Course 2BA1, Section 11: Periodic Functions and Fourier Series

Transcription

1 Course BA, 8 9 Section : Periodic Functions and Fourier Series David R. Wikins Copyright c David R. Wikins 9 Contents Periodic Functions and Fourier Series 74. Fourier Series of Even and Odd Functions Fourier Series for Genera Periodic Functions Sine Series Cosine Series i

2 Periodic Functions and Fourier Series Definition A function f: R R is said to be periodic if there exists some positive rea number such that f(x + = f(x for a rea numbers x. The smaest rea number with this property is the period of the periodic function f. A periodic function f with period satisfies f(x + m = f(x for a rea numbers x and integers m. The period of a periodic function f is said to divide some positive rea number K if K/ is an integer. If the period of the function f divides a positive rea number K then f(x + mk = f(x for a rea numbers x and integers m. Mathematicians have proved that if f: R R is any sufficienty webehaved function from R to R with the property that f(x + π = f(x for a rea numbers x then f may be represented as an infinite series of the form f(x = a + a n cos nx + b n sin nx. (33 In particuar it foows from theorems proved by Dirichet in 89 that a function f: R R which satisfies f(x + π = f(x for a rea numbers x can be represented as a trigonometrica series of this form if the function is bounded, with at most finitey many points of discontinuity, oca maxima and oca minima in the interva [, π], and if ( f(x = im f(x + h + im f(x h h + h + at each vaue x at which the function is discontinuous ( im f(x + h h + f(x h denote the imits of f(x + h and f(x h respectivey as and im h + h tends to from above. Fourier in 87 had observed that if a sufficienty we-behaved function coud be expressed as the sum of a trigonometrica series of the above form, then a = π a n = π b n = π f(x dx, (34 f(x cos nx dx, (35 f(x sin nx dx (36 74

3 for each positive integer n. These expressions for the coefficients a n and b n may readiy be verified on substituting the trigonometic series for the function f (equation (33 into the integras on the right hand side of the equation, provided that one is permitted to interchange the operations of integration and summation in the resuting expressions. Now it is not generay true that the integra of an infinite sum of functions is necessariy equa to the sum of the integras of those functions. However if the function f is sufficienty we-behaved then the trigonometric series for the function f wi converge sufficienty rapidy for this interchange of integration and summation to be vaid, so that ( π a m cos mx dx = a m cos mx dx etc. If we interchange summations and integrations in this fashion and make use of the trigonometric integras provided by Theorem., we find that ( π ( π f(x dx = a π + a m cos mx dx + b m sin mx dx = a π + = a π, a m cos mx dx + b m sin mx dx Aso f(x cos nx dx = a = cos nx dx + ( a m cos mx cos nx dx ( π + b m sin mx cos nx dx a m cos mx cos nx dx + b m = a n π, f(x sin nx dx = a sin nx dx + sin mx cos nx dx ( a m cos mx sin nx dx 75

4 = ( π + b m sin mx sin nx dx a m cos mx sin nx dx + b m = b n π. sin mx sin nx dx A trigonometric series of the form (33 with coefficients a n and b n given by the integras (34, (35 and (36 is referred to the Fourier series for the function f. The coefficients defined by the integras (34, (35 and (36 are referred to as the Fourier coefficients of the function f. Exampe Consider the function f: R R defined by if x = mπ for some integer m; f(x = if mπ < x < (m + π for some integer m; if (m π < x < mπ for some integer m. This function f has the property that f(x = f(x + mπ for a rea numbers x and integers m, and can be represented by a Fourier series. The coefficients a n and b n of the Fourier series are given by the formuae a = π a n = π b n = π f(x dx, f(x cos nx dx (n >, f(x sin nx dx (n >, Now f(x = if < x <, and f(x = if < x < π. Therefore a =, and a n = cos nx dx = π nπ [sin nx]π = (n >, b n = sin nx dx = π nπ [ cos nx]π = ( cos nπ = nπ nπ ( ( n { if n is odd and n >, = nπ if n is even and n >, 76

5 (We have here made use of the fact that sin nπ = and cos nπ = ( n for a integers n. Thus f(x = + n odd n> = + k= sin nx nπ sin ((k x (k π. Fourier Series of Even and Odd Functions Definition A function f: R R is said to be even if f(x = f( x for a rea numbers x. A function f: R R is said to be odd if f(x = f( x for a rea numbers x. Let f: R R be an integrabe function. Then f(x dx = f(x cos nx dx = f(x sin nx dx = f( x dx, (37 f( x cos nx dx, (38 f( x sin nx dx (39 (The first of these identities may be verified by making the substitution x x and then interchanging the two imits of integration. The second and the third foow from the first on repacing f(x by f(x cos nx and f(x sin nx and noting that cos( nx = cos nx and sin( nx = sin nx. It foows that the Fourier coefficients of f are given by the foowing formuae: a = π a n = π b n = π f(x dx = π f(x cos nx dx = π f(x sin nx dx = π (f(x + f( x dx, (4 (f(x + f( x cos nx dx, (4 (f(x f( x sin nx dx (4 for a positive integers n. Of course f(x + f( x = f(x and f(x f( x = for a rea numbers x if the function f: R R is even, and f(x + f( x = and f(x f( x = f(x if the function f: R R is odd. The foowing resuts foow immediatey, 77

