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

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1 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 004 g.s.mcdonald@salford.ac.uk

2 1. Theory. Exercises 3. Answers 4. Integrals 5. Useful trig results 6. Alternative notation 7. Tips on using solutions Full worked solutions Table of contents

3 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

4 Section 1: Theory 4 This property of repetition defines a fundamental spatial frequency k = L that can be used to give a first approximation to the periodic pattern f(x): f(x) c 1 sin(kx + α 1 ) = a 1 cos(kx) + b 1 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 c sin(kx + α ) = a cos(kx) + b sin(kx) c 3 sin(3kx + α 3 ) = a 3 cos(3kx) + b 3 sin(3kx) (the nd harmonic) (the 3rd harmonic) Here, symbols with subscripts are constants that determine the amplitude and phase of each harmonic contribution

5 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 + 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 + 0 h a rm o n ic s P E R IO D = L

6 Section 1: Theory 6 In this Tutorial, we consider working out Fourier series for functions f(x) with period L =. Their fundamental frequency is then k = L = 1, and their Fourier series representations involve terms like a 1 cos x, a cos x, a 3 cos 3x, b 1 sin x b sin x b 3 sin 3x We also include a constant term a 0 / 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) = a 0 / + a 1 cos x + a cos x + a 3 cos 3x b 1 sin x + b sin x + b 3 sin 3x +...

7 Section 1: Theory 7 A more compact way of writing the Fourier series of a function f(x), with period, uses the variable subscript n = 1,, 3,... f(x) = a 0 + [a n cos nx + b n sin nx] n=1 We need to work out the Fourier coefficients (a 0, a n and b n ) for given functions f(x). This process is broken down into three steps STEP ONE STEP TWO STEP THREE a 0 = 1 f(x) dx a n = 1 f(x) cos nx dx b n = 1 f(x) sin nx dx where integrations are over a single interval in x of L =

8 Section 1: Theory 8 Finally, specifying a particular value of x = x 1 in a Fourier series, gives a series of constants that should equal f(x 1 ). 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

9 Section : Exercises 9. Exercises Click on Exercise links for full worked solutions (7 exercises in total). Exercise 1. Let f(x) be a function of period such that { 1, < x < 0 f(x) = 0, 0 < x <. a) Sketch a graph of f(x) in the interval < x < b) Show that the Fourier series for f(x) in the interval < x < is 1 [sin x + 13 sin 3x + 15 ] sin 5x +... c) By giving an appropriate value to x, show that = Theory Answers Integrals Trig Notation

10 Section : Exercises 10 Exercise. Let f(x) be a function of period such that { 0, < x < 0 f(x) = x, 0 < x <. a) Sketch a graph of f(x) in the interval 3 < x < 3 b) Show that the Fourier series for f(x) in the interval < x < is 4 [cos x + 13 cos 3x + 15 ] cos 5x [sin x 1 sin x + 13 ] sin 3x... c) By giving appropriate values to x, show that (i) 4 = and (ii) 8 = Theory Answers Integrals Trig Notation

11 Section : Exercises 11 Exercise 3. Let f(x) be a function of period such that { x, 0 < x < f(x) =, < x <. a) Sketch a graph of f(x) in the interval < x < b) Show that the Fourier series for f(x) in the interval 0 < x < is 3 4 [cos x + 13 cos 3x + 15 ] cos 5x +... [sin x + 1 sin x + 13 ] sin 3x +... c) By giving appropriate values to x, show that (i) 4 = and (ii) 8 = Theory Answers Integrals Trig Notation

12 Section : Exercises 1 Exercise 4. Let f(x) be a function of period such that f(x) = x over the interval 0 < x <. 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 < is [sin x + 1 sin x + 13 ] sin 3x +... c) By giving an appropriate value to x, show that 4 = Theory Answers Integrals Trig Notation

13 Section : Exercises 13 Exercise 5. Let f(x) be a function of period such that { x, 0 < x < f(x) = 0, < x < a) Sketch a graph of f(x) in the interval < x < b) Show that the Fourier series for f(x) in the interval 0 < x < is + [cos x cos 3x + 15 ] cos 5x sin x + 1 sin x sin 3x + 1 sin 4x c) By giving an appropriate value to x, show that 8 = Theory Answers Integrals Trig Notation

