will work out the value of the infinite
series sum
1
1
1
1 + + 2 + 3 + ···
2 2
2
(If you think about it, this surely should add up to 2
...
)
In the summation notation this is displayed as
∞
X
1
...
2
2
2
2
Observe that this is very similar to SN itself: when we subtract nearly everything cancels out:
1
1
SN − SN = 1 − N +1
...
2
2
54
Ambar N
...
2
2
1 − 21
(6
...
5
1
1
+ ··· =
2
1−
1
2
= 2
...
12)
Lim with sups and infs
In this section we are going to push our grasp on sup’s and inf’s to the limit!
Consider a function f on a set S ⊂ R and a point p ∈ R∗
...
We
know of then that
inf
x∈U ∩S,x6=p
f (x) ≤
sup
f (x)
...
Both U and V contain the
neighborhood
W = U ∩ V
...
13)
x∈V ∩S,x6=p
because shrinking a set raises the inf and lowers the sup
...
(6
...
Thus, there always exists a point that lies between the sups and infs of the
values f (x) for x in neighborhoods of p, excluding x = p
...
We can reformulate the relation (6
...
Hence: the sup of f over any neighborhood V of p is
≥ the sup of the values inf x∈U ∩S,x6=p f (x) for all neighborhoods U of p:
sup
U
inf
x∈U ∩S,x6=p
f (x) ≤
sup
f (x)
(6
...
Since this holds for all
neighborhoods V of p we then conclude that
sup
U
inf
x∈U ∩S,x6=p
f (x) ≤ inf
sup
V x∈V ∩S,x6=p
f (x)
(6
...
Pondering (6
...
To summarize:
Proposition 6
...
1 Let f be a function defined on a set S ⊂ R, and p a limit
point of S
...
(6
...
Sengupta 11/6/2011
Chapter 7
Limits: Properties
In this chapter we explore the concept of limit and get familiar with some
basic properties that make the computation of limits a routine process for
many ordinary functions
...
1
Up and down with limits
Suppose f is a function on R and
lim f (x) = 5
...
Let us see how we can deduce this from the
official definition of limit
...
Thus, for instance, 4 does not lie between these sups
and infs
...
x∈U,x6=3
x∈U,x6=3
Note that this interval does contain 5 because that is actually the limit
...
58
Ambar N
...
This is exactly what we had conjectured based on common
sense intuition about limits
...
Thus
what we have really proved is this:
Proposition 7
...
1 Suppose f is a function on some subset S ⊂ R, and
L = lim f (x),
x→p
where p ∈ R∗
...
We have had to write x ∈ U ∩ S, and not just x ∈ U , because f (x) might
not be defined for all x in U
...
What we are saying is, in ordinary rough and
ready language, if f (x) → L as x → p then the values f (x) lie above b when
x is near p (but not p itself), for any given value b < L
...
1
...
If u is any value > L then there is a neighborhood U of p on
which
f (x) < u for all x ∈ U ∩ S except possibly for x = p
...
1
...
If u is any value > L and b is any value < L then there is a
neighborhood U of p on which
b < f (x) < u for all x ∈ U ∩ S except possibly for x = p
...
2
59
Limits: the standard definition
Proposition 7
...
3 can be recast and slightly broadened into the following
characterization of the notion of limit:
Proposition 7
...
1 Let f be a function on S ⊂ R and p ∈ R∗
...
Then for any neighborhood W of L there is a
neighborhood U of p such that
f (x) ∈ W for all x ∈ U ∩ S, excluding x = p
...
Suppose first that L is a real number, and not ∞ or −∞
...
Then by Proposition 7
...
3 there is
a neighborhood U of p for which
b < f (x) < u for all x ∈ U ∩ S except possibly for x = p
...
Now consider the case L = ∞
...
By Proposition 7
...
1 there is a neighborhood U of p
for which
f (x) > b for all x ∈ U ∩ S except possibly for x = p
...
Lastly, consider the case L = −∞
...
By Proposition 7
...
2 there is a neighborhood U of p for which
f (x) < u for all x ∈ U ∩ S except possibly for x = p
...
QED
The result can also be run in reverse:
60
Ambar N
...
2
...
Suppose L ∈ R∗ has the property that for any neighborhood W of
L there is a neighborhood U of p such that
f (x) ∈ W for all x ∈ U ∩ S, excluding x = p
...
1)
Then L = limx→p f (x)
...
Suppose first that L is a real number, and not ∞ or −∞
...
Suppose
this were not true
...
(7
...
(7
...
Then by the condition
(7
...
But this is impossible since, by (7
...
This
contradiction shows that (7
...
By a similar argument it also follows that
L≤
sup
f (x)
...
DRAFT Calculus Notes 11/17/2011
61
But we still need to show that L is the unique such value
...
Suppose first L0 > L
...
This means that L0 is > than all the values in W
...
1) there is a neighborhood U of p such that f (x) ∈ W for all x ∈ U ∩ S
with x 6= p
...
Thus L0 cannot be the limit of f (x) as x → p
...
The cases L = ∞ and L = −∞ are settled by different but similar
arguments, just keeping in mind that the neighborhoods of ∞ and −∞ are
‘one sided’ rays
...
2
...
A value L ∈ R∗ is said to be the limit of f (x) as x → p if for any
neighborhood W of L there is a neighborhood U of p such that f (x) ∈ W for
all x ∈ U ∩ S with x 6= p
...
3
Limits: working rules
It is an exhausting and largely pointless task to try to use the definition of
limit directly in computing actual limits such as
x2 − 9
...
There really are just two limits we have worked out directly from the
definition:
lim K = K
x→p
(7
...
The first computationally useful result for limits is simply that the limit
of a sum is the sum of the limits:
lim [f (x) + g(x)] = lim f (x) + lim g(x)
...
5)
62
Ambar N
...
Here is a formal statement:
Proposition 7
...
1 Suppose f and g are functions on S ⊂ R, and suppose
the limits limx→p f (x) and limx→p g(x) exist, where p is some point in R∗
...
x→p
x→p
(7
...
x→p
x→p
x→p
(7
...
6) is to ensure that the sum on the right in
(7
...
We will not prove this result here
...
3
...
Then the limit limx→p [f (x)g(x)] exists and
lim [f (x)g(x)] =
x→p
lim f (x)
lim g(x) ,
x→p
x→p
(7
...
More clearly, the condition is that
the limits limx→p f (x) and limx→p g(x) exist and
{lim f (x) , lim g(x)} =
6 {0, ∞}
x→p
and
x→p
{lim f (x) , lim g(x)} =
6 {0, −∞}
...
9)
x→p
It would have been easier to state this result had we never defined 0·∞ and
∞ · 0 as 0, and instead left such products as undefined
...
DRAFT Calculus Notes 11/17/2011
63
Here is a quick application of the preceding rules about limits: we can
compute the limit of x2 + 3x + 4 as x → 1:
lim x2 + 3x + 4 = lim x2 + lim (3x + 4) (if this exists)
x→1
x→1
x→1 h
i
= lim x lim x + lim 3 lim x + lim 4
(if this exists)
x→1
x→1
x→1
= 1 · 1 + [3 · 1 + 4]
= 8
...
10)
This is the kind of reasoning one should go through once but it is clearly so
routine that it is not worth mentioning all the steps every time
...
Going beyond multiplication we consider ratios
...
It is
useful, for the purposes of the next result to define
a
a
=0=
∞
−∞
if a ∈ R
...
11)
Proposition 7
...
3 Suppose f and g are functions on S ⊂ R, and p is some
(x)
exists and
point in R∗
...
12)
provided the two limits on the right and their ratio exist (this means that the
ratio must not look like something/0 or ±∞/ ± ∞)
...
x→3 x − 3
lim
64
Ambar N
...
Thus we need to be less
lazy and observe that
(x − 3)(x + 3)
x2 − 9
=
= x + 3,
x−3
x−3
from which it is clear that
x2 − 9
= 6
...
4
Limits by comparing
Sometimes we can find the limit of a function by comparing it with other
functions that are easier to understand
...
Suppose f , g, and h are
functions on a set S ⊂ R and p ∈ R∗ is such that
f (x) ≤ h(x) ≤ g(x)
(7
...
Assume
that limx→p f (x) and limx→p g(x) exist and are equal:
L = lim f (x) = lim g(x)
...
Here is a formal statement and proof:
Proposition 7
...
1 Suppose f , g, and h are functions on a set S ⊂ R, and
p ∈ R∗ is such that
f (x) ≤ h(x) ≤ g(x)
(7
...
Assume
also that limx→p f (x) and limx→p g(x) exist and are equal:
L = lim f (x) = lim g(x)
...
DRAFT Calculus Notes 11/17/2011
65
Proof
...
2
...
(Since the limit limx→p f (x) is assumed to exist, the point
p is a limit point of S
...
Note that W is an interval containing
L
...
2
...
Similarly, there is also a neighborhood U2 of p such that g(x) ∈ W for all
x ∈ U2 ∩ S with x 6= p
...
If we take
any x ∈ U ∩ S, with x 6= p, then both g(x) and f (x) lie inside W , and so
anything between g(x)) and f (x) also lies in W
...
This proves that
lim f (x) = L
...
4
...
x→p
Then
lim f (x) = 0
...
Proof
...
Hence
−|f (x)| ≤ f (x) ≤ |f (x)|
66
Ambar N
...
As x → p both |f (x)| → 0 and −|f (x)| → 0, and so, by the
‘squeeze’ theorem, f (x) → 0 as well
...
lim x sin x +
x→0
x
This follows by using the fact that | sin(·)| is ≤ 1, which shows that
3
1
≤ |x|3 ,
0 ≤ x sin x +
x
to which we can apply Proposition 7
...
2
...
5
Limits of composite functions
Suppose f and g are functions
...
For example,
√
x 7→ 1 − x2
is the√composite of the function x 7→ 1 − x2 (for all x ∈ R) and the function
u 7→ u (for u ≥ 0); its domain is [−1, 1]
...
This is an extremely useful method and mostly we use it without even noticing; for example, using the simple result
w3 − 1
(w − 1)(w2 + w + 1)
= lim
+ lim (w2 + w + 1) = 3,
w→1 w − 1
w→1
w→1
w−1
lim
DRAFT Calculus Notes 11/17/2011
67
we obtain a limit that is at first less obvious:
x1/3 − 1
w−1
1
= lim 3
= ,
x→1 x − 1
w→1 w − 1
3
lim
by using the ‘substitution’ x = w3
...
As an extreme
example, consider the functions F and G given by
(
1 for v 6= 0;
F (v) =
2 if v = 0,
and G(x) = 0 for all x
...
What has gone wrong is that we have rigged the inner function G to keep
hitting (in fact it is stuck at) the forbidden’ point v = 0 which is excluded
from consideration when defining limv→0 F (v)
...
Recall that a point p is exterior to a set B
if p has a neighborhood leying entirely outside B
...
5
...
Let S be the domain of the composite f ◦ g
...
(In other words, p
has a neighborhood U0 with the property that there is no point in U0 ,
other than p itself, that is both in the domain of f and were g takes the
value q
...
x→p
68
Ambar N
...
Notice that (ii) is automatically
satisfied in the case q = ∞, for g, being a real valued function, never takes
the value ∞
...
Proof
...
Let
W be any neighborhood of L
...
(7
...
(7
...
Now consider any point x in U1 , with x 6= p, for which f (g(x)) is defined
and g(x) 6= q; we are given that suchan x exists
...
16) we have g(x) ∈ V ,
and then by (7
...
Thus, starting with any neigborhood W of L we have produced a neighborhood U1 of p such that (f ◦ g)(x) ∈ W for all x ∈ U1 with x 6= p
...
Though we don’t really
discuss completely precise mathematical definitions for these functions, we
extract enough information about them from trigonometry to be able to do
calculus with these functions
...
8
...
C
Figure 8
...
One way to be more specific is to draw a circle, with center C at the
vertex, and think of the angle as an arc of the circle marked off by the two
rays
...
Sengupta 11/6/2011
The area of the shaded sector gives a
measure of the angle ∠P CQ (taking the
radius CP to be the unit of length)
...
2: Measuring angle using sectorial areas
center C at the vertex of the two rays, and then think of the angle as a
sector marked off in the circle by the two rays
...
Why twice? This is just to be consistent with historical practice and
convention
...
The area of this disk is what is
denoted
π
and so the full circular angle has radian measure 2π
...
4
4
This discussion has one element of haziness: what do we mean by the
area of a curved region? For this please turn back to the Introduction
...
2
Geometric specification of sin, cos and tan
We will describe the geometric meaning of the measure of an angle and also
that of sin θ and cos θ
...
3
...
The ‘semichord’ from R2 to R1 is the
segment, perpendicular to R1 , that runs from the point Q where R2 cuts the
circle to a point on R1
...
The cos of the angle is the distance from the vertex C to the semi-chord
...
1
sin θ
tan θ
θ
cos θ
Figure 8
...
4 provides more concrete formulas and also
relates visually to the measurement of the angle θ in terms of the area of the
sectorial region it cuts out of the circle
...
Let
x = CB
y = QB
...
1)
Here we take x to be negative if B is on the opposite side of C from P
...
72
Ambar N
...
4: Measuring θ and sin θ, cos θ, and tan θ
The trigonometric functions are specified by:
QB
sin θ =
;
CQ
CB
(8
...
From (8
...
3)
as long as θ 6= π/2
...
4)
tan 0 = 0
...
5)
sin π = 0;
cos π = −1;
tan π = 0
...
6)
and
If the angle θ is increased to θ+2π then geometrically the point Q remains
where it is
...
7)
again as long as tan(a + 2π), and hence tan a, is defined
...
When a ∈ (0, π), sin a is postive, and when a ∈ (π, 2π) the value sin a is
negative:
(
> 0 if a ∈ (0, π);
sin a
(8
...
For cos it is:
(
> 0 if a ∈ (−π/2, π/2);
cos a
< 0 if a ∈ (π/2, 3π/2)
...
9)
74
8
...
Sengupta 11/6/2011
Reciprocals of sin, cos, and tan
The reciprocals of sin, cos and tan also have names:
1
sin θ
1
sec θ =
cos θ
1
cot θ =
tan θ
csc θ =
(8
...
8
...
This
leads to the following identities:
π
sin − θ = cos θ
2
π
cos − θ = sin θ
(8
...
2
When an angle is replaced by its negative, it changes the sign of sin and
tan but not of cos:
sin(−a) = − sin a;
cos(−a) = cos a;
tan(−a) = − tan a,
(8
...
Pythagoras’ theorem implies the enormously useful identity
sin2 a + cos2 a = 1,
(8
...
Using this we can work out the value of sin, at least up to sign,
from the value of cos:
√
sin a = ± 1 − cos2 a
...
14)
DRAFT Calculus Notes 11/17/2011
75
The only way to decide whether it is + or whether it is − is to consider the
value of a: if a ∈ [0, π], or differs from such a value by an integer multiple of
2π, then sin a is ≥ 0
...
(8
...
For instance,
dividing
sin2 θ + cos2 θ = 1
by cos2 θ produces:
1
sin2 θ
+
1
=
,
cos2 θ
cos2 θ
which can be rewritten as
1 + tan2 θ = sec2 θ
(8
...
(For θ = ±π/2 one could define tan2 θ as
well as sec2 θboth to be ∞, and similarly for all the other trouble spots ±π/2
plus integer multiples of 2π, and this would make (8
...
)
Very clever geometric arguments can be used to prove the trigonometric
identities:
sin(a + b) = sin a cos b + sin b cos a;
cos(a + b) = cos a cos b − sin a sin b;
tan a + tan b
tan(a + b) =
,
1 − tan a tan b
(8
...
There are some consequences of these addition formulas that are also
useful
...
Special cases of these are also very useful:
sin(2a) = 2 sin a cos b;
cos(2a) = cos2 a − sin2 a = 2 cos2 a − 1 = 1 − 2 sin2 a;
2 tan a
tan(2a) =
,
1 − tan2 a
(8
...
Sengupta 11/6/2011
as long as the tan values are defined
...
Suppose we choose a and b such that
x=a+b
y =a−b
Then
sin x − sin y = sin(a + b) − sin(a − b)
= (sin a cos b + sin b cos a) − (sin a cos b − sin b cos a)
= 2 sin a cos b
...
19)
(8
...
19)
...
19) gives
x + y = 2a,
and subtracting gives
x − y = (a + b) − (a − b) = 2b
...
2
Putting these into (8
...
2
2
Following the same line of reasoning for cos instead of sin gives us
x−y
x+y
cos x − cos y = −2 sin
sin
...
21)
(8
...
23)
(8
...
5
77
Inequalities
The identity
sin2 a + cos2 a = 1
implies that neither sin a nor cos a can be bigger than 1 in magnitude:
| sin a| ≤ 1;
| cos a| ≤ 1
...
25)
For example,
|5| = 5;
|0| = 0;
and | − 4| = 4
...
26)
sin − + 2πn = −1,
2
for all integers n ∈ Z
...
27)
for all integers n ∈ Z
...
(8
...
−x
x
Moreover, we also know that
cos(−x) = cos x
...
(8
...
6
Ambar N
...
30)
lim cos x = cos p,
x→p
for all p ∈ R
...
)
In particular,
lim sin x = sin 0 = 0
x→0
and
lim cos x = cos 0 = 1
...
Recall
the bounds for (sin x)/x near 0:
cos x ≤
sin x
x
≤ 1 for all x ∈ (−π/2, π/2) with x 6= 0
...
31)
We know that as x → 0 we have cos x → 1
...
(8
...
For
instance,
lim
x→0
1 − cos x
(1 − cos x)(1 + cos x)
=
lim
x→0
x→0
x2
x2 (1 + cos x)
1 − cos2 x
= lim 2
x→0 x (1 + cos x)
sin2 x
= lim 2
x→0 x (1 + cos x)
2
sin x
1
= lim
x→0
x
1 + cos x
1
= 12 ·
1+1
1
=
...
33)
DRAFT Calculus Notes 11/17/2011
79
In summary
1 − cos x
1
=
...
32) is
lim
sin Kx
= K,
x→0
x
lim
for all K ∈ R
...
34)
(8
...
Clearly, y approaches 0 when x → 0, and so we have
sin Kx
sin y
= lim K
= K · 1 = K
...
)
lim
x→0
8
...
When x = 1/(π/2 + 2πn), for any integer n, we have sin(1/x) = 1:
1
= sin (π/2 + 2πn) = 1 for all n ∈ Z
...
1/(3π/2 + 2πn)
Now any neighborhood of 0 contains the values 1/(π/2+2πn) and 1/(3π/2+
2πn) for large enough integers n
...
Thus
inf
x∈U,x6=0
sin(1/x) = −1,
and
sup sin(1/x) = 1
x∈U,x6=0
(8
...
Sengupta 11/6/2011
for every neighborhood U of 0
...
Hence:
limx→0 sin x1 does not exist
...
37)
The problem here is that sin 1/x fluctuates too much near x = 0
...
Since sin a is at most 1 in magnitude we have
−|x| ≤ f (x) ≤ |x| for all x 6= 0
...
38)
lim x sin = 0
...
8
...
The graph for sin is
y
y = sin x
−π
π
2π
π/2
x
Figure 8
...
and
lim tan x = ∞,
x→π/2−
DRAFT Calculus Notes 11/17/2011
81
y
y = cos x
−π
π
2π
π/2
x
Figure 8
...
7: Graph of tan
8
...
For example, here are the formulas for sin and cos that can be
used to define them without using pictures:
1 3 1 5
x + x − ···
3!
5!
(8
...
It is, at this stage, impossible to see
where the identities/definitions (8
...
sin x = x −
82
Ambar N
...
limx→1 5
2
...
limx→1
x2 −9
x−3
4
...
limx→1
x4 −1
x2 −1
6
...
limx→∞
4x3 −3x+2
x2 −x+1
8
...
limx→∞
4x3 +sin
√x
2x3 + x
10
...
limx→∞ 3x2 + 1 − x2 + 1
√
√
13
...
limx→∞
√
14
...
limx→∞ x x2 + 2 − x2 + 1
√
√
√
16
...
limθ→0
sin(θ2 )
θ2
18
...
limθ→π/6
sin(θ−π/6)
θ−π/6
DRAFT Calculus Notes 11/17/2011
20
...
limx→0 x(1 − x)1Q (x)
22
...
Explain why limx→3 x(x − 1)1Q (x) does not exist
...
Explain why limx→∞ cos x does not exist
...
Explain why limx→∞ x sin x does not exist
...
Explain why limx→∞
sin x
x
= 0
...
Explain why limx→∞
sin
√x
x
= 0
...
Sengupta 11/6/2011
Chapter 9
Continuity
Continuous functions are functions that respect topological structure
...
9
...
In case p is an isolated point of S we cannot work with limx→p f (x), but
surely there is no reason to view f as being not continuous at such a point
...
