Fundamental of Fourier Series

7/28/2019 Fundamental of Fourier Series

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Series

FOURIER SERIES

Graham S McDonald

A self-contained Tutorial Module for learningthe technique of Fourier series analysis

q Table of contents

q Begin Tutorial

c 2004 [email protected]
http://www.cse.salford.ac.uk/mailto:[email protected]:[email protected]://www.cse.salford.ac.uk/


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Table of contents

1. Theory

2. Exercises

3. Answers

4. Integrals

5. Useful trig results

6. Alternative notation

7. Tips on using solutions

Full worked solutions


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Section 1: Theory 3

1. Theory

qA graph of periodic function f(x) that has period L exhibits thesame pattern every L units along the x-axis, so that f(x + L) = f(x)

for every value ofx. If we know what the function looks like over onecomplete period, we can thus sketch a graph of the function over awider interval of x (that may contain many periods)

f(x)

x

PERIOD =L

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Section 1: Theory 4

q This property of repetition defines a fundamental spatial fre-quency k = 2

Lthat can be used to give a first approximation to

the periodic pattern f(x):

f(x) c1 sin(kx + 1) = a1 cos(kx) + b1 sin(kx),

where symbols with subscript 1 are constants that determine the am-

plitude and phase of this first approximation

q A much better approximation of the periodic pattern f(x) canbe built up by adding an appropriate combination of harmonics tothis fundamental (sine-wave) pattern. For example, adding

c2 sin(2kx + 2) = a2 cos(2kx) + b2 sin(2kx) (the 2nd harmonic)c3 sin(3kx + 3) = a3 cos(3kx) + b3 sin(3kx) (the 3rd harmonic)

Here, symbols with subscripts are constants that determine the am-

plitude and phase of each harmonic contribution

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Section 1: Theory 5

One can even approximate a square-wave pattern with a suitable sumthat involves a fundamental sine-wave plus a combination of harmon-

ics of this fundamental frequency. This sum is called a Fourier series

Fundamental + 5 harmonics


x

PERIOD = L

Fundamental


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Section 1: Theory 6

q In this Tutorial, we consider working out Fourier series for func-tions f(x) with period L = 2. Their fundamental frequency is then

k =

2

L = 1, and their Fourier series representations involve terms like

a1 cos x , b1 sin x

a2 cos2x , b2 sin2x

a3 cos3x , b3 sin3x

We also include a constant term a0/2 in the Fourier series. Thisallows us to represent functions that are, for example, entirely abovethe xaxis. With a sufficient number of harmonics included, our ap-proximate series can exactly represent a given function f(x)

f(x) = a0/2 + a1 cos x + a2 cos2x + a3 cos3x + ...

+ b1 sin x + b2 sin2x + b3 sin3x + ...

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Section 1: Theory 7

A more compact way of writing the Fourier series of a function f(x),with period 2, uses the variable subscript n = 1, 2, 3, . . .

f(x) = a02

+n=1

[an cos nx + bn sin nx]

q We need to work out the Fourier coefficients (a0, an and bn) forgiven functions f(x). This process is broken down into three steps

STEP ONE

a0 =1

2

f(x) dx

STEP TWOan =

1

2

f(x)cos nx dx

STEP THREEbn =

1

2

f(x)sin nx dx

where integrations are over a single interval in x of L = 2

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Section 1: Theory 8

q Finally, specifying a particular value of x = x1 in a Fourier series,gives a series of constants that should equal f(x1). However, if f(x)

is discontinuous at this value ofx, then the series converges to a valuethat is half-way between the two possible function values

f(x)

x

Fourier series

converges to

half-way point

"Vertical jump"/discontinuity

inthefunction represented

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Section 2: Exercises 9

2. Exercises

Click on Exercise links for full worked solutions (7 exercises in total).