14 Section : Exercises 14 Exercise 6. Let f(x) be a function of period such that f(x) = x in the range < x <. a) Sketch a graph of f(x) in the interval 3 < x < 3 b) Show that the Fourier series for f(x) in the interval < x < is [sin x 1 sin x + 13 ] sin 3x... c) By giving an appropriate value to x, show that 4 = Theory Answers Integrals Trig Notation

15 Section : Exercises 15 Exercise 7. Let f(x) be a function of period such that f(x) = x over the interval < x <. a) Sketch a graph of f(x) in the interval 3 < x < 3 b) Show that the Fourier series for f(x) in the interval < x < is [cos 3 4 x 1 cos x + 13 ] cos 3x... c) By giving an appropriate value to x, show that 6 = Theory Answers Integrals Trig Notation

16 Section 3: Answers 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 =,. (i) x =, (ii) x = 0, 3. (i) x =, (ii) x = 0, 4. x =, 5. x = 0, 6. x =, 7. x =.

17 Section 4: Integrals Integrals Formula for integration by parts: f (x) x n 1 x b a u dv dx dx = [uv]b a b du a dx v dx f(x)dx f (x) f(x)dx xn+1 n+1 (n 1) [g (x)] n g (x) ln x g (x) g(x) [g(x)] n+1 n+1 (n 1) ln g (x) e x e x a x ax ln a (a > 0) sin x cos x sinh x cosh x cos x sin x cosh x sinh x tan x ln cos x tanh x ln cosh x cosec x ln tan x cosech x ln tanh x sec x ln sec x + tan x sech x tan 1 e x sec x tan x sech x tanh x cot x ln sin x coth x ln sinh x sin x sin x x 4 sinh sinh x x 4 x cos x sin x x + 4 cosh sinh x x 4 + x

18 Section 4: Integrals 18 f (x) f (x) dx f (x) f (x) dx 1 1 a +x a tan 1 x 1 1 a a x a ln a+x a x (0< x <a) 1 1 (a > 0) x a a ln x a x+a ( x > a>0) 1 a x sin 1 x a 1 a +x ( a < x < a) 1 x a ln ln x+ a +x a x+ x a a (a > 0) (x>a>0) a x a [ sin 1 ( ) x a a +x a ] x + x a x a a a [ [ sinh 1 ( x a cosh 1 ( x a ) + x a +x a ] ) + x ] x a a

19 Section 5: Useful trig results Useful trig results When calculating the Fourier coefficients a n and b n, for which n = 1,, 3,..., the following trig. results are useful. Each of these results, which are also true for n = 0, 1,, 3,..., can be deduced from the graph of sin x or that of cos x s in (x ) 1 sin n = x 3 1 c o s (x ) 1 cos n = ( 1) n 3 0 x 3 1

20 Section 5: Useful trig results 0 s in (x ) 1 c o s (x ) x x sin n 0, n even = 1, n = 1, 5, 9,... 1, n = 3, 7, 11,... cos n = 0, n odd 1, n = 0, 4, 8,... 1, n =, 6, 10,... Areas cancel when when integrating over whole periods sin nx dx = 0 cos nx dx = s in (x ) x

21 Section 6: Alternative notation 1 6. Alternative notation For a waveform f(x) with period L = k f(x) = a 0 + [a n cos nkx + b n sin nkx] n=1 The corresponding Fourier coefficients are STEP ONE a 0 = f(x) dx L L STEP TWO a n = f(x) cos nkx dx L L STEP THREE b n = f(x) sin nkx dx L L and integrations are over a single interval in x of L

22 Section 6: Alternative notation For a waveform f(x) with period L = k k = L = nx L and nkx = L f(x) = a 0 + n=1 [ a n cos nx L The corresponding Fourier coefficients are STEP ONE STEP TWO STEP THREE a 0 = 1 f(x) dx L a n = 1 L b n = 1 L L L L f(x) cos nx L f(x) sin nx L, we have that + b n sin nx ] L and integrations are over a single interval in x of L dx dx