Here is a cleaner definition of continuity at p:
Definition 9
...
1 A function f defined on a set S ⊂ R is said to be continuous at a point p ∈ S if for every neighborhood W of f (p) there is a
neighborhood U of p such that
f (x) ∈ W for all x ∈ U
...
2
Discontinuities
Sometimes a function is discontinuous (that is, not continuous) at a point p
because the value f (p) is, for whatever reason, not equal to limx→p f (x) even
85
86
Ambar N
...
For example, for the function g given by
( 2
x −9
if x 6= 3;
g(x) = x−3
4
if x = 3
...
1)
we have the limit
lim g(x) = lim (x + 3) = 6,
x→3
x→3
which is not equal to the value g(3)
...
On the other hand there are more serious discontinuties
...
x→0− x
x→0 x
lim
There is a jump from left to right, and there is no way to remove this discontinuity
...
9
...
All polynomial functions, such as
5x3 − 3x4 + 7x2 − 3x + 4
are continuous
...
3
...
You can check this easily by consulting the definition of what it means
to be continuous at a point
...
For example, consider the function 1Q on R whose value is 1 on rationals
and 0 on irrationals
...
The statement ‘f is continuous on a set T ’ can have two different meanings:
(i) f is continuous at every point of T ;
(ii) the restriction f |T is continuous
...
9
...
Hence this
function is discontinuous everywhere on R
...
has the property that
lim f (x) = 0 = f (0),
x→0
but limx→p f (x) does not exist for any p 6= 0
...
88
Ambar N
...
Can you manufacture a function that is continuous at exactly a given
set of points and nowhere else? Is there a function that continuous at every
point of (0, 1) but at no other point?
9
...
5
...
If g is
continuous at p and f is continuous at g(p) then f ◦ g is continuous at p
...
Let W be a neighborhood of L = f (q), where q = g(p)
...
Next, by continuity of g at p, there is a
neighborhood U of p such that g(x) ∈ V for all x ∈
U in the domain of g
...
This proves that
f ◦ g is continuous at p
...
6
Continuity on R∗
In calculus we work with functions defined on subsets of R and having values
in R
...
No
great additional work is needed for this; the definition remains exactly as
before:
a function F : S → R∗ is continuous at p ∈ S if either p is an isolated
point of S or if limx→p F (x) = F (p)
...
If f : (a, b) → R is continuous and limx→a f (x) exists then we can extend
f to a continuous function F : [a, b) → R∗ by setting
F (a) = lim f (x)
...
89
90
Ambar N
...
Figure 10
...
5, 2]
...
1: Graph of x3 − x2 − 2x + 1
that
f (−1) = 1
and
f (1) = −1
...
The intermediate
value theorem, which we study in this chapter, guarantees the existence of
such a point; thus, from this theorem it follows that there is a solution of the
equation
x3 − x2 − 2x + 1 = 0
91
92
Ambar N
...
Its essence remains valid in settings far beyond the real line, but even this
first glimpse of the idea, on R, is of great use
...
1
Inequalities from limits
Suppose we know that a function f has the limit
lim f (x) = 9
...
We summarize this idea in:
Proposition 10
...
1 Let f be a function on a set S, and p ∈ R∗ a point
for which limx→p f (x) exists
...
1)
for some neighborhood U of p
...
2)
for some neighborhood U of p
...
Let
L = lim f (x)
...
So if K < L then K does not lie between all
such infs and sups; thus, there is some neighborhood U of p such that K
DRAFT Calculus Notes 11/17/2011
93
does not lie between inf x∈U,x6=p f (x) and supx∈U,x6=p f (x)
...
x∈U,x6=p
This proves (10
...
The result (10
...
QED
10
...
2
...
Let t be any real number between f (a) and f (b):
f (a) ≤ t ≤ f (b)
or f (b) ≤ t ≤ f (a)
...
Proof
...
Similarly if t = f (b)
...
If f (a) < f (b) this means that f (a) < t < f (b), whereas if f (a) > f (b)
then f (b) < t < f (a)
...
Let S be the set of all x ∈ [a, b] for which f (x) < t:
S = {x ∈ [a, b] : f (x) < t}
...
Moreover, b is an upper bound for S, because S is inside
[a, b]
...
1
...
Then by the completeness property for R there is a least upper bound
s = sup S, and this, of course, also lies in [a, b]
...
Consider any neighborhood U of s of the form
U = (s − δ, s + δ),
94
Ambar N
...
Since s is an upper bound of S, any
point p of S strictly to the right of s (that is, p > s) is not in S, and so
f (p) > t,
for such p ∈ [a, b]
...
Of course
q ≤ s, since s is an upper bound of S
...
Since q ∈ S we have
f (q) < t
...
Thus t satisfies:
inf
x∈U,x6=p
f (x) < t < sup f (x)
x∈U,x6=s
for every neighborhood U of p
...
x→s
Hence t must be f (s)
...
3
Intermediate Value Theorem: a second
formulation
Here is another formulation of the intermediate value theorem:
Theorem 10
...
1 If f is continuous on an interval J then the image
def
f (J) = {f (x) : x ∈ J}
is also an interval
...
To prove that f (J) is an interval we need only to show that all the
numbers between any two distinct values y1 , y2 ∈ f (J) also lie in f (J)
...
Since y1 ∈ f (J) we have
y1 = f (a),
for some a ∈ J, and since y2 ∈ J then
y2 = f (b)
for some b ∈ J
...
Then by Theorem 10
...
1, there is a point s ∈ [a, b] for which
f (s) = t
...
Thus any point t between y1 and y2 is of the form t = f (s), with
s ∈ J, which just means that t ∈ J
...
QED
10
...
But how do we know that such a real number exists? We can obtain existence
by using the intermediate value theorem
...
This is clearly continuous, and from
q(0) = 0 and q(7) = 74
96
Ambar N
...
Hence, by the intermediate value theorem, there is a real number s ∈ (0, 7)
for which
s4 = 73
...
Thus there is a unique positive real number whose
4-th power is 73
...
In this was one can see that
xy
exists for all positive real x and all rational y
...
But we can sharpen this much further
...
Hence 73/4 actually lies between 4 and 5
...
It is clear that not much is special about the numbers 7 and 3/4 in this
discussion
...
10
...
There is no systematic way to work out exact solutions of equations such as
this
...
Here let us
see how the intermediate value theorem shows that there are solutions of the
equation and helps localize them somewhat
...
This is clearly continuous
...
Since f is continuous and 0 lies between f (−2) and f (−1):
f (−2) = −123 < 0 < 1 = f (−1),
it follows by the intermediate value theorem that
there is a point p ∈ (−2, −1) where f (p) = 0
...
3)
has a solution on the interval (−2, −1)
...
3) also has a solution
in (−1, 0) and and a solution lying in (1, 2)
...
3) would
be to divide, say the interval (1, 2) into ten pieces, each of width
...
1), f (1
...
, f (1
...
We can calculate
f (1
...
35
f (1
...
98
and this tells us there is a root (which means the same as ‘solution’) of the
equation (10
...
1, 1
...
Next, repeating the same strategy
by dividing (1
...
2) into ten pieces and calculating the values
f (1
...
12), f (1
...
14), f (1
...
, f (1
...
2),
98
Ambar N
...
13) ' −0
...
14) ' 0
...
Thus, there is a root in the interval
(1
...
14)
...
Later, with the use of calculus,
we can study much faster methods for locating roots
...
The solution is
then displayed as
x = f −1 (y),
and f −1 is the called the inverse of the function f
...
1
Inverse trigonometric functions
The graph of y = sin x oscillates between −1 and 1
...
5
−π
π/6
π
5π/6
−1
Figure 11
...
If we try to solve an equation such as
sin φ =
...
Sengupta 11/6/2011
there are infinitely many values for φ:
π 5π π
5π
π
5π
π
5π
,
, + 2π,
+ 2π,
− 2π,
− 2π,
+ 4π,
+ 4π,
...
5 in
the sense that the sin of each of these is
...
However, to avoid ambiguity we
can focus on just the value π/6: what makes it unique is that it is the only
value between −π/2 and π/2 whose sin is
...
y
−π/2
1
...
2: Graph of sin over [−π/2, π/2]
...
5) to be π/6:
def
def
arcsin(
...
5) = π/6
...
1)
that is,
sin sin−1 w = w
and sin−1 w ∈ [−π/2, π/2]
...
2)
y = sin−1 x means that y ∈ [−π/2, π/2] and sin y is x
...
3)
Thus,
Since the values sin always lie between −1 and 1, there is no value whose
sin is 2; thus
sin−1 x is not defined, as a real number, if x is not in [−1, 1]
...
1
...
Proof
...
Therefore, by the intermediate value theorem, for any A ∈ [−1, 1] there is a
B ∈ [−π/2, π/2] for which
A = sin B
...
QED
Thus,
sin−1 x is defined for all x ∈ [−1, ]1]
...
We have to be careful to observe that y = cos x is not strictly
increasing on [−π/2, π/2]
...
5:
cos(−π/3) = cos(π/3) =
...
y
y = cos x
1
...
3: Graph of cos
...
So, instead we use the interval
[0, π]
x
102
Ambar N
...
Running through the argument we conclude that
for every A ∈ [−1, 1] there is a unique B ∈ [0, π] for which cos B = A
...
4)
or, equivalently,
y = cos−1 x means that y ∈ [0, π] and cos y is x
...
5)
We can run the same reasoning for tan as well and see that
for every A ∈ R there is a unique B ∈ [−π/2, π/2] for which tan B = A
...
6)
or, equivalently,
y = tan−1 x means that y ∈ [−π/2, π/2] and tan y is x
...
2
(11
...
We say that f is strictly
increasing if
f (s) < f (t)
for all s, t in the domain of f for which s < t
...
We say that f is strictly
decreasing if
f (s) > f (t)
DRAFT Calculus Notes 11/17/2011
103
for all s, t in the domain of f for which s < t
...
For this reason we will often state or prove results just for increasing functions, it being generally understood from context that the corresponding result for decreasing functions also holds
...
We say f is strictly monotone if f is stricly increasing or strictly decreasing
...
8)
is an increasing function that is not strictly increasing
...
3
Inverse functions
Here is a somewhat strange result, guaranteeing continuity of certain types
of strictly monotone functions:
Proposition 11
...
1 If g is a strictly monotone function defined on a set
S ⊂ R such that the range g(S) is an interval then g is continuous
...
3
...
(11
...
The inverse function also satisfies
f f −1 (y) = y
for all y ∈ V
...
(11
...
Sengupta 11/6/2011
Proof
...
To show that V is an interval we have to show that if c, d ∈ V , with c < d,
then every point between c and d is also in the range V of f
...
Now consider a point q strictly between c and d:
c < q < d
...
By the intermediate value theorem (keep in mind f is continuous there is a
point p ∈ (a, b) for which
q = f (p)
...
Thus V is an interval
...
If y ∈ V then there is a point x ∈ U
with f (x) = y
...
Set
f −1 (y) = x
if f (x) = y
...
9), by simply writing in the value f (x) in place of y in
f −1 (y) = x
...
10)
...
3
...
QED
Suppose f is a strictly increasing function defined on a set S
...
DRAFT Calculus Notes 11/17/2011
105
Thus, f (S) contains a largest element if and only if S contains a largest
element, and f (S) contains a smallest element if and only if S contains a
smallest element
...
3
...
106
Ambar N
...
12
...
Theorem 12
...
1 Let f be a continuous function on [a, b], where a, b ∈ R
with a < b
...
(12
...
x∈[a,b]
Let us follow a point t, starting at a and moving to the right towards b and
keep track of the ‘running supremum’
Sf (t) = sup f (x)
...
Sengupta 11/6/2011
If f (a) itself is already the maximum M then we are done; assuming then
that f (a) < M , surely the ‘first exit time’ t when Sf (t) escapes from below
the value M is where f actually takes the value M
...
2)
where BM is thet set of all t for which Sf (t) is below M :
BM = {t ∈ [a, b] : Sf (t) < M }
...
3)
(We are assuming the initial value f (a) isn’t already M
...
It is intuitively clear that for t to the left of d∗ the value Sf (t) is < M
whereas for any t to the right of d∗ the value of Sf (t) is M ; this would imply
that that the supremum of f on any neighborhood of d∗ is in fact M
...
Observe that if Sf (x) < M then, since d∗ is an upper bound of BM , it
follows that d∗ ≥ x
...
Hence:
Sf (x) = M
for all x > d∗
...
4)
Proof of Theorem 12
...
1
...
The argument for minimum is exactly similar
(or we can use the trick of applying the maximum result to −f in place of f
to find where f is minimum
...
If f (a) happens to be equal to M then, of course, we are done, on taking
d to be a
...
This implies, in particular,
that the set BM is not empty, containing at least the point a
...
Choose any r ∈ U ∩ [a, b] with
r > d∗ ; what if d∗ = b? We will deal with that case later, assuming for
now that d∗ < b
...
4), Sf (r) = M
...
x∈U ∩[a,b]
DRAFT Calculus Notes 11/17/2011
109
Hence M satisfies
inf
x∈U ∩[a,b]
f (x) ≤ M ≤
sup
f (x),
x∈U ∩[a,b]
wth the second ≤ being actually an equality
...
Therefore, by our definition of limit,
lim f (x) = M
...
Hence
f (d∗ ) = M,
and we are done
...
Then taking any q ∈ [a, b] withq < b, we know that
q is not an upper bound of BM (for d∗ = b is the least upper bound of BM )
...
Therefore also
sup f (x) < M
...
Thus the
supremum of f over every neighborhood of d∗ (which is b) is M
...
The result for inf x∈[a,b] f (x) is obtained similarly or just applying the
result for sup to the function −f instead of f
...
But what of functions defined on other types
of intervals? For example, for the function
1
x
for x ∈ (0, ∞)
it is clear that the function is trying to reach its supremum ∞ at the left
endpoint 0 and its infimum 0 at the right endpoint ∞
...
1 shows
the graph of the function given on (0, ∞) by x2 + x2 − 2
...
110
Ambar N
...
1: Graph of x2 +
2
x
− 2, for x > 0
...
1
...
Then either f attains a maximum value in the interior of U or supx∈U f (x)
is the larger of the endpoint limits La and Lb
...
We will not work through the proof but sketch the ideas
...
Then f is defined on the interval [a, b] ⊂ R∗ (denoting the left endpoint of
U by a), and f is allowed to take the values ±∞ at the endpoints a and b
...
1
...
12
...
2
...
Thus, there exist points c, d ∈ [a, b] such that
F (c) = inf F (x)
x∈[a,b]
F (d) = sup F (x)
...
5)
x∈[a,b]
12
...
Such a set is closed and bounded
...
3
...
Thus, there exist points c, d ∈ K such that
f (c) = inf f (x)
x∈K
f (d) = sup f (x)
...
6)
112
Ambar N
...
A line
is tangent to a circle at a point if that is the only point where the line and
the circle meet
...
Both of these ideas are illuminating and reflect our intuition of what a
tangent line ought to be
...
For example, visually it is perfectly clear that the line y = 1 is
tangent to the graph y = sin x, yet it meets this graph at infinitely many
points
...
Since the graph
y = sin x has no natural notion of ‘center’, it is also useless to try to define
tangent as a line perpendicular to a ‘radius
...
Think of a circle C, and a tangent line l, at a
point P , to the circle, drawn on a transparent sheet of paper
...
This curve
C 0 is called an ellipse
...
This notion goes back to the greek study of conic
sections
...
Sengupta 11/6/2011
to provide a definition of tangent that works for more general curves
...
This
is formalized in the next section
...
1: Tangent line and secant segment
...
1
Secants and tangents
Consider a function f defined on R and think of the graph of f :
{(x, y) : y = f (x), x ∈ R}
...
A secant line is a straight line through P and any other point Q on the
graph
...
There are possibly
many such lines, each with a different slope: (y − y∗ )/(x − x∗)
...
1)
x∈U,x6=x∗
x − x∗
x − x∗
x∈U,x6=x∗
DRAFT Calculus Notes 11/17/2011
115
for all neighborhoods U of x∗
...
We interpet an
infinite value, ∞ or −∞, of slope to mean that the tangent line is ‘vertical’,
parallel to the y-axis
...
2)
slope of tangent at point P = lim
x→x∗
x − x∗
with the existence of this limit signifying the existence of a tangent line
...
2)
...
For some functions there may be multiple lines l with slope satisfying the
condition (13
...
For example, for the graph of
y = |x|
any line with slope ∈ [−1, 1] satisfies the condition (13
...
Here is an illustration of a graph with a whole range of slopes satisfying the condition (13
...
Though this is not a standard notion, let us agree that by a quasi-tangent
line at P (p, f (p)) to the graph y = f (x) for a function f we mean a line
through P of slope satisfying the bounds
f (w) − f (p)
f (w) − f (p)
inf
≤ slope of l ≤
sup
,
(13
...
(Note that, as with tangent lines, this notion is meaningful only when p ∈ S
is not an isolated point of S
...
116
Ambar N
...
2
Derivative
Consider a function f defined on a set S ⊂ R and let p be a point of S that
is not an isolated point
...
x→p
x−p
f 0 (p) = lim
(13
...
Of course, if the graph fails to have a tangent line then
it fails to have a derivative
...
First consider the constant function
K whose value everywhere is 5:
K(x) = 5
for all x
...
We can check this readily
from the official definition
0
K(x) − K(p)
= lim
= lim 0 = 0
...
If the function f is constant near p, then f (x) = f (p), for x in a neighborhood of p, and so the derivative f 0 (p) is 0
...
Thus,
If a function is constant on a neighborhood of a point p then the
derivative of the function at p is 0
...
g(x) = x
Then for any real number p we have
lim
x→p
x−p
g(x) − g(p)
= lim
= lim 1 = 1
...
We can check readily that for the function x 7→ M x + C, where M and
C are real numbers (constants) the derivative is M everywhere:
lim
x→p
13
...
x→p
x→p
x−p
x−p
Notation
The derivative if f at p is denoted
f 0 (p)
...
If a formula is given for f (x) we denote the derivative of f at x by
df (x)
= f 0 (x)
...
Instead we should write
df (x)
dx
at x = 3, or
df (x)
dx
x=3
If we are writing y = f (x) then the derivative of f at x is
dy
...
5)
118
Ambar N
...
For algebraic calculations the notation f 0 (p) is inconvenient
...
For example, for the function s given by
s(x) = x2
for all x ∈ R
we denote the derivative s0 (x) by
s0 (x) =
dx2
...
The entire expression dx2 /dx shoud
be viewed (at least at this stage) as one object, the derivative
...
4
The derivative of x2
Let us work out the derivative of the function given by f (x) = x2 for all
x ∈ R at x = 3
...
x→3 x − 3
lim
We can work this out easily
...
2: A secant segment to y = x2 at P (3, 32 )
...
6)
x→3
= 6
...
If you trace through the calculations above for a general point (p, p2 ) on
y = x2 you see that the slope of the tangent at (p, p2 ) is 2p:
(x − p)(x + p)
x2 − p 2
= lim
x→p
x→p x − p
x−p
= lim (x + p)
lim
(13
...
Thus the derivative of the function given by f (x) = x2 at x = p is 2p
...
dx
(13
...
Sengupta 11/6/2011
y
y = x3
Q(w, w3 )
slope P Q =
P (x, x3 )
w3 −x3
w−x
x
Figure 13
...
13
...
Following the method used for y = x2 we have first the picture
We can see that
w 3 − x3
...
The slope of the tangent at P is then
w 3 − x3
...
9)
DRAFT Calculus Notes 11/17/2011
13
...
This leads to
dxn
= nxn−1
...
10)
dx
Thus, for example, the slope of the curve
y = x7
at the point (1, 1) is
7 · 16 = 7
...
7
Derivative of x−1 = 1/x
For 1/x we have the graph in Figure 13
...
P (x, 1/x)
slope P Q =
1
− x1
w
w−x
Q(w, 1/w)
Figure 13
...
The derivative at x is the slope of the tangent at P (x, 1/x):
d x1
= lim (slope of P Q)
...
w−x
122
Ambar N
...
x
B−A
)
AB
(13
...
dx
x
Observe that this follows the formula dxn /dx = nxn−1 :
dx−1
= −1 · x−2 ,
dx
even though n = −1 is not a positive integer
...
13
...
xk
DRAFT Calculus Notes 11/17/2011
123
We can calculate its derivative:
d(1/xk )
(1/wk ) − (1/xk )
= lim
w→x
dx
w−x
k
(x − wk )/(xk wk )
= lim
w→x
w−x
(using A1 − B1 = B−A
)
AB
xk − w k
w→x xk w k (w − x)
w k − xk
1
· k k
= lim (−1) ·
w→x
w−x x w
1
= (−1) · kxk−1 · 2k ,
x
(13
...
w→x w − x
lim
Thus
d(1/xk )
1
1
= −k 2k−k+1 = −k k+1
...
