Euler's Formulas Examples Convergence Arbitrary Periods Integral formulas for Fourier coefficients Ryan C. Daileda Trinity University Partial Differential Equations Lecture 6 Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods Recall: Relative to the inner product Z π hf ; gi = f (x)g(x) dx −π the functions 1; cos(x); cos(2x); cos(3x);::: sin(x); sin(2x); sin(3x);::: satisfy the orthogonality relations hcos(mx); sin(nx)i = 0; 8 0 if m 6= n; <> hcos(mx); cos(nx)i = π if m = n 6= 0; :>2π if m = n = 0; ( 0 if m 6= n; hsin(mx); sin(nx)i = π if m = n: Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods By the linearity of the inner product, if 1 X f (x) = a0 + (an cos(nx) + bn sin(nx)) ; n=1 then =0 unless m=0 =0 unless m=n 1 z }| { X z }| { hf (x); cos(mx)i = a0 h1; cos(mx)i + (an hcos(nx); cos(mx)i n=1 + bn hsin(nx); cos(mx)i ) | {z } =0 = am hcos(mx); cos(mx)i ( 2πa if m = 0; = 0 πam if m 6= 0: Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods Likewise, one can show that hf (x); sin(mx)i = bm hsin(mx); sin(mx)i = πbm: Solving for the coefficients gives: Theorem (Euler's Formulas) If f is 2π-periodic and piecewise smooth, then its Fourier coefficients are given by hf (x); 1i 1 Z π a0 = = f (x) dx; h1; 1i 2π −π hf (x); cos(nx)i 1 Z π an = = f (x) cos(nx) dx (n 6= 0); hcos(nx); cos(nx)i π −π hf (x); sin(nx)i 1 Z π bn = = f (x) sin(nx) dx: hsin(nx); sin(nx)i π −π Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods Remarks f (x+)+f (x−) Technically we should have used 2 . However, the integrals cannot distinguish between this and f (x). Because all the functions in question are 2π-periodic, we can integrate over any convenient interval of length 2π. If f (x) is an odd function, so is f (x) cos(nx), and so an = 0 for all n ≥ 0. If f (x) is an even function, then f (x) sin(nx) is odd, and so bn = 0 for all n ≥ 1. Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods Example Find the Fourier series for the 2π-periodic function that satisfies f (x) = x for −π < x ≤ π. The graph of f (a sawtooth wave): Because f is odd, we know an = 0 (n ≥ 0): Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods According to Euler's formula: 1 Z π bn = x sin(nx) dx π −π π 1 −x cos(nx) sin(nx) = + 2 π n n −π 1 −π cos(nπ) π cos(−nπ) = − π n n −2 cos(nπ) (−1)n+12 = = : n n Therefore, the Fourier series of f is 1 1 X (−1)n+12 X (−1)n+1 sin(nx) sin(nx) = 2 : n n n=1 n=1 Remark: Except where it is discontinuous, this series equals f (x). Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods Example Find the Fourier series of the 2π-periodic function satisfying f (x) = jxj for −π ≤ x < π. The graph of f (a triangular wave): This time, since f is even, bn = 0 (n ≥ 1): Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods By Euler's formula we have Z π Z π 2 π 1 1 1 x π a0 = jxj dx = x dx = = 2π −π π 0 π 2 0 2 and for n ≥ 1 1 Z π 2 Z π an = jxj cos(nx) dx = x cos(nx) dx π −π | {z } π 0 even π 2 x sin(nx) cos(nx) 2 cos(nπ) 1 = + 2 = 2 − 2 π n n 0 π n n 8 −4 > 2 if n is odd; 2 n <πn = 2 ((−1) − 1) = πn > :>0 if n is even. Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods In the Fourier series we may therefore omit the terms in which n is even, and assume that n = 2k + 1, k ≥ 0: 1 1 X π X −4 a + a cos(nx) = + cos((2k + 1)x) 0 n 2 π(2k + 1)2 n=1 k=0 1 π 4 X cos((2k + 1)x) = − : 2 π (2k + 1)2 k=0 Remarks: Since k is simply an index of summation, we are free to replace it with n again, yielding 1 π 4 X cos((2n + 1)x) − : 2 π (2n + 1)2 n=0 Because f (x) is continuous everywhere, this equals f (x) at all points. Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods Example Use the result of the previous exercise to show that 1 1 1 1 X 1 π2 1 + + + + ··· = = : 9 25 49 (2n + 1)2 8 n=0 If we set x = 0 in the previous example, we get 1 1 π 4 X cos(0) π 4 X 1 0 = f (0) = − = − 2 π (2n + 1)2 2 π (2n + 1)2 n=0 n=0 Solving for the series gives the result. Remark: In Calculus II you learned that this series converges, but were unable to obtain its exact value. Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods Example Find the Fourier series of the 2π-periodic function satisfying f (x) = 0 for −π ≤ x < 0 and f (x) = x2 for 0 ≤ x < π. The graph of f : Because f is neither even nor odd, we must compute all of its Fourier coefficients directly. Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods Since f (x) = 0 for −π ≤ x < 0, Euler's formulas become Z π 3 π 2 1 2 1 x π a0 = x dx = = 2π 0 2π 3 0 6 and for n ≥ 1 Z π 2 π 1 2 1 x sin(nx) 2x cos(nx) 2 sin(nx) an = x cos(nx) dx = + 2 − 3 π 0 π n n n 0 1 2π cos(nπ) 2(−1)n = · = ; π n2 n2 Z π 2 π 1 2 1 x cos(nx) 2x sin(nx) 2 cos(nx) bn = x sin(nx) dx = − + 2 + 3 π 0 π n n n 0 1 π2 cos(nπ) 2 cos(nπ) 2 (−1)n+1π 2((−1)n − 1) = − + − = + : π n n3 n3 n πn3 Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods Therefore the Fourier series of f is 1 π2 X 2(−1)n (−1)n+1π 2((−1)n − 1) + cos(nx) + + sin(nx) : 6 n2 n πn3 n=1 This will agree with f (x) everywhere it's continuous. Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods Convergence of Fourier series Given a Fourier series 1 X a0 + (an cos(nx) + bn sin(nx)) (1) n=1 let its Nth partial sum be N X sN (x) = a0 + (an cos(nx) + bn sin(nx)) : (2) n=1 According to the definition of an infinite series, the Fourier series (1) is equal to lim sN (x): N!1 Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods According to the definition of the limit: We can approximate the (infinite) Fourier series (1) by the (finite) partial sums (2). The approximation of (1) by (2) improves (indefinitely) as we increase N. Because sN (x) is a finite sum, we can use a computer to graph it. In this way, we can visualize the convergence of a Fourier series. Let's look at some examples... Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods These examples illustrate the following results. In both, f (x) is 2π-periodic and piecewise smooth. Theorem (Uniform convergence of Fourier series) If f (x) is continuous everywhere, then the partial sums sN (x) of its Fourier series converge uniformly to f (x) as N ! 1. That is, by choosing N large enough we can make sN (x) arbitrarily close to f (x) for all x simultaneously. Theorem (Wilbraham-Gibbs phenomenon) If f (x) has a jump discontinuity at x = c, then the partial sums sN (x) of its Fourier series always \overshoot" f (x) near x = c. More precisely, as N ! 1, the the ratio between the peak of the overshoot and the height of the jump tends to 1 Z π sint 1 dt − = 0:08948 ::: (about 9% of the jump): π 0 t 2 Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods The Wilbraham-Gibbs phenomenon A function f (x) (in blue) with a jump discontinuity and a partial sum sN (x) (in red) of its Fourier series: j N h j lim N = 0:08948 ::: N!1 h Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods General Fourier series If f (x) is 2p-periodic and piecewise smooth, then f^(x) = f (px/π) 2p has period p/π = 2π, and is also piecewise smooth. It follows that f^(x) has a Fourier series: 1 f^(x+) + f^(x−) X = a + (a cos(nx) + b sin(nx)): 2 0 n n n=1 Since f (x) = f^(πx=p), we find that f also has a Fourier series: 1 f (x+) + f (x−) X nπx nπx = a + a cos + b sin : 2 0 n p n p n=1 Daileda Fourier Coefficients Euler's Formulas Examples Convergence Arbitrary Periods We can use Euler's formulas to find an and bn.