23 Section 6: Alternative notation 3 For a waveform f(t) with period T = ω f(t) = a 0 + [a n cos nωt + b n sin nωt] n=1 The corresponding Fourier coefficients are STEP ONE a 0 = f(t) dt T T STEP TWO a n = f(t) cos nωt dt T T STEP THREE b n = f(t) sin nωt dt T T and integrations are over a single interval in t of T

24 Section 7: Tips on using solutions 4 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

25 Solutions to exercises 5 Full worked solutions Exercise 1. f(x) = { 1, < x < 0 0, 0 < x <, and has period a) Sketch a graph of f(x) in the interval < x < 1 f(x ) 0 x

26 Solutions to exercises 6 b) Fourier series representation of f(x) STEP ONE a 0 = 1 f(x)dx = 1 0 f(x)dx + 1 f(x)dx 0 = dx dx 0 = 1 0 dx = 1 [x]0 = 1 (0 ( )) = 1 () i.e. a 0 = 1.

27 Solutions to exercises 7 STEP TWO a n = 1 cos nx dx = f(x) 1 0 cos nx dx + f(x) 1 f(x) cos nx dx 0 = cos nx dx cos nx dx 0 = 1 0 cos nx dx = 1 [ ] 0 sin nx = 1 n n [sin nx]0 = 1 (sin 0 sin( n)) n = 1 (0 + sin n) n i.e. a n = 1 (0 + 0) = 0. n

28 Solutions to exercises 8 STEP THREE b n = 1 f(x) sin nx dx = 1 0 f(x) sin nx dx + 1 f(x) sin nx dx 0 = sin nx dx sin nx dx 0 0 [ cos nx ] 0 i.e. b n = 1 i.e. b n = sin nx dx = 1 = 1 n [cos nx]0 = 1 (cos 0 cos( n)) n = 1 1 (1 cos n) = n n (1 ( 1)n ), see Trig { { 0, n even 1, n even n, n odd n, since ( 1) n = 1, n odd

29 Solutions to exercises 9 We now have that f(x) = a 0 + [a n cos nx + b n sin nx] n=1 with the three steps giving { 0, n even a 0 = 1, a n = 0, and b n =, n odd n It may be helpful to construct a table of values of b n n 1 3 ( 4 5 b n 0 1 ( 3) 0 1 5) Substituting our results now gives the required series f(x) = 1 [sin x + 13 sin 3x + 15 ] sin 5x +...

30 Solutions to exercises 30 c) Pick an appropriate value of x, to show that 4 = Comparing this series with f(x) = 1 [sin x + 13 sin 3x + 15 ] sin 5x +..., we need to introduce a minus sign in front of the constants 1 3, 1 7,... 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 sin sin sin sin i.e Looking at the graph of f(x), we also have that f( ) = 0.

31 = 4. Return to Exercise 1 Solutions to exercises 31 Picking x = thus gives [ 0 = 1 sin sin sin 5 ] sin i.e. 0 = 1 [ ] A little manipulation then gives a series representation of 4 [ ] +... = 1

32 Solutions to exercises 3 Exercise. f(x) = { 0, < x < 0 x, 0 < x <, and has period a) Sketch a graph of f(x) in the interval 3 < x < 3 f(x ) 3 3 x

33 Solutions to exercises 33 b) Fourier series representation of f(x) STEP ONE 0 a 0 = 1 f(x)dx = 1 f(x)dx = 1 0 dx + 1 = 1 [ ] x 0 = 1 ( ) 0 i.e. a 0 =. 0 0 f(x)dx x dx

