bre; fourier

8/20/2019 Bre; Fourier

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Table of contents

Page

Acknowledgments………………………………………………….ii

Objective…………………………………………………………………….i

1. Chapter one: Fourier series

1.1 Introduction…………………………………………………1

1.2 Periodic function…………………..…..………………..2

1.3 Periodic function of odd and even function…2

1.4 Fourier sine series.........................................3

1.5 Fourier cosine series…………………………………..6

1.6 Fourier series of odd and even function…….9

1.7 Fourier series……………………………………………11

1.8 Convergence theorem of Fourier series…….14

1.9 Dirichlet’s theorem of Fourier series…………15

1.10 Half rang of Fourier series……………….……….16

1.11 Half rang of Fourier sine and cosine series..17

1.12 Reference…………………………………………………20


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[mathematics project] [2014]

[By Birhanu] Page i

Acknowledgments

First, I would like to thank my advisor Ato Waredewu

Worku for this advisor and Guide me how to prepare this

paper. Next, I would like to thank my parents and all of my

sisters and brothers for their financial support and for their

advice.

Finally, I would like to thank all of my friends who helped me

in any time during the preparation of this paper.


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[By Birhanu] Page ii

The objective of this project

The main object of Fourier series is:

. To find the Fourier series representation of period

function

. To express any periodic function in the finite number of

sequence of function where the sequence of function are

convergence or non convergence


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[By Birhanu] Page 1

Chapter one: Fourier series

1.1 Introduction

In, mathematical, A Fourier series decomposes a periodic functions or periodic

signals in to the sum of (a possibly infinitely) set of simple oscillating functions

namely the sine function and cosine function.

The study of Fourier series is a branch of pure mathematics which is called Fourier

analysis.

The idea of a Fourier series is that any function f(x) that is periodic on the interval

(- ≤ ≤ )of period 2that means f(x +2πn) = f(x) for all n. can be decomposedinto the general form of sin() and cos() then generally the concept ofFourier series is that representing the periodic function f(x) in the form of:

f(x) = + ∑ cos + ∑ sin where L is the period. And, are called the Fourier coefficients of the function.


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[By Birhanu] Page 2

1.2 periodic functions

The first topic we need to discuss is that of a periodic function. Function f is said

to be periodic with period T (T being anon –zero constant) if we have.

f(X+T) =f(X) for all x.

Let’s do that

(f + g)(X +T) = f(X + T) + g(X + T) = f(X) + g(X) = (F + g)(X)( )(X + T) = f(X + T) g(X + T) = f(X)g(X) = (fg)(X)EXAMPLE1: For example, the sine function is periodic with period 2, sinceSin(x +2π) =

sin

For all values of x this function rerates on intervals of length 2π.

1.3. Periodic function of odd and even function

1.3.1 Even functions

Let f(x) is a real valued function of areal variable .the f is even if the following

equation holds for all x and –x in the domain of f:

f(X) =f(-X)

Examples of even function are||, , cos and and cosh(x)1.3.2 Odd function

Again let f(x) is area valued function of area variable. Then f is odd if the following

equation holds for all x and –x in the domain of f:

f(-x) =-f(x) or f (-x) + f(x) = 0

Example of odd functions are x,, s i n , sinh(x).


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[By Birhanu] Page 3

1.4 Fourier sin series

Let’s instead think back to our calculus class where we looked at Taylor series.

With Taylor series we wrote a series representation of a function, f(x), as a serieswhose terms where powers of x – a for some x = a with some condition we were

able to show that,

() = ()()!

( − ) And that the series will converge to f(x) on | − | < for some R that will bedependent upon the function itself. The is nothing wrong with this, but it does

require the derivatives of all orders exist at x = a or in other words ()() exists forn = 0, 1, 2, 3, . . . . . . . We’ll start thing of f by assuming that the function ,f(x) ,wewant to write a series representation for is an odd function (i.e. f(-x) = -f(x)

).Because f(x) is odd it make sense that should be able to write a series

representation for this in terms of sine’s only (since they are also odd function ).

What we’ll try to do here is write f(x) as the following series s representation,

called a Fourier sin series, on

– L ≤ ≤ .

sin () The question now is how to determine the coefficients, in the series. Let’s startwith the series above and multiply both sides by sin () where m is a fixedinteger m the range {1, 2, 3 . . .}.Inn other words we multiply both sides by any of

the sines in the set of sine’s that we are working with here .doing this gives,

f(x)sin () = ∑ sin () sin () Now let’s integrate both sides of this from x = -L to x =L

∫ () sin d x


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[By Birhanu] Page 4

= ∫ sin sin dxDoing this gives and factoring the constant,

out of the integral gives.