Exercise 1.Let f(x) be a function of period 2 such that

f(x) =

1, < x < 00, 0 < x < .

a) Sketch a graph of f(x) in the interval 2 < x < 2b) Show that the Fourier series for f(x) in the interval < x < is

1

2 2

sin x +1

3

sin3x +1

5

sin5x + ...c) By giving an appropriate value to x, show that

4= 1 1

3+

1

5 1

7+ . . .

q Theory q Answers q Integrals q Trig q Notation

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Exercise 2.

Let f(x) be a function of period 2 such that

f(x) =

0, < x < 0x, 0 < x < .

a) Sketch a graph of f(x) in the interval 3 < x < 3b) Show that the Fourier series for f(x) in the interval < x < is

4 2

cos x +

1

32cos3x +

1

52cos5x + ...

+ sin x 1

2

sin2x +1

3

sin3x

...

c) By giving appropriate values to x, show that

(i) 4 = 1 13 + 15 17 + . . . and (ii) 2

8 = 1 +1

32 +1

52 +1

72 + . . .


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Exercise 3.


f(x) =

x, 0 < x < , < x < 2 .

a) Sketch a graph of f(x) in the interval 2 < x < 2b) Show that the Fourier series for f(x) in the interval 0 < x < 2 is

3

4 2

cos x +

1

32cos3x +

1

52cos5x + . . .

sin x + 12 sin2x + 13 sin3x + . . .c) By giving appropriate values to x, show that

(i) 4 = 1 13 + 15 17 + . . . and (ii) 2

8 = 1 +1

32 +1

52 +1

72 + . . .


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Exercise 4.


f(x) =x

2over the interval 0 < x < 2.

a) Sketch a graph of f(x) in the interval 0 < x < 4

b) Show that the Fourier series for f(x) in the interval 0 < x < 2 is

2

sin x +1

2sin2x +

1

3sin3x + . . .

c) By giving an appropriate value to x, show that

4= 1 1

3+

1

5 1

7+

1

9 . . .


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Exercise 5.


f(x) =

x, 0 < x <

0, < x < 2


b) Show that the Fourier series for f(x) in the interval 0 < x < 2 is

4+

2

cos x +

1

32cos3x +

1

52cos5x + . . .

+ sin x +

1

2sin2x +

1

3sin3x +

1

4sin4x + . . .


2

8= 1 +

1

32+

1

52+ . . .


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Exercise 6.


f(x) = x in the range < x < .


b) Show that the Fourier series for f(x) in the interval < x < is2

sin x 1

2sin2x +

1

3sin3x . . .


4= 1 1

3+

1

5 1

7+ . . .


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Exercise 7.


f(x) = x2 over the interval < x < .


b) Show that the Fourier series for f(x) in the interval < x < is2

3 4

cos x 1

22cos2x +

1

32cos3x . . .


2

6= 1 +

1

22+

1

32+

1

42+ . . .


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A


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Section 3: Answers 16

3. Answers

The sketches asked for in part (a) of each exercise are given within

the full worked solutions click on the Exercise links to see thesesolutions

The answers below are suggested values of x to get the series ofconstants quoted in part (c) of each exercise

1. x = 2 ,

2. (i) x = 2 , (ii) x = 0,

3. (i) x = 2 , (ii) x = 0,

4. x =

2 ,5. x = 0,

6. x = 2 ,

7. x = .

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S i 4 I l 17


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Section 4: Integrals 17

4. Integrals

Formula for integration by parts: b

au dvdx

dx = [uv]ba

b

adudx

v dx

f(x)

f(x)dx f(x)

f(x)dx

xn xn+1

n+1 (n = 1) [g (x)]n g (x) [g(x)]n+1

n+1 (n = 1)1x

ln |x| g(x)g(x) ln |g (x)|

ex

ex

ax a

x

ln a (a > 0)sin x cos x sinh x cosh xcos x sin x cosh x sinh xtan x ln |cos x| tanh x ln cosh xcosec x ln tan x2 cosech x ln tanh x2 sec x ln |sec x + tan x| sech x 2tan1 exsec2 x tan x sech2 x tanh xcot x ln |sin x| coth x ln |sinh x|sin2 x x2 sin 2x4 sinh2 x sinh 2x4 x2cos2 x x