34 Solutions to exercises 34 STEP TWO a n = 1 cos nx dx = f(x) 1 0 cos nx dx + f(x) 1 f(x) cos nx dx 0 = cos nx dx + 1 x cos nx dx 0 i.e. a n = 1 x cos nx dx = 1 {[ ] sin nx } sin nx x dx 0 n 0 0 n (using integration by parts) i.e. a n = 1 {( ) sin n 0 1 [ cos nx ] } n n n 0 = 1 { ( 0 0) + 1 } n [cos nx] = {cos n cos 0} = n n {( 1)n 1} { 0, n even i.e. a n =, see Trig. n, n odd

35 Solutions to exercises 35 STEP THREE b n = 1 f(x) sin nx dx = 1 0 f(x) sin nx dx + 1 f(x) sin nx dx 0 = 1 0 sin nx dx x sin nx dx 0 i.e. b n = 1 x sin nx dx = 1 { [ ( cos nx )] ( cos nx ) } x dx 0 n 0 0 n (using integration by parts) = 1 { 1 n [x cos nx] } cos nx dx n 0 = 1 { 1 n ( cos n 0) + 1 [ ] } sin nx n n 0 = 1 n ( 1)n + 1 (0 0), see Trig n = 1 n ( 1)n

36 Solutions to exercises 36 { 1 n, n even i.e. b n = + 1 n, n odd We now have f(x) = a 0 + [a n cos nx + b n sin nx] n=1 { where a 0 = 0, n even, a n = n, n odd, b n = Constructing a table of values gives { 1 n 1 n, n even, n odd n a n b n

37 Solutions to exercises 37 This table of coefficients gives f(x) = 1 ( ) ( ) cos x + 0 cos x ( ) 1 3 cos 3x + 0 cos 4x ( ) 1 5 cos 5x sin x 1 sin x + 1 sin 3x... 3 [cos x + 13 cos 3x + 15 ] cos 5x +... i.e. f(x) = 4 + [sin x 1 sin x + 13 ] sin 3x... and we have found the required series!

38 Solutions to exercises 38 c) Pick an appropriate value of x, to show that (i) 4 = Comparing this series with [cos x + 13 cos 3x + 15 ] cos 5x +... f(x) = 4 + [sin x 1 sin x + 13 ] sin 3x..., the required series of constants does not involve terms like 1 3, 1 5, 1 7,... So we need to pick a value of x that sets the cos nx terms to zero. The Trig section shows that cos n = 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 =, i.e. cos = cos 3 = cos 5 =... = 0

39 Solutions to exercises 39 Setting x = in the series for f(x) gives ( ) f = 4 [cos + 13 cos cos 5 ] [sin 1 sin + 13 sin 3 14 sin sin 5 ]... = 4 [ ] sin }{{} ( 1) 1 sin 4 } {{ } + 1 (1)... 5 =0 =0 The graph of f(x) shows that f ( ) =, so that i.e. 4 = =

40 Solutions to exercises 40 Pick an appropriate value of x, to show that (ii) 8 = Compare this series with [cos x + 13 cos 3x + 15 ] cos 5x +... f(x) = 4 + [sin x 1 sin x + 13 ] sin 3x.... 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 x = 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

41 Solutions to exercises 41 So, picking x = 0 gives 0 = 4 [cos cos cos cos ] + sin 0 sin 0 + sin i.e. 0 = 4 [ ] We then find that [ ] +... = 4 and = 7 8. Return to Exercise

42 Solutions to exercises 4 Exercise 3. f(x) = { x, 0 < x <, < x <, and has period a) Sketch a graph of f(x) in the interval < x < f(x ) 0 x

43 Solutions to exercises 43 b) Fourier series representation of f(x) STEP ONE a 0 = 1 0 f(x)dx = 1 = [ x f(x)dx + 1 xdx + 1 ] [ x = = 1 ( ) 0 + = + ] dx ( f(x)dx ) i.e. a 0 = 3.

44 Solutions to exercises 44 STEP TWO a n = 1 0 f(x) cos nx dx = 1 x cos nx dx + 1 cos nx dx 0 [ [ = 1 ] ] sin nx sin nx x n 0 0 n dx + [ sin nx n } {{ } using integration by parts [ ] = 1 1 n ( ) sin n 0 sin n0 [ cos nx n ] 0 ] + 1 (sin n sin n) n

45 Solutions to exercises 45 i.e. a n = 1 [ 1 n ( ) ( cos n n cos 0 n ) ] + 1 ( ) 0 0 n = = 1 n (cos n 1), 1 ( ( 1) n n 1 ), see Trig i.e. a n = { n, n odd 0, n even.

