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Title: Civil engineering easy ways
Description: Formulas question and answers will set in short and easy methods
Description: Formulas question and answers will set in short and easy methods
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Applied II
DEBREMARKOSUNIVERSITY
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CHAPTER ONE
REAL SEQUENCES
1
...
For instance, 1, 2, 3, 4,
...
...
The numbers in the sequence are called terms of the sequence and described as 1 stterm,
2ndterm, 2rdterm etc
...
Consider A=
i
...
The set A is also described by defining a function f by the rule
where the domain f is a positive integer
...
In general we have the following definition
...
, an ,
...
, n th terms & so on
Or
equivalently we can write as a1 , a2 , a3 ,
...
Example-1:-a) If
b) If
, then
, then
c) If
, then
for all n N
for all n N
for all n N
Example-2:-Find a formula an for each of the following
a)
b)
Solution: -
c)
a)
b)
for all positive integers n
for all positive integer n
c)
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Example-3:-Write down the first few terms of each of the following sequences
a)
b)
Solution:-a) If
then
=
b) If
then
If
then
If
If
then
then
If
etc
...
=
Note:-Since the terms of the sequence in (b) alternate in signs it sometimes called
alternating sequence
...
2 Convergence and Divergence of real Sequences
Definition:-A Sequence
said to have a limit L, written as
there exists a natural number
such that for every n > , we have | an L|< Є
...
)
If such a number L does not exist, we say that the sequence
that
if for each Є>0,
diverges or
does not exist
...
Example-1:-show that the sequence
convergences and that
Solution:- Let
Choose
]
Then by definition of sequence, the sequence
Note:-The sequence
is known as the Harmonic Sequence
...
We need to find N>0 such that
< whenever
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DEPARTMENT OF MATHEMATICS
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Applied II
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But
<
=
Choose =
=
<
]
Then
<
<
<
Hence
n
1
2n 1 2
Example-3:- show that the sequence
diverges
Solution:Assume that
is convergent Sequence
Choose
Case 1)
, then for odd values of , we have
Case 2)
then even values of , we have
This is contradiction our assumption
...
e
...
Note:-The sequence
is an example of oscillating sequence
...
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DEPARTMENT OF MATHEMATICS
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Example-4 Show that the sequence
Solution:-
As
is arbitrarily large
Hence
Definition :-Let
also diverges
also arbitrarily large i
...
diverges
...
natural number
If for each positive number
for all n
such that
, there is a
...
In this case we say that
diverges to and we write as
Example-4:-prove that
Solution:-Let >0 be given
...
whenever n
Choose
Then
n
whenever n
Example-5:- Show that
Solution:- Let <0 be given
...
But
Choose
Then
n
whenever n
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DEPARTMENT OF MATHEMATICS
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Theorem:-Let
be a sequence, a number and f a function defined for every positive
integer such that
a) If
, then
b) If
( or
convergences and
) , then
diverges and
...
Then there is some integer
If n >N, then | an
b) If
such that if x > , then|
|=|
|< ε
...
So that, if n > , then
...
e
...
etc
...
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Example-8:- Show that
...
Example-9:- Let be any number
...
Solution:- Case 1) if
Let
so that
...
From (1)
converge for
In particular,
and diverges for
converges to 0
converges to 1
diverges
...
is the initial term
...
Example:10 Show that
...
Then by the above theorem
Example:11 Show that
Solution:-Note that
Let
so that f is continuous and
and
Then
using the above theorem
...
3Properties of convergent real Sequences
Since Sequences are functions, we may add, subtract, multiply, and divide as we do for
functions
...
Example-1:-Find a)
b)
Solution a) The highest power of
in both in the numerator and the denominator is 1
...
i
...
Then by applying convergence properties, we have
b)
Theorem:-(squeezing theorem for sequences)
Given three sequences
,
and such that
=
for every
and
, then
=
Proof:-For every ε>0, there is N1 and N2 such that|
|< ε for n > N1
and|
|< ε for n > N2
...
For n > N, we have|
|< ε which implies - ε <
<ε
- ε < < L+ε and hence - ε <
...
(**)
Since
by hypothesis, from (*)and(**) it follows that
-ε<
< +ε
=
Therefore
| bn
|< ε, for every n > N
...
Example-2:-show that
Solution:-We know that
, since
and
Then by show that squeezing theorem we have
1
...
b) The sequence
is bounded below if there is a number M2 such that
an M2 for every positive integer n m
...
Otherwise the sequence is unbounded
...
So it is bounded
...
are unbounded because no matter what
number M we choose, there is value
Theorem: -
converges
is unbounded
Proof
and
...
diverges
...
q)
i
...
part (b) is logically equivalent to part(a)
Example:-4) the sequences
and
are unbounded and hence by the above
theorem the sequences diverge
...
e
...
Example:-5) the sequence
is bounded but it diverges
...
