FOURIER SERIES

Series FOURIER SERIES Graham S McDonald A self-contained Tutorial Module for learning the technique of Fourier series analysis

● Table of contents ● Begin Tutorial

c 2004 [email protected]

Table of contents 1. 2. 3. 4. 5. 6. 7.

Theory Exercises Answers Integrals Useful trig results Alternative notation Tips on using solutions Full worked solutions

Section 1: Theory

3

1. Theory ● A graph of periodic function f (x) that has period L exhibits the same pattern every L units along the x-axis, so that f (x + L) = f (x) for every value of x. If we know what the function looks like over one complete period, we can thus sketch a graph of the function over a wider interval of x (that may contain many periods) f(x )

x

P E R IO D = L

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Section 1: Theory

4

● This property of repetition defines a fundamental spatial frequency k = 2π L that can be used to give a first approximation to the periodic pattern f (x): f (x) ' c1 sin(kx + α1 ) = a1 cos(kx) + b1 sin(kx), where symbols with subscript 1 are constants that determine the amplitude and phase of this first approximation ● A much better approximation of the periodic pattern f (x) can be built up by adding an appropriate combination of harmonics to this fundamental (sine-wave) pattern. For example, adding c2 sin(2kx + α2 ) = a2 cos(2kx) + b2 sin(2kx) c3 sin(3kx + α3 ) = a3 cos(3kx) + b3 sin(3kx)

(the 2nd harmonic) (the 3rd harmonic)

Here, symbols with subscripts are constants that determine the amplitude and phase of each harmonic contribution Toc

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Section 1: Theory

5

One can even approximate a square-wave pattern with a suitable sum that involves a fundamental sine-wave plus a combination of harmonics of this fundamental frequency. This sum is called a Fourier series F u n d a m e n ta l F u n d a m e n ta l + 2 h a rm o n ic s

x

F u n d a m e n ta l + 5 h a rm o n ic s F u n d a m e n ta l + 2 0 h a rm o n ic s Toc

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P E R IO D = L

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Section 1: Theory

6

● In this Tutorial, we consider working out Fourier series for functions f (x) with period L = 2π. Their fundamental frequency is then k = 2π L = 1, and their Fourier series representations involve terms like a1 cos x ,

b1 sin x

a2 cos 2x ,

b2 sin 2x

a3 cos 3x ,

b3 sin 3x

We also include a constant term a0 /2 in the Fourier series. This allows us to represent functions that are, for example, entirely above the x−axis. With a sufficient number of harmonics included, our approximate series can exactly represent a given function f (x)

f (x) = a0 /2

Toc

+ a1 cos x + a2 cos 2x + a3 cos 3x + ... + b1 sin x + b2 sin 2x + b3 sin 3x + ...

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Section 1: Theory

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A more compact way of writing the Fourier series of a function f (x), with period 2π, uses the variable subscript n = 1, 2, 3, . . . ∞ a0 X f (x) = + [an cos nx + bn sin nx] 2 n=1 ● We need to work out the Fourier coefficients (a0 , an and bn ) for given functions f (x). This process is broken down into three steps STEP ONE

a0

=

1 π

=

1 π

=

1 π

Z f (x) dx 2π

STEP TWO

an

Z f (x) cos nx dx 2π

STEP THREE

bn

Z f (x) sin nx dx 2π

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

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Section 1: Theory

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● 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 of x, then the series converges to a value that is half-way between the two possible function values " V e rtic a l ju m p " /d is c o n tin u ity in th e fu n c tio n re p re s e n te d

f(x )

x

F o u rie r s e rie s c o n v e rg e s to h a lf-w a y p o in t Toc

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Section 2: Exercises

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2. Exercises Click on Exercise links for full worked solutions (7 exercises in total). Exercise 1. Let f (x) be a function of period 2π such that 1, −π < x < 0 f (x) = 0, 0 < x < π . a) Sketch a graph of f (x) in the interval −2π < x < 2π b) Show that the Fourier series for f (x) in the interval −π < x < π is 1 2 1 1 − sin x + sin 3x + sin 5x + ... 2 π 3 5 c) By giving an appropriate value to x, show that π 1 1 1 = 1 − + − + ... 4 3 5 7 ● Theory ● Answers ● Integrals ● Trig ● Notation Toc

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Exercise 2. Let f (x) be a function of period 2π such that 0, −π < x < 0 f (x) = x, 0 < x < π . a) Sketch a graph of f (x) in the interval −3π < x < 3π b) Show that the Fourier series for f (x) in the interval −π < x < π is π 2 1 1 − cos x + 2 cos 3x + 2 cos 5x + ... 4 π 3 5 1 1 + sin x − sin 2x + sin 3x − ... 2 3 c) By giving appropriate values to x, show that (i) π4 = 1 − 31 + 15 − 17 + . . .