46 Solutions to exercises 46 STEP THREE b n = 1 f(x) sin nx dx 0 = 1 x sin nx dx + 1 sin nx dx 0 [ = 1 [ ( cos nx )] ( ) ] cos nx x dx + [ ] cos nx n 0 0 n n } {{ } using integration by parts [ ( = 1 ) [ ] ] cos n sin nx n n 1 (cos n cos n) 0 n [ = 1 ( 1) n ( ) ] sin n sin 0 + n n 1 ( 1 ( 1) n ) n = 1 n ( 1)n n ( 1 ( 1) n )

47 Solutions to exercises 47 We now have i.e. b n = 1 n ( 1)n 1 n + 1 n ( 1)n i.e. b n = 1 n. f(x) = a 0 + [a n cos nx + b n sin nx] where a 0 = 3, a n = n=1 { Constructing a table of values gives 0, n even n, n odd, b n = 1 n n 1 3 ( 4 5 a n 0 1 ) ( ) 5 b n

48 Solutions to exercises 48 This table of coefficients gives f(x) = 1 ( ) ( ) [ ] 3 + cos x + 0 cos x cos 3x ( ) [ 1 sin x + 1 sin x sin 3x +... ] i.e. f(x) = 3 4 [ ] cos x cos 3x cos 5x +... [ ] sin x + 1 sin x sin 3x +... and we have found the required series.

49 Solutions to exercises 49 c) Pick an appropriate value of x, to show that (i) 4 = Compare this series with f(x) = 3 4 [cos x + 13 cos 3x + 15 ] cos 5x +... [sin x + 1 sin x + 13 ] sin 3x +... Here, we want to set the cos nx terms to zero (since their coefficients 1 1 are 1, 3, 5,...). Since cos n = 0 when n is odd, we will try setting x = in the series. Note also that f( ) = This gives = 3 4 [ cos cos cos ] [ sin + 1 sin sin sin sin ]

50 Solutions to exercises 50 and = 3 4 [ ] then [ (1) + 1 (0) ( 1) (0) (1) +...] = 3 4 ( ) = = 4, as required. To show that (ii) 8 = , We want zero sin nx terms and to use the coefficients of cos nx

51 Solutions to exercises 51 Setting x = 0 eliminates the sin nx terms from the series, and also gives cos x cos 3x cos 5x cos 7x +... = (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 at x = 0 i.e. = 3 4 [cos cos cos ] cos [sin sin ] sin and = 3 4 [ ] [ ]

52 Solutions to exercises 5 Finally, this gives ( ) and 4 8 = = Return to Exercise 3

53 Solutions to exercises 53 Exercise 4. f(x) = x, over the interval 0 < x < and has period a) Sketch a graph of f(x) in the interval 0 < x < 4 f(x ) x

54 Solutions to exercises 54 b) Fourier series representation of f(x) STEP ONE a 0 = 1 = 1 i.e. a 0 =. 0 0 [ x = = 1 [ () 4 f(x) dx x dx ] ] 0

55 Solutions to exercises 55 STEP TWO a n = 1 = 1 i.e. a n = { [ = 1 x f(x) cos nx dx x cos nx dx ] } sin nx 1 sin nx dx n 0 n 0 } {{ } using integration by parts { ( = 1 sin n 0 sin n 0 ) } 1n n n 0 { } = 1 (0 0) 1n 0, see Trig

56 Solutions to exercises 56 STEP THREE b n = 1 = 1 = 1 0 f(x) sin nx dx = 1 0 ( x ) sin nx dx x sin nx dx 0 { [ ( )] cos nx ( ) } cos nx x dx n 0 0 n } {{ } using integration by parts { } = 1 1 n ( cos n + 0) + 1 n 0, see Trig = n cos(n) = 1 n cos(n) i.e. b n = 1 n, since n is even (see Trig)

57 Solutions to exercises 57 We now have f(x) = a 0 + [a n cos nx + b n sin nx] n=1 where a 0 =, a n = 0, b n = 1 n These Fourier coefficients give f(x) = + (0 1n ) sin nx n=1 i.e. f(x) = { sin x + 1 sin x sin 3x +... }.