13)
dxn
= nxn−1 ,
dx
correct again, even though n is now a negative integer
...
9
Derivative of x1/2 =
Consider the function
s(x) =
√
x
√
x = x1/2
defined on all x ≥ 0
...
Then the derivative of this
√ function
√
at p is the slope of the tangent at P (p, p) to the graph y = x, and we
know that
slope of tangent at P = lim (slope of P Q)
...
Sengupta 11/6/2011
√
Q(w, w)
√
slope P Q =
P (p,
√
√
w− x
w−x
p)
Figure 13
...
√
√
w− p
s (p) = lim
w→p
w−p
√
√ √
√
( w − p)( w + p)
√
= lim
√
w→p
(w − p)( w + p)
0
(using (A − B)(A + B) = A2 − B 2 )
w−p
√
= lim
√
w→p wp( w +
p)
1
= lim √
√
w→p 2( w +
p)
1
= √ ,
2 p
(13
...
If p = 0 we can just set p = 0 in the preceding calculations
except for the very last line, being careful to note that the values w we work
with are > 0 (because s(w) is not defined for w < 0; thus,
1
√ = ∞
...
(13
...
15) is
1 −1/2 1 1 −1
x
= x2
...
16)
dxn
= nxn−1 ,
dx
now with n = 1/2
...
10
Derivatives of powers of x
Let r = p/q be a rational number, where p and q are integers, and q 6= 0
...
The existence of such a real number xr follows from the intermediate value
theorem, and ultimately is a consequence of the completeness of the real line
...
17)
dx
and can be proved by extension of the methods used before for negative
powers of x and for x1/2
...
11
Derivatives with infinities
It does not seem useful to bother defining the derivative F 0 (∞), for a function
F , defined at ∞
...
(13
...
Sengupta 11/6/2011
If F (x) approaches a finite limit F (∞), as x → ∞, then F 0 (p) is 0, which
conforms to intuition: the tangent line at x = ∞ is the ‘horizontal’ line
y = F (∞)
...
1
tan θ
= 1
...
Thus it is:
sin0 x = lim (slope of P Q),
w→x
where Q is the point (w, sin w)
...
w−x
We have then
sin w − sin x
...
Sengupta 11/6/2011
Q(w, sin w)
slope P Q =
P (x, sin x)
sin w−sin x
w−x
y = sin x
Figure 14
...
To work this out there are two possible routes
...
22), which implies:
sin w − sin x = 2 sin
w+x
w−x
cos
...
1)
Using this we have
2 sin w−x
cos w+x
2
2
w→x
w−x
2 sin w−x
w+x
2
= lim
cos
...
2)
To make this look more like something involving sin θ/θ (whose limit we
understand), we write this as
2 sin w−x
w+x
2
cos
w−x
w→x
2
2 2
w−x
sin
w+x
= lim w−x2 cos
w→x
2
2
x+x
= 1 · cos
...
3)
Thus we have found the derivative of sin:
d sin x
= sin0 x = cos x
...
4)
DRAFT Calculus Notes 11/17/2011
14
...
w→x
w−x
cos0 x = lim
Recall the relation (8
...
2
2
(14
...
= −1 · sin
2
0
(14
...
dx
14
...
w→x
w−x
tan0 x = lim
Recall the relation (8
...
1 + tan a tan b
(14
...
Sengupta 11/6/2011
From this we have
tan a − tan b = (1 + tan a tan b) tan(a − b)
...
We can proceed to the derivative now:
tan w − tan x
w→x
w−x
(1 + tan w tan x) tan(w − x)
= lim
w→x
w−x
tan(w − x)
= lim (1 + tan w tan x)
w→x
w−x
= (1 + tan x tan x) · 1,
tan0 x = lim
because
(14
...
θ→0
θ
lim
Thus,
tan0 x = 1 + tan2 x
...
Hence
d tan x
= tan0 x = sec2 x
...
9)
Chapter 15
Differentiability and Continuity
15
...
For example,
the absoute value function is continuous but is not differentiable at 0:
y
y = |x|
x
However, if a function is differentiable then it is continuous:
Theorem 15
...
1 Suppose f is a function defined on a set S ⊂ R and is
differentiable at a point p ∈ S
...
Proof
...
x→p
x−p
f 0 (p) = lim
Note that this is meaningful only when p ∈ S is not an isolated point of S
...
For this we first write f (x)
as f (p) plus the amount it deviates from f (p):
f (x) = f (p) + f (x) − f (p)
...
Sengupta 11/6/2011
Since we have information about
f (x)−f (p)
x−p
let us bring this in:
f (x) − f (p)
f (x) = f (p) + (x − p)
...
QED
We have allowed infinite values for derivatives
...
But note that f is not continuous at 0
...
They are hard to visualize and have severely zig-zag
graphs
...
In summary the basic algebraic rules are
dU
dV
d(U + V )
=
+
dx
dx
dx
d(kU )
dU
=k
dx
dx
dU
dV
d(U V )
=
V +U
dx
dx
dx
(product rule)
(16
...
We
have retained some redundancy here; for example, that d(kU )/dx = kdU/dx
can be deduced from the ‘product rule’ for d(U V )/dx, keeping in mind that
dk/dx = 0
...
1
Ambar N
...
For example
d(5 sin x + 4x3 − 3)
d(5 sin x) d(4x3 − 3)
=
+
dx
dx
dx
(assuming these derivatives exist and are finite)
d sin x d(4x3 ) d(−3)
=5
+
+
dx
dx
dx
(assuming these derivatives exist and are finite)
dx3
+0
dx
= 5 cos x + 12x2
...
Mostly we can write down the
derivative of a sum directly:
d(5 sin x − 3x5 + 2x − 4)
= 5 cos x − 15x4 + 2
...
2
Using the product rule
The product rule gets us to more complicated functions
...
Using the product rule we find the derivative very easily:
√
√ d sin x
√
dy
d x
1
=
sin x + x
= √ sin x + x cos x
...
dx
dx
dx
(16
...
Here is another example of the product rule
√
d(t + tan t)(2t2 + t)(4t − cos t)
dt
√
(2t2 + t)
d(t + tan t) 2 √
(2t + t)(4t − cos t) + (t + tan t)
=
dt
dt
√
d(4t
−
cos t)
+ (t + tan t)(2t2 + t)
dt
√
1
= (1 + sec2 t)(2t2 + t)(4t − cos t) + (t + tan t)(4t + √ )(4t − cos t)
2 t
√
2
+ (t + tan t)(2t + t)(4 + sin t)
...
3
Using the quotient rule
Let us first see quickly how to use the formula
(1/V )0 = −
1 0
V
...
=−
(sin w + cos w + 1)2
Now let us do an example of the full quotient rule
(U/V )0 =
V U0 − UV 0
...
(sin w + cos w + 1)2
=
136
Ambar N
...
17
...
As a first example, look at
d sin(x2 + 3x)
= cos(x2 + 3x) (2x + 3)
dx
The sin has become cos (which is the derivative of sin), and then we have a
multiplier 2x + 3 which we recognize to be the derivative of x2 + 3x
...
dt
3
We recognize the factor on the right as the derivative of the function (·)1/3 ,
evaluated at 4t3 − 2 sin t; next, this is multiplied by the derivative of 4t3 −
2 sin t
...
4
dx
2 x + x2
137
138
Ambar N
...
17
...
This means that to calculate the value H(x) we must first work out the value
G(x)
and then apply the function F to it:
F G(x)
...
As another example,
√
sin w
√
is the cmomposite of sin with ·
...
For
example, the composite function given by
√
1+x
is defined only for x ≥ −1
...
Returning to examples, we have then
√
i 1
d tan x h 2 √ 4
= sec x + 3x2 √ ,
dx
2 x
because we recognize that
√
tan x
√
x:
√
√
tan x = (tan ◦ )(x)
...
1)
140
Ambar N
...
18
...
Suppose now that p ∈ S is a point where the derivatives f 0 (p) and g 0 (p) exist
...
1)
if this sum is defined (that is not ∞ + (−∞) or (−∞) + ∞)
...
18
...
141
142
Ambar N
...
Notice that if a rectangle, whose sides are A
and B, gets enlarged so that it becomes a C by D rectangle then its area
increases by
CD − AB = CD − AD + AD − AB
= (C − A)D + A(D − B)
= (C − A)(D − B + B) + A(D − B)
= (C − A)(D − B) + (C − A)B + A(D − B)
...
2)
Proposition 18
...
1 Let f and g be functions on a set S ⊂ R, and at p ∈ S
is a point where f and g are both differentiable (that is, the derivatives f 0 (p)
and g 0 (p) exist and are finite)
...
(18
...
Recall that for any function h the derivative h0 (p) is defined to be
h(w) − h(x)
...
Then
f (w)g(w) − f (x)g(x)
...
4)
Now we split the numerator following the idea of (18
...
Now divide by w − x to obtain
f (w)g(w) − f (x)g(x)
g(w) − g(x)
= [f (w) − f (x)]
w−x
w−x
f (w) − f (x)
g(w) − g(x)
+
g(x) + f (x)
...
w−x
w−x
Now just let w → x:
f (w)g(w) − f (x)g(x)
= 0 · f 0 (x) · g 0 (x)
w→x
w−x
+ f 0 (x)g(x) + f (x)g 0 (x),
lim
which works because the derivatives f 0 (x) and g 0 (x) have been assumed to
exist and be finite
...
18
...
5)
valid whenever f and g are functions defined on some common domain S
and x is a point of S where f and g are both differentiable and g(x) 6= 0
...
6)
as w → x
...
Sengupta 11/6/2011
f (w) − g(w) and g(w) − g(x):
f (w)g(x) − f (x)g(w)
f (w) f (x)
−
=
g(w)
g(x)
g(w)g(x)
[f (w) − f (x) + f (x)] g(x) − f (x) [g(w) − g(x) + g(x)]
=
g(w)g(x)
[f (w) − f (x)] g(x) + f (x)g(x) − f (x) [g(w) − g(x)] − f (x)g(x)
=
g(w)g(x)
[f (w) − f (x)] g(x) − f (x) [g(w) − g(x)]
=
...
Thus
h
i
h
i
f (w)−f (x)
f (w)
f (x)
g(w)−g(x)
−
g(x)
−
f
(x)
g(w)
g(x)
w−x
w−x
=
...
7)
w−x
g(w)g(x)
Now we let w → x
...
Applying this to the identity (18
...
QED
If in the quotient rule we take the numerator to be the constant function
1 we obtain
(1/V )0 =
a useful formula in itself
...
But, as we shall see,
turning this easy understanding into a proof runs into a snag
...
19
...
To work out the derivative dy/dx, let us introduce some notation:
y = F (u)
where
u = G(x)
...
∆x
∆u ∆x
(19
...
dx
du dx
(19
...
Sengupta 11/6/2011
we take u = 1 + sin x, and then
√
and
y= u
u = 1 + sin x,
so that
dy
dy du
1
1
=
= √ cos x = √
cos x
...
The argument used above is natural but has one technical gap
...
1) is undefined and the argument breaks down
...
Either we can try to
tread through the obstacle very carefully, or we can try an entirely different
pathway, one that is less natural but one that gets us to the destination faster
...
This is a situation one
sometimes faces in trying to construct a proof out of a reasonable idea
...
(In fact, the idea does provide a proof in the case G0 (x) is
not 0
...
2
Proof the chain rule
A proof of the chain rule can be built out of the following useful observation:
Lemma 19
...
1 Let f be a function, with domain S ⊂ R, differentiable at a
point p ∈ R
...
3)
and
fp (p) = f 0 (p)
...
3):
f (x) = f (p) + (x − p)f 0 (p) + p (x)
where
p (x)
= 0,
x→p x − p
lim
for all x ∈ S,
(19
...
5)
DRAFT Calculus Notes 11/17/2011
147
because p (x) = (x − p)[fp (x) − f 0 (p)]
...
4) observe that the first two terms of the right describe the y-value of the
tangent line to y = f (x) at p, and so condition (19
...
6)
where
def
y = Tp f (x) = f (p) + (x − p)f 0 (p)
is the equation of the tangent to the graph of f at p, f (p)
...
1: The tangent as an approximation to the graph
Proof
...
We want to denote the ratio
f (x)−f (p)
x−p
However, we need to say what fp (p) is, for at x = p the ration
not defined
...
So let us define the function fp on S by
(
fp (x) =
f (x)−f (p)
x−p
0
f (p)
f (x)−f (p)
x−p
if x ∈ S and x 6= p;
if x = p
...
7)
by fp (x)
...
8)
148
Ambar N
...
Moreover, from (19
...
But putting in x = p shows that this is also valid
when x = p
...
Proposition 19
...
1 Let f and g be functions on subsets of R, and let S0 be
the set of all x for which the composite
f ◦ g : x 7→ f g(x)
is defined
...
Assume also that p is not an isolated point of
S0
...
(19
...
Thus, if f ◦ g is defined in a neighborhood
of p, g is differentiable at p and f is differentiable at g(p) then f ◦ g is
differentiable at p and (19
...
Proof
...
Recall from Lemma 19
...
1 the functions fq and gp
...
Then
f g(x) − f g(p)
= gp (x)fq g(x)
x−p
(19
...
Now letting x → p (recall that p has been assumed
to be a limit point of S0 ) we have
f g(x) − f g(p)
lim
= lim gp (x)fq g(x)
x→p
x→p
x−p
(19
...
150
Ambar N
...
For example, we will see soon how to find maximum and minimum values of
g(x) = 2x3 − 6x2 + 1
of x ∈ [−1, 1]
...
To see how this works let us apply it to the function g(x) = 2x3 − 6x2 + 1
for x ∈ [−1, 1]
...
This is 0 at x = 0 and at x = 2
...
Now we compute the values of g at x = 0 and at the endpoints
1 and −1:
g(0) = 1
g(1) = −3,
g(−1) = −7
...
Sengupta 11/6/2011
The largest value of g(x) is therefore 1, occurring at x = 0, and the smallest
value is −7, occurring at x = −1
...
If the function f is defined on the whole real line R or intervals such as
(2, ∞), we have to modify step (ii) above to:
(ii)’ the supremum (infimum) of a function f , defined on an interval U , is
the largest (smallest) of the values of f at the points where f 0 is 0 and
the endpoint limit values limx→a f (a) and limx→b f (b), where a and b
are the endpoints of the interval U and the limits here are assumed to
exist
...
The term ‘maximum’ is used when the supremum is actually attained at
a point in the domain of the function; similarly, we speak of the ‘minimum’
value of a function if the infimum is actually attained in the domain of the
function
...
20
...
The value of y here is
y(1) = 13
...
So we need to work out the endpoint limits:
lim y(x) = ∞
and
x→∞
lim y(x) = ∞
...
x∈R
20
...
This is by using the ancient method of
‘completing the square’:
3x2 −6x+16 = 3(x2 −2x)+16 = 3(x2 −2x+1 −1)+16 = 3(x2 −2x+1)−3+16
which shows that
3x2 − 6x + 16 = 3(x − 1)2 + 13
...
Hence 3x2 − 6x + 16 has minimum value
0 + 13 = 13,
and this value is attained when x = 1
...
Completing the square we
have
B
2
2
Ax + Bx + C = A x + x + C
A
"
2
2 #
B
B
B
= A x2 + 2 x +
−
+C
2A
2A
2A
"
2 #
2
B
B
B
2
(20
...
Sengupta 11/6/2011
Thus,
2
B
(B 2 − 4AC)
Ax + Bx + C = A x +
(20
...
(20
...
2) is always ≤ 0 and
the largest it gets is 0, this happening when x = −B/(2A); so
max(Ax2 + Bx + C) = −
x∈R
(B 2 − 4AC)
4A
if A < 0
...
4)
This is a clean and nice solution, but in practice it is faster to simply observe
that
(Ax2 + Bx + C)0 = 2Ax + B
is 0 when x = −B/(2A) and this point corresponds to the maximum/minimum
value of Ax2 + Bx + C
...
5)
Using the completed square form this reads
2
B
(B 2 − 4AC)
=0
−
A x+
2A
4A
from which we have
B
x+
2A
Taking square roots shows that
2
=
(B 2 − 4AC)
...
Thus the two solutions of the
quadratic equation (20
...
2A
2A
DRAFT Calculus Notes 11/17/2011
Observe that
155
√
α−β =
B 2 − 4AC
...
A2
The square of this is
The quantity
A2 (α − β)2 = B 2 − 4AC
(20
...
If the discriminant is 0 then (20
...
On the other hand if
the discriminant is not 0 then the roots α and β are distinct
...
20
...
This distance is, by definition, the shortest distance from P to any point on
the line:
def
d(P, L) = inf d(P, Q),
Q∈L
where
d(P, Q) = distance between the points P and Q
...
156
Ambar N
...
We
can avoid unpleasant calculations by minimizing the distance squared:
d(P, Q)2 = (x − xP )2 + (y − yP )2
...
7)
Clearly if we can find the minimum value of this then we can just take the
square-root to find the minimum distance
...
7) (x, y)
is on the line L and so
y = mx + k
...
This is clearly quadratic in x:
d(P, Q)2 = x2 − 2xP x + x2P + m2 x2 + 2m(k − yP )x + (k − yP )2
= (1 + m2 )x2 + 2[−xP + (k − yP )]x + x2P + (k − yP )2
...
From our study of quadratic functions we know then that
d(P, Q)2 does attain a minimum value and at the point Q0 where it attains
minimum the derivative
d
d(P, Q)2
dx
is 0
...
dx
dx
This is 0 if and only if (x, y) is the special point
Q0 (x0 , y0 )
which satisfies
(x0 − xP ) + (y0 − yP )m = 0
...
8)
DRAFT Calculus Notes 11/17/2011
157
It is worth observing that this implies
1
y0 − yP
=− ,
x0 − xP
m
(20
...
The geometric significance of (20
...
Geometrically this makes perfect sense
...
8), we subsitute in the value of y as mx+k to obtain
(x0 − xP ) + (mx0 + k − yP )m = 0,
which is
(1 + m2 )x0 + km − xP − yP m = 0
...
1 + m2
(20
...
(20
...
Sengupta 11/6/2011
We can now work out the distance between P and Q0 :
d(P, Q0 )2 = (x0 − xP )2 + (y0 − yP )2
= [−m(y0 − yP )]2 + (y0 − yP )2
= m2 (y0 − yP )2 + (y0 − yP )2
= (m2 + 1)(y0 − yP )2
on using (20
...
12)
We need to work oit y0 − yP from (20
...
12) for d(P, Q0 )2 we have:
2
2
d(P, Q0 ) = (m + 1)
mxP + k − yP
1 + m2
2
=
(mxP + k − yP )2
1 + m2
Taking the square root produces at last the distance of P from the line L:
d(P, L) =
|mxP + k − yP |
√
...
13)
Taking a concrete example, let us work out
the distance of the point (1, 2) from the line y = 5x − 2
...
2
1+5
26
The absolute value in the numerator erases a piece of information
...
In fact,
mxP + k − yP
DRAFT Calculus Notes 11/17/2011
159
measures how far ‘below’ (in the vertical y-direction) the line L the point P
lies
...
1 + m2
1 + tan2 α
sec2 α
Thus
|mxP + k − yP |
√
= |(mxP + k − yP ) cos α|
...
Now consider a different way of writing the equation of a line:
Ax + By + C = 0,
where at least one of A and B is not 0
...
Then we can rewrite the
equation as
−C
A
...
m=− , k=
B
B
Then the distance between P (xP , yP ) and L is
C
A
xP − B
− yP |
|− B
|mxP + k − yP |
√
q
=
A2
1 + m2
1+ B
2
Simplifying the algebra this produces the formula
d(P, L) =
|AxP + ByP + C|
√
...
14)
You can check that this formula works even when B is 0, for then the line
L has constant x value −C/A and the x-coordinate of P is xP , so that the
distance is
|AxP + C|
|xP − (−C/A)| =
,
|A|
which matches (20
...
160
Ambar N
...
1: Distance of a point P from points on a line
...
4
Other geometric examples
A straight piece of wire of length L units is to be cut into two pieces, one
of which will be bent into a square and the other a circle
...
This intuition (where does it come
from?) is verified to be correct by the mathematical solution we work out
...
Let x units be the length of the piece that is bent into a circle
...
2π
4π
DRAFT Calculus Notes 11/17/2011
161
The remaining piece is of length
L−x
and when this is bent to form a square, each side of the square has length
L−x
,
4
and its area is
L−x
4
2
...
4π
16
(20
...
(Taking x = 0 means we just form a large square and no circle, and taking
x = L means we form a circle out of the full length of wire and no square at
all
...
8π
Thus
the solution of A0 (x) = 0 is x0 =
π
L
...
π+4
π+4
The total area enclosed is
1
A(x0 ) =
4π
π
L
π+4
2
1
+
16
2
4
L
π+4
(20
...