2

and (ii) π8 = 1 + 312 + 512 + 712 + . . .

● Theory ● Answers ● Integrals ● Trig ● Notation Toc

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Exercise 3. Let f (x) be a function of period 2π such that x, 0 < x < π f (x) = π, π < x < 2π . a) Sketch a graph of f (x) in the interval −2π < x < 2π b) Show that the Fourier series for f (x) in the interval 0 < x < 2π is 2 1 1 3π − cos x + 2 cos 3x + 2 cos 5x + . . . 4 π 3 5 1 1 − sin x + sin 2x + sin 3x + . . . 2 3 c) By giving appropriate values to x, show that (i)

π 4

= 1−

1 3

+

1 5

−

1 7

+...

and (ii)

π2 8

= 1+

1 32

+

1 52

+

1 72

+...


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Exercise 4. Let f (x) be a function of period 2π such that f (x) =

x over the interval 0 < x < 2π. 2

a) Sketch a graph of f (x) in the interval 0 < x < 4π b) Show that the Fourier series for f (x) in the interval 0 < x < 2π is 1 π 1 − sin x + sin 2x + sin 3x + . . . 2 2 3 c) By giving an appropriate value to x, show that π 1 1 1 1 = 1 − + − + − ... 4 3 5 7 9


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Exercise 5. Let f (x) be a function of period 2π such that π − x, 0 < x < π f (x) = 0, π < x < 2π a) Sketch a graph of f (x) in the interval −2π < x < 2π b) Show that the Fourier series for f (x) in the interval 0 < x < 2π is π 1 2 1 + cos x + 2 cos 3x + 2 cos 5x + . . . 4 π 3 5 1 1 1 + sin x + sin 2x + sin 3x + sin 4x + . . . 2 3 4 c) By giving an appropriate value to x, show that π2 1 1 = 1 + 2 + 2 + ... 8 3 5 ● Theory ● Answers ● Integrals ● Trig ● Notation Toc

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Exercise 6. Let f (x) be a function of period 2π such that f (x) = x in the range − π < x < π. a) Sketch a graph of f (x) in the interval −3π < x < 3π b) Show that the Fourier series for f (x) in the interval −π < x < π is 1 1 2 sin x − sin 2x + sin 3x − . . . 2 3 c) By giving an appropriate value to x, show that π 1 1 1 = 1 − + − + ... 4 3 5 7


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Exercise 7. Let f (x) be a function of period 2π such that f (x) = x2 over the interval − π < x < π. a) Sketch a graph of f (x) in the interval −3π < x < 3π b) Show that the Fourier series for f (x) in the interval −π < x < π is 1 π2 1 − 4 cos x − 2 cos 2x + 2 cos 3x − . . . 3 2 3 c) By giving an appropriate value to x, show that π2 1 1 1 = 1 + 2 + 2 + 2 + ... 6 2 3 4


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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 these solutions The answers below are suggested values of x to get the series of constants quoted in part (c) of each exercise 1. x =

π 2,

2. (i) x = 3. (i) x = 4. x =

π 2, π 2,

(ii) x = 0, (ii) x = 0,

π 2,

5. x = 0, 6. x =

π 2,

7. x = π.

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Section 4: Integrals

17

4. Integrals R b dv Rb b Formula for integration by parts: a u dx dx = [uv]a − a R R f (x) f (x)dx f (x) f (x)dx xn 1 x x

e sin x cos x tan x cosec x sec x sec2 x cot x sin2 x cos2 x

xn+1 n+1

(n 6= −1)

ln |x| ex − cos x sin x − ln |cos x| ln tan x2 ln |sec x + tan x| tan x ln |sin x| x sin 2x 2 − 4 x sin 2x 2 + 4 Toc

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n

[g (x)] g 0 (x) 0

g (x) g(x) x

a sinh x cosh x tanh x cosech x sech x sech2 x coth x sinh2 x cosh2 x II

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

du dx v dx

(n 6= −1)

ln |g (x)| ax (a > 0) ln a cosh x sinh x ln cosh x ln tanh x2 2 tan−1 ex tanh x ln |sinh x| sinh 2x − x2 4 sinh 2x + x2 4 I