Show that whether the following sequences are convergent or divergent
a)
b)
c)
Definition: - A sequence
a)
b)
c)
d)
e)
is said to be
Monotone increasing if
...
Monotonic if it is either monotone increasing or monotone decreasing
...
Strictly decreasing if
...
Example:-6 a)
b)
c)
and
are increasing sequences
...
is neither increasing nor decreasing because it oscillates between -1 and 1
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Theorem:-Every bounded monotonic sequence
Proof:-Assume that
Since
is convergent
...
is bound, the set
has an upper bound
...
in such that
for some integer
Since the given sequence is increasing, we have
for
Hence
Thus
–
...
be the sequence defined by
Show that
converges
...
(i)
To prove boundedness
Claim that
Assume that
...
Using (i) and (ii) and applying the above theorem the given sequence converges
...
Activity
Show that whether the following sequences are convergent or divergent
a)
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CHAPTER 2
2
...
Definition:-consider a sequence with terms
we may form in succession
The sum
are called partial sums
...
It is denoted by
is called
Therefore, an infinite series or just series is equal to the limit of the partial sums, that is,
Example:-1) If
, Then
,
,
Sum is
Example:-2) If
,Then
,
is
,
Sums
...
2 Convergence and Divergence real series
Consider the sequence of partial sums
to a number
then we say that the series
of the series
converges to
...
diverges, then we say that the
or
is not unique ,
then
Note:-If
is not unique then
oscillates
...
If it converges determine its value
Solution:-The sequence of partial sum is
This is known series and its value can be shown to be
So, to determine if the series is convergent we will first need to see if the sequence of partial
sums,
is convergent or divergent
...
converges and find the sum
...
Hence
Hence
converges and
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DEPARTMENT OF MATHEMATICS
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Note:-The series
is called a telescoping series because when we write the partial
sums all but the first and the last terms cancel
...
If it converges
determine its value
Solution: - The sequence of partial sum
So, to determine if the series is convergent we will first need to see if the sequence of partial
sums,
is convergent or divergent
...
The value of the series is
Example:-4) consider the series
Here
Hence
is not unique
...
Activities
Determine whether the following series is convergent or divergent
...
(where c is a constant)
ii
...
Example:- 5) Find the sum of the series
Solution:- The series
is a geometric series with
and
so
-------------------- (1)
We have
and so
Therefore, the given series is convergent and
----------------- (2)
So, by properties of convergent series
Theorem: -If
is convergent and
Proof:-Let
Since
is divergent, then
is divergent
then
is convergent by hypothesis,
otherwise
cannot be convergent for
would be convergent by the above theorem which contradicts the
hypothesis
...
and
We have
,
which is convergent
...
Proof:-Consider
Let
(the partial sum of the 1st (k-1) terms
...
are two series differing only in their first m terms ( i
...
Theorem:- If the series
is convergent, then
(the
term of a
convergent series tends to zero)
Proof:-Let
Then
Since
is convergent, the sequence
is convergent
...
i
...
does not converge but
Theorem :-( The Test for Divergence)
If
then the series
Proof:-Suppose
is diverges, then
Hence by contra positive
is diverges
...
Solution:So the series
diverges by the Divergence Test theorem
...
a)
b)
Some types of infinite series
1
...
is divergent (see example 7 above)
2
...
is called the ratio of the geometric series
...
Case (iii) if
Then
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DEPARTMENT OF MATHEMATICS
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DEBREMARKOSUNIVERSITY
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Therefore,
does not exist since it oscillates between 0 and
...
Case (iii) If
,then
Then
The series is diverges
...
Hence theorem proved
...
In case of
convergence find the sum
a)
b)
Solution:-a)
=
which is a geometric series with
Since
b)
c)
and
the series converges
is a geometric series with
Since
c)
and
the series converges
Which is a geometric series with
Since
and
the series diverges
...
Example:-10) Express decimals as a rational number
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DEPARTMENT OF MATHEMATICS
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a)
b)
Solution:-a)
is a geometric series with
Since
and
the series converges
b)
is a geometric series with
Since
and
the series converges
2
...
This is true when we can’t find any
simpler expression for the sum of the first terms of a series as a function but several tests
are developed which enable us to say whether the series is Convergent or divergent without
explicitly finding its sum
...
The Integral Test
Suppose is a continuous, positive, decreasing function on
then the series
and let
converges iff the improper integral
Example:1)Test the series
for
converges
...
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Solution:-The function
is continuous, positive, and decreasing on
...
Example:-2) Determine whether the series
Solution:-The function
converges or diverges
...
But it is not obvious whether or not is decreasing,
so we compute its derivative:
Thus
when
that is,
It follows that is decreasing when
and so we can apply the Integral Test:
Since this improper integral is divergent, the series
is also divergent by the Integral
Test
...
Assume
Let
on
...
