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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 Toc JJ II J I Back 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 JJ II J I Back 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 JJ II P E R IO D = L J I Back 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 + ... JJ II J I Back Section 1: Theory 7 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 JJ II J I Back Section 1: Theory 8 ● 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 JJ II J I Back 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 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 JJ II J I Back Section 2: Exercises 10 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 JJ II J I Back Section 2: Exercises 11 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 +... ● Theory ● Answers ● Integrals ● Trig ● Notation Toc JJ II J I Back Section 2: Exercises 12 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 ● Theory ● Answers ● Integrals ● Trig ● Notation Toc JJ II J I Back Section 2: Exercises 13 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 JJ II J I Back Section 2: Exercises 14 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 ● Theory ● Answers ● Integrals ● Trig ● Notation Toc JJ II J I Back Section 2: Exercises 15 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 ● Theory ● Answers ● Integrals ● Trig ● Notation Toc JJ II J I Back 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. Toc JJ II J I Back 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 JJ 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 J [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 Back Section 4: Integrals f (x) 1 a2 +x2 √ 1 a2 −x2 √ a2 − x2 18 R f (x) dx f (x) R 1 a tan−1 1 a2 −x2 1 2a ln a+x a−x (0 < |x| < a) (a > 0) 1 x2 −a2 1 2a ln x−a x+a (|x| > a > 0) sin−1 x a √ 1 a2 +x2 ln √ x+ a2 +x2 a (a > 0) (−a < x < a) √ 1 x2 −a2 ln √ x+ x2 −a2 a (x > a > 0) a2 2 −1 sin √ +x Toc x a x a a2 −x2 a2 JJ √ i √ a2 +x2 a2 2 a2 2 x2 −a2 II f (x) dx J h h sinh−1 − cosh−1 I x a i √ x a2 +x2 2 a i √ 2 2 + x xa2−a + x a Back 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 Toc JJ II J I Back 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 Toc JJ II J I Back 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 JJ II J I Back Section 6: Alternative notation 22 ● 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 Toc JJ II J I Back Section 6: Alternative notation 23 ● 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 Toc JJ II J I Back 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 Toc JJ II J I Back Solutions to exercises 25 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 Toc −π JJ 0 II 2π π J x I Back Solutions to exercises 26 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 Toc π 1 π JJ II Z 0 J I Back Solutions to exercises 27 STEP TWO an = 1 π Z π f (x) cos nx dx = −π = = = = = i.e. an Toc JJ = 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 J I Back Solutions to exercises 28 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 Toc Z 0 JJ II J I Back Solutions to exercises 29 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 JJ II J I Back Solutions to exercises 30 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 JJ II J I Back 1 7 sin 7 π2 1 7 Solutions to exercises 