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Section 4: Integrals

f (x) 1 a2 +x2

√ 1 a2 −x2

√

a2 − x2

18

R

f (x) dx

f (x)

1 a

tan−1

1 a2 −x2

R

(a > 0)

1 x2 −a2

f (x) dx a+x 1 2a ln a−x (0 < |x| < a) x−a 1 ln 2a x+a (|x| > a > 0)

sin−1

x a

√ 1 a2 +x2

√ 2 2 ln x+ aa +x (a > 0)

(−a < x < a)

√ 1 x2 −a2

√ 2 2 ln x+ xa −a (x > a > 0)

a2 2

−1 sin √

+x

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x a

x a

a2 −x2 a2

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√ i √

a2 +x2

a2 2

x2 −a2

II

a2 2

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h h

sinh−1

− cosh−1

I

x a

i √ x a2 +x2 2 a i √ 2 2 + x xa2−a

+

x a

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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 from the graph of sin x or that of cos x 1

● sin nπ = 0

s in (x )

x −3π

−2π

−π

0

π

2π

3π

π

2π

3π

−1

1

● cos nπ = (−1)n

c o s(x )

x −3π

−2π

−π

0 −1

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Section 5: Useful trig results 1

20

s in (x )

1

c o s(x )

x −3π

−π

−2π

0

π

3π

2π

x −3π

−2π

−π

−1

 π  ● sin n =  2

π

0

3π

2π

−1

 0 , n even 0 , n odd π  1 , n = 1, 5, 9, ... ● cos n = 1 , n = 0, 4, 8, ...  2 −1 , n = 3, 7, 11, ... −1 , n = 2, 6, 10, ...

Areas cancel when when integrating overR whole periods ● sin nx dx = 0 2π R ● cos nx dx = 0

1

+ −3π

−2π

s in (x )

+ +

−π

0

π

−1

2π

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2π

x 3π

Section 6: Alternative notation

21

6. Alternative notation ● For a waveform f (x) with period L =

2π k

∞

f (x)

=

a0 X + [an cos nkx + bn sin nkx] 2 n=1

The corresponding Fourier coefficients are STEP ONE

a0

Z

=

2 L

Z

=

2 L

=

2 L

f (x) dx L

STEP TWO

an

f (x) cos nkx dx L

STEP THREE

bn

Z f (x) sin nkx dx L

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

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● For a waveform f (x) with period 2L = 2π k , we have that 2π π k = 2L =L and nkx = nπx L ∞ nπx a0 X h nπx i an cos f (x) = + + bn sin 2 L L n=1 The corresponding Fourier coefficients are STEP ONE

a0

=

Z

1 L

f (x) dx 2L

STEP TWO

an

=

1 L

=

1 L

Z f (x) cos

nπx dx L

f (x) sin

nπx dx L

2L

STEP THREE

bn

Z 2L

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

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● For a waveform f (t) with period T =

2π ω

∞

f (t)

a0 X + [an cos nωt + bn sin nωt] 2 n=1

=

The corresponding Fourier coefficients are STEP ONE

a0

Z

=

2 T

Z

=

2 T

=

2 T

f (t) dt T

STEP TWO

an

f (t) cos nωt dt T

STEP THREE

bn

Z f (t) sin nωt dt T

and integrations are over a single interval in t of T

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Section 7: Tips on using solutions

24

7. Tips on using solutions ● When looking at the THEORY, ANSWERS, INTEGRALS, TRIG or NOTATION pages, use the Back button (at the bottom of the page) to return to the exercises

● Use the solutions intelligently. For example, they can help you get started on an exercise, or they can allow you to check whether your intermediate results are correct

● Try to make less use of the full solutions as you work your way through the Tutorial

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Solutions to exercises

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Full worked solutions Exercise 1. 1, −π < x < 0 f (x) = 0, 0 < x < π, and has period 2π a) Sketch a graph of f (x) in the interval −2π < x < 2π

f(x ) 1

−2π

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−π

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2π

π

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b) Fourier series representation of f (x) STEP ONE

1 a0 = π

Z 1 π f (x)dx = f (x)dx + f (x)dx π 0 −π −π Z Z 1 0 1 π = 1 · dx + 0 · dx π −π π 0 Z 1 0 = dx π −π 1 0 = [x] π −π 1 = (0 − (−π)) π 1 = · (π) π i.e. a0 = 1 .