2+ sin 2x

4cosh2 x sinh 2x

4+ x

2

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S ti 4 I t l 18


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Section 4: Integrals 18

f(x)

f(x) dx f(x)

f(x) dx

1

a2+x21

a tan1 x

a

1

a2x21

2a ln a+xax (0 < |x|< a)(a > 0) 1

x2a21

2a lnxax+a (|x| > a > 0)

1a2x2 sin

1 x

a

1a2+x2 ln x+

a2+x2

a (a > 0)(a < x < a) 1

x2a2 lnx+x2a2a (x > a > 0)

a2 x2 a22 sin1xa a2 +x2 a22 sinh1 xa + xa2+x2a2 +x

a2x2a2

x2a2 a22

cosh1 x

a

+ x

x2a2a2

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S ti 5 U f l t i lt 19


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Section 5: Useful trig results 19

5. Useful trig results

When calculating the Fourier coefficients an and bn , for which n =

1, 2, 3, . . . , the following trig. results are useful. Each of these results,which are also true for n = 0,1,2,3, . . . , can be deduced fromthe graph of sin x or that of cos x

q sin n = 0

1

0

1

2 33 2

sin(x)

x

q cos n = (1)n

1

0

1

2 33 2

cos(x)

x

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1

0

1

2 33 2

sin(x)

x

1

0

1

2 33 2

cos(x)

x

q sin n 2

=

0 , n even1 , n = 1, 5, 9,...

1 , n = 3, 7, 11,...q cos n

2=

0 , n odd1 , n = 0, 4, 8,...

1 , n = 2, 6, 10,...

Areas cancel whenwhen integrating

over whole periodsq

2

sin nx dx = 0

q

2cos nx dx = 0

++1

0

1

2 33 2

sin(x)

x

+

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Section 6: Alternative notation 21


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6. Alternative notation

q For a waveform f(x) with period L = 2k

f(x) =a02

+n=1

[an cos nkx + bn sin nkx]

The corresponding Fourier coefficients are

STEP ONEa0 =

2L

L

f(x) dx

STEP TWOan =

2

L L f(x)cos nkx dxSTEP THREE

bn =2

L

L

f(x)sin nkx dx

and integrations are over a single interval in x of L

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q For a waveform f(x) with period 2L = 2k

, we have thatk = 22L =

L

and nkx = nxL

f(x) = a02

+n=1

an cos nx

L+ bn sin nx

L


STEP ONE a0 = 1L

2L

f(x) dx

STEP TWOan =

1

L 2Lf(x)cos

nx

Ldx

STEP THREEbn =

1

L

2L

f(x)sinnx

Ldx

and integrations are over a single interval in x of 2L

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q For a waveform f(t) with period T = 2

f(t) =

a0

2+

n=1

[an

cosnt

+bn

sinnt

]


STEP ONE

a0 =2

T T

f(t) dt

STEP TWOan =

2

T

T

f(t)cos nt dt

STEP THREE bn = 2T

T

f(t)sin nt dt

and integrations are over a single interval in t of T

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Section 7: Tips on using solutions 24


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Section 7: Tips on using solutions 24

7. Tips on using solutions

q When looking at the THEORY, ANSWERS, INTEGRALS, TRIGor NOTATION pages, use the Back button (at the bottom of thepage) to return to the exercises

q Use the solutions intelligently. For example, they can help you getstarted on an exercise, or they can allow you to check whether yourintermediate results are correct

q Try to make less use of the full solutions as you work your waythrough the Tutorial

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Solutions to exercises 26


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b) Fourier series representation of f(x)

STEP ONE

a0 = 1

f(x)dx = 1

0

f(x)dx + 1

0

f(x)dx

=1

0

1 dx + 1

0

0 dx

=1

0

dx

=1

[x]0

=1

(0 ())

=1

()

i.e. a0 = 1 .