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 Toc JJ II J I Back Solutions to exercises 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 JJ π −π II J 2π 3π I Back x Solutions to exercises 33 b) Fourier series representation of f (x) STEP ONE a0 = 1 π Z π f (x)dx = −π = = = Toc 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 J I Back Solutions to exercises 34 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 f (x) cos nx dx 0 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 = JJ II J I Back Solutions to exercises 35 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 JJ 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 + J I Back Solutions to exercises ( 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 I Back , n even , n odd Solutions to exercises 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 JJ II J I Back Solutions to exercises 38 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 JJ II J when x = I π 2, Back Solutions to exercises 39 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 JJ = = π 2 = π 2, so that π 1 1 1 + 1 − + − + ... 4 3 5 7 1 1 1 1 − + − + ... 3 5 7 II J I Back Solutions to exercises 40 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 JJ II J I Back Solutions to exercises 41 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 JJ II J I Back Solutions to exercises 42 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 J I 2π x Back Solutions to exercises 43 b) Fourier series representation of f (x) STEP ONE 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 J I Back Solutions to exercises 44 STEP TWO an = = = Z 1 π Z 1 π " f (x) cos nx dx 0 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 Solutions to exercises 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 Back Solutions to exercises 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 JJ II J I Back Solutions to exercises 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 Solutions to exercises 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 JJ II J I Back Solutions to exercises 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 Solutions to exercises 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 J I Back Solutions to exercises 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 J I Back Solutions to exercises 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 JJ II J I Back Solutions to exercises 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 Solutions to exercises 54 b) Fourier series representation of f (x) STEP ONE 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 Back Solutions to exercises 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 using integration by parts } ) 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 Back Solutions to exercises 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 using integration by parts = −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 Solutions to exercises 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 J I Back Solutions to exercises 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 I Back Solutions to exercises 59 Exercise 5. π−x , 0<x<π f (x) = 0 , 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 J I 2π x Back Solutions to exercises 60 b) Fourier series representation of f (x) STEP ONE a0 = = i.e. a0 Toc 1 π Z 1 π Z f (x) dx 0 π 1 (π − x) dx + π 0 π 1 πx − x2 + 0 2 0 2 π π2 − −0 2 = 1 π = 1 π = π . 