Z

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π

1 π

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Z

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STEP TWO

an =

1 π

Z

π

f (x) cos nx dx = −π

= = = = = i.e. an

Toc

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=

1 π 1 π

Z

0

f (x) cos nx dx + −π Z 0

1 · cos nx dx + −π Z 0

1 π Z

1 π

π

Z

f (x) cos nx dx 0 π

0 · cos nx dx 0

1 cos nx dx π −π 0 1 sin nx 1 0 = [sin nx]−π π n nπ −π 1 (sin 0 − sin(−nπ)) nπ 1 (0 + sin nπ) nπ 1 (0 + 0) = 0. nπ II

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STEP THREE

bn

1 π

=

1 π

=

1 π

=

i.e. bn

= = =

i.e. bn

=

Z

π

f (x) sin nx dx −π Z 0

f (x) sin nx dx + −π Z 0

1 · sin nx dx + −π

1 π

1 π Z

π

Z

f (x) sin nx dx 0 π

0 · sin nx dx 0

0 1 − cos nx sin nx dx = π n −π −π 1 1 − [cos nx]0−π = − (cos 0 − cos(−nπ)) nπ nπ 1 1 − (1 − cos nπ) = − (1 − (−1)n ) , see Trig nπ nπ 0 , n even 1 , n even n , since (−1) = 2 −1 , n odd − nπ , n odd 1 π

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0

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We now have that ∞

a0 X f (x) = + [an cos nx + bn sin nx] 2 n=1 with the three steps giving a0 = 1, an = 0 , and bn =

0 2 − nπ

, n even , n odd

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

1 − π2

2 3 0 − π2 13

4 5 0 − π2 15

Substituting our results now gives the required series 1 1 2 1 f (x) = − sin x + sin 3x + sin 5x + . . . 2 π 3 5 Toc

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c) Pick an appropriate value of x, to show that π 4

=1−

1 3

+

1 5

−

1 7

+ ...

Comparing this series with 1 2 1 1 f (x) = − sin x + sin 3x + sin 5x + . . . , 2 π 3 5 we need to introduce a minus sign in front of the constants 13 , 17 ,. . . So we need sin x = 1, sin 3x = −1, sin 5x = 1, sin 7x = −1, etc The first condition of sin x = 1 suggests trying x = This choice gives i.e.

sin π2 1

+ −

1 3

sin 3 π2 1 3

+ +

1 5

π 2.

sin 5 π2 1 5

+ −

Looking at the graph of f (x), we also have that f ( π2 ) = 0. Toc

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1 7

sin 7 π2 1 7


Picking x =

31

π 2

thus gives h 0 = 21 − π2 sin π2 +

i.e. 0 =

1 2

−

2 π

h

1

−

1 3

sin 3π 2 +

1 5

sin 5π 2

+

1 7

sin 7π 2 + ...

1 3

+

1 5

−

1 7

+ ...

i

i

A little manipulation then gives a series representation of π4 2 1 1 1 1 1 − + − + ... = π 3 5 7 2 1 1 1 π 1 − + − + ... = . 3 5 7 4 Return to Exercise 1

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Exercise 2. 0, f (x) = x,

32

−π < x < 0 0 < x < π, and has period 2π

a) Sketch a graph of f (x) in the interval −3π < x < 3π

f(x ) π

−3π

Toc

−2π

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π

−π

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2π

3π

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x


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a0 =

1 π

Z

π

f (x)dx = −π

=

=

=

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i.e. a0

=

JJ

II

1 π

Z

1 π

Z

0

f (x)dx + −π 0

0 · dx + −π

1 π

1 π Z

Z

π

f (x)dx 0 π

x dx 0

π 1 x2 π 2 0 1 π2 −0 π 2 π . 2

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STEP TWO

an =

1 π

Z

π

f (x) cos nx dx = −π

1 π 1 π

Z

0

f (x) cos nx dx + −π Z 0

1 π Z

Z

π


1 0 · cos nx dx + x cos nx dx π −π 0 π Z π Z π 1 sin nx sin nx 1 x cos nx dx = x − dx i.e. an = π 0 π n n 0 0 =