∫ () sin = ∑ ∫ sin sin dx=∑ ∫ sin sin dx

We will use the fact that {sin } do to the orthogonal set on the interval -L≤ ≤ and that,∫ sin sin d x = 0 =≠

Compute the value of the integral all but one of the integrals will be zero. The

only none zero integral will come when we have n=m in which case the integral

has the value of L. Therefore, the only none zero term in the series will come

when we have n=m and our equation becomes,

∫ () sin d x = LFinally all we need to do is dividing by L and we know have an equation for each

of the coefficients.

= ∫ () sin d x m = 1, 2, 3…Next, note that because we are integrating two odd functions the integrand of

this integral is even and so we also know that,

= ∫ () sin d x m = 1, 2, 3…


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[By Birhanu] Page 5

Summarizing all this work up the Fourier sine series of an odd function f(x) on -

L≤ ≤ is gives by,f(x) =

∑ sin

= ∫ () sin d x n = 1, 2, 3=∫ () sin d x n = 1, 2, 3….

Example1. Find the Fourier sine series for f(x) = L –x on 0≤ ≤ Solution

: the really isn’t much to do have other than computing the coefficients

so here they are, = ∫ () sin d x=

∫ ( − ) sin d x=

sin − ( − ) cos =

(−sin())

=

In the simplification process don’t that n is an integer. So, with the coefficient we

get the following Fourier sine series for t his function.

F(x) = ∑ sin( ) Example2. Find the Fourier sin series for f(x) = 1 + on 0≤ ≤ Solution: in this case the coefficients are liable to be somewhat messy given thefact that the integrals will involve integration by part twice.

= ∫ () sin


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[By Birhanu] Page 6

= ∫ (1 + ) sin

=

[(2

−

](1 +

))cos

+2sin ]0 =

[2 − (1 + ))cos() + 2sin() − (2 − )] =

2 − (1 + )(−1) − 2 + The Fourier sine series for this function is then,

F(x) = ∑ 2 − (1 + )(−1) − 2 + sin 1.5 Fourier cosine series

Let’ start by assuming g that the function, f(x) we will be working with initially is

an even function (i.e. f (-x) = f(x)) and that we want to write a series

representation for this function on –L ≤ ≤ in terms of cosines (which are alsoeven) .In other words we are going to look for the following ,

F(x) = ∑ cos This series is called Fourier cosine series and note that in this case (unlike with

Fourier sine series) we able to start the series representation at n=0 since that

term will not be zero as it was with sine’s. Note as well that we are assuming that

the series will in fact converge to f(x) on -L≤ ≤ at this point. So to determinea formula for the coefficients,

,we will use the fact that {

cos

} do from

an orthogonal set on the interval -L≤ ≤ and that,∫ cos cos = 20

= = 0 = ≠ 0 ≠


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[By Birhanu] Page 7

We’ll start with the representation above and multiply both sides by cos where m is affixed integer in the range {0, 1, 2...}.Doing this gives,

F(x)cos = ∑ cos cos

Next, we integrate both sides from X =-L to =L and as we were able to do with the

Fourier sine series we can again interchange the integral and the series.

∫ () cos = ∫ ∑ cos cos =∑ ∫ cos cos

We now that the all of the integrals on the right side will be zero except when

n=m because the set of cosines from an orthogonal set on the interval L≤ ≤ however we need to be careful about the value of m. so, after evaluating all of

the integrals we arrive at the following set of formulas for the coefficients.

m = 0: ∫ () = (2L) → = ∫ () m ≠ 0: ∫ () cos = () → = ∫ () cos summarizing everything up then, the Fourier cosine series of an even of function,

f(x) on -L≤ ≤ is given by.f(x) =

∑ cos

= ∫ () = ∫ () cos ≠ 0 Example1: find the Fourier cosine series for f(x) = on - ≤ ≤


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[By Birhanu] Page 8

Solution: we clearly have an even function here and so all we really to do is

compute the coefficients and we’ll need to do integration by parts twice. We’ll

leave most of the actual integration details you to verify.