So we can apply the Integral Test
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DEPARTMENT OF MATHEMATICS
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And
=
Hence
=
And
Converges if
Note:-the series
and diverges if
is called the
series
Activities
Determine whether the following series are convergent or divergent
...
The comparison Test
Suppose that
and
are series with positive terms
...
If
is convergent and
ii
...
is also divergent
...
Solution:-For large the dominant term in the denominator is
series with the series
...
Example:-4) Test the series
for convergence or divergence
...
Therefore, the given series
is divergent by the comparison test
...
a)
b)
3
...
i
...
ii
...
If
and
diverges, then
Example:-5) Test the series
also converges
...
for convergence or divergence
...
Example:-6) Test the series
is a convergent geometric series, the given series converges by
for convergence or divergence
Solution:- We use the limit comparison test with
,
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(finite and non zero)
Then by limit comparison test either both series converge or both diverge
...
Activities
Test the following series for convergence or divergence
...
The Ratio Test
The Ratio Test determines if the terms of the given series are approaching 0 at a rate sufficient
for convergence by considering the ratio between successive terms of the series
...
The series converges if
ii
...
The test provides no conclusion if
then
and
Example:-7) Test the convergence of
Solution: - Let
and
Therefore, the series converges if
and diverges if
Put
diverges by
Therefore, the given series converges if
since
and diverges if
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Example:-8) Test the series for convergence
Solution: - Let
and
Therefore, the series is convergent
...
converges
Solution:-Here
Therefore, the given series
converges
Activities
Investigate the following series for convergence or divergence
...
The Root Test
As Ratio Test, The Root Test can be applied without considering other series whose convergence
is known
...
Let
be a series of positive terms such that
i
...
The series diverges if
iii
...
Example:-12) Show that
converges
Solution:-Here
Thus
Therefore, the series
converges by Root Test
...
Example:-14) Show that the convergence or divergence of
Solution:-
1
and
and
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DEPARTMENT OF MATHEMATICS
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Hence we can’t conclude any thing about the convergence or divergence of both series by a root
test
...
Activities
Investigate the following series for convergence or divergence
...
4 Alternating Series and Alternating Series Test
Alternating Series
An Alternating Series is a series whose terms are alternately positive and negative
...
ii
...
The Alternating Series Test
If the Alternating Series
Satisfies
i
...
Then the series is convergent
...
for convergence or divergence
...
It satisfies
because
ii
...
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Example:-Discuss the convergence of the series
Solution:The given series is an alternating series since the terms are alternatively positive and negative
...
because
ii
...
Activities
Test the following series for convergence or divergence
...
5 Absolute and Conditional Convergence
Absolute Convergence
Definition: - Let
be a series of positive and negative terms
...
Solution:-The given series
is absolutely convergent
because
is a convergent series by
where
Example:-Show that the series
is absolutely convergent
...
is absolutely convergent
...
Remark: - If
diverges, we can’t say any thing about
Theorem: - If
and
...
Conditionally Convergent
Definition: - Let
be a series of positive and negative terms, then
conditionally Convergent if
is Convergent and
is said to be
is divergent
...
It is divergent by
So
since
is conditionally convergent
...
It is divergent by
So
since
is conditionally convergent
...
6 Generalized Convergence Tests
Let
be a series and
a positive series then
a
...
Generalized limit comparison test
If
converges, then
where L is a positive number, and if
converges (absolutely)
...
c
...
d
...
No conclusion about the convergence of the series
...
Activities
Test the following series for convergence or divergence
...
If
or
,
then
Example: - Show that
=0,
Solution: -
,
Hence by the above Corollary
=0,
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CHAPTER THREE
Introduction
In Chapter two, we studied series with constant terms
...
Such series are of fundamental importance in many branches of
Mathematics and the Physical Sciences
...
1 Definition of power series at any
and
Power Series at any
A Series of the form
------------- (1)
is called a Power Series in
or a power series centered at
or a Power Series about
Power Series at
A Series of the form
--------------- (2)
is called a power series, where
coefficients of the series
...
ii) All tests can be used for power series
...
2 Convergence and divergence, Radius and interval of convergence
i)
For each fixed x, the series
is a series of constants that we can test for
convergence or divergence
...
The sum of the series is a function
whose
domain is the set of all for which the series converges
...
In Power Series centered at
even when
all of the terms are for
and so
the power series in equation (1) always converges when
...
Example:-1) For what values of x is the Series
, we get a geometric series
converges?
Solution:-To find the values of , let’s use the Ratio test
...
and
denote the
term of
...
then
Hence, by the Ratio test, the given series converges when
...
, the harmonic series, which is divergent
...
...
The Series converges for all
iii
...
By convention, the radius of convergence is
in case (i) and
in case (ii)
Interval of convergence
The interval of convergence of a power series is the interval that consists of all values of
which the series converges
...