(using integration by parts) 1 sin nπ 1 h cos nx iπ π −0 − − π n n n 0 1 1 ( 0 − 0) + 2 [cos nx]π0 π n 1 1 {cos nπ − cos 0} = {(−1)n − 1} 2 2 πn πn 0 , n even , see Trig. − πn2 2 , n odd

i.e. an

= = =

i.e. an Toc

=

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STEP THREE

bn =

1 π

Z

π

f (x) sin nx dx = −π

= i.e. bn =

1 π

Z

π

x sin nx dx = 0

= = = = Toc

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Z 1 π f (x) sin nx dx π 0 −π Z Z 1 π 1 0 0 · sin nx dx + x sin nx dx π −π π 0 Z π 1 h cos nx iπ cos nx x − − − dx π n n 0 0 (using integration by parts) Z 1 1 π 1 π − [x cos nx]0 + cos nx dx π n n 0 π 1 1 sin nx 1 − (π cos nπ − 0) + π n n n 0 1 1 − (−1)n + (0 − 0), see Trig n πn2 1 − (−1)n n 1 π

II

Z

0

f (x) sin nx dx +

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( i.e.

bn =

36

− n1

, n even

+ n1

, n odd

We now have ∞

f (x)

a0 X + [an cos nx + bn sin nx] 2 n=1

=

(

π where a0 = , 2

an =

0

(

, n even

− πn2 2

,

, n odd

bn =

− n1 1 n

Constructing a table of values gives

Toc

n

1

an

− π2

0

bn

1

− 12

JJ

2

3 − π2

II

·

4 1 32

1 3

0

5 − π2

− 14 J

·

1 52

1 5

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, n even , n odd


37

This table of coefficients gives f (x) =

1 π 2 2

i.e. f (x) =

π 4

2 π 2 + − π 2 + − π

cos x + 0 · cos 2x 1 · 2 cos 3x + 0 · cos 4x 3 1 · 2 cos 5x + ... 5 1 1 + sin x − sin 2x + sin 3x − ... 2 3 2 1 1 − cos x + 2 cos 3x + 2 cos 5x + ... π 3 5 1 1 + sin x − sin 2x + sin 3x − ... 2 3 +

−

and we have found the required series! Toc

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

π 4

=1−

1 3

+

1 5

−

1 7

+ ...

Comparing this series with π 2 1 1 f (x) = − cos x + 2 cos 3x + 2 cos 5x + ... 4 π 3 5 1 1 + sin x − sin 2x + sin 3x − ... , 2 3 the required series of constants does not involve terms like 312 , 512 , 712 , .... So we need to pick a value of x that sets the cos nx terms to zero. The Trig section shows that cos n π2 = 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

i.e.

cos π2 = cos 3 π2 = cos 5 π2 = ... = 0 Toc

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when x =

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π 2,

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Setting x = π2 in the series for f (x) gives π π 2 π 1 1 3π 5π f = − cos + 2 cos + 2 cos + ... 2 4 π 2 3 2 5 2 2π 1 3π 1 4π 1 5π π 1 + sin − sin + sin − ... + sin − sin 2 2 2 3 2 4 2 5 2 π 2 = − [0 + 0 + 0 + ...] 4 π  1 1 1 1 π + · (−1) − sin 2π + · (1) − ... + 1 − sin 2 | {z } 3 4 | {z } 5 =0

=0

The graph of f (x) shows that f π 2 π i.e. 4 Toc

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= =

π 2

=

π 2,

so that

π 1 1 1 + 1 − + − + ... 4 3 5 7 1 1 1 1 − + − + ... 3 5 7 II

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Pick an appropriate value of x, to show that (ii)

π2 8

=1+

1 32

+

1 52

+

1 72

+ ...

Compare this series with f (x) =

π 4

2 1 1 cos x + 2 cos 3x + 2 cos 5x + ... π 3 5 1 1 + sin x − sin 2x + sin 3x − ... . 2 3 −

This time, we want to use the coefficients of the cos nx terms, and the 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 Toc

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So, picking x = 0 gives 0

i.e. 0

π 2 1 1 1 − cos 0 + 2 cos 0 + 2 cos 0 + 2 cos 0 + ... 4 π 3 5 7 sin 0 sin 0 + sin 0 − + − ... 3 2 π 2 1 1 1 = − 1 + 2 + 2 + 2 + ... + 0 − 0 + 0 − ... 4 π 3 5 7 =