Sengupta 11/6/2011
Simplifying this we have
A(x0 ) =
=
=
=
=
π2
16
L2 +
L2
2
4π(π + 4)
16(π + 4)2
π
1
L2 +
L2
2
2
4(π + 4)
(π + 4)
4
π
L2 +
L2
4(π + 4)
4(π + 4)2
π+4
L2
4(π + 4)2
1
L2
4(π + 4)
(20
...
15) with x = 0 and x = L:
A(0) =
L2
16
and
A(L) =
L2
...
Thus the largest area is enclosed when x = L, which means we take the
entire length of wire and bend it into a circle
...
A rectangle of sides L units by W units
has four little squares, each having side x units, cut out of the four corners;
the edges are now folded to form a box (with no cover)
...
What
should x be to maximize the volume of the box?
The volume V cubic units is given by
V (x) = x(L − 2x)(W − 2x) = xLW − 2(L + W )x2 + 4x3
The value of x is ≥ 0 but cannot be more than W/2 (we assume W is the
shorter edge, that is: W ≤ L)
...
DRAFT Calculus Notes 11/17/2011
163
The derivative V 0 (x) is
V 0 (x) = LW − 4(L + W )x + 12x2
For this to be 0 we have
12x2 − 4(L + W )x + LW = 0
...
Since
L ≥ W the numerator for the + sign is
p
√
(L + W ) + L2 + W 2 − LW = L + W + L(L − W ) + W 2
√
≥ W + W + 0 + W2
= 3W
(L + W ) ±
which makes the ratio
(L + W ) +
√
L2 + W 2 − LW
3W
W
≥
=
,
6
6
2
falling outside the allowed range, unless we have the extreme case L = W
for which the ratio is W/2
...
6
6
3
164
Ambar N
...
Clearly this choice of x must produce the
maximum value of the volume, for the value of V (x) at the endpoints x = 0
and x = W/2 is 0
...
54
If we start with a square, for which L = W , this simplifies to
2 3
L,
27
with x0 being L/6
...
Find the maximum and minimum values of x2 for x ∈ [−1, 2]
...
Find the maximum and minimum values of
x(6 − x)(3 − x)
for x ∈ [0, 2]
...
A wire of length 12 units is bent to form an isosceles triangle
...
A piece of wire is bent into a rectangle of maximum area
...
5
...
What is the value of x which would make the total area enclosed
by the pieces maximum, and what is the value of x which would make
this area minimum
...
Here are some practice problems on straight lines and distances:
(i) Work out the distance from (1, 2) to the line 3x = 4y + 5
(ii) Work out the distance from (2, −2) to the line 4x − 3y − 5 = 0
...
What is the angle between P0 P and
the line L?
(iv) Let P0 be the point on the line L, with equation 3x + 4y − 11 = 0,
closest to the point P (1, 3)
...
Find the equation of the line through
P and P0
...
Prove the inequality
x3 k 3/2
+
≥ kx,
3
3/2
(20
...
Explain when ≥ is =
...
Note that Φ(0) = 0 and limx→∞ Φ(x) = −∞; so you have to
find a point p ∈ (0, ∞) where Φ0 (p) is 0 and compare the value Φ(p)
with Φ(0) and choose the larger
...
Prove the inequality
x6 + 5k 6/5 ≥ 6kx,
for all x, k ∈ (0, ∞)
...
(20
...
Sengupta 11/6/2011
Chapter 21
Local Extrema and Derivatives
21
...
Suppose u ∈ S is such that the value
of f at u is maximum:
f (u) = sup f (x)
...
Then, in any neighborhood U of u, there are points of S to the right of u
that are in U and there are points of S to the left of d that are also in U
...
On the other hand, for x ∈ U to the left of u we have
f (x) − f (u)
≤0
x−u
because the numerator is ≥ 0 but the denominator is < 0
...
x∈U,x6=u
x−u
x−u
x∈U,x6=u
inf
167
168
Ambar N
...
1) for quasi-tangents
...
In fact these arguments easily establish:
Proposition 21
...
1 Suppose f is a function on a set S ⊂ R, and b ∈ S is
a point in the interior of S
...
If, moreover, f 0 (b) exists then
f 0 (b) = 0
...
If, moreover, f 0 (u) exists then
f 0 (u) = 0
...
A point u is said to be a local maximum of f if the value f (u) is ≥ all other
values over U :
f (u) ≥ f (x) for all x ∈ U ∩ S
...
1
...
Recalling the term ‘quasi-tangent’ we introduced back at the end of section (13
...
DRAFT Calculus Notes 11/17/2011
169
Review Exercises
1
...
Answer and explain briefly:
(i) If 4 < sup T is 4 an upper bound of T ?
(ii) In (i), is there a point of T that is > 4?
(iii) If inf T < 3 is 3 a lower bound of T ?
(iv) In (iii), is there a point of T that is < 3?
3
...
Sengupta 11/6/2011
(ix) If G0 (3) = 4 then
G(3 + h) − G(3)
=
h→0
h
lim
4
...
Using the definition of the derivative, show that
√
1
d(1 x)
=− √
...
This result shows that for any differentiable function f , the slope
of a secant line P Q is actually equal to the slope of a suitable tangent line
to the graph y = f (x)
...
1: A tangent line parallel to a secant
...
1
Rolle’s Theorem
The following is a version of Rolle’s theorem, which is a key step towards
proving the mean value theorem:
Theorem 22
...
1 Let f be a function, continuous on the interval [a, b], where
a, b ∈ R with a < b, and suppose
f (a) = f (b)
...
Sengupta 11/6/2011
Suppose also that the derivative f 0 (x) exists for all x ∈ (a, b)
...
Proof
...
Let us first see that at least one of these values must occur at a point p
in (a, b), the interior of the interval
...
Thus
in all cases there is a p ∈ (a, b) such that f attains either its maximum or its
minimum value at p
...
1
...
QED
There is a small sharpening of Rolle’s theorem we could note, just to see
how proofs can be tweaked to sharpen results
...
Thus, we could drop the requirement in Rolle’s theorem that
f is differentiable and conclude that there is a point p ∈ [a, b] where the
graphy = f (x) has a flat quasi-tangent line
...
2
Mean Value Theorem
The following is the enormously useful Mean Value Theorem:
Theorem 22
...
1 Let f be a function, continuous on the interval [a, b], where
a, b ∈ R with a < b
...
Then there is a point on the graph of f over (a, b) where there is
a tangent line that has slope equal to
f (b) − f (a)
...
(22
...
Consider the secant line passing through the points A = (a, f (a)) and
B = (b, f (b))
...
b−a
(22
...
Consider now how high f rises above this line:
H(x) = f (x) − [M (x − a) + f (a)]
for all x ∈ [a, b]
...
3)
y
y = f (x)
P x, f (x)
B(b, f (b))
H(x)
Equation of line AB is
y = M (x − a) + f (a)
H(x) = f (x)−[M (x−a)+f (a)]
A(a, f (a))
x
Figure 22
...
This function is continuous, being the sum of two continuous functions
...
Hence there is a point p ∈ (a, b) where the tangent line to the graph of H is
flat:
H 0 (p) = 0
...
3) we have
H 0 (x) = f 0 (x) − M
...
Sengupta 11/6/2011
(This makes geometric sense: the slope of H is the slope of f minus the slope
of the line M (x − a) + f (a)
...
QED
22
...
1
...
3
...
Suppose also that the derivative F 0 (x) exists for all x ∈ (a, b)
...
We will use this result in establishing one case of l’Hospital’s rule (Proposition 28
...
2)
...
1
...
Recall that for a function f defined on a set S ⊂ R, and a point p ∈ S, if
U is a neighborhood of p then part of U might not be inside S
...
23
...
In this section we make this
idea precise
...
1
...
If f is increasing on S, that is if f (s) ≤ f (t) for all s, t ∈ S with s ≤ t,
then f 0 (p) ≥ 0 for all p ∈ S where f 0 (p) exists
...
175
176
Ambar N
...
This follows directly from the definition of the derivative:
f (x) − f (p)
...
Hence the ratio f (x)−f
is
x−p
0
≥ 0, and so the limit f (p) is also ≥ 0
...
Hence in this case f 0 (p) ≤ 0
...
1
...
Then there is a neighborhood U of p such that the f (x) > f (p) for x ∈ U ∩ S
to the right of p and f (x) < f (p) for x ∈ U ∩ S to the left of p:
f (x) > f (p) for all x ∈ U ∩ S for which x > p
f (x) < f (p) for all x ∈ U ∩ S for which x < p
(23
...
Proof
...
x→p
x−p
f 0 (p) = lim
If this is > 0 then the ratio
f (x) − f (p)
x−p
DRAFT Calculus Notes 11/17/2011
177
y = f (x)
y
p, f (p)
p
)
(
x
U
Figure 23
...
is also > 0 when x is near p, but 6= p (see Proposition 10
...
1 for this)
...
If we take an x ∈ U ∩ S to the right of p, we have x − p > 0 and so the
f (x) − f (p)
f (x) − f (p) = (x − p)
> 0
...
On the other hand, if x ∈ U ∩ S is to the left of p, we have x − p < 0 and
so the
f (x) − f (p)
< 0
...
QED
Using this we can step up to another result going in the converse direction
to Proposition 23
...
1:
Proposition 23
...
3 Suppose f is a continuous function defined on an interval U , and suppose f 0 (p) exists and is positive (this means > 0) for all p
in the interior of U
...
Sengupta 11/6/2011
f (x1 ) < f (x2 )
for all x1 , x2 ∈ U with x1 < x2
...
Proof
...
By the mean value theorem there
is a point c ∈ (x1 , x2 ) where the derivative f 0 (c) is given by
f 0 (c) =
f (x2 ) − f (x1 )
...
If we assume only that f 0 ≥ 0 then the same argument shows that f (x2 ) ≥
f (x1 )
...
1
...
Then f is strictly increasing
on U
...
Proof
...
By Proposition 23
...
3 f is an increasing function on U in the
sense that f (a) ≤ f (b) for all a, b ∈ U with a ≤ b
...
This proves the result for f 0 ≥ 0
...
QED
DRAFT Calculus Notes 11/17/2011
23
...
Proposition 23
...
1 Let f be a function on a set S ⊂ R, and p a point in
S where f 0 (p) exists and is negative, that is
f 0 (p) < 0
...
2)
If f slopes downward along an interval then it is decreasing:
Proposition 23
...
2 If f is defined on an interval [a, b], where a, b ∈ R with
a < b, and if f 0 (p) exists and is negaive, that is < 0, for all p ∈ [a, b] then f
is strictly decreasing on [a, b] in the sense that:
f (x1 ) > f (x2 )
for all x1 , x2 ∈ [a, b] with x1 < x2
...
23
...
One can run this also in the converse direction,
but with just a bit of care
...
180
Ambar N
...
On the other hand it is also clear that
G really is constant, separately on each interval on which it is defined
...
3
...
Then f is constant on
[a, b]
...
One can tinker with this as usual
...
Proof
...
Then by the mean value
theorem
f (x2 ) − f (x1 )
= f 0 (c),
x 2 − x1
for some c ∈ (x1 , x2 )
...
Thus the values of f at any two different points
are equal; that is, f is constant
...
If f is differentiable then
formal common sense suggests that the derivative of the inverse function
should be
1
1
dx
= dy = 0
...
Of course, we need
to avoid the points where dy/dx is 0 (or undefined)
...
Sengupta 11/6/2011
Note that we have been referring to ‘an’ inverse function
...
Things could be really made messy by choosing an inverse function that
√
√
switches wildly back and forth between the branches y and − y
...
24
...
Let V denote the range of f :
V = f (U ) = {f (x) : x ∈ U }
...
f −1 f (x) = x
for all x ∈ U
...
1
...
Then f −1 (y) exists for all y ∈ V , the range of f , and
0
f −1 (y) =
1
f 0 (x)
(24
...
39) we take the right side 1/f 0 (x) to be 0 in case
f 0 (x) is ∞, and ∞ if f 0 (x) is 0
...
As usual, there is a corresponding result if f 0 is ≤ 0, being
< 0 at all but finitely many points
...
Suppose f 0 ≥ 0 on U and is actually > 0 except possibly at finitely
many points
...
1
...
By Proposition 11
...
2 the range V of f is an interval and the inverse
function f −1 is defined on V
...
Then
f −1 (y) − f −1 (q)
...
2)
Writing x for f −1 (y) and p for f −1 (q) we see that the difference ratio on the
right here is
x−p
...
Thus (24
...
(24
...
4)
this being taken to be 0 if f 0 (p) = ∞ and to be ∞ if f 0 (p) is 0 (these extreme
cases require some care)
...
3
...
Then by Proposition 7
...
1
we have
lim D f −1 (y) = lim D(x)
...
5)
y→q
x→p
184
Ambar N
...
3) and the limit value
in (24
...
(24
...
QED
Chapter 25
Analyzing local extrema with
higher derivatives
25
...
As we have seen, if f 0 (p) exists it must
be 0
...
The basic idea is simple: if a continuous graph is sloping downward just
to the left of a point p and sloping upward just to the right then it must have
a local minimum at p
...
Proposition 25
...
1 Suppose f is defined and continuous on a neighborhood
U of p ∈ R, and the derivative f 0 is ≥ 0 to the right of p and ≤ 0 to the
left of p; more precisely, suppose f is differentiable on U except possibly at
p, and f 0 (x) ≥ 0 for x ∈ U with x > p and f 0 (x) ≤ 0 for x ∈ U with x < p
...
Sengupta 11/6/2011
Then p is a local minimum for f
...
Let us look at the point x = 3
...
Then Proposition 25
...
1 implies that x = 3 gives a local minimum for g
...
Thus g(x) continues to decrease in value as x passes from the left of 0 to the
right of 0, and 0 is not a local maximum or minimum
...
1
...
Consider any x ∈ U to the right of p, that is
x > p; the mean value theorem (Theorem 22
...
1) says that
f (x) − f (p) = (x − p)f 0 (c) for some c ∈ (p, x)
...
Thus
f (x) ≥ f (p) for x ∈ U , with x > p
...
Thus,
f (x) ≥ f (p) for x ∈ U , with x < p
...
This means that f has a local minimum at p
...
1
...
Then p is a local maximum for f
...
2
Ambar N
...
If this is positive we have a local minimum at p, while if it
is negative then we have a local maximum at p
...
(You should,
however, see that since sin θ ' θ near θ = 0, the graph y = sin(x2 ) looks
about like that of y = x2 , near x = 0, and since this clearly has a local
minimum at x = 0, it is a good guess that sin(x2 ) has a local minimum at
x = 0
...
2
...
Assume also that the second derivative f 00 (p) exists
...
If
f 0 (p) = 0
and
f 00 (p) < 0
then f has a local maximum at p
...
1
...
Proof
...
Thus the derivative f 0 (x) exists and is finite at all x ∈ W
...
DRAFT Calculus Notes 11/17/2011
189
Suppose f 00 (p) > 0
...
Then by
Proposition 23
...
2, the value f 0 (x) for x immediately to the left of p is less
than the value f 0 (p), whereas the value f 0 (x) for x immediately to the right
of p is greater than the value f 0 (p)
...
Since we are given that f 0 (p) is 0 this means
f 0 (x) > 0) for x > p and x ∈ U ;
< 0 for x < p and x ∈ U
...
1
...
The argument for f 00 (p) < 0 is very similar, using Proposition ?? to see
first that f 0 is positive to the left of p and negative to the right of p and
concluding then that p is a loca maximum for f
...
For example,
y = x4
is ‘very flat’ at x = 0 since both its first derivative, which is 4x3 , and its
second derivative 12x3 are 0 at x = 0
...
As another example of what can go wrong, consider
g(x) = x4 − 4x3 + 2
The derivative is
g 0 (x) = 4x3 − 12x2 = 4x2 (x − 3)
and the second derivative is
g 00 (x) = 12x2 − 24x = 12x(x − 2)
...
Sengupta 11/6/2011
Thus
g 0 (0) = 0,
and g 0 (3) = 0
and
g 00 (0) = 0,
and g 00 (3) = 36 > 0
...
Chapter 26
Exp and Log
The exponential function is one of the most useful functions in mathematics,
and is expressed through the amazing formula
x
1
x2 x3
1
+
+ ···
...
1)
1 + 1 + + + ··· = 1 + x +
2! 3!
2!
3!
Its inverse function is the natural logarithm log
...
The history of the discovery/understanding of these two functions is an
entertaining example of how mathematical concepts develop trough unexpected twists and turns and near-misses [1, 2, 4]
...
This approach is fast but gives little insight on how or why these
ideas were developed historically
...
1
Exp summarized
The function exp is defined on R by
x2 x3
+
+ ···
exp(x) = 1 + x +
2!
3!
for all x ∈ R
...
The number exp(1) is denoted e:
191
192
Ambar N
...
718281
...
2)
exp(x) = ex ,
which is another way of writing (26
...
What exactly it means to raise e to
the power x, which is a real number, will be explored more carefully later
...
Moreover, the derivative of exp is again exp:
exp0 = exp,
(26
...
dx
(26
...
Figure 26
...
The graph of the exponential function shows rapid increase as x → ∞
and rapid decay to 0 as x → −∞:
lim ex = ∞
x→∞
lim ex = 0
...
5)
x→−∞
Among the early studies of the exponential function is its use in describing
the growth of money under compound interest
...
(26
...
1: Graph of the exponential function
...
2
Log summarized
The natural logarithm is the inverse function of exp: thus log(A) is the real
number with the property that
exp (log(A)) = A,
or, equivalently,
elog A = A
...
7)
It is defined for all A > 0
...
Since e1 = 1 we have
log 1 = e
...
An alternative way to state the fact that log is inverse to the function
exp is
log ex = x for all x ∈ R
...
8)
The basic algebraic properties of log are:
log(AB) = log(A) + log(B)
log(A/B) = log(A) − log(B)
k
log(A ) = k log(A),
(26
...
Sengupta 11/6/2011
y
y = ex
A
eB = A is equivalent to B = log A
B = log(A)
x
Figure 26
...
for all A, B > 0 and k ∈ R
...
In
fact log (up to a scaling) was studied and used well before the exponential
function was identified
...
We have the following limits of interest:
lim log(x) = ∞
x→∞
(26
...
x→−∞
The derivative of log is the reciprocal:
1
d log(x)
=
dx
x
(26
...
Figure 26
...
3: The graph of log
...
(26
...
26
...
13)
Real Powers
For any real number y we have the positive integer powers
y 1 = 1, y 2 = y · y, y 3 = y 2 · y,
...
The 0-th power is
y 0 = 1
...
Of course, this makes sense only for y 6= 0
...
For example,
y 1/2 =
√
y
is defined to be the unique real number ≥ 0 whose square is y:
(y 1/2 )2 = y
...
)
More generally, for any positive real number A > 0 and integers p and q,
with q 6= 0, the power
Ap/q
is defined to be the unique positive real number whose q-th power is Ap :
(Aq )p/q = Ap
...
Sengupta 11/6/2011
Thus, we have a definition for
Ar
for all positive real A and all rational r
...
14)
for all positive real A, B and rationals r and s
...
r→x,q∈Q
(26
...
Perhaps the shortest way to see that Ax exists is by using
log: for rational r we have
r
Ar = (elog(A) ) = e(log(A))r = er log(A) = exp (r log(A))
...
(26
...
We can also verify the algebraic relations:
Ax+y = Ax Ay
(Ax )y = Axy
(AB)x = Ax B x ,
(26
...
Using the derivative of the exponential function we obtain immediately
that
dAx
= ex log A · log(A)
...
18)
dx
Note that on the right we have ex log A which is, in fact, Ax
...
dx
DRAFT Calculus Notes 11/17/2011
26
...
We have already observed in (26
...
Thus, for example,
x)
d2(2
dx
(26
...
Next consider the derivative of the function xx
...
Now we can differentiate this:
dex log(x)
dxx
=
dx
dx
dx log(x)
= ex log(x)
dx
(by the chain rule)
1
x log(x)
=e
1 · log(x) + x ·
x
x
= x [log x + 1]
(recognizing ex log(x) as xx )
...
Similarly,
dx1/x
1/x 1 − log x
=x
...
20)
dx
x2
Observe that this is > 0 when x < e and is < 0 when x > e
...
So x1/x is maximum at
e:
max x1/x = e1/e
...
5
Ambar N
...
We assume that there
exists a function exp defined on R that is differentiable and satisfies the
following conditions:
exp(x) = 1 + x +
exp0 = exp
exp(0) = 1
...
21)
We will prove that there can be at most one such function and then prove
the crucial relation
exp(x) = ex ,
where e = exp(1)
...
In particular,
d exp(−x)
= − exp(−x)
...
5
...
(26
...
(26
...
First we take the derivative of exp(x) exp(−x) and show that it is 0:
exp(−x) =
d exp(x)
d exp(−x)
d exp(x) exp(−x)
=
exp(−x) + exp(x)
dx
dx
dx
= exp(x) exp(−x) − exp(x) exp(−x)
= 0
...