If the radius of convergence
for
is zero, then the interval of convergence of a power
series consists of just a single point
ii
...
If the radius of convergence is R, there are four possibilities for the interval of
convergence
...
Solution:-a) Let
, applying the ratio test, we have
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DEPARTMENT OF MATHEMATICS
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Now, by ratio test the series converges for
...
, which is a -series with
hence the
series diverges
...
...
3 Algebraic operations on convergent power series
We may also add, subtract, multiply and divide the series just like polynomials
...
Theorem:-The Algebra of Power Series
Assume that
Then
, we have
(i)
(ii)
(iii)
(iv)
is obtained by long division, if
3
...
Representation of functions as a power series is useful for integrating functions that do not have
elementary anti-derivatives, for solving differential equations and for approximating functions
by polynomials
...
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Example:-1) Express
as the sum of a power series and find the interval of convergence
...
Solution:-Given
We know that
This series converges for
Therefore, the power series representation of
is –
and the interval of
convergence is
Example:-Express
as the sum of a power series and find the interval of convergence
...
5 Differentiation and integration of power series
A power series
with a nonzero radius of convergence is always differentiable
and the derivative is obtained from
by differentiating term by term, the way
we differentiate polynomials
...
Solution:-Differentiating both sides of the equation
We get
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DEPARTMENT OF MATHEMATICS
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Therefore,
The radius of convergence
since the radius of convergence of the differentiated series is
the same as the radius of convergence of the original series
...
Solution: - Let
then –
------------- (i)
But
Now
By substituting in (i), we get
To find the value of we put
, we get –
Therefore,
Hence,
Then
and It converges for
The radius of convergence
Example:-3) Find the power series representation for
and determine the radius
of convergence
...
Example:-4) Evaluate the indefinite integral
as a power series
Solution:-We know that
Now
Activity
1
...
Find the power series representation of
3
...
Here we investigate more general problems: which functions have
power series representations? How can we find such representations?
We start by supposing that is any function that can be represented by a power series
---------- (i)
and let us try to determine what the coefficients must be in terms of To begin, notice that
if we put
in equation (i), then all terms after the first one are 0 and we get
...
Differentiate of the series in equation (iii) gives
---------------(iv)
and substitute
in equation (iv), we get
In general,
Solving this equation for the
coefficient
, we get
Theorem:- If has a power series representation (expansion) at , that is, if
Then its coefficients are given by the formula
Taylor series of the function at
If
has a power series expansion at
then it must be the following form
Maclaurin Series
If
, the Taylor series becomes
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DEPARTMENT OF MATHEMATICS
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DEBREMARKOSUNIVERSITY
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Note:-If can be represented as a power series about (such functions are called analytic at ),
then is equal to the sum of its Taylor series
...
so
Therefore, the Taylor series for at 0 (that is, Maclaurin series) is
To find the radius of convergence, let
Then
So by the ratio test, the series converges for all and the radius of convergence is
Taylor Polynomial of at
If
where
polynomial of
is the
degree Taylor
at and
for
that is,
then
is equal to the sum its Taylor series on the interval
is analytic at
Note: - In trying to show that
for a specific function
we usually use the
expression in the next theorem
...
then
Where z lies between
If
so the remainder term in Taylor’s Formula is
(Note, however, that z depends on )
then
Therefore
So
If
by the squeeze theorem
then
Again
and
Thus, by Taylor polynomial,
Note: - In particular if we put
is equal to the sum of its Taylor series, that is,
we obtain the following expression for the number as a
sum of an infinite series:
Applications of Taylor Polynomials
Suppose that
is equal to the sum of its Taylor series at
We know that
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DEPARTMENT OF MATHEMATICS
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DEBREMARKOSUNIVERSITY
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Since is the sum of its Taylor series, we know that
be used as an approximation to
Example:-3) Approximate the function
and so
can
by a Taylor polynomial of degree 2
...
1 Periodic Functions, trigonometric series
Definition:-A function
is said to be periodic if
in the domain of
the period of
for all real number
and some positive number
...
also has a period 2 but this is not a fundamental
period
...
If
is a period of , then
that is,
2
...
3
...
4
...
Thus there exists no smallest period, so does not have the fundamental period
...
The sum of the number of trigonometric functions in
and
is also a
periodic function with period the least common multiple of the periods of each
term of the sum
...
and respectively
...
Trigonometric series
Definition: - the series the form
is called a trigonometric series
...
4
...
5
...
6
...
a)
b)
c)
4
...
To determine the coefficients
assume the series
...
(ii)
To determine
to
multiply both sides of (i) by
and then integrate from
=
(Remaining terms vanish
...
(iv)
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Note:-1
...