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STEP TWO

an =1

f(x)cos nx dx =

1

0

f(x)cos nx dx +

1

0

f(x)cos nx dx

=1

0

1 cos nx dx + 1

0

0 cos nx dx

=1

0

cos nx dx

=1

sin nx

n

0

=1

n[sin nx]

0

=1

n(sin0

sin(

n))

=1

n(0 + sin n)

i.e. an =1

n(0 + 0) = 0.

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STEP THREE

bn =1

f(x)sin nx dx

=1

0

f(x)sin nx dx +1

0

f(x)sin nx dx

=1

0

1

sin nx dx +

1

0

0

sin nx dx

i.e. bn =1

0

sin nx dx =1

cos nxn

0

= 1

n [cos nx]0 =

1

n (cos0 cos(n))= 1

n(1 cos n) = 1

n(1 (1)n) , see Trig

i.e. bn = 0 , n even

2

n

, n odd, since (

1)n =

1 , n even

1 , n odd

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We now have that

f(x) =a0

2

+

n=1[an cos nx + bn sin nx]with the three steps giving

a0 = 1, an = 0 , and bn = 0 , n even

2n

, n odd

It may be helpful to construct a table of values of bn

n 1 2 3 4 5bn 2 0 2

13

0 2

15

Substituting our results now gives the required series

f(x) =1

2 2

sin x +

1

3sin3x +

1

5sin5x + . . .

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c) Pick an appropriate value of x, to show that

4= 1

1

3+ 1

5 1

7+ . . .

Comparing this series with

f(x) =1

2 2

sin x +

1

3sin3x +

1

5sin5x + . . .

,

we need to introduce a minus sign in front of the constants 13 ,17 , . . .

So we need sin x = 1, sin3x = 1, sin5x = 1, sin 7x = 1, etc

The first condition of sin x = 1 suggests trying x =

2 .

This choice gives sin 2 +13 sin3

2 +

15 sin5

2 +

17 sin7

2

i.e. 1 13 + 15 17Looking at the graph of f(x), we also have that f(2 ) = 0.

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Picking x = 2 thus gives

0 = 1

2 2 sin 2 + 13 sin 32 + 15 sin 52

+ 17 sin72 + . . .

i.e. 0 = 12 2

1 13 + 15

17 + . . .

A little manipulation then gives a series representation of 4

2

1 1

3+

1

5 1

7+ . . . = 12

1 13

+1

5 1

7+ . . . =

4.

Return to Exercise 1

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Exercise 2.

f(x) = 0, < x < 0x, 0 < x < , and has period 2a) Sketch a graph of f(x) in the interval 3 < x < 3

2 33 2

f(x)

x

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STEP ONE

a0 =1

f(x)dx =1

0

f(x)dx +1

0

f(x)dx

=1

0 0 dx +

1

0x dx

=1

x2

2

0

= 1

22 0

i.e. a0 =

2.

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STEP THREE

bn =1

f(x)sin nx dx =

1

0

f(x)sin nx dx +

1

0

f(x)sin nx dx

=1

0

0 sin nx dx + 1

0

x sin nx dx

i.e. bn =1

0

x sin nx dx =1

xcos nx

n

0

0 cos nx

n dx(using integration by parts)