This proves (26
...
QED
The stragegy used in the proof above, showing that the derivative of a
function is 0 and then concluding that its value is constant, equal to the
value at x = 0, will be used several times
...
5
...
Then F is a constant multiples of exp:
F (x) = F (0) exp(x)
for all x ∈ R
...
Consider the function F (x)/ exp(x), which is defined for all x ∈ R
since the denominator exp(x) is never 0 (Proposition 26
...
1)
...
=
(by the quotient rule)
Thus, F (x)/ exp(x) is constant, and so
F (x)
F (0)
F (0)
=
=
= F (0),
exp(x)
exp(0)
1
for all x ∈ R
...
Next we can prove a key algebraic property for exp:
QED
200
Ambar N
...
5
...
(26
...
Consider any a ∈ R and let F be the function
F (x) = exp(a + x)
for all x ∈ R
...
Then by Proposition 26
...
2 we conclude that
F (x) = F (0) exp(x) for all x ∈ R
...
Recalling that F (x) is exp(a + x) we are done
...
5
...
(26
...
This follows from writing x as x/2 + x/2 and using the previous
Proposition:
x x
exp(x) = exp
+
= exp(x/2) exp(x/2) = [exp(x/2)]2
...
Moreover, we know from Proposition 26
...
1
that exp(x/2) is not 0
...
QED
From the exponential multiplicative property in Proposition 26
...
3 we
have
exp(2a) = exp(a + a) = [exp(a)]2
and
exp(3a) = exp(2a + a) = exp(2a) exp(a) = [exp(a)]2 exp(a) = [exp(a)]3
...
26)
for all a ∈ R and N ∈ {1, 2, 3,
...
23))
[exp(a)]N
= [exp(a)]−N
...
27)
for a ∈ R and all integers n ∈ Z (you can check the case n = 0 directly: both
sides are 1 in that case)
...
Hence exp( 12 a) is the positive square-root of exp(a):
exp
1
a
2
= [exp(a)]1/2
...
Then,
by defininition of (·)p/q , we have
exp(ra) = [exp(a)]p/q
...
Sengupta 11/6/2011
Thus,
exp(ra) = [exp(a)]r
(26
...
Finally, we can proceed to an arbitary real power
...
The definition of the x-th power is that
[exp(a)]x =
lim [exp(a)]r
...
r→x,r∈Q
Putting everything together we have:
Proposition 26
...
5 For any real numbers a and x we have
exp(ax) = [exp(a)]x
...
29)
Now we specialize this to a = 1 to obtain the crucial formula:
exp(x) = ex
for all x ∈ R,
(26
...
(26
...
dx
Proposition 26
...
6 The exponential function is strictly increasing:
ea < eb
for all a, b ∈ R with a < b
...
32)
DRAFT Calculus Notes 11/17/2011
203
y
y = ex
y =1+x
x
Figure 26
...
Proof
...
Hence exp is
strictly increasing
...
(26
...
5
...
34)
with = holding only when x = 0
...
4
...
First observe that
d (ex − (1 + x))
= ex − 1
...
Thus, ex − (1 + x)
attains its minmum value when x = 0:
ex − (1 + x) > e0 − (1 + 0) = 1 − 1 = 0
for all x > 0 and all x < 0
...
34)
...
34) is the exp(x) goes to ∞ when
x → ∞:
lim ex = ∞
...
35)
x→∞
204
Ambar N
...
x→−∞
(26
...
37)
exp (log(A)) = A for all A ∈ (0, ∞);
log (exp(B)) = B for all B ∈ R
...
38)
satisfying
If a function f , defined on an open interval U , has an inverse f −1 , and if
f is differentiable at p ∈ U with f 0 (p) 6= 0 then the derivative of the inverse
function is given by
1
(f −1 )0 (q) = 0 ,
(26
...
Applying this to the exponential
function and its inverse, log, we have first
log0 q =
1
1
= ,
p
e
q
for all p ∈ R and q = ep
...
(26
...
It is useful in establishing results about
maxima and minima, and useful in many different applications
...
27
...
y
y = f (x)
B b, f (b)
A
a
b
x
Figure 27
...
Sengupta 11/6/2011
might touch or run along each other
...
A function f is strictly convex on an interval U if the graph of f between
any two points in U lies strictly below the corresponding secant segment
...
It
is strictly concave of the graph lies strictly above all secant segments
...
y
B
A
x
Figure 27
...
27
...
Let us take three points a, p, b ∈ U
with
a
...
Next let B be (b, f (b)) and P the point (p, f (p))
...
Thus Q is of the
form
(p, L(p))
where L(p) is the y-cordinate of Q
...
DRAFT Calculus Notes 11/17/2011
207
This means that the slope of AP is ≤ the slope of AQ
...
Thus the condition that P is below AB is
equivalent to
slope AP ≤ slope AB
...
Thus, convexity of f is equivalent to the relations
slope AP ≤ slope AB ≤ slope P B,
for all a, p, b ∈ U , with p ∈ [a, b]
...
3: Convex function: secant slopes increase
Just the condition
slope AP ≤ slope P B
(27
...
For example, since
the slope of AP is ≤ the slope of P B, the segment AP produced beyond P
would end up below B, and this would mean that the slope of AP is ≤ the
slope of AB
...
208
27
...
Sengupta 11/6/2011
Checking convexity/concavity
In this section we will see how to check convexity/concavity of specific functions and how to work with such functions
...
A function f is convex on an interval U if its derivative f 0 is an increasing
function on U (this means f (s) ≤ f (t) if s ≤ t); it is strictly convex if f 0 is
strictly increasing (f (s) < f (t) if s < t)
...
For concavity we have the analogous results
...
If f 00 is ≤ 0 on U
then f is concave on U , and if f 00 < 0 on U then f is strictly concave on U
...
Its second derivative is 2:
(x2 )0 = 2x and (x2 )00 = 2 > 0,
and so x2 is strictly convex on R
...
(In fact it is strictly convex even though 12x2 hits 0 at the single
point x = 0
...
The derivative is pxp−1
and the second derivative is
p(p − 1)xp−2
...
In fact, xp is strictly convex on (0, ∞) if
p > 1
...
Thus
1/x is strictly convex for x ∈ (0, ∞),
and is strictly concave for x < 0
...
The second derivative is ex , which is positive
...
Lastly consider
log x
for x > 0
...
Hence log x is strictly concave
...
4
Inequalities from convexity/concavity
Recall that a function Φ, on an inerval U , is convex if its graph lies below
the secant segments, with strict convexity meaning that the graph always
lies strictly below the secant segments
...
2)
210
Ambar N
...
Strict convexity means that here ≤ would be replaced by
< except in the trivial case where a = b
...
2) is
altered by replacing ≤ by ≥
...
For example, if Φ is convex on an interval U then
1
8
1
1
8
1
a + b + c ≤ Φ(a) + Φ(b) + Φ(c),
Φ
10
10
10
10
10
10
for all a, b, c ∈ U
...
Thus,
2
1
3
1
a+ b+ c+ d
7
7
7
7
is a convex combination of a, b, c, and d
...
, pN ∈ R is of the form
w1 p1 + · · · + wN pN ,
where the weights w1 ,
...
Thus, (27
...
3)
for all p1 ,
...
, wN ∈ [0, 1] (adding to 1), for
every choice of N ∈ {1, 2, 3,
...
3)
holds with ≤ replaced by < if the points p1 ,
...
For concavity we simply reverse the inequalities
...
3) to the functions we looked at in the preceding section
...
(27
...
Note
that since x2 is actually strictly convex the inequality ≤ can be replaced by
<, unless a = b
...
3
3 3
(27
...
, xN ∈ R we have
x1 + · · · + xN
N
2
≤
1 2
x1 + · · · + x2N
...
6)
Here the ≤ can be replaced by < except in the case all the xi ’s are equal
...
6) as:
x21 + · · · + x2N ≥
(x1 + · · · + xN )2
,
N
(27
...
Here is a quick application: suppose a length L of wire is cut into N
pieces, of lengths x1 ,
...
7) we see that this is
≥
(x1 + · · · + xN )2 /N
L2
=
,
16
16N
with ≥ being = if and only if all the xi are equal
...
Now we turn to another example, the strictly concave function log, defined
on (0, ∞)
...
(27
...
Sengupta 11/6/2011
The right side simplifies to
√
1
1
(log a + log b) = log(ab) = log ab
...
9)
with equality if and only if a = b
...
10)
2
with equality if and only if a = b
...
11)
3
with equality if and only if a, b and c are equal
...
9) and (27
...
(The first inequaity (27
...
However, both express a common
idea: the arithmetic mean (AM) is greater or equal to the geometric mean
...
xN ) N ,
N
(27
...
This follows from the strict
concavity of log
...
Here
are a few convex functions that are not differentiable at all points:
|x|,
x+ = max{x, 0},
for any constant K ∈ R
...
4: The convex function |x|
y
y = (x − 1)+
x
Figure 27
...
5
Convexity and derivatives
We have seen in section 27
...
From this we arrive at a condition for
convexity for differentiable functions:
Proposition 27
...
1 Suppose f is a function on an interval U on which f
is differentiable
...
The function f is concave if and only if f 0 is an decreasing function in
the sense that f 0 (s) ≥ f 0 (t) for all s, t ∈ U with s ≤ t
...
Suppose f is convex on U , and s, t ∈ U with s < t
...
Sengupta 11/6/2011
we have the increasing slope condition
f (x) − f (s)
f (t) − f (w)
≤
x−s
t−w
Now taking the limits x → s and w → t we have
f 0 (s) ≤ f 0 (t)
...
Consider
a, p, b ∈ U with
a < p < b
...
Then
slope of AP =
f (p) − f (a)
= f 0 (s) for some s ∈ (a, p),
p−a
by the mean value theorem
...
Hence
slope of AP ≤ slope of P B
...
The argument for concavity is exactly similar
...
Similarly, if f 0 is strictly decreasing then f is strictly concave
...
5
...
Then
(i) f is convex if and only if f 00 is ≥ 0 on U ;
(ii) f is concave if and only if f 00 is ≤ 0 on U
...
Note that in (iii) and (iv) we don’t have the ‘only if’ parts
...
Proof
...
1
...
5
...
Conversely,
if f is convex then by Proposition 27
...
1 the derivative f 0 is an increasing
function on U and so, by Proposition 23
...
1, its derivative f 00 is ≥ 0 on U
...
QED
4
27
...
We have seen in
(27
...
(27
...
Consider now the line L through (p, Φ(p)) whose slope is m
...
But to the left of p the slope of L
is greater than the secant slopes and so again the line L lies below the graph
of Φ to the left of p
...
Sengupta 11/6/2011
Proposition 27
...
1 Suppose Φ is a convex function on an open interval U ,
and let p be anypoint in U
...
y = Φ(x)
y = L(x)
y
x
Figure 27
...
If Φ0 (p) exists at p then there is only one choice for the line L: it is the
line through (p, Φ(p)) with slope m = Φ0 (p), that is, it is the tangent line
to the graph of Φ at (p, Φ(p))
...
Proof
...
(27
...
13) holds for every a ∈ U to the left of p we see that the ‘right
slope’ Φ(b)−Φ(p)
is an upper bound for all the ‘left slopes’, and so
b−p
Φ(a) − Φ(p)
Φ(b) − Φ(p)
≤
,
a−p
b−p
a∈U,a
because sup is the least upper bound
...
b−p
a∈U,a
DRAFT Calculus Notes 11/17/2011
217
Thus,
Φ(b) − Φ(p)
for all b ∈ U with p < b
...
15)
Φ(b) − Φ(p)
...
Now we can simply take m to lie between these values:
m− leqm ≤ m+
QED
The secant slope inequalities for a convex function have the following
remarkable consequence:
Proposition 27
...
2 Any convex function on an open interval U is continuous on U
...
6
...
6
...
16)
L≤Φ
where L denotes any function of the form L(x) = M x + k, with constant
M, k ∈ R, for all x ∈ U
...
We have already seen in Proposition 27
...
1 that the graph of Φ has
a supporting line at every point
...
This proves (27
...
QED
218
Ambar N
...
7
Convex combinations
If a, b ∈ R are points with a < b, then any point p ∈ [a, b] can be reached by
starting at a and moving towars b, covering a fraction
p−a
b−a
of the interval [a, b]
...
Let µ denote the
fraction
p−a
µ=
...
We can write p as
p = a + µ(b − a) = a + µ(b) − µa = a − µa + µb = (1 − µ)a + µb
...
17)
where λ and µ are weights, in the sense that they are non-negative and add
up to 1 (which forces λ and µ to be ≤ 1):
λ, µ ∈ [0, 1],
λ + µ = 1
...
17) is at most b (the weight on a,
which is < b, would draw the value of p down below b) and at least a:
a ≤ any convex combination of a and b ≤ b
...
5
5
5
A convex combination of points p1 ,
...
, wN all lie in [0, 1] and sum to 1:
w1 + · · · + wN = 1
...
For a given set of points p1 ,
...
Similarly, the least value a convex combintation of p1 ,
...
A multiple convex combination can be built out of convex combinations
of pairs
...
More generally, for N ∈ {2, 3,
...
, pN ∈ R, and any
weights w1 ,
...
18)
1 − w1
220
Ambar N
...
, pN −1
...
7
...
19)
for all p1 ,
...
, wN ∈ R with w1 + · · · + wN = 1
...
19) we have just a linear combination
w 1 p1 + · · · + w N pN
and the coefficients wi need not be in [0, 1] nor have to sum to 1
...
It is more convenient to start with the right side of (27
...
Thus L maps linear combinations to linear combinations
...
7
...
20)
for all a, b ∈ R and all weights λ, µ ∈ [0, 1] with λ + µ = 1
...
20) holds for all a, b, λ, µ as above but
with ≤ replaced by < whenever the three points a, λa + µb and b are distinct
(no two are equal to each other)
...
We can work with a, b ∈ U , with a < b (if a = b then (27
...
Let A be the point (a, Φ(a)) and B the
point (b, Φ(b))
...
Consider now any point p ∈ [a, b]; we can write this as
p = λa + µb
for some λ, µ ∈ [0, 1] with λ + µ = 1 (see (27
...
The condition that the
graph of Φ is below the graph of L is
Φ(p) ≤ L(p)
for all such p
...
7
...
Since y = L(x) passes through A and B, on the graph
y = Φ(x), we have
L(a) = Φ(a),
and
L(b) = Φ(b)
...
21)
Combining all these observations we have
Φ(λa + µb) ≤ λL(a) + µL(b) = λΦ(a) + µΠ(b),
which establishes (27
...
For strict convexity, the point p, Φ(p) lies strictly below (p, L(p)), which
means Φ(p) < L(p) when p is strictly between a and b
...
21) we obtain the condition for
strict convexity of Φ
...
20) to an inequality for convex
combinations for multiples points
...
222
Ambar N
...
22)
for every convex function Φ on any interval U , any N ∈ {1, 2,
...
, pN ∈ U , and all weights w1 ,
...
DRAFT Calculus Notes 11/17/2011
223
Exercises on Maxima/Minima, Mean Value Theorem,
Convexity
1
...
Explain your reasoning
fully and present all calculations clearly
...
Find the distance of the point (1, 2) from the line whose equation is
3x + 4y − 5 = 0
...
Suppose f is a twice differentiable function on [1, 5], with f (1) = f (3) =
f (5)
...
4
...
01
...
Prove the inequality
1
≤
a+b 2
2
for any a, b > 0
...
Sengupta 11/6/2011
Chapter 28
L’Hospital’s Rule
L’Hospital’s rule makes it possible to compute weird limits such as
lim x1/x ,
x→∞
and is worth studying just for that reason
...
Briefly, l’Hospital’s rule says
f (x)
f 0 (x)
= lim 0
x→p g(x)
x→p g (x)
lim
(28
...
Assuming that f and g both have domain a set S, here
is a more detailed statement of the conditions:
• there is a neighborhood U of p ∈ R∗ such that that the part of U in
S, with p added in if necessary, that is the set W = (S ∩ U ) ∪ {p}, is
either U or a one-sided neighborhood of p of the form (a, p] or [p, a),
for some a ∈ R;
• g(x) 6= 0 and g 0 (x) 6= 0 for all x ∈ W , with x 6= p;
• limx→p f (x) and limx→p g(x) are either both 0 or are both in {−∞, ∞};
• the limit limx→p
f 0 (x)
g 0 (x)
exists
...
1
Ambar N
...
1 3
x→0
x
3!
lim
The first thing to observe is that both numerator and denominator go to 0
as x → 0
...
2)
provided the limit on the right exists
...
3)
and we do know that this limit exists and its value is −1
...
3) holds, and this shows that the right
side of (28
...
2) by l’Hospital’s
rule
...
cos x − 1
= lim 1 2
(if the right side exists)
1 3
x→0
x→0
x
3x
3!
3!
− sin x
l0 H
...
(28
...
Now we carry this a step beyond, finding an estimate for the difference
1 3
sin x − x − x
...
cos x − 1 − 3!1 3x2
lim
= lim
(if the right side exists)
1 5
1 5
x→0
x→0
x
x
5!
5!
cos x − 1 − 2!1 x2
= lim
(by algebraic simplification)
1 4
x→0
x
4!
− sin x − − 2!1 2x
l0 H
...
(28
...
We turn now to a different example:
1
1
lim x x = lim elog x x
x→∞
x→∞
= lim e
log x
x
x→∞
= elimx→∞
l0 H
...
log x
x
1/x
limx→∞ 1
(the exponent here is formally ∞/∞)
(28
...
2
Ambar N
...
This is formalized in
the following version of the mean value theorem:
Proposition 28
...
1 Suppose F and G are continuous functions on a closed
interval [a, b], where a, b ∈ R∗ and a < b, with values in R
...
Then
F 0 (c)
F (b) − F (a)
= 0
G(b) − G(a)
G (c)
(28
...
Since G0 is never 0 on (a, b) it follows by Rolle’s theorem that G(b)−G(a) 6= 0
...
Consider the function H defined on [a, b] by
H(x) = [G(b) − G(a)] [F (x) − F (a)] − [F (b) − F (a)] [G(x) − G(a)]
(28
...
This is clearly continuous on [a, b] and differentiable on (a, b) with derivative
given by
H(0 x) = [G(b) − G(a)] F 0 (x) − [F (b) − F (a)] G0 (x)
for all x ∈ (a, b)
...
(28
...
(28
...
7)
...
2
...
x→p
f (x)
f 0 (x)
lim
= lim 0
x→p g(x)
x→p g (x)
(28
...
11) exists
...
Let W be a closed interval, one of whose endpoints is p and for which
all points of W , except possibly for p, lie inside U
...
What this does is make f and g defined and continuous at p in case it wasn’t
to start with; more precisely, F and G are continuous on U ∪ {p}
...
Then,
for such x, we have by Proposition 28
...
1
f (x)
F 0 (cx )
f 0 (cx )
= 0
= 0
,
g(x)
G (cx )
g (cx )
230
Ambar N
...
Letting x → p makes cx → p and so
f (x)
f 0 (cx )
f 0 (w)
= lim 0
= lim 0
,
x→p g(x)
x→p g (cx )
w→p g (w)
lim
since we assume that the right side here exists
...
2
...
x→p
f 0 (x)
f (x)
= lim 0
x→p g (x)
x→p g(x)
lim
(28
...
12) exists
...
From
"
f (a) #
f (x) 1 − f (x)
f (x) − f (a)
=
g(x) − g(a)
g(x) 1 − g(a)
g(x)
(28
...
This is the motivation for
the strategy we use
...
Choose a smaller neighborhood W1 of L such
that
cW1 ⊂ W
for all c ∈ (1/2, 2)
...
14)
There is an interval V that is a neighborhood of p such that
f 0 (x)
∈ W1
g 0 (x)
(28
...
Pick any two distinct points a, x ∈ V ∩ U ,
neither equal to p; then
f (x) − f (a)
f 0 (c)
= 0
g(x) − g(a)
g (c)
for some c between a and x, and hence in V (note that g(x) 6= g(a) because
by Rolle’s theorem if g(x) = g(a) then g 0 would be 0 at a point between a
and x, contrary to the given assumption that g 0 is never 0 on U )
...
15) we have
f (x) − f (a)
∈ W1 ,
g(x) − g(a)
for distinct a, x ∈ V ∩ U , neither equal to p
...
13) we have
"
1−
f (x) − f (a)
f (x)
=
g(x)
g(x) − g(a)
1−
g(a)
g(x)
f (a)
f (x)
#
and so, on using (28
...
Since this is true for any neighborhood W of L, we
conclude that f (x)/g(x) → L as x → p
...
Sengupta 11/6/2011
Exercises on l’Hosptal’s rule
1
...
2
...
Work out the limit
g(2 + w) + g(2 + 3w)
w→0
w
lim
clearly justifying each step
...