2 JJ 2π II J Z 2π 0 · dx π I Back Solutions to exercises 61 STEP TWO an = = i.e. an = 1 π Z 2π f (x) cos nx dx 0 Z Z 1 π 1 2π (π − x) cos nx dx + 0 · dx π 0 π π π Z π sin nx sin nx 1 (π − x) − (−1) · dx +0 π n n 0 0 | {z } using integration by parts = = = i.e. an = Toc Z π 1 sin nx (0 − 0) + dx , see Trig π n 0 π 1 − cos nx πn n 0 1 − 2 (cos nπ − cos 0) πn 1 − 2 ((−1)n − 1) , see Trig πn JJ II J I Back Solutions to exercises i.e. an 62 = 0 , n even 2 πn2 , n odd STEP THREE bn = = Z 1 π Z 2π f (x) sin nx dx 0 π Z 2π (π − x) sin nx dx + 0 0 · dx π cos nx iπ Z π cos nx (π − x) − − (−1) · − dx + 0 n n 0 0 π 1 1 = 0− − − · 0 , see Trig π n n 1 = . n Toc JJ II J I Back = i.e. bn 1 π 1 π h Solutions to exercises In summary, a0 = π 2 63 and a table of other Fourier cofficients is n an = 2 πn2 bn = 1 n ∴ f (x) = (when n is odd) 1 2 3 4 5 2 π 0 2 1 π 32 0 2 1 π 52 1 1 2 1 3 1 4 1 5 ∞ X a0 + [an cos nx + bn sin nx] 2 n=1 π 2 2 1 2 1 + cos x + cos 3x + cos 5x + . . . 4 π π 32 π 52 1 1 1 + sin x + sin 2x + sin 3x + sin 4x + . . . 2 3 4 π 2 1 1 i.e. f (x) = + 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 Toc JJ II J I Back = Solutions to exercises 64 π2 8 c) To show that =1+ 1 32 + 1 52 + ... , note that, as x → 0 , the series converges to the half-way value of and then π π = 2 4 + + giving Toc π 2 π 4 π2 8 JJ = = = π 2, 1 1 cos 0 + cos 0 + . . . 32 52 1 1 sin 0 + sin 0 + sin 0 + . . . 2 3 π 2 1 1 + 1 + 2 + 2 + ... + 0 4 π 3 5 2 1 1 1 + 2 + 2 + ... π 3 5 1 1 1 + 2 + 2 + ... 3 5 Return to Exercise 5 2 π II cos 0 + J I Back Solutions to exercises 65 Exercise 6. f (x) = x, over the interval −π < x < π and has period 2π a) Sketch a graph of f (x) in the interval −3π < x < 3π π −3π −2π −π f(x ) x π 0 2π 3π −π Toc JJ II J I Back Solutions to exercises 66 b) Fourier series representation of f (x) STEP ONE a0 = = = = i.e. a0 Toc JJ = II 1 π Z 1 π Z π f (x) dx −π π x dx −π π 1 x2 π 2 −π 1 π2 π2 − π 2 2 0. J I Back Solutions to exercises 67 STEP TWO an = = = Z 1 π f (x) cos nx dx π −π Z π 1 x cos nx dx π −π ( π ) Z π 1 sin nx sin nx x − dx π n n −π −π | {z } using integration by parts Z 1 1 1 π (π sin nπ − (−π) sin(−nπ)) − sin nx dx π n n −π 1 1 1 (0 − 0) − · 0 , π n n Z since sin nπ = 0 and sin nx dx = 0, i.e. an = = 2π i.e. an = 0. Toc JJ II J I Back Solutions to exercises STEP THREE bn = = = = = = = i.e. bn = Toc Z 1 π f (x) sin nx dx π −π Z π 1 x sin nx dx π −π ( π ) Z π 1 −x cos nx − cos nx − dx π n n −π −π Z 1 1 1 π π cos nx dx − [x cos nx]−π + π n n −π 1 1 1 − (π cos nπ − (−π) cos(−nπ)) + · 0 π n n π − (cos nπ + cos nπ) nπ 1 − (2 cos nπ) n 2 − (−1)n . n JJ II J I Back 68 Solutions to exercises 69 We thus have f (x) = ∞ i a0 X h an cos nx + bn sin nx + 2 n=1 with a0 = 0, an = 0, bn = − n2 (−1)n and n 1 2 bn 2 −1 3 2 3 Therefore f (x) i.e. f (x) = b1 sin x + b2 sin 2x + b3 sin 3x + . . . 1 1 = 2 sin x − sin 2x + sin 3x − . . . 2 3 and we have found the required Fourier series. Toc JJ II J I Back Solutions to exercises 70 c) Pick an appropriate value of x, to show that π 4 =1− 1 3 + 1 5 − 1 7 + ... Setting x = π2 gives f (x) = π2 and π π 1 2π 1 3π 1 4π 1 5π = 2 sin − sin + sin − sin + sin − ... 2 2 2 2 3 2 4 2 5 2 This gives π = 2 π = 2 π i.e. = 4 1 1 1 2 1 + 0 + · (−1) − 0 + · (1) − 0 + · (−1) + . . . 3 5 7 1 1 1 2 1 − + − + ... 