58 Solutions to exercises 58 c) Pick an appropriate value of x, to show that Setting x = gives f(x) = [ ]... 4 = and = [ ] = [ ] = 4 i.e = 4. Return to Exercise 4

59 Solutions to exercises 59 Exercise 5. f(x) = { x, 0 < x < 0, < x <, and has period a) Sketch a graph of f(x) in the interval < x < f(x ) 0 x

60 Solutions to exercises 60 b) Fourier series representation of f(x) STEP ONE a 0 = 1 i.e. a 0 =. 0 f(x) dx = 1 ( x) dx = 1 [ x 1 ] x = 1 ] [ 0 0 dx

61 Solutions to exercises 61 STEP TWO a n = 1 0 f(x) cos nx dx = 1 ( x) cos nx dx dx 0 i.e. a n = 1 {[ ] sin nx } sin nx ( x) ( 1) dx +0 n 0 0 n } {{ } { (0 0) + = 1 = 1 [ cos nx n n using integration by parts } 0 ] 0 sin nx n dx = 1 (cos n cos 0) n i.e. a n = 1 n (( 1)n 1), see Trig, see Trig

62 Solutions to exercises 6 STEP THREE b n = 1 = 1 i.e. b n = 1 n. 0 0 i.e. a n = = 1 { [ ( x) f(x) sin nx dx 0, n even n ( x) sin nx dx + ( cos nx )] n 0, n odd 0 dx = 1 { ( ( 0 )) 1 } n n 0, see Trig 0 ( cos nx ) } ( 1) dx + 0 n

63 Solutions to exercises 63 In summary, a 0 = and a table of other Fourier cofficients is n a n = n (when n is odd) b n = 1 n f(x) = a 0 + [a n cos nx + b n sin nx] n=1 = 4 + cos x + i.e. f(x) = cos 3x + 1 cos 5x sin x + 1 sin x sin 3x + 1 sin 4x [cos x + 13 cos 3x + 15 ] cos 5x sin x + 1 sin x sin 3x + 1 sin 4x

64 Solutions to exercises 64 c) To show that 8 = , note that, as x 0, the series converges to the half-way value of, and then = 4 + (cos cos cos ) + sin sin sin giving 4 8 = 4 + ( ) = ( ) +... = Return to Exercise 5

65 Solutions to exercises 65 Exercise 6. f(x) = x, over the interval < x < and has period a) Sketch a graph of f(x) in the interval 3 < x < 3 f(x ) 3 0 x 3

66 Solutions to exercises 66 b) Fourier series representation of f(x) STEP ONE a 0 = 1 f(x) dx = 1 x dx [ ] x i.e. a 0 = 0. = 1 = 1 ( )

67 Solutions to exercises 67 STEP TWO a n = 1 f(x) cos nx dx = 1 x cos nx dx { [ = 1 ] sin nx ( ) } sin nx x dx n n } {{ } using integration by parts i.e. a n = 1 { 1 n ( sin n ( ) sin( n)) 1 } sin nx dx n = 1 { 1 n (0 0) 1 } n 0, since sin n = 0 and sin nx dx = 0, i.e. a n = 0.

68 Solutions to exercises 68 STEP THREE b n = 1 f(x) sin nx dx = 1 x sin nx dx { [ x = 1 ] cos nx ( ) } cos nx dx n n = 1 { 1 n [x cos nx] + 1 } cos nx dx n = 1 { 1n ( cos n ( ) cos( n)) + 1n } 0 = (cos n + cos n) n = 1 ( cos n) n i.e. b n = n ( 1)n.