Find
1
lim y y
y→∞
Chapter 29
Integration
The development of calculus has two original themes: (i) the notion of tangent to a curve, (ii) computing areas of curved regions
...
29
...
This simple and perfectly natural idea fails to produce a completely
satisfactory and usable measure of area when the region S is ‘unintuitive’
(for example S consists of all points (x, y) ∈ R2 with irrational coordinates),
but it is meaningful, intuitive and computable for regions bounded by well
behaved curves
...
See Figure 29
...
For this discussion we
assume f ≥ 0
...
An overestimate of A will surely be obtained by the sum of areas of ‘upper
rectangles’ obtained by slicing the region into vertical pieces
...
Sengupta 11/6/2011
partition [a, b] into N intervals marked off by
a = t0 < t1 <
...
By the k-th ‘upper rectangle’ we mean the rectangle whose base runs along
the x-axis from x = tk−1 to x = tk and whose height is
Mk =
sup
f (x)
...
1)
x∈[tk−1 ,tk ]
The area of this upper rectangle is
Mk (tk − tk−1 )
...
(29
...
The sum of all the upper rectangles is
N
X
Mk ∆tk = M1 ∆t1 + · · · + MN ∆tN
...
k=1
Similarly, working with lower rectangles we have an understimate of the
area:
N
X
mk ∆tk ≤ A,
k=1
where
mk =
inf
f (x)
...
3)
x∈[tk−1 ,tk ]
Thus the actual area A lies between the overestimates given by the upper
sums and underestimates given by the lower sums:
N
X
k=1
mk ∆tk ≤ A ≤
N
X
Mk ∆tk
...
4)
k=1
Surely, the area A is the unique value that lies between all the upper sums
and all the lower sums
...
DRAFT Calculus Notes 11/17/2011
235
y
y = f (x)
a
x
b
Figure 29
...
2: Area below y = f (x) overestimated by upper recangles
29
...
Let us first extract some helpful terminology from our earlier
discussions
...
Typically we denote
a partition by
P = {x0 , x1 ,
...
< xN = b
...
Sengupta 11/6/2011
The width of the k-th interval is denoted
∆xk = xk − xk−1
...
5)
For a function
f : [a, b] → R,
and the partition P , the upper sum is
U (f, P ) =
N
X
Mk (f )∆xk ,
(29
...
7)
k=1
and the lower sum is
L(f, P ) =
N
X
k=1
where
Mk (f ) =
sup
f (x)
x∈[tk−1 ,tk ]
mk (f ) =
inf
f (x)
...
8)
x∈[tk−1 ,tk ]
In the degenerate case where b = a, the only partition of [a, a] is just the
one-point set {a}, and the upper and lower sums are taken to be 0
...
9)
for every partition P of [a, b], then A is called the Riemann integral of f , and
denoted
Z
b
f
...
Rb
We say that f is integrable if a f exists and is finite (in R)
...
1) and the concept of tangent line in (13
...
Rb
From (29
...
10)
DRAFT Calculus Notes 11/17/2011
237
where x∗k is any point in [xk−1 , xk ], for each k ∈ {1,
...
The sum on the
right in (29
...
The relation (29
...
Note that if supx∈[a,b] f (x) is ∞ then at least one Mk (f ) is ∞, for any
P , and so the upper sums are all ∞, and in this case the integral
Rb
Rpartition
b
f
,
if
it
exists, must also be ∞
...
Thus, if f is integrable then it is bounded, in the sense that both its
supremum and its infinmum are finite
...
, N }, and so
U (f, P ) = C∆x1 + · · · + C∆xN = C(∆x1 + · · · + ∆xN ) = C(b − a),
and
L(f, P ) = C∆x1 + · · · + C∆xN = C(∆x1 + · · · + ∆xN ) = C(b − a)
...
(29
...
3
Refining partitions
Consider a function f on an interval [a, b], and a partition P = {x0 ,
...
< xN = b
...
Let us start by adding one
point s ∈ (xj−1 , xj ) to the j-th interval
...
Sengupta 11/6/2011
all terms remain the sum except for the j-th term: for this
Mj (f )∆xj
is replaced by
A(s − xj−1 ) + B(xj − s)
where A is the sup of f over [xj−1 , s] and B is the sup of f over [s, xj ]:
A=
sup
f (x)
and
B = sup f (x)
...
Hence
A(s − xj−1 ) + B(xj − s) ≤ Mj (f )(s − xj−1 ) + Mj (f )(xj − s),
and observe that the right side here adds up to Mj (f )∆xj ; thus:
A(s − xj−1 ) + B(xj − s) ≤ Mj (f )∆xj
...
Thus, adding a point to a partition (that is, splitting one of the intervals into
two) lowers the upper sum
...
Adding points one by one enlarges a given partition P to any given larger
partition P 0 , and at each stage in this process the upper sum is lowered and
the lower sum is raised:
DRAFT Calculus Notes 11/17/2011
239
Proposition 29
...
1 Let f : [a, b] → R be a function, where a, b ∈ R and
a ≤ b, and P and P 0 any partitions of [a, b] with P ⊂ P 0 ; then
L(f, P ) ≤ L(f, P 0 )
U (f, P 0 ) ≤ U (f, P )
...
12)
This implies the following natural but strong observation:
Proposition 29
...
2 Let f : [a, b] → R be a function, where a, b ∈ R and
a ≤ b, and P and Q any partitions of [a, b]; then
L(f, P ) ≤ U (f, Q)
...
13)
Thus, every upper sum of f is ≥ every lower sum of f
...
14)
...
Let
P 0 = P ∪ Q
...
3
...
Combining this with the fact that L(f, P 0 ) ≤ U (f, P 0 ) produces the inequality
(29
...
QED
29
...
, xN } be
a partition of [a, b] with
a = x0 <
...
We know that the integral of f , if it exists, lies between the upper sum
U (f, P ) and the lower sum L(f, P )
...
Let us find how far from each other the upper and
lower sums are:
N
N
X
X
U (f, P ) − L(f, P ) =
Mk (f )∆xk −
mK (f )∆xk
=
k=1
N
X
k=1
k=1
[Mk (f ) − mk (f )] ∆xk ,
(29
...
Sengupta 11/6/2011
where Mk (f ) is the sup of f over [xk−1 , xk ] and mk (f ) is the inf of f over
[xk−1 , xk ]
...
(29
...
01 then
U (f, P ) − L(f, P ) <
...
01∆xN
=
...
01)(b − a)
...
16)
Thus we can shrink U (f, P ) − L(f, P ) down by choosing the partition P such
that the fluctuation of f over each interval [xk−1 , xk ] is very small
...
5
Continuous functions are integrable
The discussions of the preceding section lead to the following important result:
Theorem 29
...
1 If f : [a, b] → R is continuous, where a, b ∈ R with a ≤ b,
then f is integrable
...
5
...
, xN } of [a, b], with
a = x0 <
...
This property is called uniform continuity of f
...
We work with a < b, as the case a = b is trivial
...
Since f is continuous at a there is some x1 > a = x0 such that
f (x0 ) −
< f (x) < f (x0 ) +
4
4
DRAFT Calculus Notes 11/17/2011
241
for all x ∈ [x0 , x1 ]
...
Thus
sup f (x) ≤ f (x0 ) +
4
x∈[x0 ,x1 ]
and
f (x) ≥ f (x0 ) −
...
Now we can start at x1 , if it isn’t already b, and produce a point x2 > x1
for which
f (x1 ) − < f (x) < f (x1 ) +
4
4
for all x ∈ [x1 , x2 ]
...
It might seem that in this
way we could produce the desired partition P
...
Fortunately, we can
show that this will not happen
...
, xK } for which the fluctuations of f over every interval of the
partition is <
...
By continuity of f at s there is an interval
(p, q), centered at s, such that the fluctuation of f over (p, q) ∩ [a, b] is <
...
, xK }
of [a, t] such that the fluctuation of f over each interval [xj−1 , xj ] is <
...
Since the fluctuation of f over (p, q) is < , the fluctuation of f over the
subinterval [t, r] is <
...
, xK+1 } of [a, r] for which the
fluctuations of f are all <
...
QED
Now we can prove Theorem 29
...
1
...
Sengupta 11/6/2011
Proof of Theorem 29
...
1
...
Let > 0 and set, for convenience
...
5
...
, xN },
of [a, b] such that
a = x0 <
...
, N }
...
16), we
have
U (f, P ) − L(f, P ) < 1 (b − a)
...
QED
29
...
17)
We focus on the restriction of 1Q over any interval [a, b] ⊂ R, with a < b
...
, xN } be any partition of [a, b], with
a = x0 < x1 <
...
Observe that 1Q attains both the value 1 and the value 0 on each interval
[xk−1 , xk ] (because every such interval contains both rational and irrational
points)
...
DRAFT Calculus Notes 11/17/2011
243
This makes the upper sum large:
U (1Q , P ) = 1 · ∆x1 +
...
+ 0 · ∆xN = 0
...
Hence,
Z b
1Q does not exist
...
7
Basic properties of the integral
Integration of a larger function produces a larger number:
Proposition 29
...
1 If f and g are functions on an interval [a, b], where
Rb
Rb
a, b ∈ R and a ≤ b, for which the integrals a f and a g exist, and if f ≤ g
then
Z b
Z b
g
...
Suppose a f > a g
...
a
Rb
Again, since a f is the unique value lying between all upper and lower sums
for f there is a partition Q of [a, b] for which
L(f, Q) > U (g, P )
...
Since Q ⊂ P 0 and f ≤ g we have
L(f, Q) ≤ L(g, Q) ≤ L(g, P 0 )
...
18)
244
Ambar N
...
Combining these inequalities produces
L(f, Q) ≤ L(g, P ),
contradicting (29
...
QED
Next we have linearity of the integral:
Proposition 29
...
2 Suppose f and g are functions on an interval [a, b],
Rb
Rb
where a, b ∈ R and a ≤ b, for which the integrals a f and a g exist
...
Then the integral
and
Z b
Z b
Z b
(f + g) =
f+
g
...
19)
a
kf of the function kf also exists and
b
Z
kf = k
a
Rb
b
f
...
20)
a
If a function is integrable over an interval then it is integrable over any
subinterval:
Proposition 29
...
3 Suppose f is integrable over [a, b], where a, b ∈ R and
a ≤ b
...
Moreover,
Z
b
Z
f=
a
c
Z
f+
a
b
f
...
21)
Chapter 30
The Fundamental Theorem of
Calculus
The fundamental theorem of calculus connects differential calculus and integral calculus, and makes it possible to compute integrals by running the
derivative process in reverese
...
1
Fundamental theorem of calculus
Here is one form of the fundamental theorem:
Theorem 30
...
1 Suppose f is an integrable function on [a, b], where a, b ∈
R, with a < b
...
Thus,
if f is continuous on [a, b] then
Rx
d a f
= f (x)
(30
...
Notice that (30
...
Here is another version of the result:
245
246
Ambar N
...
1
...
Then
Z b
b
0
(30
...
a
a
b
where g means g(b) − g(a):
a
b
g = g(b) − g(a)
...
2
Differentials and integrals
Consider a function f , differentiable at a point
p ∈ R
...
y
y = f (x)
tangent line
df |p (h) = f 0 (p)h
P
h
p
x
Figure 30
...
The differential of
f at p is denoted df |p and its value on any h ∈ R is thus
df |p (h) = f 0 (p)h
...
3)
DRAFT Calculus Notes 11/17/2011
247
For example, for the function sin we have
d sin |π (3) = cos(π) ∗ 3 = −3
...
The slope of y = x is 1 everywhere and so
dx|p (h) = 1 ∗ h = h
...
Thus, for a differentiable function f on some set S ⊂ R, the differential
df
is a differential form
...
(30
...
5)
Working with a differentiable function f we have
(df (x))|p (h) = f 0 (p)h
and this is exactly the same as
f 0 (x)dx |p (h) = f 0 (p)h
...
(30
...
Sengupta 11/6/2011
All of this notation has been designed to produce this notational consistency:
df (x)
= f 0 (x),
dx
(30
...
Using equation (30
...
8)
where f and g are differentiable functions on some common domain except
that in the last identity we assume the composite f (g(x)) is defined on some
open interval
...
)
As an example, we have
d log(sin x2 ) =
1
cos(x2 ) ∗ 2x dx
...
(30
...
All of what we have done in this section is essentially just notation
...
10)
dg(x) =
g = g(b) − g(a) =
a
a
a
Thus,
b
Z
Z
0
g (x) dx =
b
g0
...
11)
a
a
If f is any continuous function on [a, b] then by the fundamental theorem of
calculus there is a function F on [a, b] for which
F 0 = f
...
a
Thus we have, finally, a complete identification of the Riemann integral as
the integral of a differential form:
Z b
Z b
f (x) dx
...
12)
f=
a
a
The Riemann integral is rooted in ideas going back to Archimedes’ computation of areas
...
The equality
sions) was done only in the early 20th century by Elie
(30
...
12) is viewed simply
as different notation for the same integral
...
3
Using the fundamental theorem
Let us start with a simple example
...
0
250
Ambar N
...
1
...
We can easily find a function g for which
g 0 (x) = x
for all x ∈ R
...
dx
2
Then by Theorem 30
...
2 we have
Z 1
Z 1
x2 1 12 02
1
x dx =
(x2 /2)0 dx = =
−
=
...
This is just a rght angled triangle with base 1 and height 1,
and so its area is indeed (1/2)1 · 1 = 1/2
...
The area is
Z
b
x2 dx
...
Recall that
(x3 )0 = 3x2
and so
(x3 /3) = x2
...
2: Area below y = x2 for x ∈ [a, b]
Z
b
2
Z
x dx =
a
b
0
3
(x /3) dx
b 3 a3
x3 b
−
...
30
...
2
=
a
Archimedes amazing determination of areas assocated with parabolas has
thus been reduced to a simple routine calculation
...
y = sin x
Rπ
0
sin x dx = 2
π
x
Figure 30
...
(30
...
4
Ambar N
...
The indefinite integral of a function f is the general function
g for which
g 0 = f
...
For example, we know that
(x3 /3)0 = x2
...
3
Denoting this ‘arbitrary constant’ by C we have
g(x) =
x3
+ C
...
The presence of this arbitrary constant
ensures that we have the ‘general’ solution to the problem of finding a function whose derivative is x2 , not just one special choice
...
In general, in mathematics it is not good practice to use ambiguous notation, but writing
Z
f (x) dx
for not one function but for the class of all functions with derivative f (x) is
worth the occasional discomfort caused by the ambiguity
...
Thus an integral of a differential form Φ is a function g for which
dg = Φ
...
For any integer n, except for n = −1,
Z
xn+1
+ C,
xn dx =
n+1
(30
...
n+1
What if n = −1? Then we are seeking a function g(x) whose derivative is
1/x:
1
g 0 (x) =
...
t
254
Ambar N
...
15)
for any arbitary constant C, provided we restrict the functions all to the
positive ray (0, ∞)
...
Thus sometimes it is convenient to use the combined formula
log |x| + C
for the indefinite integral of 1/x
...
A full exploration of this would require going into the complex plane,
a subject well beyond our objectives
...
5
Revisiting the exponential function
Historically the logarithm log was invented before the exponential function
...
This is essentially
what we have done, except that we never really defined ex in a logically
connected way
...
5 we devloped the properties of ex by assuming
that there is a function exp on R with the following two properties:
exp0 = exp
exp(0) = 1
...
(30
...
We defined log to be the inverse function of exp, defining
log a to be the unique real number for which
elog a = a,
with log a being defined for all a ∈ (0, ∞)
...
Having developed integration theory we are not at a point where we can
turn this logical development on its head (thereby regaining the correct, if
strange, historical development) by defining the function log directly as
Z a
1
dx
for all a ∈ (0, ∞)
...
17)
log a =
1 x
Since 1/x is continuous on (0, ∞), the fundamental theorem of calculus
guarantees that the integral defining log a exists and is finite; moreover it
also assures us that
log0 a =
1
a
for all a ∈ (0, ∞)
...
18)
This derivative being strictly positive, log has an inverse function, call it exp
...
1
...
1/x
Moreover, the definition of exp as inverse of log along with the simple fact
that
log(1) = 0
shows that
exp(0) = 1
...
16), making
the development of both exp and log logically complete
...
Sengupta 11/6/2011
Chapter 31
Riemann Sum Examples
In this chapter we work out some Riemann sums and compare them with the
corresponding integrals
...
1
Riemann sums for
RN
1
dx
x2
Let N be a an integer > 1
...
, N }
...
, [N − 1, N ]
...
We will work out the upper and lower Riemann sums for 1/x2 relative to
P
...
The area
of the k-th lower rectangle is therefore
area of k-th lower rectangle =
257
1
· 1,
(k + 1)2
258
Ambar N
...
Similarly
area of k-th upper rectangle =
1
· 1
...
2
(k + 1)
2
3
N
Similarly, the upper sum is
U (1/x2 , P ) =
N
−1
X
k=1
1
1
1
1
·
1
=
+
+
·
·
·
+
...
x2
Z
1
lies between these value:
2
Z
L(1/x , P ) ≤
1
(31
...
We can work out the integral of 1/x2 using the fundamental theorem of
calculus:
Z N
Z N
dx
1
1
1
1
=
d −
=− − −
=1−
...
1) produces:
1
1
1
1
1
1
1
+ 2 + ··· + 2 ≤ 1 −
≤ 2 + 2 + ··· +
...
2)
DRAFT Calculus Notes 11/17/2011
259
We can extract some information from this by focusing on the first inequality:
1
1
1
1
+ 2 + ··· + 2 ≤ 1 −
...
3)
This is true for all integers N ∈ {2, 3,
...
increases in value as additional terms are added on:
s1 < s2 < s3 < · · ·
Therefore, there is a limit
lim sN =
N →∞
sup
sN
...
}
From (31
...
N →∞
Thus,
lim
N →∞
1
1
1
+
+ · · · + 2 ≤ 1
...
One says that the series
X 1
1
1
1
=
+
+
+ ···
2
2
2
2
n
1
2
3
n
(31
...
The convergence of theR series above can be seen in other ways, but the
method using the integral 1/x2 dx is useful for other similar sums as well
...
4) can be computed by more advanced
methods; the amazing result is
1
1
1
π2
+
+
+
·
·
·
=
...
5)
260
Ambar N
...
Observe that
Z t
dx
1
lim
= 1
...
31
...
6)
on the left side to mean the limit of
Rt
1
Riemann sums for 1/x
For the partition
P = {1, 2,
...
1
2
N −1
The exact integral, which lies between these sums, is
Z
1
N
N
dx
= log(x) = log(N ) − log(1) = log(N )
...
2
N
1
2
N −1
(31
...
(31
...
7) we have
1 1
lim
+ + · · · ≥ lim log(N ) = ∞,
N →∞ 1
N →∞
2
and so
∞
X
1
1 1 1
= + + + · · · = ∞
...
9)
P
The series ∞
n=1 is called the harmonic series and the fact that the sum is
∞ is expressed by saying that series is divergent
...
k
It turns out that this has a finite limit as N → ∞:
1
1 1
γ = lim
+ + ··· +
− log(N ) ,
N →∞ 1
2
N
(31
...
31
...
Let N be a an integer > 1
...
,
...
,
,
...
Sengupta 11/6/2011
The k-th interval has the form
k−1 k
,
...
On the interval
k−1 k
,
N N
the sup of x is just Nk and the inf is k−1
...
Similarly
area of k-th upper rectangle =
k 1
·
...
N
(31
...
N
The integral
Z
1
x dx
0
(31
...
L(x, P ) ≤
(31
...
This implies that there can be only one number lying between the upper
sums and the lower sums and hence that the integral
Z
1
x dx
0
exists and
Z
1
1
[1 + 2 + · · · + N ]
...
24) below):
1 + 2 + ··· + N =
N (N + 1)
...
14)
Hence
Z
1
1 N (N + 1)
1 N · N (1 + 1/N )
(1 + 1/N )
= lim 2
= lim
...
2
(31
...
(31
...
4
Ambar N
...
Let N be a an integer > 1
...
,
...
N N
On the interval
k−1 k
,
N N
2
2
...
Similarly
area of k-th upper rectangle =
k2 1
·
...
N
Similarly, the upper sum is
N
X
k2 1
·
2 N
N
k=1
1 12
22
N2
+
+ ··· + 2
=
N N2 N2
N
1 2
= 3 1 + 22 + · · · + N 2
...
17)
U (x2 , P ) =
(31
...
L(x , P ) ≤
(31
...
N
Hence there can be only one number lying between the upper sums and the
lower sums and so the integral
Z 1
x2 dx
U (x2 , P ) − L(x2 , P ) =
0
exists and
Z
1
1 2
1 + 22 + · · · + N 2
...
31) below):
x2 dx = lim
12 + 22 + · · · + N 2 =
N (N + 1)(2N + 1)
...