=1

1

n[x cos nx]0 +

1

n

0

cos nx dx

=1

1n ( cos n 0) + 1n sin nxn

0

= 1

n(1)n + 1

n2(0 0), see Trig

=

1

n(

1)n

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i.e. bn =

1n

, n even

+ 1n

, n odd

We now have

f(x) =a02

+

n=1


where a0 =

2, an =

0 , n even 2

n2, n odd

, bn = 1n , n even

1n

, n odd

Constructing a table of values gives

n 1 2 3 4 5

an 2 0 2 132 0 2 152

bn 1

1

2

1

3 1

4

1

5

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This table of coefficients gives

f(x) =1

2

2 + 2 cos x + 0 cos2x

+

2

1

32

cos3x + 0 cos4x

+ 2 152 cos5x + ...+ sin x 1

2sin2x +

1

3sin3x ...

i.e. f(x) =

4 2

cos x + 132 cos3x + 152 cos5x + ...+

sin x 1

2sin2x +

1

3sin3x ...

and we have found the required series!

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(i)

4

= 1

1

3

+ 1

5 1

7

+ ...

Comparing this series with

f(x) =

4 2

cos x +

1

32cos3x +

1

52cos5x + ...

+

sin x 12

sin2x +1

3sin3x ...

,

the required series of constants does not involve terms like 132 ,1

52 ,1

72 ,....So we need to pick a value of x that sets the cos nx terms to zero.

The Trig section shows that cos n2 = 0 when n is odd, and note also

that cos nx terms in the Fourier series all have odd n

i.e. cos x = cos 3x = cos 5x = ... = 0 when x = 2 ,

i.e. cos 2 = cos 32 = cos 5

2 = ... = 0

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Setting x = 2 in the series for f(x) gives

f

2 =

4 2

cos

2

+1

32

cos3

2

+1

52

cos5

2

+ ...+

sin

2 1

2sin

2

2+

1

3sin

3

2 1

4sin

4

2+

1

5sin

5

2 ...

=

4 2

[0 + 0 + 0 + ...]

+

1 12

sin =0

+1

3 (1) 1

4sin2

=0

+1

5 (1) ...

The graph of f(x) shows that f2 = 2 , so that

2=

4+ 1 1

3+

1

5 1

7+ ...

i.e.

4

= 1

1

3

+1

5 1

7

+ ...

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Pick an appropriate value of x, to show that

(ii)

2

8 = 1 +

1

32

+

1

52

+

1

72

+ ...

Compare this series with

f(x) =

4 2

cos x +1

32cos3x +

1

52cos5x + ...

+

sin x 12

sin2x +1

3sin3x ...

.

This time, we want to use the coefficients of the cos nx terms, andthe same choice of x needs to set the sin nx terms to zero

Picking x = 0 gives

sin x = sin 2x = sin 3x = 0 and cos x = cos 3x = cos 5x = 1

Note also that the graph of f(x) gives f(x) = 0 when x = 0

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So, picking x = 0 gives

0 =

4 2

cos 0 + 132 cos 0 + 152 cos 0 + 172 cos 0 + ...+sin0 sin0

2+

sin0

3 ...

i.e. 0 =

4 2

1 +1

32+

1

52+

1

72+ ... + 0 0 + 0 ...

We then find that

2

1 + 132 + 152 + 172 + ... = 4and 1 +

1

32+

1

52+

1

72+ ... =

2

8.


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i


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Exercise 3.

f(x) = x, 0 < x < , < x < 2, and has period 2a) Sketch a graph of f(x) in the interval 2 < x < 2

0 22

f(x)

x

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b) F i i i f f ( )


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STEP ONE

a0 =1

20

f(x)dx =1

0

f(x)dx +1

2

f(x)dx

=1

0

xdx +1

2

dx

=1

x2

2

0

+

x

2

=1

2

2 0 + 2 =

2+

i.e. a0 = 32 .