If
, we obtain the Fourier series in
as
Where
Example:-1) Determine the Fourier series for
in
Solution: - let
Then the Fourier series for
in interval
is given by
Where
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DEPARTMENT OF MATHEMATICS
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In order to evaluate this integral we use the following formula
Since
and
Here we apply the formula
Since
Substituting the values of
in
and
we get
Example:-2) Determine the Fourier series for
in
Solution: - let
Then
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DEPARTMENT OF MATHEMATICS
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Substituting the values of
in
we get
Example:-3) let
Then, obtain the Fourier series for
Solution: - Then the Fourier series for
obtain the Fourier series for
in –
in interval –
is given by
Where
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DEPARTMENT OF MATHEMATICS
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And
By Substituting the values of
in
we get the Fourier series of the function
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Activity
Determine the Fourier series for each of the following functions
...
3 FUCTIONS OF ANY PERIOD
In the previous section we expanded a function
in a Fourier series of period
In most of the Engineering problems the period to be expanded is not
but some
other
...
Let
be a periodic function of period
such that the period becomes
...
we convert
into
Let
Where
,
When
defined in
is equivalent to
Hence we now write the Fourier series for
Where the coefficients
,
in
in
as
are given by
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DEPARTMENT OF MATHEMATICS
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We now make the inverse substitutions
The Fourier series for
in
is
Where
from
from
from
Hence the Fourier series for
in
is
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DEPARTMENT OF MATHEMATICS
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Where
Example:-1) obtain the Fourier series for
in
Solution:- Fourier series is
Where
Since
is even
...
a) Find the Fourier series period 3 to represent
b) Find the Fourier series period 2 to represent
in
in
4
...
2) The product of two even functions is even
3) If
is an even function
4 ) the graph of an even function is symmetric about the Y-axis
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Definition: - A function
is said to be odd if
Example:-2)
for all
is odd function because
for all
b)
is an odd function since
for all
Note: - 1) The sum of two odd functions is odd
...
4) If
is an odd function
5) The graph of an even function is symmetric about the orgin
...
However there are functions which are neither odd nor even
...
Then
Where
and
Since
And
Example:-3)
is neither odd nor even but by putting
and
we get
and
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Fourier series for even and odd function in the interval
1
...
If
by
is an odd function in the interval
we get the
given
Where
Here we also need not calculate
Note: - If
is neither an even nor an odd function in the interval
calculate
using the formula
Example:-4) Find the Fourier series for
, then
in
Solution: - let
is an even function
Hence the Fourier series in
is
...
(*)
Where
Substituting for
in (*), we get
Activity
Find the Fourier series for each of the following functions in the interval
a)
c)
e)
b)
d)
f)
...
e
...
Suppose a periodic function
of period is defined only on half interval
...
Define
is an even function in
only cosine terms
...
Hence the Fourier series of
So that
contains
Where
Since
in
In this case the Fourier series is called Half-Range Fourier Cosine series which is defined
as
Where
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Note: - The Half-Range cosine series in
is obtained by putting
Half-Range sine series
Let
be defined in the range
in order to obtain Fourier series we construct a
new function in such a way that the function
plus its extension yield an odd
function
...
Hence the Fourier series of
contains only
sine terms
...
5
...
But, in the real world,
physical quantities usually depend on two or more variables, so in this chapter we turn our
attention to functions of several variables and extend the basic ideas of differential calculus to
such functions
...
We can think of as being a function of the two
variables and or as a function of the pair
...
Example:-The volume of a circular cylinder depends on its radius and height
...
We say that V is a function of and and we write
Definition
Let
pair
of
Note
i
...
The variables and
takes on, that is,
are independent variables and is the dependent variable
...
is the domain
to make explicit the value taken on by at the general point
[Compare this with the notation
ii
...
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Example:-Find the domain of each of the following functions and evaluate
Solution:- a)Given that
The expression for makes sense if the denominator is not 0 and the quantity under the
square root sign is nonnegative
...
b
...
1 Notations, Examples, level Curves and graphs
Graph of a function of two variables
The graph of a function of two variables is a surface with equation
the graph of as lying directly above or below its domain in the
If
is a function of two variables with domain
, the graph of
We can see
plane
...
The portion of this graph that lies in the first octant is sketched
...
The range of is
Since z is positive square root,
also
So the range is
...
But, since
the graph of is just the top half of this sphere
...
A
third method, borrowed from mapmakers,
is a contour map on which points of
constant elevation are joined to form
contour curves, or level curves
...
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Example
Sketch the level curves of the function
Solution:-The level curves are
for the values
This is a family of lines with slope
The four particular level curves with
are
They are sketched in above figure
...
Contour map of
5
...
The following figure illustrates limit by means of
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an arrow diagram
...
Note
The above limit definition refers only to the distance between
and
It does not refer to the direction of approach
...
Thus, if
we can find two different paths of approach along which
has different limits, then it
follows that
does not exist
...
if it exists
...
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Example:-Find
if it exists
...