20)
Hence
Z 1
x2 dx = lim
0
1 N (N + 1)(2N + 1)
1 N · N (1 + 1/N )N (2 + 1/N )
= lim 3
3
N →∞ N
N →∞ N
6
6
(1 + 1/N )(2 + 1/N )
= lim
N →∞
6
1
=
...
21)
Hence
Z
1
1
x2 dx =
...
22)
3
0
This, of course, agrees with the result using the fundamental theorem of
calculus:
0
Z 1
Z 1
1 3
x3 1 1
2
x dx =
x
dx = =
...
23)
3
3 0 3
0
0
266
31
...
Sengupta 11/6/2011
Power sums
We will work out a few power sum formulas, starting with
N (N + 1)
...
Let
1 + 2 + ··· + N =
(31
...
Writing the same sum backwards we have:
S1 = 1 + 2 + · · · + (N − 1) + N
S1 = N + (N − 1) + · · · + 2 + 11
(31
...
Adding
them we have
2S1 = (N + 1) + (N + 1) + · · · + (N + 1) = N (N + 1)
...
2
Here is a second, longer and far less appealing method
...
Now use a = 1, through a + N in this to obtain:
22 − 12 = 2 ∗ 1 + 1
32 − 22 = 2 ∗ 2 + 1
...
2
2
(N + 1) − N = 2 ∗ N + 1
(31
...
{z
}
|
S1
DRAFT Calculus Notes 11/17/2011
267
Hence
2S1 = (N +1)2 −1−N = (N +1)2 −(N +1) = (N +1 −1)(N +1) = N (N +1),
which implies
S1 =
N (N + 1)
...
27)
The advantage of this method is that it works for sums of higher powers
...
(31
...
Using a = 1 through a = N in this produces
23 − 13 = 3 ∗ 12 + 3 ∗ 1 + 1
33 − 23 = 3 ∗ 22 + 3 ∗ 2 + 1
...
3
3
(N + 1) − N = 3 ∗ N 2 + 3 ∗ N + 1
...
29)
When we add these up on the left everything cancels except the very first
term −13 and the very last term (N + 1)3 , and we have:
(N + 1)3 − 13 = 3 ∗ 12 + 22 + · · · + N 2 +3 [1 + 2 + · · · + N ] +N ∗ 1
...
Sengupta 11/6/2011
Using the formula (31
...
= (N + 1)N
2
(31
...
6
Applied to the sum of the cubes this strategy produces
2
N (N + 1)
3
3
3
1 + 2 + ··· + N =
...
31)
(31
...
(31
...
What this means is that we are given a function f (x) and we have to find a
function g(x) for which
dg(x) = f (x)dx
...
We write the general such function as the indefinite integral
Z
g(x) = f (x) dx
...
32
...
Instead we substitute y for x + 1 and rewrite the integral in terms
of y:
y =x+1
269
270
Ambar N
...
The essence of the idea behind the substitution method is simple
...
The main challenge
is in identifying the functions F and p which express f (x) as F (p(x))p0 (x)
...
For example, it may be easier to write f (x) as a constant multiple of
F (p(x))p0 (x) or, in some cases, we can break up f (x) into a sum of pieces,
each of which is easier to work out separately
...
We substitute
z = 2x − 5
which gives
dz = 2dx
and so
1
dx = dz
2
DRAFT Calculus Notes 11/17/2011
271
and so
Z
(2x − 5)
3/5
Z
dx =
z
3/5
Z
3
1
1
dz =
z 5 dz
2
2
3
1
1
z 5 +1 + C
= ∗ 3
2 5 +1
5
= (2x − 5)8/5 + C,
16
where C is an arbitrary constant
...
Sometimes the integrand should be reworked a bit before or after the
substitution is made
...
1)
1
dx = − dy,
4
272
Ambar N
...
xy
4
We need to replace the x in the integrand with its expression in terms of y
by solving (32
...
4
Then our integral becomes
Z
Z
1
1
1
2/5
(3 − y)y
(3 − y)y 2/5 dy
...
1)
...
Moving on to more complicated integrands, consider
Z
(x2 + 1)2/3 x dx
Observe that xdx is about the same as d(x2 + 1), aside from a constant
multiple
...
DRAFT Calculus Notes 11/17/2011
273
With this we have
1
dy = 2xdx and so xdx = dy
2
which converts the given integral as follows:
Z
Z
Z
2
1
1 1
2
2/3
2/3 1
y 3 +1 + C,
dy =
(x + 1) xdx = y
y 2/3 dy = 2
2
2
2 3 +1
where C is any constant
...
10
Here is a faster display of this method:
Z
Z
2 1
2
2/3
(x + 1) x dx = (x2 + 1) 3 d(x2 + 1)
2
1 1
= 2
(x2 + 1)2/3+1 + C
2 3 +1
3
= (x2 + 1)5/3 + C
...
For example,
for
Z
(3x2 − 12x + 2)−7/3 (x − 2) dx
observe again that (x − 2)dx is a constant multiple of d(3x2 − 5x + 2); so we
use
y = 3x2 − 12x + 2
for which
dy = (6x − 12) dx = 6(x − 2)dx
and this transforms the given integral as follows:
Z
Z
Z
1
2
−7/3
−7/3 1
dy =
y −7/3 dy,
(3x − 12x + 2)
(x − 2) dx = y
6
6
which integrates out to
Z
7
4
1 1
1
(3x2 −12x+2)−7/3 (x−2) dx =
y − 3 +1 +C = − (3x2 −12x+2)− 3 +C,
7
6 −3 + 1
8
274
Ambar N
...
Sometimes the substitution is not as obvious as in the preceding (manufactured) example
...
2
x +1
Observe that the numerator is x, which is constant times the dervative of
x2 + 1
...
x2 + 1
Z
Similarly,
Z
(32
...
3)
1√
=−
u+C
2
1√
=−
3 − 2x2 + C,
2
where C is an arbitrary constant
...
For
Z
dx
x log x
we observe the x in the denominator and recall that d log x = 1/x; so we
rewrite the integral as
Z
Z 1
Z
dx
dx
d(log x)
x
=
=
= log(log x) + C,
x log x
log x
log x
x
√
dx =
3 − 2x2
Z
DRAFT Calculus Notes 11/17/2011
275
for any arbitrary constant C
...
Here is an example with trigonometric functions:
Z
Z
1
sin x cos x dx = sin x d(sin x) = sin2 x + C,
2
(32
...
We can do the same integral using the
substitution cos x instead:
Z
Z
Z
1
sin x cos x dx = cos x sin x dx = − cos x d(cos x) = − cos2 x + C1 ,
2
(32
...
For the first time we have now denoted
the arbitrary constant by C1 instead of C and this is the first time we face
a possible pitfall of the ambiguity in the meaning of the indefinite integral
...
4 ) for sin x cos x dx into
1
1
1
(1 − cos2 x) + C = − cos2 x + + C,
2
2
2
and this agrees with (32
...
Here is a slightly more involved example of this type of substitution:
Z
Z
1
6
sin x cos x dx = sin6 x d(sin x) = sin7 x + C,
(32
...
We can take this a step beyond, with
Z
sin4 x cos7 x dx
...
276
Ambar N
...
This brings us to
Z
Z
4
7
sin x cos x dx = sin4 x(1 − sin2 x)3 d(sin x)
...
Here we have the integral of a polynomial and though it is a lengthy write-up
the integration is routine:
Z
Z
4
2 3
y (1 − y ) dy = y 4 (1 − 3y 2 + 3y 4 − y 6 ) dy
Z
= (y 4 − 3y 6 + 3y 8 − y 10 ) dy
3
3
1
1
= y 5 − y 7 + y 9 − y 11 + C,
5
7
9
11
where C is the arbitrary constant; substituting back sin x for y produced the
complete integral:
Z
1
3
3
1
sin4 x cos7 x dx = sin5 x − sin7 x + sin9 x −
sin11 x + C
...
2
Some trigonometric integrals
As starting point we have the integrals
Z
sin x dx = − cos x + C1
Z
cos x dx = sin x + C2
(32
...
Next up we have the simplest substitutions:
Z
Z
Z
1
1
1
d(3x) =
sin(3x) dx = sin(3x)
sin(3x) d(3x) = − cos(3x) + C
3
3
3
and, similarly,
Z
1
sin(5x) + C 0 ,
5
with C and C 0 being arbitrary constants
...
The key strategy is the use of the trigonometric sum formulas such as
sin(a + b) = sin a cos b + sin b cos a
sin(a − b) = sin a cos b − sin b cos a
(32
...
2
(32
...
2
2
Integrating, we obtain
Z
1
1
1
sin(3x) cos(7x) dx =
− cos(10x) + cos(4x) + C,
2
10
4
where C is any constant
...
For
these, recall
cos(a + b) = cos a cos b − sin a sin b
cos(a − b) = cos a cos b + sin a sin b
...
10)
278
Ambar N
...
Hence
1
[cos(a + b) + cos(a − b)]
2
1
sin a sin b = [cos(a − b) − cos(a + b)]
...
11)
Using these we have
Z
Z
1
[cos(8x) + cos(2x)] dx (note that cos(−2x) = cos(2x))
cos(3x) cos(5x) dx =
2
1 1
1
=
sin(8x) + sin(2x) + C
2 8
2
(32
...
Similarly,
Z
Z
1
[cos(4x) − cos(8x)] dx (note that cos(−4x) = cos(4x))
sin(2x) sin(6x) dx =
2
1 1
1
=
sin(4x) + sin(8x) + C
2 4
8
(32
...
We can apply this even to
Z
sin2 x dx,
viewing the integrand as the product sin x sin x:
sin2 x = sin x sin x =
1
1
[cos(0) − cos(2x)] = [1 − cos(2x)] ,
2
2
DRAFT Calculus Notes 11/17/2011
from which we have
Z
Z
1
2
sin x dx =
[1 − cos(2x)] dx
2
1
1
x − sin(2x) + constant
...
14)
where C is an arbitrary constant
...
2 2
Similarly for cos2 x we have
cos2 x = cos x cos x =
which leads to
Z
1
1
[cos(0) + cos(2x)] = [1 + cos(2x)],
2
2
1
1
cos x dx =
x + sin(2x)
...
15)
We can use the sum-formula strategy multiple times
...
16)
sin(5x) sin(3x) cos(4x) =
from which we have
Z
1
1
1
1 1
sin(5x) sin(3x) cos(4x) dx =
sin(6x) + sin(2x) −
sin(12x) − sin(4x)
4 6
2
12
4
280
32
...
Sengupta 11/6/2011
Summary of basic trigonometric integrals
Let us recall the following derivatives
sin0 x = cos x
cos0 x = − sin x
tan0 x = sec2 x
csc0 x = − csc x cot x
sec0 x = sec x tan x
cot0 x = − csc2 x
...
17)
These invert to give the following integrals:
Z
cos x dx = sin x + C
Z
sin x dx = − cos x + C
Z
sec2 x dx = tan x
(32
...
dx = − cot x + C
...
For instance,
tan x dx
...
So we
use the substitution
y = cos x
DRAFT Calculus Notes 11/17/2011
281
which gives
dy = − sin x dx
and so
Z
tan x dx =
and so
−dy
=−
y
Z
dy
= − log y + C,
y
Z
tan x dx = − log cos x + C
...
(32
...
(In complex analysis the correct integral is (32
...
)
The integral
Z
sec x dx
is also of interest
...
The following tricky method is based on
already happening on the correct answer by accident:
Z
Z
sec x(sec x + tan x)
dx
sec x dx =
sec x + tan x
(32
...
Sengupta 11/6/2011
and so we use the subtitution
y = sec x + tan x
...
y
Thus
Z
sec x dx = log(sec x + tan x) + C
...
21)
Again, we can use log | · · · | if sec x + tan x is negative
...
4
Using trigonometric substitutions
For the integral
Z
dx
√
1 − x2
the best substitution is
x = sin θ
...
Then
Z
Z
dx
cos(θ) dθ
√
p
=
1 − x2
1 − sin2 θ
Z
cos(θ) dθ
√
=
(which shows why we use x = sin θ)
2θ
cos
Z
cos θ dθ
=
(here we use cos θ ≥ 0 for θ ∈ [−π/2, π/2]
...
22)
where C is an arbitrary constant
...
A similar integral is
Z
dx
...
More involved is
Z √
1 − x2 dx
...
Sengupta 11/6/2011
When we substitute
x = sin θ
and
dx = cos θ dθ
we obtain
Z p
Z √
Z √
2
2
1 − x dx =
1 − sin θ cos θ dθ =
cos2 θ cos θ dθ
(32
...
We get past this bit of unpleasantness by requiring that
θ = sin−1 x ∈ [−π/2, π/2],
which ensures that
cos θ ≥ 0
and so
√
cos2 θ = cos θ
...
23) we have
Z
Z
Z √
2
1 − x dx = cos θ cos θ dθ = cos2 θ dθ
...
15) we have then
Z √
1
1
1 − x2 dx =
θ + sin(2θ) + C,
2
2
where C is an arbitrary constant
...
Before doing this observe that
sin(2θ) = 2 sin θ cos θ
and so, since
sin θ = x and
cos θ =
√
1 − x2 ,
DRAFT Calculus Notes 11/17/2011
we conclude that
Z √
1 − x2 dx =
285
i
√
1 h −1
sin x + x 1 − x2 + C
...
24)
Sometimes we need to do some algebra before using a trigonometric substitution
...
√
It is best to work the term inside the · · · into a ‘completed squares’ form;
for convenience we work with the negative so that the coefficient of x2 is
positive:
4x2 − 12x + 6 = (2x)2 − 2 ∗ (2x) ∗ 3 + 32 − 32 + 6
= (2x − 3)2 + [6 − 32 ] (using (A − B)2 = A2 − 2AB + B 2 )
= (2x − 3)2 − 3
...
25)
Thus
−4x2 + 12x − 6 = 3 − (2x − 3)2 ,
and so
Z √
Z p
−4x2 + 12x − 6 dx =
3 − (2x − 3)2 dx
...
23)
...
Alternatively, we can make
the similarity with (32
...
26)
2x − 3 = 3y,
which ensures that
− 4x2 + 12x − 6 = 3 − (2x − 3)2 = 3 − 3y 2 = 3(1 − y 2 ),
and
2 dx =
So
Z √
(32
...
√
√
Z p
Z p
3
3√
2
2
−4x + 12x − 6 dx =
3(1 − y )
dy =
3
1 − y 2 dy
...
Sengupta 11/6/2011
Using (32
...
22
Substituting in the value of y from (32
...
The
algebra can be made nicer by observing that
i 1h
i
p
p
3 1 h −1
sin y + y 1 − y 2 =
sin−1 y + 3y 1 − y 2
22
4
i
√ p
1 h −1
sin y + 3y 3(1 − y 2 )
=
4
√
1
2x − 3
−1
2
√
=
sin
+ (2x − 3) −4x + 12x − 6
4
3
(32
...
27) and the value of y from (32
...
32
...
Then
Z
Z
U dV = U V − V dU
...
For example,
Z
Z
log x dx = (log x)x − x d(log x)
Z
dx
= x log x − x
x
(32
...
DRAFT Calculus Notes 11/17/2011
287
Thus,
Z
log x dx = x log x − x + C,
(32
...
Sometimes the choice of U and V requires some planning ahead:
Z
x log x dx =
=
=
=
=
Z
1
log x d(x2 )
2
Z
1
1
2
(log x)x −
x2 d(log x)
2
2
Z
1
1
dx
2
(log x)x −
x2
2
2
x
Z
1
1
(log x)x2 −
x dx
2
2
1
1
(log x)x2 − x3 + C,
2
6
(32
...
We can apply this method to
Z
x sin x dx
as follows:
Z
Z
x sin x dx = −
x d(cos x)
Z
= − x(cos x) − cos x dx
Z
= −x cos x + cos x dx
= −x cos x + sin x + C
where C is an arbitrary constant
...
32)
288
Ambar N
...
33)
R
and at this stage it looks bad at first because we have ended up with sec3 x dx
again, on the right
...
21) for
Z
sec3 x dx =
R
sec x dx
...
34)
where C is an arbitrary constant
...
One other example of this type is
Z
eAx sin(Bx) dx
DRAFT Calculus Notes 11/17/2011
289
where A and B are non-zero constants
...
35)
B
Z
1 Ax
Ax
e cos(Bx) − A cos(Bx) e dx
=−
B
Z
1 Ax
A
= − e cos(Bx) +
cos(Bx) eAx dx
...
36)
with our original integral reappearing on the right side, with a negative sign
...
1+ 2
B
B
B
Multiplying both sides by B 2 gives
Z
2
2
(A + B ) eAx sin(Bx) dx = −BeAx cos(Bx) + AeAx sin(Bx) + constant
= eAx [cos(Bx) + A sin(Bx) − B cos(Bx)] + constant
(32
...
Sengupta 11/6/2011
Dividing by A2 + B 2 produces, at last, the formula
Z
Ax
Ax A sin(Bx) − B cos(Bx)
e sin(Bx) dx = e
+ C,
(32
...
We assumed A and B are both nonzero
...
Exercises on Integration by Substitution
1
...
1
Paths
A path c in the plane R2 is a mapping
c : I → R2 : t 7→ c(t) = xc(t), yc(t) ,
(33
...
We can think of c(t) as being the position of
a point at time t
...
1) we are denoting the x-coordinate of a point p by x(p):
x(p) = x-coordinate of a point p,
so that the x-coordinate of c(t) is x c(t) , which we write briefly as
xc(t)
...
As our first example, consider
c(t) = (t, 2t + 1)
for t ∈ R
...
1
...
Sengupta 11/6/2011
y
t 7→ c(t) = (t, 2t + 1)
c(1
...
1: The path c : R → R2 : t 7→ (t, 2t + 1)
This is a point traveling at a uniform speed along a straight line
...
dt
dt
This is called the velocity of the path c at time t
...
Here is a different path that also travels along the same line, but with
increasing speed:
R → R2 : t 7→ c(t) = (t2 , 2t2 + 1)
...
2
...
5) = (0
...
5)
x
Figure 33
...
The path
[0, 2π] → R2 : t 7→ cos t, sin t)
traces out the unit circle counterclockwise exactly once, starting out at
(cos 0, sin 0) = (1, 0)
and ending also at
(cos 2π, sin 2π) = (1, 0)
...
3: The path [0, 2π] → R2 : t 7→ (cos t, sin t)
...
In this example we have a different interpretation of t also possible: t is
simply the measure of the angle between the x-axis and the line from the
origin to the location of the point
...
It is said to be differentiable at t if its x and y components are
differentiable at t
...
(33
...
3)
294
33
...
Sengupta 11/6/2011
Lengths of paths
Consider a path
c : [a, b] → R2 ,
where a, b ∈ R and a < b
...
, tN }
be a partition of [a, b], with
a = t0 < t1 <
...
We think of these as time instants when we ‘observe’ the path and note
where it is
...
The length of this polygonal
approximation is
l(c; P ) = d (c(t0 ), c(t1 )) + d (c(t1 ), c(t2 )) +
...
4)
(33
...
If we refine the partition P by adding one more point, say s, which lies
between tj−1 and tj , then for the new partition P1 we have a corresponding
length
l(c; P1 )
...
DRAFT Calculus Notes 11/17/2011
295
Thus adding points to a partition increases the length of the corresponding polygonal approximation
...
(33
...
Hence we define the length of c to be
l(c) =
sup
l(c; P )
...
7)
partitions P of [a,b]
If c has a continuous derivative then we have a clean and simple formula
for the length of c;
Z bp
(xc)0 (t)2 + (yc)0 (t)2 dt
...
8)
l(c) =
a
Let us apply this to the circle
c(t) = (cos t, sin 2) t ∈ [0, 2π]
...
33
...
Briefly, a curve is a path but without worrying
about the specific speed at which the path moves
...
Sengupta 11/6/2011
R → R2 : t 7→ (t, sin t)
are the same curve, the only difference between them being that the second
path is always a bit ‘ahead’ of the first one
...
We say that a path
c1 : [a, b] → R2
is a reparametrization of a path
c2 : [a0 , b0 ] → R2
if
c2 (t) = c1 φ(t)
for all t ∈ [a0 , b0 ],
for some clock-changing mapping
φ : [a0 , b0 ] → [a, b] : t 7→ φ(t)
that is continuous and satisfies
φ(a0 ) = a and φ(b0 ) = b,
and
φ(p) < φ(q) if
p
for all p, q ∈ [a, b]
...
We can say that two paths which are reparametrizations of each other
correspond to the same curve
...
Here is an important observation:
Proposition 33
...
1 If a path c1 is a reparametrization of a path c2 then
they have the same lengths:
l(c1 ) = l(c2 )
...
This is also called arc length
...
One could insist, for example,
that only differentiable reparametrizations be allowed
...
4
297
Lengths for graphs
Consider now a function
f : [a, b] → R
where a, b ∈ R with a < b
...