= ∫ () = ∫ =

= ∫ () = ∫ () cos = ∫ cos = ∫ 2cos + ( − 2) sin ]0

= (2cos() + () sin( )) = () n =1, 2, 3… TheFourier cosine series is the, =

= () n = 1, 2, 3…Fourier cosine series is then,

∑ cos = ∑ cos =

+∑ () cos f(x) =

+ ∑ () cos


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[By Birhanu] Page 9

Example2: find the Fourier cosine series for f(x) = L - x on –L ≤ ≤ Solution: all we need to do is compute the coefficients so here is the work for that

= ∫ () =

∫ − =

= ∫ () cos

= ∫ (−)cos =

( − ) sin −cos ]0 =

(−cos() + ) =

(1+(−1)) , n = 1, 2, 3…The Fourier cosine series is then,

f(x) = + ∑ 1+(−1) cos 1.6 Fourier series for Even and odd function

1.6.1 Even function

If y = f(x) is even, then the f(-x) = f(x)

Example1: y = The graph of this function is symmetric w.r.t the y-axis

If y = g(x) is even, the ∫ () =2∫ ()


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[By Birhanu] Page 10

∫ (() = ∫ () + ∫ () =

− ∫ (−)

+

∫ ()

=∫ () + ∫ () = 2 ∫ () Example 2: ∫ cos = 2 ∫ c os= 0 If f is even, then the integrand f(x) sin x is odd, = ∫ () sin = 0 For Fourier series of an even function f having period 2L is a Fourier cosine series

f(x) + ∑ cos ( ) With coefficients = ∫ () and = ∫ () cos n=1, 2 …

1.6.2odd function

If y = f(x) is odd, then f(-x) = -f(x)

Example 1: y =

If h(x) is odd function, then ∫ ℎ() = ∫ ℎ() + ∫ ℎ() = − ∫ ℎ(−) + ∫ ℎ() = − ∫ ℎ() + ∫ ℎ()

The product q = gh of an even function g and an odd function h is an odd function,

because q(-x) =g(-x).-h(x) = g(-x)h(x) then q(-x) = -q(x)

If f is odd, then the integrand f(x) is odd, = ∫ () = ∫ () = 0 = ∫ () cos = ∫ () cos = 0 The Fourier series of odd function f having period = 2L is a Fourier sine series.


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f(x) = ∑ sin ℎ = 0, = ∫ () sin

1.7 Fourier series

Fourier series is specific types of infinite mathematical series involving

trigonometric functions. In mathematics an infinite series used to solve special

types of differential equation. It consists of an infinite sum of sine and cosine and

because, it is periodic (i.e. its values over affixed intervals). With Fourier we are

going to try to write aeries representation for f(x) on –L≤ ≤ ℎ , f(x) = ∑ cos + ∑ sin

So Fourier series is in some way combination of Fourier sine series and Fourier

cosine series.

Determining formulas for the coefficient, and, will be done in exactly thesome manner. We will betake advantage of the fact that cos ∞ = 0 and

sin

∞ = 1 are mutually orthogonal on -L

≤ ≤ . We will also need the

following formulas that we derived when we proved the two were mutually

orthogonal.

∫ cos cos = 2 = = 0 = ≠ 00 ≠

∫ sin

sin

= 0

So, let’s start off by multiplying both sides of the series above by cos andintegrating form –L to L. doing this gives.


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∫ () cos =∫ ∑ cos cos +∫ ∑ Bsin cos Now, jest as we have been able to do in the last two sections we can interchangethe integral and the summation .doing this gives.

∫ () sin = ∑ ∫ cos cos +∑ ∫ sin cos

The integral in the second series will always be zero and in the first series theintegral will be zero if n≠ and so this reduces to,

Solve for gives, = ∫ () = ∫ () cos , m = 1, 2, 3…

Now, do it all again only this time multiply both sides by sin , integrate bothsides from –L to L and inter change the integral and summation to get,∫ () sin

= ∑ ∫ cos sin + ∑ ∫ sin () sin () In this case the integral in the first series will always be zero and the second will

be zero if n=m and so we get,

∫ () sin = (L)

Finally, solving for, gives.


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= ∫ () sin , m = 1, 2, 3…However in this case we don’t know anything about whether f (x) will be even,

odd or more likely neither even or odd .Therefore, this the only form of thecoefficients for the Fourier series.

Example1: find the Fourier series for f(x) =L-x on - ≤ ≤ Solution: so, let’s as go ahead and just run through formulas for the coefficients.