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Example: - Let
We know is continuous for
have
Therefore, is continuous at
Theorem
If
is continuous at
continuous at
since it is equal to a rational function there, also we
and so it is continuous on
and
is a function of a single variable that is
then the composite function
defined by
is continuous at
Example: - On what set is the function
Solution:-Let
So
domain
and
continuous?
Then
Now is continuous everywhere since it is a polynomial, and is continuous on its
Thus by above theorem, is continuous on its domain
This consists of all points outside the circle
5
...
Then we really considering a function of a single variable
namely,
...
To find
2
...
4 Higher Derivatives
If is a function of two variables, then its partial derivatives
are also functions of two
variables, so we can consider their partial derivatives
which
are called the second partial derivatives of
...
Example: - Find the second partial derivatives of
Solution:We know that
Therefore
Clairaut’s Theorem
Suppose
is defined on a disk
that contains the point
are both continuous on
If the functions
then
5
...
Then the differential , also called the total differential, is defined by
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Example:- If
, find the differential
...
Let
be a
be the curves obtained by intersecting
the vertical planes
with the surface
...
Then the tangent
plane to the surface at the point is defined to be the plane that
contains both of the tangent lines
The following table shows the equation of the tangent plane to the surface
An equation of the tangent plane to the surface
point
at the
is
Application: tangent plane approximation of values of a function
Example:-Find the tangent plane to the elliptic paraboloid
point
Solution:-Let
at the
...
6 The Chain Rule and Implicit differentiation
The Chain Rule
Case (i)
Suppose that
and
is a differentiable function of
are both differentiable functions of
...
Then
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Example:-If
, where
and
, find
and
...
Solution:-The given equation can be written as
So by Implicit differentiation, we get
Definition
Example:-Find
and
if
Solution:-Let
...
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Then, from the above definition, we have
5
...
Theorem
If f is a differentiable function of x and y, then f has a directional derivative in the
direction of any unit vector
and
Note:-If the unit vector u makes an angel with the positive
, then we can write
and the formula becomes
Example:-Find the directional derivative
unit vector given by angle
if
and u is the
what is
Solution:We know that
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DEPARTMENT OF MATHEMATICS
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Therefore
Gradient
If f is a function of two variables x and y, then the gradient of f is the vector
function
defined by
Example:-
If
then
Note:Example:-Find the directional derivative of the function
at the point
in the direction of the vector
Solution:-We first compute the gradient vector at
Note that v is not a unit vector, but since
...
Solution:- (a )The gradient of is
(b) At (1, 3, 0) we have
...
8 Relative extreme of functions of two variables
Definition
A function of two variables has a relative (local) maximum at
for all points
in some disk with center
called a relative (local) maximum value
...
The number
is
for all
in
is a relative (local) minimum value
...
Example:-Let
Then
These partial derivatives are equal to 0 when
and
,
so the only critical point is (1,3)
...
Therefore
is a local minimum, and in fact it is the absolute minimum of
...
5
...
Let
(a) If
and
then
is a local minimum
(b) If
and
, then
is a local maximum
...
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Example:-Find the local extrema of
...
from the first equation into the second one
...
The three critical points are
and
Next we calculate the second partial derivatives and
Since
, it follows from case (c) of the Second Derivatives Test that the origin
is a saddle point; that is, has no local extremum at
...
Similarly, we have
and
so
is
also a local minimum
...
To find the absolute maximum and minimum values of a continuous function
a closed, bounded set
on
:
1
...
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2
...
3
...
Example:-Find the absolute maximum and minimum values of the function
on the rectangle
...
According to step 1
we first find the critical points
...
On we have
and
This is an increasing function of , so its minimum value is
...
On we have
and
and its minimum value is
By the methods of Chapter 4, or simply by observing that
minimum value of this function is
and the maximum value is
Finally, on
we have
, we see that the
and
With maximum value
and minimum value
minimum value of is and the maximum is 9
...
is
...
10 Extreme values under constraint conditions: Lagrange’s multiplier
Let and differentiable at
...
If grad
and has an extreme value on at
then there is a number such that
...
Example:- Let
...
Solution:-Let
so that the constraint is
Since grad
Grad
Use these equations to find
constraint
------------- (1)
---------------- (2)
--------------- (3)
By equation (2), either
If
from (1)
If
from (3)
If
from (1)
If
from (1)
The extreme values of occurs at
Now,
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DEPARTMENT OF MATHEMATICS
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The maximum value
occur at
and minimum value occur at
at
Example:- Let
Find the extreme values of as
Solution:-Let
...
1Double Integrals
Before starting on double integrals let’s do a quick review of the definition of a definite integrals
for functions of single variables
...
For these integrals we can say that we
are integrating over the interval
...
Now, when we derived the definition of the definite integral we first thought of this as an area
problem
...
To get the exact area we then take the limit as n goes to infinity and this is also the definition of
the definite integral
...