3 5 7 1 1 1 1 − + − + ... 3 5 7 Return to Exercise 6 Toc JJ II J I Back Solutions to exercises 71 Exercise 7. f (x) = x2 , over the interval −π < x < π and has period 2π a) Sketch a graph of f (x) in the interval −3π < x < 3π f(x ) π 2 −3π Toc −2π JJ −π II π 0 J 2π I 3π x Back Solutions to exercises 72 b) Fourier series representation of f (x) STEP ONE 1 a0 = π Z π f (x)dx = −π = = = i.e. a0 Toc JJ II = 1 π Z π x2 dx −π π 1 x3 π 3 −π 1 π3 π3 − − π 3 3 1 2π 3 π 3 2π 2 . 3 J I Back Solutions to exercises 73 STEP TWO an = = = Z 1 π f (x) cos nx dx π −π Z π 1 x2 cos nx dx π −π ( π ) Z π 1 sin nx sin nx 2x x2 − dx π n n −π −π | {z } using integration by parts = = = ) Z 2 π 1 2 2 x sin nx dx π sin nπ − π sin(−nπ) − n n −π ) ( Z 1 1 2 π x sin nx dx , see Trig (0 − 0) − π n n −π Z π −2 x sin nx dx nπ −π 1 π ( Toc JJ II J I Back Solutions to exercises i.e. an 74 −2 nπ ( π ) Z π − cos nx − cos nx x − dx n n −π −π | {z } = −2 nπ ( = −2 nπ = −2 nπ = −2 nπ = using integration by parts again Toc ) Z 1 1 π π − [x cos nx]−π + cos nx dx n n −π ) ( 1 1 − π cos nπ − (−π) cos(−nπ) + · 0 n n ( ) 1 − π(−1)n + π(−1)n n ( ) −2π (−1)n n JJ II J I Back Solutions to exercises i.e. an 75 ( = −2 nπ = +4π (−1)n πn2 = 4 (−1)n n2 ( i.e. an = 2π − (−1)n n 4 n2 , n even −4 n2 , n odd. Toc JJ ) II J I Back Solutions to exercises 76 STEP THREE bn = = π Z 1 π 2 f (x) sin nx dx = x sin nx dx π −π −π ( π ) Z π 1 − cos nx − cos nx x2 − 2x · dx π n n −π −π | {z } 1 π Z 1 π ( using integration by parts = = = π 2 1 2 x cos nx −π + − n n Z ) π x cos nx dx −π ) Z π 1 2 2 − π cos nπ − π 2 cos(−nπ) + x cos nx dx n n −π ) ( Z 2 π 1 1 2 2 − π cos nπ − π cos(nπ) + x cos nx dx π n| {z } n −π 1 π ( =0 = 2 πn Z Toc π x cos nx dx −π JJ II J I Back Solutions to exercises i.e. bn = 2 πn 77 ( sin nx x n | = = = i.e. bn = 2 πn π Z π − −π −π {z ) sin nx dx n } using integration by parts ( 1 1 (π sin nπ − (−π) sin(−nπ)) − n n ( ) Z 2 1 1 π sin nx dx (0 + 0) − πn n n −π Z −2 π sin nx dx πn2 −π Z ) π sin nx dx −π 0. Toc JJ II J I Back Solutions to exercises 78 ∞ a0 X + [an cos nx + bn sin nx] 2 n=1 ( 4 , n even 2π 2 n2 a0 = 3 , an = , −4 , n odd n2 ∴ f (x) = where n an 1 i.e. f (x) = 2 2π 2 3 1 2 −4(1) 4 1 22 3 −4 bn = 0 4 1 32 4 1 42 1 1 1 − 4 cos x − 2 cos 2x + 2 cos 3x − 2 cos 4x . . . 2 3 4 + [0 + 0 + 0 + . . .] π2 1 1 1 i.e. f (x) = − 4 cos x − 2 cos 2x + 2 cos 3x − 2 cos 4x + . . . . 3 2 3 4 Toc JJ II J I Back Solutions to exercises 79 π2 6 = 1 + 212 + 312 + 412 + . . . , ( 1 , n even use the fact that cos nπ = −1 , n odd c) To show that cos x − 1 22 cos 2x + 1 32 cos 3x − 1 42 cos 4x + . . . with x = π gives cos π − 1 22 cos 2π + 1 32 cos 3π − 1 42 cos 4π + . . . 1 32 · (−1) − i.e. i.e. (−1) − i.e. −1 − 1 22 · (1) + 1 22 1 32 − − 1 42 · (1) + . . . 1 42 +... 1 1 1 = −1 · 1 + 2 + 2 + 2 + . . . 2 3 4 | {z } (the desired series) Toc JJ II J I Back Solutions to exercises 80 The graph of f (x) gives that f (π) = π 2 and the series converges to this value. Setting x = π in the Fourier series thus gives π2 1 1 1 2 π = − 4 cos π − 2 cos 2π + 2 cos 3π − 2 cos 4π + . . . 3 2 3 4 2 π 1 1 1 π2 = − 4 −1 − 2 − 2 − 2 − . . . 3 2 3 4 π2 1 1 1 2 + 4 1 + 2 + 2 + 2 + ... π = 3 2 3 4 2 2π 1 1 1 = 4 1 + 2 + 2 + 2 + ... 3 2 3 4 π2 1 1 1 i.e. = 1 + 2 + 2 + 2 + ... 6 2 3 4 Return to Exercise 7 Toc JJ II J I Back

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