69 Solutions to exercises 69 We thus have f(x) = a 0 + [ ] a n cos nx + b n sin nx with n=1 a 0 = 0, a n = 0, b n = n ( 1)n and n 1 3 Therefore b n 1 3 f(x) = b 1 sin x + b sin x + b 3 sin 3x +... i.e. f(x) = [sin x 1 sin x + 13 ] sin 3x... and we have found the required Fourier series.

70 Solutions to exercises 70 c) Pick an appropriate value of x, to show that 4 = Setting x = gives f(x) = and [ = sin 1 sin + 13 sin 3 14 sin sin 5 ]... This gives i.e. 4 = [ ( 1) (1) ] ( 1) +... = [ ] +... = Return to Exercise 6

71 Solutions to exercises 71 Exercise 7. f(x) = x, over the interval < x < and has period a) Sketch a graph of f(x) in the interval 3 < x < 3 f(x ) x

72 Solutions to exercises 7 b) Fourier series representation of f(x) STEP ONE a 0 = 1 f(x)dx = 1 x dx [ ] x 3 = 1 3 = 1 ( 3 3 ( ) 3 ( 3 3 )) = 1 3 i.e. a 0 = 3.

73 Solutions to exercises 73 STEP TWO a n = 1 f(x) cos nx dx = 1 x cos nx dx { [ = 1 ] sin nx ( ) } sin nx x x dx n n } {{ } using integration by parts { = 1 1 ( sin n sin( n) ) } x sin nx dx n n { = 1 1 n (0 0) } x sin nx dx, see Trig n = n x sin nx dx

74 Solutions to exercises 74 i.e. a n = n = n = n = n = n { [ ( )] cos nx ( ) } cos nx x dx n n } {{ } { using integration by parts again 1 n [x cos nx] + 1 cos nx dx n { 1 ( ) } cos n ( ) cos( n) + 1n n 0 { 1 ) (( 1) } n + ( 1) n n { } n ( 1)n }

75 Solutions to exercises 75 i.e. a n = n { n ( 1)n } = +4 n ( 1)n = 4 n ( 1)n i.e. a n = { 4 n, n even 4 n, n odd.

76 Solutions to exercises 76 STEP THREE b n = 1 f(x) sin nx dx = 1 x sin nx dx { [ = 1 ( )] cos nx ( ) } cos nx x x dx n n } {{ } using integration by parts { = 1 1 [ x cos nx ] n + } x cos nx dx n { = 1 1 ( cos n cos( n) ) + } x cos nx dx n n { = 1 1 ( cos n cos(n) ) + } x cos nx dx n } {{ } n =0 = x cos nx dx n

77 Solutions to exercises 77 { [ i.e. b n = x n i.e. b n = 0. ] } sin nx sin nx dx n n } {{ } using integration by parts { = 1 n n ( sin n ( ) sin( n)) 1 sin nx dx n { = 1 n n (0 + 0) 1 } sin nx dx n = n sin nx dx }

78 Solutions to exercises 78 where f(x) = a 0 + [a n cos nx + b n sin nx] n=1 { 4 a 0 = 3, a n = n, n even 4 n, n odd, b n = 0 i.e. f(x) = 1 n a n 4(1) 4 ( ) 1 4 ( ) ( ) 1 4 ( 3 ) [ 4 cos x 1 cos x + 13 cos 3x 14 ] cos 4x... + [ ] i.e. f(x) = [cos 3 4 x 1 cos x + 13 cos 3x 14 ] cos 4x +....

79 Solutions to exercises 79 c) To show that 6 = , { 1, n even use the fact that cos n = 1, n odd i.e. cos x 1 cos x cos 3x 1 4 cos 4x +... with x = gives cos 1 cos cos cos i.e. ( 1) 1 (1) ( 1) 1 4 (1) +... i.e = 1 ( ) +... } {{ } (the desired series)

80 Solutions to exercises 80 The graph of f(x) gives that f() = and the series converges to this value. Setting x = in the Fourier series thus gives = (cos cos + 13 cos 3 14 ) cos = ( )... = ( ) +... = 4 ( ) +... i.e. 6 = Return to Exercise 7