With functions of one
variable we integrated over an interval (i
...
a one-dimensional space) and so it makes some
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sense then that when integrating a function of two variables we will integrate over a region of
(two dimensional space)
...
Also, we will initially assume that
although this doesn’t really have to be the case
...
Now, just like with functions of one variable let’s not worry about integrals quite yet
...
We will first
approximate the volume much as we approximated the area above
...
This will divide R into a
series of smaller rectangles and from each of these we will choose a point
...
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Now, over each of these smaller rectangles we will construct a box whose height is given
by
...
Each of the rectangles has a base area of
and a height of
so the volume of each of
these boxes is
...
To get a better estimation of the volume we will take n and m larger and larger and to get the
exact volume we will need to take the limit as both n and m go to infinity
...
This looks a lot like the definition of the integral of a function of
single variable
...
Here is the official definition of a double integral of a function of two variables over a
rectangular region R as well as the notation that we’ll use for it
...
We have two integrals to
denote the fact that we are dealing with a two dimensional region and we have a differential
here as well
...
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Note as well that we don’t have limits on the integrals in this notation
...
Note that one interpretation of the double integral of f (x, y) over the rectangle R is the volume
under the function f (x, y) (and above the xy – plane)
...
We can do this by choosing
to be the midpoint of each rectangle
...
This leads to the Midpoint Rule,
Iterated Integrals
In the previous section we gave the definition of the double integral
...
We will continue to assume that
we are integrating over the rectangle
...
The following theorem tells us how to compute a double integral over a rectangle
...
Note that there are in fact two ways of computing a double integral and also notice that the
inner differential matches up with the limits on the inner integral and similarly for the out
differential and limits
...
Now, on some level this is just notation and doesn’t really tell us how to compute the double
integral
...
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We will compute the double integral by first computing
and we compute this by
holding x constant and integrating with respect to y as if this were a single integral
...
We’ve done a similar process with partial derivatives
...
Double integrals work in the same manner
...
Let’s take a look at some examples
...
2 Double integrals over rectangular regions
Example1
...
To prove that let’s work this one with each order to make
sure that we do get the same answer
...
So, the iterated integral that we
need to compute is,
When setting these up make sure the limits match up to the differentials
...
e
...
To compute this we will do the inner integral first and we typically keep the outer
integral around as follows,
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Remember that we treat the x as a constant when doing the first integral and we don’t do any
integration with it yet
...
Method 2:-In this case we’ll integrate with respect to x first and then y
...
= 84
Sure enough the same answer as the first solution
...
(a)
In this case we will integrate with respect to x first
...
e
...
We’ll also rewrite the integrand to help
with the first integration
...
Remember that when integrating with respect to y all x’s are treated as constants and so as far
as the inner integral is concerned the 2x is a constant and we know that when we integrate
constants with respect to y we just tack on a y and so we get 2xy from the first term
...
However,
sometimes one direction of integration is significantly easier than the other so make sure that you
think about which one you should do first before actually doing the integral
...
Fact
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So, if we can break up the function into a function only of x times a function of y then we can do
the two integrals individually and multiply them together
...
Example 2 :-Evaluate
Solution:-Since the integrand is a function of x times a function of y we can use the fact
...
Let R be the rectangular region bounded by the lines x = -1, x = 2, y = 0 and y = 2 then
find
b
...
Evaluate
d
...
Evaluate
f
...
Evaluate
Answers
6
...
For instance, we might have a region that is a disk, ring, or a portion of a disk or
ring
...
For instance
let’s suppose we wanted to do the following integral,
To this we would have to determine a set of inequalities for x and y that describe this region
...
However, a disk of radius 2 can be defined in polar coordinates by the following inequalities,
These are very simple limits and, in fact, are constant limits of integration which almost always
makes integrals somewhat easier
...
The problem is that we can’t just convert the dx and the dy into
a dr and a d
...
Once we’ve moved into polar
coordinates
and so we’re going to need to determine just what dA is under polar coordinates
...
So, our general region will be defined by inequalities,
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Now, to find dA let’s redo the figure above as follows,
As shown, we’ll break up the region into a mesh of radial lines and arcs
...
The area of this piece is
...
Basic geometry then tells us
that the length of the inner edge is
while the length of the out edge is
where
is
the angle between the two radial lines that form the sides of this piece
...
This is not
an unreasonable assumption
...
In fact,
as the mesh size gets smaller and smaller the formula above becomes more and more accurate and
so we can say that,
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Now, if we’re going to be converting an integral in Cartesian coordinates into an integral in polar
coordinates we are going to have to make sure that we’ve also converted all the x’s and y’s into
polar coordinates as well
...
It is important to not forget the added r and don’t forget to convert the Cartesian coordinates in
the function over to polar coordinates
...
Example1 Evaluate the following integrals by converting them into polar coordinates
...