6 Theorem. Let f: R R be an even periodic function whose period divides π. Suppose that the function f may be represented by a Fourier series. Then the Fourier series of f is of the form and for a positive integers n. f(x = a + a = π a n = π a n cos nx, f(x dx f(x cos nx dx Theorem. Let f: R R be an odd periodic function whose period divides π. Suppose that the function f may be represented by a Fourier series. Then the Fourier series of f is of the form for a positive integers n. f(x = b n = π b n sin nx, f(x sin nx dx. Fourier Series for Genera Periodic Functions Let f: R R be a periodic function whose period divides, is some positive rea number. Then f(x + = f(x for a rea numbers x. Let ( ( x πx g(x = f so that f(x = g. π Then g: R R is a periodic function, and g(x + π = g(x for a rea numbers x. If the function f is sufficienty we-behaved (and, in particuar, if the function f is bounded, with ony finitey many oca maxima and minima and points of discontinuity in any finite interva, and if f(x at each point of discontinuity is the average of the imits of f(x + h and f(x h 78

7 as h tends to zero from above then the function g may be represented by a Fourier series of the form g(x = a + a n cos nx + b n sin nx. The coefficients of this Fourier series are then given by the formuae a = π a n = π b n = π g(u du, g(u cos nu du, g(u sin nu du for each positive integer n. If we make the substitution u = πx in these integras, we find that f(x = a + ( nπx a n cos + ( nπx b n sin, (43 a = = a n = = b n = = g ( πx dx f(x dx, ( ( πx nπx g cos dx ( nπx f(x cos dx, ( ( πx nπx g sin dx ( nπx f(x sin dx for a positive integers n. Note that these integras are taken over a singe period of the function, from to +. It foows from the periodicity of 79

8 the integrand that these integras may be repaced by integras from c to c+ for any rea number c, and thus a = a n = b n = c+ c c+ c c+ c f(x dx, (44 ( nπx f(x cos dx, (45 ( nπx f(x sin dx (46 for a positive integers n. (Indeed, if h: R R is any integrabe function with the property that h(x + = h(x for a rea numbers x, and if p and q are rea numbers with p q p + then p+ p h(x dx = = q p q+ p+ h(x dx + h(x dx + p+ q p+ q h(x dx Repeated appications of this identity show that p+ for a rea numbers p and q. p h(x dx = q+ q h(x dx = h(x dx q+ q h(x dx. Exampe Let k be a positive rea number. Consider the function f: R R defined by { e k(x m if m < x < m + for some integer m; f(x = (ek + if x is an integer. This function is periodic, with period, and may be expanded as a Fourier series f(x = a + a n cos nπx + b n sin nπx, a = a n = b n = f(x dx, f(x cos nπx dx (n >, f(x sin nπx dx (n >. 8

9 Note that f(x = e kx if < x <. We see therefore that a = e kx dx = k [ e kx ] = k (ek. Now if k and ω a positive rea numbers then e kx k ω cos ωx dx = k + ω ekx cos ωx + k + ω ekx sin ωx + C, e kx k ω sin ωx dx = k + ω ekx sin ωx k + ω ekx cos ωx + C, C is a constant of integration. (These identities may be verified by differentiating the expressions on the right hand side. We find therefore that a n = e kx cos nπx dx = [ ] k 4nπ k + 4n π ekx cos nπx + k + 4n π ekx sin nπx = k k + 4n π (ek b n = e kx sin nπx dx, [ ] k 4nπ = k + 4n π ekx sin nπx k + 4n π ekx cos nπx 4nπ = k + 4n π (ek, for each positive integer n, since cos nπ = and sin nπ = when n is an integer. Thus e kx = k (ek + k k + 4n π (ek cos nπx 4nπ k + 4n π (ek sin nπx for a rea numbers x satisfying < x <..3 Sine Series Let f: [, ] R be a function defined on the interva [, ], [, ] = {x R : x }. Suppose that f( = f( =. Let f: R R be the 8