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S P WO


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STEP TWO

an =1

2

0

f(x)cos nx dx

=1

0

x cos nx dx +1

2

cos nx dx

= 1 x sin nxn

0

0sin nxn dx

using integration by parts

+ sin nxn

2

=

1

1n sin n 0 sin n0 cos nxn2

0

+1

n(sin n2 sin n)

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STEP THREE


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STEP THREE

bn =1

2

0

f(x)sin nx dx

=1

0

x sin nx dx +1

2

sin nx dx

=1

xcos nx

n

0

0 cos nx

n dx using integration by parts

+

cos nx

n 2

=1

cos nn

+ 0

+

sin nx

n2

0

1

n(cos2n cos n)

=1

(1)nn

+

sin n sin0

n2

1

n

1 (1)n

= 1n

(1)n + 0 1n1 (1)

n

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i b1

( 1)n1 1

( 1)n


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i.e. bn = n

(1)n n

+n

(1)n

i.e. bn =

1

n.

We now have

f(x) =a02

+

n=1


where a0 =32 , an =

0 , n even 2

n2, n odd

, bn = 1n

Constructing a table of values gives

n 1 2 3 4 5an 2 0 2

1

32

0 2

1

52

bn 1 12 13 14 15

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f(x) = 12 3

2

+ 2

cos x + 0 cos2x + 132 cos3x + . . .

+ 1

sin x + 12 sin2x +

13 sin3x + . . .

i.e. f(x) = 34 2

cos x + 132 cos3x +1

52 cos5x + . . .

sin x +12 sin2x +

13 sin3x + . . .

and we have found the required series.

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c) Pick an appropriate value of x to show that


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(i) 4 = 1

13 +

15

17 + . . .

Compare this series with

f(x) =3

4 2

cos x +

1

32cos3x +

1

52cos5x + . . .

sin x +1

2sin2x +

1

3sin3x + . . .

Here, we want to set the cos nx terms to zero (since their coefficientsare 1, 132 ,

152 , . . .). Since cos n

2 = 0 when n is odd, we will try setting

x =

2 in the series. Note also that f(

2 ) =

2

This gives2 =

34 2

cos 2 +

132 cos3

2 +

152 cos5

2 + . . .

sin 2 + 12 sin22 + 13 sin32 + 14 sin42 + 15 sin52 + . . .Toc Back


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Setting x = 0 eliminates the sin nx terms from the series, and also


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Setting x 0 eliminates the sin nx terms from the series, and alsogives

cos x +

1

32 cos3x +

1

52 cos5x +

1

72 cos7x + . . . = 1 +

1

32 +

1

52 +

1

72 + . . .

(i.e. the desired series).

The graph of f(x) shows a discontinuity (a vertical jump) at x = 0

The Fourier series converges to a value that is half-way between thetwo values of f(x) around this discontinuity. That is the series willconverge to 2 at x = 0

i.e.

2=

3

4 2

cos 0 +1

32cos 0 +

1

52cos 0 +

1

72cos 0 + . . .

sin 0 +1

2sin 0 +

1

3sin 0 + . . .

and

2 =

3

4 2

1 + 132 + 152 + 172 + . . . [0 + 0 + 0 + . . .]Toc Back



52/80

Finally, this gives

4

= 2

1 + 1

32+ 1

52+ 1

72+ . . .

and

2

8= 1 +

1

32+

1

52+

1

72+ . . .


Toc Back


Exercise 4.


53/80

f(x) = x2 , over the interval 0 < x < 2 and has period 2


3 40 2

f(x)

x

Toc Back




54/80

) p f ( )

STEP ONE

a0 =1

20

f(x) dx

=

1

2

0

x

2 dx

=1

x2

4

20

= 1

(2)24

0i.e. a0 = .

Toc Back


STEP TWO


55/80

an =1

2

0

f(x)cos nx dx

=1

20

x

2cos nx dx

=1

2x sin nxn

2

0

1

n2

0sin nx dx


=

1

22 sin n2n 0 sin n 0n 1n 0=

1

2

(0 0) 1

n 0

, see Trig

i.e. an = 0.