The length of this path
is
Z bp
1 + f 0 (x)2 dx
...
9)
a
Here is an example
...
10)
had four parts, one in the positive quadrant and the others obtained by
reflections across the two axes
...
This length is
Z 1p
1 + (y 0 )2 dx
...
10) we have on taking d/dx of both sides:
2 −1/3 2 −1/3 0
x
+ y
y = 0,
3
3
and so, on doing the algebra, we have
y0 = −
y 1/3
...
− 13 + 1 0 2
298
Ambar N
...
2
Next consider the curve
y = x2
x ∈ [0, a],
for any a > 0
...
4
...
4: Arc length for y = x2
The length is
1
Z
Z
p
0
2
1 + (y ) dx =
1+
w2
√
1 + 4x2 dx
0
0
Recall that
Z √
a
i
√
1h √
2
2
dw =
w 1 + w + log(w + 1 + w + C,
2
where C is an arbitrary constant
...
Hence
Z a√
i
√
1h √
1 + 4x2 dx =
2a 1 + 4a2 + log(2a + 1 + 4a2
...
Sengupta 11/6/2011
Chapter 34
Selected Solutions
Solutions for Exercise Set 8
...
1
...
2
...
limx→1
x2 −9
x−3
4
...
= −8/(−2) = 4
...
x→1
x+1
lim
5
...
x→1
x+1
lim
301
302
Ambar N
...
limx→∞
1
x2
7
...
x3 4 − x32 + x23
4x3 − 3x + 2
lim
= lim 2
x→∞ x2 − x + 1
x→∞ x 1 − 1 + 12
x
x
4 − x32 + x23
= lim x
x→∞
1 − x1 + x12
4−0+0
=∞·
= ∞
...
x6 5 −
5x6 − 7x + 2
=
lim
lim
x→∞ x6 3 +
x→∞ 3x6 + x + 2
7
x5
1
x5
+
+
2
x6
2
x6
5 − x75 + x26
x→∞ 3 + 15 + 26
x
x
5−0+0
=
= 5/3
...
x
x3 4 + sin
4x3 + sin x
3
x
√
√ = lim
lim
x→∞ 2x3 +
x→∞ x3 2 + x
x
x3
= lim
x→∞
sin x
x3
√
+ x3x
4+
2
4+0
=
= 2,
2+0
where we used (sin x)/x3 → 0 as x → ∞ by using, for example, the
‘squeeze theorem’:
sin x 1
x3 ≤ x3 → 0 as x → ∞,
√
1
and also x/x3 = x3−1/2
→ 0 as x → ∞
...
3
x5 7 + x14 + cosx5x
7x5 + x + cos(x3 )
lim
=
lim
x→∞
x→∞ x5 2 − 53 + 15
2x5 − 5x2 + 1
x
x
7 + x14 + cosx5x
= lim
x→∞ 2 − 53 + 15
x
x
7
7+0+0
= ,
=
2−0+0
2
11
...
In order to deterime
whether one of the two terms wins out over the other we need some
trick
...
x→∞
x+1+ x
√ i
x + 1 − x = lim
√
304
Ambar N
...
lim
h√
x→∞
3x2
√
√
√
√
3x2 + 1 − x2 + 1
3x2 + 1 + x2 + 1
√
√
x→∞
3x2 + 1 + x2 + 1
3x2 + 1 − (x2 + 1)
√
= lim √
x→∞
3x2 + 1 + x2 + 1
(using (A − B)(A + B) = A2 − B 2 )
2x2
= lim q
p
x→∞
x2 3 + 1/x2 + x2 (1 + 1/x2 )
i
√
+ 1 − x2 + 1 = lim
= lim
2x2
p
3 + 1/x2 + x 1 + 1/x2
2x2
i
= lim hp
p
x→∞
2
2
x
3 + 1/x + 1 + 1/x
x→∞
x
p
= lim p
x→∞
2x
3 + 1/x2 +
p
1 + 1/x2
∞
3+1
= ∞
...
limx→∞
√
4x4 + 2 −
√
x4 + 1
Use the same method as for the previous problem and reach
DRAFT Calculus Notes 11/17/2011
305
h√
i
√
4
4
lim
4x + 2 − x + 1 = stuff
x→∞
3x4 + 1
hp
i
p
x→∞ 2
x
4 + 2/x4 + 1 + 1/x4
= lim
x4 (3 + 1/x4 )
hp
i
p
x→∞ 2
4
4
x
4 + 2/x + 1 + 1/x
= lim
x2 (3 + 1/x4 )
p
= lim p
x→∞
4 + 2/x4 + 1 + 1/x4
∞(3 + 0)
=
= ∞
...
limx→∞
√
x+1
√
x
15
...
306
Ambar N
...
=
1+1
2
x→∞
1
2/x2
+
p
1 + 1/x2
DRAFT Calculus Notes 11/17/2011
16
...
1+1
2
= lim
17
...
18
...
limθ→π/6
= limθ→0
sin(θ−π/6)
θ−π/6
sin θ 2
θ
= limx→0
= 12 = 1
sin x
x
= 1, on setting
x = θ − π/6
308
Ambar N
...
20
...
21
...
22
...
23
...
Sol: Near x = 3 the supremum of the values x(x − 1)1Q (x) is around 3(3 −
1) = 6 (actually, more than this, because if x > 3, with x rational, then
x(x − 1)1Q (x) = x(x − 1) > 3(3 − 1)), whereas the inf is 0 on taking x
irrational
...
Explain why limx→∞ cos x does not exist
...
) and the next higher such multiple
...
x∈(t,∞)
DRAFT Calculus Notes 11/17/2011
309
Since there is no unique value lying between 1 and −1, the limit
limx→∞ cos x does not exist
...
Explain why limx→∞ x sin x does not exist
...
) and the next higher such multiple
...
x∈(t,∞)
x∈(t,∞)
Since there is no unique value lying between 1 and −1, the limit
limx→∞ sin x does not exist
...
Explain why limx→∞
sin x
x
= 0
...
x ≤ x → 0
27
...
Sol: This follows, for instance, by the ‘squeeze’ theorem on observing that
sin x 1
√ ≤ √ → 0
as x → ∞
...
1
...
Sengupta 11/6/2011
(i) an interior point: −2
(ii) a limit point: −5
...
(v) the interior S 0 = [−∞, −1) ∪ (1, 2) ∪ (9, ∞]
(vi) the boundary ∂S =
Sol: ∂S = {−1, 1, 2, 6, 8, 9}
...
2
...
(iv) In (iii), is there a point of T that is < 3?
Sol: Since 3 is not a lower bound of T there must be a point of T that
is < 3
...
Answer the following concerning limits, with brief explanations:
(i) If limx→1 F (x) = 2 does it follow that F (1) = 2?
DRAFT Calculus Notes 11/17/2011
311
Sol: No, the definition of the limit limx→1 F (x) contains no information
about the value of F at 1
...
has limx→1 F (x) = 2 but F (1) = 0
...
For example, the function g given by
g(x) = |x − 3|
for all x ∈ R,
is continuous everywhere but is not differentiable at 3
...
(iv) If h0 (5) = 4 and h(5) = 8 then limx→5 h(x) =
Sol: Since h0 (5) = 4 the function h is differentiable at 5
...
Hence limx→5 h(x) is equal to h(5), which is
given to be 8
...
=
(v) If H 0 (2) = 5 and H(2) = 3 then limw→2 H(w)−3
w−2
Sol: The limit here is
H(w) − H(2)
H(w) − 3
= lim
...
Thus the value of the limit is 5
...
y→5
y→2
y−5
y−5
We recognize this to be the derivative G0 (5), which is given to be
1
...
312
Ambar N
...
(viii) limw→π/3
sin w−sin(π/3)
w−π/3
=
Sol: We recognize this limit as the derivative of sin at π/3, and so its
value is sin0 (π/3) = cos π/3 = 1/2
...
4
...
(v)
√
d sin(cos(tan( x)))
dx
Sol: Using the chain rule repeatedly we have
√
√
√
d sin (cos (tan ( x)))
= cos cos tan x
· − sin tan x
dx
√
1
· sec2 x · √
2 x
314
Ambar N
...
Using the definition of the derivative, show that
√
d(1 x)
1
=− √
...
1)
Solutions for Exercise Set 20
...
1
...
Sol: The deriative of f (x) = x2 is 2x, which is 0 at x = 0
...
2
...
Sol: For f (x) = x(6 − x)(3 − x) = x3 − 9x2 + 18x, we have f 0 (x) = 3x2 −
18x + 18, and this is 0 when x2 − 6x + 6 = 0, the solutions of which
DRAFT Calculus Notes 11/17/2011
315
√
are 3 ± 3 (instead of using the formula for solutions of quadratic
equations it is easier to observe that
x2 − 6x + 6 = x2 − 2 ∗ 3x√
+9 = 3
√
2
3 is in
and this is (x − 3) = 3
...
√We compare
the
values
f
(0)
=
0,
f
(2)
=
8,
and
f
(3
−
3) =
√ √
√
(3√− 3)(3 + 3)
√ 3 = 6 3 ' 10
...
So the maximum of f on [0, 2] is
6 3 at x = 3 − 3, and the minimum is 0 at x = 0
...
A wire of length 12 units is bent to form an isosceles triangle
...
Then 2x + 2y = 12, so x + y = 6
...
x2 = y 2 + h2
h=
x
x
h
y
y
p
x2 − y 2
p
1
A = 2yh = y x2 − y 2
2p
p
= y 6 − y)2 − y 2 = y 36 − 12y
Thus y ∈ [0, 3]
...
The derivative is (A2 )0 =
72y − 36y 2 = 36y(2 − y)
...
When y = 0 or y = 3 then
√ A is 0
...
4
...
Show that
this maximal area rectangle is a square
...
Then the perimeter is 2x + 2y = L, thus x + y = L/2
...
Sengupta 11/6/2011
area is A = xy
...
We write A as A = x(L/2 − x) =
(L/2)x−x2
...
We know that
the maximum value of a quadratic can be obtained without calculus,
but it is simple to work out the derivative A0 = L/2 − 2x and observe
that this is 0 when x = L/4
...
Thus
the rectangle has equal sides, which means that it is a square
...
5
...
What is the value of x which would make the total area enclosed
by the pieces maximum, and what is the value of x which would make
this area minimum
...
2
The area is π x/(2π)
...
Thus the total area is
2
2
1
1 2
x +
(L − x)2
...
The derivative of A is
A0 (x) =
1
1
1
x+
(L − x)(−1) =
(2x − L)
...
The area enclosed is then
1
1
L2
2
2
A(L/2) =
(L/2) +
(L − L/2) =
...
Thus the minimum are is obtained by splitting the wire into two equal
pieces, each rolled into a circle, and the maximum area is obtained by
rolling up the entire length of wire into a circle
...
Here are some practice problems on straight lines and distances:
DRAFT Calculus Notes 11/17/2011
317
(i) Work out the distance from (1, 2) to the line 3x = 4y + 5
(ii) Work out the distance from (2, −2) to the line 4x − 3y − 5 = 0
...
What is the angle between P0 P and
the line L?
(iv) Let P0 be the point on the line L, with equation 3x + 4y − 11 = 0,
closest to the point P (1, 3)
...
Find the equation of the line through
P and P0
...
5
...
Use
u = 4 − 3x
and
du = −3dx
...
3
Z
(4 − 3x)
2/3
Z
dx =
2/3
1
− du
3
u
Z
1
=−
u2/3 du
3
2
1 1
=− 2
u 3 +1 + C
3 3 +1
1
= − (4 − 3x)5/3 + C
...
Use
y = 2 + 5x
and
dy = 5dx
...
2)
318
Ambar N
...
15
Z
2 + 5x dx =
=
=
=
=
(34
...
Use
w = 2 − 3x
So
and
dw = −3dx
...
3
Then
Z
1
1
√
−
dw
3
w
Z
1
1
√ dw
=−
3
w
Z
1
=−
w−1/2 dw
3
1
1 1
=−
w− 2 +1 + C
1
3 −2 + 1
2
= − w1/2 + C
3
2
= − (2 − 3x)1/2 + C
...
Use
y = 3 − 2x
...
2
(34
...
5)
v
...
Then
dy = 5dx,
and
1
x = (y − 2)
...
Sengupta 11/6/2011
So
Z
x
dx =
(2 + 5x)3/5
=
=
=
=
=
=
vi
...
25 7
(34
...
y
2 y
R
2
vii
...
Use y = −x2 /2, for which
dy = −xdx
and so
Z
Z
Z
2
−x2 /2
y
e
x dx = e (−dy) = − ey dy = −ey +C = −e−x /2 +C
...
R √log(2x+5)
2x+5
dx
DRAFT Calculus Notes 11/17/2011
321
Use y = log(2x + 5), for which
dy =
1
2dx
2x + 5
and then
Z p
Z
Z
log(2x + 5)
1
√ 1
dx =
y dy =
y 1/2 dy
2x + 5
2
2
1 3/2
y + C and so
and this equals 12 3/2
Z p
log(2x + 5)
1
dx = (log(2x + 5))3/2 + C
...
R
1
x log(x) log(log x)
dx
Using y = log x converts the integral to
Z
1
dy
y log y
R
and then w = log y converts this to w1 dw = log w + C, and
so
Z
1
dx = log log log x + C
...
Z
Z
1
[sin(5x + 2x) + sin(5x − 2x)] dx
2
1
1
1
=
− cos(7x) − cos 3x + C
2
7
3
(34
...
Z
Z
1
[cos(5x − 2x) − cos(5x + 2x)] dx
2
1 1
1
=
sin 3x − sin 7x + C
2 3
7
(34
...
Sengupta 11/6/2011
xiii
...
R
cos(5x) cos(2x) dx =
1
2
1
7
sin 7x + 31 sin 3x + C
sin3 x dx
sin3 x = sin x sin x sin x
1
= [cos 0 − cos(2x)] sin x
2
1
= [1 − cos 2x] sin x
2
1
= [sin x − sin x cos 2x]
2
1
1
=
sin x − [sin 3x + sin(−x)]
2
2
1
1
= sin x − [sin 3x − sin x]
2
4
1
1
1
= sin x + sin x − sin 3x
2
4
4
1
3
= sin x − sin 3x
4
4
Integration gives
Z
3
1
sin3 x dx = − cos x +
cos 3x + C
...
R
(34
...
10)
DRAFT Calculus Notes 11/17/2011
323
Integration gives
Z
cos3 x dx =
3
1
sin x +
sin 3x + C
...
Z
2
Z
sin (5x) dx =
Z
sin(5x) sin(5x) dx =
1
[cos 0 − cos(5x + 5x)] dx
2
and so
Z
xvii
...
x−
2
10
R√
3 − 6x − x2 dx
Completing the square for x2 + 6x − 3 we have
x2 + 6x − 3 = x2 + 2 ∗ 3 ∗ x + 32 − 32 − 3 = (x + 3)2 − 12
and so the integral is
Z p
12 − (x + 3)2 dx
Substitute
x+3=
√
12 sin θ
for which
dx =
√
12 cos θ dθ
324
Ambar N
...
11)
Now substitute back in
x+3
sin θ = √
12
and
cos θ =
p
12 − (x + 3)2
to get
Z p
x + 3p
−1 x + 3
2
2
12 − (x + 3) dx = 6 sin √
12 − (x + 3) +C
...
R
√
1
3−6x−x2
dx
Bibliography
[1] Havil, Julian, Gamma: Exploring Euler’s Constant
...
[2] Maor, Eli, e: The Story of a Number, Princeton University Press (2009)
...
, When Least is Best: How Mathematicians Discovered
Many Clever Ways to Make Things as Small (or as Large) as Possible,
Princeton University Press (2004)
...
325
Index
arc length, 296
absolute value
definition, 41
larger of x and −x, 42
AM-GM inequality, 212
angle
as a pair of rays, 69
Archimedes, 12
boundary
notation ∂S, 37
boundary points, 36
Cartesian product, 19
chain rule
initiating examples, 137
proof, 146
statement, 139
chainrule
dy/dx form, 145
closed sets, 39
complements of open sets, 40
codomain
of a function, 20
complement, 17
completeness
existence of suprema, 31
of R, 31
of R∗ , 31
completing squares, 153
composite function, 138
composite functions, 66
composites
of continuous functions, 88
concave function, 206
strictly concave, 206
continuity
at a point, 85
continuous
at a point, 85
at exactly one point, 87
at exactly two points, 88
nowhere, 87
on a set, ambiguity, 87
continuous functions, 86
composites of, 88
polynomials, 86
convergence
P
of n 1/n2 , 259
convex combination, 210, 218
convex function, 205
strict convexity, 206
cosecant function csc, 74
cosine
geometric meaning, 70
cotangent function cot, 74
curve
definition, 296
decreasing functions, 102
dense
326
DRAFT Calculus Notes 11/17/2011
irrationals in R, 51
rationals in R, 51
dense subset
Q in R, 26
Qc in R, 26
derivative
as slope of tangent line, 116
at a point, 116
definition, 116
finiteness implies continuity, 131
(x)
notation dfdx
, 118
notation df (x)/dx, 117
notation f 0 (p), 116, 117
of constant is 0, 116
sign of, 175–179
zero and constancy, 180
derivatives
algebraic rules, 133
differential df , 246
differential form, 247
differential forms
working rules, 248
discontinuity
removable, 86
discriminant, 155
distance
on R, 43
triangle inequality, 43
divergence
P
of n 1/n, 261
domain
of a function, 20
dummy variable, 252
elements, 13
empty set
as subset, 16
empty set ∅, 14
327
extended real line R∗ , 26
exterior points, 36
factorials, 23
functions, 19
definition, 20
graph
of a function, 20
of a function f , 114
of unit circle, 22
greatest lower bound, 30
harmonic series
divergence, 261
Hausdorff property, 35
Havil, Julian, 325
increasing functions, 102
indefinite integral, 252
indicator function, 51
infimum
larger for larger set, 32
notation inf S, 30
of ∅, 31
integers, 18
integrability
and continuity, 240
integral
of a differential, 248
of a differential form, 253
integrand, 269
integration
by parts, 286
by substitution, 269
interior
notation S 0 , 37
interior point, 35
intermediate value theorem, 93
328
constructing rational powers, 95
with intervals, 94
intersections, 17
interval, 33
inverse sin: sin−1 or arcsin, 100
inverse function, 99, 103
derivative of, 204
inverse trigonometric functions, 99
irrational numbers, 18
irrationals, 26
L’Hospital’s rule, 225
least upper bound, 30
length
of a path, 295
lHospital’s rule
proof, 228
limit
as unique value between suprema
and infima, 47
notation, 49
of 1/x as x → 0+, 51
of 1/x as x → 0−, 51
of 1/x as x → −∞, 50
of 1/x as x → ∞, 49
limits
‘squeeze’ theorem, 65
‘squeezing’, 64
and ratios, 63
between suprema and infima, 54
by comparision, 64
definition with neighborhoods, 61
of composites, 66
of sums, 61
products, 62
the notion, 46
with neighborhoods, 59
linear combination, 220
Ambar N
...
, 325
Napier, John, 325
neigborhoods, 34
neighborhoods
and distance, 43
of ±∞, 34
open sets, 38
complements of closed sets, 40
finite intersections, 39
unions are open, 39
ordered pairs, 18
DRAFT Calculus Notes 11/17/2011
polygonal approximation
to paths, 294
powers
rational, definition, 196
real, definition, 196
product rule
and integration, 286
329
sets, 13
equality, 14
intersections of, 17
unions of, 17
sin
geometric meaning, 70
squeeze theorem, 64
strictly decreasing functions, 102
quadratic equations
strictly increasing functions, 102
solutions, 154
subsets, 15
quasi-tangents
properties, 16
definition, 115
sum of cubes 13 + · · · + N 3 , 268
flat at interior maxima/minima, sum of squares 12 + · · · + N 2 , 267
168
summation notation, 53
Rolle’s theorem, 172
supporting line, 216
uniqueness and tangents, 115
and tangent line, 216
supremum
radian measure, 70
notation sup S, 30
Ramanujan formula, 12
of ∅, 31
range
smaller for larger set, 32
of a function, 21
rational numbers, 18
tan
notation Q, 18
geometric meaning, 70
real line, 26
tangent
real numbers, 18, 26
and maxima/minima, 168
notation R, 26
tangent
line
Riemann integral
and supporting line, 216
definition, 236
definition, 114
Riemann sum, 237
triangle
inequality
Rolle’s Theorem, 171
for distance, 43
Rolle’s theorem
for magnitudes, 42
with derivatives, 172
trigonometric functions
sin(1/x) and cos(1/x), 79
secan function sec , 74
sin2 + cos2 = 1, 74
secant
addition formulas, 75
to a graph, 114
bounds, 77
semi-chord
bounds for (sin x)/x, 77
and sin, 71
continuity, 78
set theory, 13
330
Ambar N