Continuing from the last lecture.... Example. (ii) Let f be a periodic function of period 2π such that f(x) = π2 − x2 for − π ≤ x < π Solution: So f is periodic with period 2π and its graph is: We first check if f is even or odd. f(−x) = π2 − (−x)2 = π2 − x2 = f(x), so f(x) is even. Since f is even, bn = 0 2 Z π an = f(x) cos(nx)dx π 0 Using the formulas for the Fourier coefficients we have 2 Z π an = f(x) cos(nx)dx π 0 2 Z π = (π2 − x2) cos(nx)dx π 0 Z π 2 2 2 sin nxπ sin nx = ( (π − x ) 0 − −2x dx) π n 0 n 2 sin nπ sin 0 Z π sin nx = ([(π2 − π2) − (π2 − 0) ] + 2x dx) π n n 0 n 2 2 Z π = x sin(nx)dx π n 0 Z π 2 2 cos nxπ cos nx = ( − x 0 − (− )dx) π n n 0 n 2 2 1 sin nx = ([−π cos nπ] + [ ]π) π n n n2 0 4 = − cos nπ n2 −4/n2 if n is even = 4/n2 if n is odd It remains to calculate a0. Z π 2 2 2 a0 = (π − x )dx π 0 2 x3 = [π2x − ]π π 3 0 4π2 = 3 The Fourier Series of f is therefore 1 f(x) = a + a cos x + a cos 2x + a cos 3x + ... 2 0 1 2 3 b1 sin x + b2 sin 2x + b3 sin 3x + ... 2π2 1 1 1 1 = + 4(cos x − cos 2x + cos 3x − cos 4x + cos 5x + ....) 3 4 9 16 25 10.6 Functions with period 2L If a function has period other than 2π, we can find its Fourier series by making a change of variable. Suppose f(x) has period 2L, that is f(x + 2L) = f(x) for all x. If we let πx Lt t = and g(t) = f(x) = f( ) L π then, g has period 2π and x = ±L corresponds to t = ±π. We know the Fourier series of g(t): ∞ 1 X a + (a cos nt + b sin nt). 2 0 n n n=1 where 1 Z π a0 = f(t)dt π −π 1 Z π an = f(t) cos(nt)dt π −π 1 Z π bn = f(t) sin(nt)dt π −π Lt πx π If we now use the Substitution Rule with x = π , then t = L , dt = ( L )dx, we have the following: Definition. If f(x) is a piecewise continuous function on [−L, L], its Fourier Series is ∞ 1 X nπx nπx a + a cos( ) + b sin( ), 2 0 n L n L n=1 where 1 Z L a0 = f(x)dx L −L 1 Z L nπx an = f(x) cos( )dx L −L L 1 Z L nπx bn = f(x) sin( )dx L −L L Example. Find the Fourier series of the triangular wave function defined by f(x) = |x| − a for −a ≤ x ≤ a and f(x + 2) = f(x) for all x. Solution: So f is periodic with period 2a and its graph is: We first check if f is even or odd. f(−x) = | − x| − a = |x| − a = f(x), so f(x) is even. Therefore, bn = 0 2 Z a nπx an = f(x) cos( )dx a 0 a Using the formulas for the Fourier coefficients we have 2 Z a nπx an = (|x| − a) cos( )dx a 0 a 2 Z a nπx = (x − a) cos( )dx a 0 a nπx Z a nπx 2 sin( a )a sin( a ) = ( (x − a) nπ 0 − a nπ dx) a a 0 a 2 a2 nπx = ( cos( )a) a n2π2 a 0 2a = (cos nπ − 1) n2π2 ( 0 if n is even = 4a − if n is odd n2π2 It remains to determine a0. 2 Z a a0 = |x| − adx a 0 2 Z a = x − adx a 0 2 x2 = [ − ax]a a 2 0 = −a The Fourier Series of f is therefore f(x) = a0 + a1 cos x + a2 cos 2x + a3 cos 3x + ... a 4a πx 1 3πx 1 5πx = − − (cos( ) + cos( ) + cos( ) + ...) 2 π2 a 9 a 25 a 10.7 Parseval’s Result The folowing result can be used to find the sum of certain series from calcu- lated Fourier series. Theorem Let f be a bounded function defined in the interval [−L, L] with at most a finite number of maxima and minima and at most a finite number of discontinuities in the interval [−L, L], and associated to the Fourier series ∞ 1 X a + (a cos nx + b sin nx). 2 0 n n n=1 Then ∞ 1 Z L 1 X (f(x))2dx = a2 + (a2 + b2 ). L 2 0 n n −L n=1 Example. Let f be a periodic function defined by f(x) = x for all −π ≤ x ≤ π. Use Parseval’s result to show that ∞ π2 X 1 = . 6 n2 n=1 Solution: We have shown that 2 2 2 2 f(x) = 2 sin x − sin 2x + sin 3x − sin 4x + sin 5x + ... 2 3 4 5 Using Parseval’s Identity ∞ 1 Z π 1 X (f(x))2dx = a2 + (a2 + b2 ) π 2 0 n n −π n=1 1 x3 1 1 π = 4(1 + + + ...) π 3 −π 4 9 2π2 1 1 = 4(1 + + + ...) 3 4 9 π2 1 1 = 1 + + + ... 6 4 9 ∞ π2 X 1 = 6 n2 n=1.