D is the portion of the region between the circles of radius 2 and radius 5
centered at the origin that lies in the first quadrant
...
D is the unit circle centered at the origin
...
First let’s get D in terms of polar coordinates
...
We want the region between them so we will have the following
inequality for r
...
Now that we’ve got these we can do the integral
...
Now, let’s simplify and make use of
the double angle formula for sine to make the integral a little easier
...
In this case we can’t do this integral in terms of Cartesian coordinates
...
First, the region D is defined by,
0
In terms of polar coordinates the integral is then,
Notice that the addition of the r gives us an integral that we can now do
...
There is one more type of example that we need to look at before moving on to the next
section
...
We need to see an example of
how to do this kind of conversion
...
Solution
First, notice that we cannot do this integral in Cartesian coordinates and so converting to polar
coordinates may be the only option we have for actually doing the integral
...
Let’s first determine the region that we’re integrating over and see if it’s a region that can be
easily converted into polar coordinates
...
Now, the upper limit for the x’s is,
and this looks like the right side of the circle of radius 1 centered at the origin
...
The range for the y’s however, tells us that we are only going to have positive y’s
...
So, we know that the inequalities that will define this region in terms of polar coordinates are
then,
Finally, we just need to remember that,
and so the integral becomes,
Note that this is an integral that we can do
...
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Exercise
a
...
Evaluate
by changing into polar coordinates
...
If D is the part of the annulus
lying in the first quadrant and
below the line y = x, evaluate
d
...
e
...
Suppose D is the region bounded by the circles r = 0, r = 2 and
, Find
then evaluate
Answers
(
6
...
The problem
with this is that most of the regions are not rectangular so we need to now look at the following
double integral,
Where D is any region
There are two types of regions that we need to look at
...
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We will often use set builder notation to describe these regions
...
This notation is really just a fancy way of saying we are going to use all the points, ( x, y) , in
which both of the coordinates satisfy the two given inequalities
...
In Case 1 where
the integral is defined to be
In Case 2 where
the integral is defined to be
Here are some properties of the double integral that we should go over before we actually do
some examples
...
Properties
Let us take a look at some examples of double integrals over general regions
Example 1 Evaluate each of the following integrals over the given region D
...
b
...
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Solutions:- a
...
b
...
The best way to do this is the graph the two curves
...
So, from the sketch we can see that the two inequalities are
We can now do the integral
=
We got even less information about the region this time
...
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Since we have two points on each edge it is easy to get the equations for each edge and so we’ll
leave it to you to verify the equations
...
If we use functions of x, as shown in the image
we will have to break the region up into two different pieces since the lower function is different
depending upon the value of x
...
If we do this then we’ll need to do two separate integrals, one for each of the regions
...
Either way should give the same answer and so we can get an example in the notes of splitting a
region up let’s do both integrals
...
Exercise
Answers
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6
...
We found the area of a very special type of surface – a surface of revolution – by the method of
single-variable calculus
...
Let S be a surface with equation z = f(x, y), where f has
continuous partial derivatives
...
We consider a partition
P of D into small rectangles
with areas
...
The tangent plane to S at
is an approximation to S
near
...
Thus the sum
is an
approximation to the total area of S, and this approximation appears to improve as
Therefore, we define the surface area of S to be
...
Then
and
and
are the slopes of the
tangent lines through in the direction of a and b
...
that lies above the
Solution
Figure 1
Figure 2
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The region shown in figure 1 and is described by T =
Using surface area formula with
, we get
dA =
dy dx
dx =
Figure 2 shows the portion of the surface whose area we have just computed
...
Solution:-The plane intersects the paraboloid in the circle
, z = 9
...
Using the surface area formula, we obtain
Converting to polar coordinates, we obtain
Exercise
i
...
ii
...
iii
...
Find the surface area S of the portion of the paraboloid
plane
above xy plane
...
6 TRIPLE INTEGRALS
Fubini’s Theorem for Triple Integrals
If f is continuous on the rectangular box B =
The iterated integral on the right side of Fubini’s Theorem means that we integrate first with
respect to x (keeping y and z fixed), then we integrate with respect to y (keeping z fixed), and
finally we integrate with respect to z
...
Evaluate
ii
...
Evaluate
where
Answers
iii
...
7Applications of Triple Integrals
Volume
We know that if f(x)
, then the single integral
y = f(x) from a to b, and if
represents the area under the curve
then the double integral
represents the
volume under the surface z = f(x, y) and above D
...
(Remember E is just the domain of the function f; the graph of f lies in four dimensional space )
...
Let us begin with the special case where f(x, y, z) = 1 for all points in E
...
The
lower boundary of T is the plane z = 0 and the upper boundary is the plane
, that
is,
Therefore, we have
Figures 12, 13 page 851
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Title: Civil engineering easy ways
Description: Formulas question and answers will set in short and easy methods
Description: Formulas question and answers will set in short and easy methods