= ∫ () =

∫ −

= L

= ∫ () cos =

∫ ( − ) cos =

( − ) sin

−cos

]

−

(−2sin(−)) = 0 = ∫ () sin

= ∫ ( − ) sin

=

sin

− ( − ) cos

]

−

= (2cos() −2sin()) =

() , n = 1, 2, 3……Note that in this case we have ≠ 0 =0, n = 1, 2, 3....


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The Fourier series is then,

f(x) = ∑ cos +∑ sin = +∑ cos +∑ sin = L+∑ () sin

1.8 convergent theorems of Fourier series

As I have aforementioned, we showed that if the Fourier series,

+ ∑ cos + sin , . . . . . . . . . . . . . . . ; (1) And there by defines a function f, then f is period with period 2L and the

coefficients are related to f(x by the Euler Fourier formulas: = ∫ () cos , m = 0, 1, 2. . . (2) =

∫ () sin

, m = 0, 1, 2. . . (3)

In this section we adopt a somewhat different point of view. Suppose that a

function f is given .if this function is periodic with period 2L and integral on the

interval– , , then a set of coefficients can be computed formequation (2) (3) and a series of the form (1) can be formally constructed.The equation is whether this series converges of each value of x and, if so,

whether its sum is f(x). a Fourier series corresponding to a function f may not

converge to f(x), or may even diverge, functions whose Fourier series do not

converge to the value of the function at isolated points are easily constructed,Functions whose Fourier series diverge at one or more points are more

pathological.

To guarantee convergence of a Fourier series to the function from which its

coefficients were computed it is essential to place additional conditions on the


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function. From a practical point of view, such condition should be broad enough

to cover all situations of interest, yet simple enough to be easily checked for

particular functions.

Before stating convergence theorem for Fourier series, we define a term that

appears in the theorem. A function f said to be piece wise continuous on an

interval a ≤ ≤ if the interval can be partitioned by a finite number of points a=


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Where the notation,

f(x+) =() →

f( x-) = () →

Denotes the right or left limits of f,

A function satisfying Dirichlet condition must have right and limits at each point of

discontinuity or else the function would need to oscillate at that point violating

the condition of maximum and minimum. Note that any point where f is

continuous.

(f () + f()) = f(xThus Dirichlet theorem says in a particular that under the Dirichlet condition the

Fourier series for f converges and is equal to f where very f is continuous.

1.9 Half rang Fourier series

Definition: it is a Fourier series defined on an interval [0, ] with the implicationthat the analyzed function f(x, x ∈ [0 , ]), should be extended to – , 0 as eitheran even (f (-x) = f(x)) or odd function (f(-x) =-f(x). this allows the expansion of the

function in aeries solely o f sine (odd) or cosines (even). The choice between odd

and even is typically motivated by boundary conditions, associated with a

differential equation satisfied by f(x)

Example: calculate the half rang Fourier sin series for the function f(x):

cos() =cos() where 0


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1.10 Half rang of Fourier sine and cosine series

1.10.1 Half rang Fourier sine series

Expanding the odd periodic exploration F(X) of the function f(x) in to a Fourier

series, we find

F(x) = ∑ sin where

=

∑ ∫ ()

sin

.

So the half rang Fourier sine series representation of f(x) is

f(x) = ∑ sin where = ∫ ( ) sin

Example1: find the half rang Fourier sine series of the function

f(x) =

4 0 < <

0 < < Solution: L = π, so thatf(x) =∑ sin()

= ∫ () sin() = ∫ () sin()

= ∫ sin() =

∫ sin()


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= [cos]0 = 1−cos

1.10.2 Half rang Fourier cosine series

Expanding the even periodic extrapolation F(x) of function f(x) into a function in a

Fourier series, we find

F(x) = = ∑ cos With

=

∫ ()

= ∫ ( ) cos F(x) = + ∑ cos

With = ∫ ()

=

∫ ()

cos

Example1: find the half rang Fourier cosine series of the function

f(x) = 4 0 < < 0 < < Solution: L = , so that

F(x) = +∑ cos() , = ∫ () = ∫ () cos() = ∫ ( )


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= ∫ 4

= 2

= ∫ () cos() =

∫ cos() =

∫ cos =

[sin]

0

= sin


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References

1. Erwin Krayzing advanced engineering mathematics,10 Edition, 1989.2.

W.A Strauss partial Differential equation, 1 Edition, 1992.

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