We then find that 2 1 1 1 1 + 2 + 2 + 2 + ... = π 3 5 7 1 1 1 and 1 + 2 + 2 + 2 + ... = 3 5 7

π 4 π2 . 8

Return to Exercise 2 Toc

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Exercise 3. x, 0 < x < π f (x) = π, π < x < 2π,

and has period 2π

a) Sketch a graph of f (x) in the interval −2π < x < 2π

f(x ) π

−2π

Toc

−π

JJ

0

II

π

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2π

x

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a0 =

1 π

Z

2π

f (x)dx = 0

=

=

=

Toc

Z 1 2π f (x)dx π π 0 Z Z 1 π 1 2π xdx + π · dx π 0 π π π 2π 1 x2 π + x π 2 0 π π 2 1 π − 0 + 2π − π π 2 1 π

Z

π

f (x)dx +

=

π +π 2

i.e. a0

=

3π . 2

JJ

II

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44

STEP TWO

an

=

=

=

Z

1 π

Z

1 π

"


0

1 x cos nx dx + π

sin nx x n

| =

2π

1 π

π

Z − 0

0

{z

π

Z

2π

π · cos nx dx π

# 2π sin nx π sin nx dx + n π n π }

using integration by parts

" π # − cos nx 1 1 π sin nπ − 0 · sin n0 − π n n2 0 +

Toc

JJ

II

J

1 (sin n2π − sin nπ) n

I

Back


i.e. an

45

=

" # 1 1 cos nπ cos 0 1 0−0 + 0−0 − 2 + π n n2 n n

=

1 (cos nπ − 1), see Trig n2 π

=

1 (−1)n − 1 , 2 n π

(

− n22 π

, n odd

0

, n even.

i.e. an =

Toc

JJ

II

J

I

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46

STEP THREE

bn

= = =

1 π

Z

1 π

Z

1 π

"

2π

f (x) sin nx dx 0 π

x sin nx dx + 0

1 π

Z

2π

π · sin nx dx π

# 2π h cos nx iπ Z π − cos nx π − cos nx x − − dx + n n π n 0 0 π | {z } using integration by parts

=

= =

π # −π cos nπ sin nx 1 +0 + − (cos 2nπ − cos nπ) n n2 0 n " # 1 −π(−1)n sin nπ − sin 0 1 + − 1 − (−1)n 2 π n n n 1 1 − (−1)n + 0 − 1 − (−1)n n n 1 π

"

Toc

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47

i.e. bn i.e. bn

1 1 1 = − (−1)n − + (−1)n n n n 1 = − . n

We now have ∞

f (x) =

where a0 =

3π 2 ,

a0 X + [an cos nx + bn sin nx] 2 n=1 ( 0 , n even an = , bn = − n1 − n22 π , n odd

Constructing a table of values gives n an bn

Toc

1 − π2 −1

JJ

2 0 − 21

3 − π2 312 − 13

II

J

4 0 − 14

5 − π2 512 − 15

I

Back


48

This table of coefficients gives

f (x) =

1 2

i.e. f (x) =

3π 2

3π 4

+

−

+

h −1 sin x +

− −

2 π

2 π

h

cos x + 0 · cos 2x +

1 32

cos 3x + . . .

i

1 2

sin 2x +

1 3

sin 3x + . . .

i

h

cos x +

1 32

cos 3x +

1 52

cos 5x + . . .

i

h

sin x +

1 2

sin 2x +

1 3

sin 3x + . . .

i

and we have found the required series.

Toc

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II

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49

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

π 4

=1−

1 3

+

1 5

−

1 7

+ ...

Compare this series with 3π 2 1 1 f (x) = − cos x + 2 cos 3x + 2 cos 5x + . . . 4 π 3 5 1 1 − sin x + sin 2x + sin 3x + . . . 2 3 Here, we want to set the cos nx terms to zero (since their coefficients are 1, 312 , 512 , . . .). Since cos n π2 = 0 when n is odd, we will try setting x = π2 in the series. Note also that f ( π2 ) = π2 This gives π 2

=

3π 4

− −

2 π

cos π2 +

1 32

sin π2 +

1 2

Toc

JJ

cos 3 π2 +

sin 2 π2 + II

1 3

1 52

cos 5 π2 + . . .

sin 3 π2 + J

1 4

sin 4 π2 + I

1 5

sin 5 π2 + . . .