Toc Back


STEP THREE


56/80

bn =1

2

0

f(x)sin nx dx =1

2

0 x

2 sin nx dx=

1

2

20

x sin nx dx

=1

2 x cos nx

n 2

0 2

0 cos nx


=1

2

1

n(2 cos n2 + 0) + 1

n 0

, see Trig

= 22n

cos(n2)

= 1n

cos(2n)

i.e. bn = 1

n , since 2n is even (see Trig)Toc Back


We now have


57/80

f(x) =a02

+

n=1 [an cos nx + bn sin nx]where a0 = , an = 0, bn = 1n

These Fourier coefficients give

f(x) =

2+

n=1

0 1

nsin nx

i.e. f(x) = 2sin x + 1

2sin2x + 1

3sin3x + . . .

.

Toc Back




58/80

4 = 1

13 +

15

17 +

19

. . .

Setting x = 2 gives f(x) =4 and

4

=

2 1 + 0 1

3

+ 0 +1

5

+ 0

. . .

4=

2

1 13

+1

5 1

7+

1

9 . . .

1 1

3+

1

5 1

7+

1

9 . . . =

4

i.e. 1 13

+1

5 1

7+

1

9 . . . =

4.


Toc Back


Exercise 5.


59/80

f(x) = x , 0 < x < 0 , < x < 2, and has period 2a) Sketch a graph of f(x) in the interval 2 < x < 2

0 22

f(x)

x

Toc Back




60/80

STEP ONE

a0 =1

20

f(x) dx

=1

0

( x) dx + 1

2

0 dx

=1

x 1

2x2

0

+ 0

=

1

2 2

2 0i.e. a0 =

2.

Toc Back


STEP TWO


61/80

an =1

2

0

f(x)cos nx dx

=1

0

( x)cos nx dx + 1

2

0 dx

i.e. an =1

( x)sin nx

n

0

0

(1) sin nxn

dx using integration by parts

+0

=1

(0 0) +

0

sin nx

ndx

, see Trig

=

1

n cos nxn

0

= 1n2

(cos n cos0)

i.e. an =

1

n2

((

1)n

1) , see Trig

Toc Back



62/80

i.e. an = 0 , n even

2n2 , n odd

STEP THREE

bn =1

2

0

f(x)sin nx dx

=1

0

( x)sin nx dx +2

0 dx

=1

( x)cos nx

n

0

0

(

1)

cos nx

n dx + 0=

1

0

n

1

n 0

, see Trig

i.e. bn =

1

n .Toc Back


63/80


c) To show that 2

8 = 1 +1

32 +1

52 + . . . ,


64/80

note that, as x 0 , the series converges to the half-way value of 2 ,

and then

2=

4+

2

cos 0 +

1

32cos 0 +

1

52cos 0 + . . .

+ sin 0 +1

2 sin 0 +1

3 sin 0 + . . .

2=

4+

2

1 +

1

32+

1

52+ . . .

+ 0

4

= 21 + 1

32+ 1

52+ . . .

giving2

8= 1 +

1

32+

1

52+ . . .


Toc Back


Exercise 6.


65/80

f(x) = x, over the interval < x < and has period 2


0

2 33 2

f(x)

x

Toc Back




66/80

STEP ONE

a0 =1

f(x) dx

=1

x dx

=1

x2

2

=1

2

2 2

2 i.e. a0 = 0.

Toc Back


STEP TWO

1


67/80

an =1

f(x)cos nx dx

=1

x cos nx dx

=1

xsin nx

n

sin nx


i.e. an =1

1

n( sin n () sin(n)) 1

n

sin nx dx

= 1 1n (0 0) 1n 0 ,since sin n = 0 and

2

sin nx dx = 0,

i.e. an = 0.