10 function defined such that and f(x = f(x n if n x (n + for some integer n f(x = f((n + x if (n + x (n + for some integer n. The function f: R R is an odd function with the property that f(x+ = f(x for a rea numbers x. Indeed it is easiy seen that f: R R is the unique odd function with this property which agrees with the function f on the interva [, ]. If the function f is sufficienty we-behaved (and, in particuar, if the function f is bounded, with at most finitey many oca maxima and minima and points of discontinuity, and has the property that f(x at each point of discontinuity is the average of the imits of f(x + h and f(x h as h tends to zero from above then the function f may be represented as a Fourier series. This Fourier series is of the form f(x = ( nπx b n sin, b n = = ( nπx f(x sin dx ( nπx f(x sin dx for a positive integers n. (This foows from equations (43 and (46 on repacing by, and then using the fact that f( x = f(x for a rea numbers x. Therefore every sufficienty we-behaved function f: [, ] R which satisfies f( = f( = may be represented in the form f(x = ( nπx b n sin, (47 b n = for each positive integer n. ( nπx f(x sin dx (48 8

11 Exampe Let be a positive rea numbers, and et f: [, ] R be the function defined by f(x = x( x ( x. This function can be expanded in a sine series of the form f(x = b n = ( nπx b n sin, f(x sin nπx Using the method of integration by parts, and the resut that sin nπ = and cos nπ = ( n for a integers n, we find then Thus b n = x( x sin nπx dx = nπ = [ x( x cos nπx ] + nπ nπ = ( x cos nπx dx nπ = ( x d ( sin nπx dx n π dx = [ ( x sin nπx ] n π = 4 sin nπx [ dx = 4 n π n 3 π 3 dx. ( n π cos nπx = 4 ( cos nπ = 4 n 3 π3 n 3 π ( 3 ( n { 8 = if n is odd; n 3 π 3 if n is even. f(x = n odd n> 8 n 3 π x( x d ( cos nπx dx ( x cos nπx dx sin nπx dx ] nπx sin. 3 or (setting n = k for each positive integer k, dx x( x = k= 8 (k πx sin (k 3 π3 ( x. 83

12 .4 Cosine Series Let f: [, ] R be a function defined on the interva [, ], [, ] = {x R : x }. Let g: R R be the function defined by and g(x = f(x n if n x (n + for some integer n g(x = f((n + x if (n + x (n + for some integer n. The function g: R R is an even function with the property that g(x+ = g(x for a rea numbers x. Indeed it is easiy seen that g: R R is the unique even function with this property which agrees with the function f on the interva [, ]. If the function f is sufficienty we-behaved then the function g may be represented as a Fourier series. This Fourier series is of the form g(x = a + ( nπx a n cos, a n = = ( nπx g(x cos dx ( nπx g(x cos dx for a non-negative integers n. (This foows from equations (43, (44 and (45 on repacing by, and then using the fact that g( x = g(x for a rea numbers x. Therefore every sufficienty we-behaved function f: [, ] R may be represented in the form f(x = a + ( nπx a n cos (49 a n = for each positive integer n. ( nπx f(x cos dx (5 84

13 Exampe Consider the function f: [, ] R defined by f(x = x if x. This function may be represented as a cosine series of the form and a = f(x = a + a n = a n cos nπx f(x dx = f(x cos nπx dx x dx =, for a positive integers n. Using the method of integation by parts, and making use of the fact that sin nπ = and cos nπ = ( n for a integers n, we find that Thus a n = x cos nπx dx = x d (sin nπx dx nπ dx = nπ [x sin nπx] sin nπx dx = nπ nπ = n π [cos nπx] = n π ( ( n { = 4 if n is odd; n π if n is even. x = n odd n> Remark The function g defined by 4 cos nπx when x. n π g(x = n odd n> 4 cos nπx n π sin nπx dx for a rea numbers x is an even periodic function, with period equa to, which coincides with the function f: [, ] R on the interva [, ], f(x = x for a rea numbers x satisfying x. It foows that g(x = x m whenever m is an integer and m x m +. 85

14 Remark Setting x = in the identity we find that and thus But n odd n> x = n odd n> n = = n odd n> n 4 cos nπx when x, n π 4 n π cos nπ = + n odd n> n odd n> n = π 8. ( (n = 4 4 n π n = 3 4 n. Therefore n = π 6. Probems. Let f: R R be the periodic function with period π given for rea numbers x satisfying x π by the formua x if f(x = π π x π; if x π or π x π. (Here x, the absoute vaue of x, is defined by x = x if x, and x = x if x <. The function f can be expanded as a Fourier series of the form f(x = a + a n cos nx. (The terms invoving sin nx are zero since the given function is even. Find the coefficients a n of this series, and hence write down the Fourier series for the function f. 86

15 . Cacuate the Fourier series of the function f: R R which is periodic, with period π, and which is defined on the interva x π by the foowing formuae: + x π if x π; f(x = if π x π; x π if π x π. 3. Let f: R R be defined such that f(x = 4(x m if m x m + for some integer m; f(x = 4(m + x if m + x m + for some integer m. Express the function f as a Fourier series of the form f(x = a + a n cos πnx. 87