Back


50

and π 2

=

3π 4

2 π

−

[0 + 0 + 0 + . . .] (1) +

−

1 2

· (0) +

1 3

· (−1) +

1 4

· (0) +

1 5

· (1) + . . .

then π 2

=

3π 4

− 1−

1 3

+

1 5

−

1 7

1−

1 3

+

1 5

−

1 7

+ ... =

3π 4

1−

1 3

+

1 5

−

1 7

+ ... =

π 4,

(ii)

π2 8

+ ... −

π 2

as required.

To show that =1+

1 32

+

1 52

+

1 72

+ ... ,

We want zero sin nx terms and to use the coefficients of cos nx Toc

JJ

II

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51

Setting x = 0 eliminates the sin nx terms from the series, and also gives 1 1 1 1 1 1 cos x + 2 cos 3x + 2 cos 5x + 2 cos 7x + . . . = 1 + 2 + 2 + 2 + . . . 3 5 7 3 5 7 (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 the two values of f (x) around this discontinuity. That is the series will converge to π2 at x = 0 π i.e. 2

=

π 2

=

and

Toc

3π 2 1 1 1 − cos 0 + 2 cos 0 + 2 cos 0 + 2 cos 0 + . . . 4 π 3 5 7 1 1 − sin 0 + sin 0 + sin 0 + . . . 2 3 3π 2 1 1 1 − 1 + 2 + 2 + 2 + . . . − [0 + 0 + 0 + . . .] 4 π 3 5 7 JJ

II

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52

Finally, this gives π 4 π2 8

− and

=

1 1 1 + + + . . . 32 52 72 1 1 1 1 + 2 + 2 + 2 + ... 3 5 7

= −

2 π

1+

Return to Exercise 3

Toc

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II

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53

Exercise 4. f (x) = x2 , over the interval 0 < x < 2π and has period 2π a) Sketch a graph of f (x) in the interval 0 < x < 4π

π

f(x )

0

Toc

JJ

3π

2π

π

II

J

I

4π

Back

x


54


a0

i.e. a0

Toc

JJ

1 π

Z

Z

=

1 π

=

1 π

=

1 π

=

2π

f (x) dx 0 2π

x dx 2 0 2 2π x 4 0 (2π)2 −0 4

= π.

II

J

I

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55

STEP TWO

an

=

=

=

1 π

Z

1 π

Z

2π

f (x) cos nx dx 0 2π

x cos nx dx 2

0

(

1 2π

sin nx x n

| =

= i.e. an Toc

=

2π 0

1 − n {z

Z

)

2π

sin nx dx 0


}

) sin n2π sin n · 0 1 2π −0· − ·0 n n n ( ) 1 1 (0 − 0) − · 0 , see Trig 2π n

1 2π

(

0. JJ

II

J

I

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56

STEP THREE

bn

= = =

1 π

Z

2π

f (x) sin nx dx = 0

1 π

Z 0

2π

x 2

sin nx dx

2π

1 2π

Z

1 2π

( 2π Z 2π ) − cos nx − cos nx x − dx n n 0 0 | {z }

1 2π

(

x sin nx dx 0


=

−2π cos(n2π) 2πn 1 = − cos(2nπ) n 1 = − , since 2n is even (see Trig) n Toc JJ II J I Back =

i.e. bn

) 1 1 (−2π cos n2π + 0) + · 0 , see Trig n n


57

We now have ∞

a0 X f (x) = + [an cos nx + bn sin nx] 2 n=1 where a0 = π, an = 0, bn = − n1 These Fourier coefficients give

f (x)

=

i.e. f (x)

=

Toc

∞ π X 1 + 0 − sin nx 2 n=1 n π 1 1 − sin x + sin 2x + sin 3x + . . . . 2 2 3

JJ

II

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58

c) Pick an appropriate value of x, to show that π 4

Setting x =

π 2

=1−

gives f (x) =

1 3 π 4

+

1 5

JJ

1 7

+

1 9

− ...

and

π = 4 π = 4 1 1 1 1 1 − + − + − ... = 3 5 7 9 1 1 1 1 i.e. 1 − + − + − . . . = 3 5 7 9

Toc

−

II

π 1 1 − 1 + 0 − + 0 + + 0 − ... 2 3 5 π 1 1 1 1 − 1 − + − + − ... 2 3 5 7 9 π 4 π . 4 Return to Exercise 4

J

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59

Exercise 5. π−x , 0