Toc Back


STEP THREE

1


68/80

bn =1

f(x)sin nx dx

=1

x sin nx dx

=1

x cos nx

n

cos nxn dx

=1

1

n[x cos nx] +

1

n

cos nx dx

=1

1

n( cos n ()cos(n)) + 1

n 0

= n

(cos n + cos n)

= 1n

(2cos n)

i.e. bn = 2

n (1)n

.Toc Back


We thus have

a


69/80

f(x) =a02

+ n=1 an cos nx + bn sin nx

with a0 = 0, an = 0, bn = 2n(1)n

and

n 1 2 3

bn 2 1 23Therefore

f(x) = b1 sin x + b2 sin2x + b3 sin3x + . . .i.e. f(x) = 2

sin x 1

2sin2x +

1

3sin3x . . .

and we have found the required Fourier series.

Toc Back




70/80

4 = 1

13 +

15

17 + . . .

Setting x = 2 gives f(x) =2 and

2= 2

sin

2 1

2sin

2

2+

1

3sin

3

2 1

4sin

4

2+

1

5sin

5

2 . . .

This gives

2= 2

1 + 0 +

1

3 (1) 0 + 1

5 (1) 0 + 1

7 (1) + . . .

2 = 21 13 + 15 17 + . . .

i.e.

4= 1 1

3+

1

5 1

7+ . . .

Return to Exercise 6Toc Back


Exercise 7.


71/80

f(x) = x2, over the interval < x < and has period 2


0

2

233 2

f(x)

x

Toc Back


72/80


STEP TWO

1


73/80

an =1

f(x)cos nx dx=

1

x2 cos nx dx

=1

x2 sin nx

n

2xsin nx


=1

1

n

2 sin n 2 sin(n)

2

n

x sin nx dx

=1

1n

(0 0) 2n

x sin nx dx

, see Trig

=2

n

x sin nx dx

Toc Back


2 cos nx cos nx


74/80

i.e. an =2

n xcos nx

n cos nx

n dx using integration by parts again

=2

n

1

n[x cos nx] +

1

n

cos nx dx

= 2n

1

n

cos n ()cos(n)

+ 1

n 0

=2

n

1

n(1)n + (1)n

=2

n

2

n(1)n

Toc Back


75/80


STEP THREE

1 ( )

1 2


76/80

bn =1

f(x)sin nx dx =

1

x2 sin nx dx

=1

x2 cos nx

n

2x cos nx

n

dx


= 1

1n

x2 cos nx

+

2n

x cos nx dx

=1

1

n 2 cos n 2 cos(n)

+

2

n

x cos nx dx

=1

1

n

2 cos n 2 cos(n)

=0

+2

n

x cos nx dx

=2

n

x cos nx dx

Toc Back


i b2

sin nx sin nx

d


77/80

i.e. bn =n

x n

ndx


=2

n

1

n( sin n ()sin(n)) 1

n

sin nx dx

=2

n

1

n(0 + 0) 1

n

sin nx dx

=2n2

sin nx dx

i.e. bn = 0.

Toc Back


78/80


c) To show that 2

6 = 1 +1

22 +1

32 +1

42 + . . . ,


79/80

use the fact that cos n = 1 , n even1 , n oddi.e. cos x 122 cos2x + 132 cos3x 142 cos4x + . . . with x =

gives cos 1

22 cos2 +1

32 cos3 1

42 cos4 + . . .

i.e. (1) 122 (1) + 132 (1) 142 (1) + . . .

i.e.

1

1

22

1

32

1

42 + . . .

= 1

1 +1

22+

1

32+

1

42+ . . .

(the desired series)Toc Back


The graph of f(x) gives that f() = 2 and the series converges tothis value.


80/80

Setting x = in the Fourier series thus gives

2 =2

3 4

cos 1

22cos2 +

1

32cos3 1

42cos4 + . . .

2 =2

3 41

1

2

2

1

3

2

1

4

2

. . .

2 =2

3+ 4

1 +

1

22+

1

32+

1

42+ . . .

22

3= 41 +

1

22+

1

32+

1

42+ . . .

i.e.2

6= 1 +

1

22+

1

32+

1

42+ . . .


Toc Back

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Fundamental of Fourier Series

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