Question

3. Evaluating a Fourier series at a point: You may use any of the Fourier series we have de- rived in class, you have obtained in the homework or any in the Table of Fourier series in MyCourses (a) By evaluating a Fourier series at some point, show that 9 25 49 n (2n+1)2 Page 1 of 2 (b) Use another Fourier series different from the one used in class to show that 4 2n+1 (c) Use a Fourier series to find the sum-+-+-+-+-+ … =? (d) Use a Fourier series to find the sum-+-+-+ -+--? (e) Use a Fourier series to find the sum 3 15 35 63 99 12 22 3 4252

Answer 1

(a) f(x) = x^2 - pi lessthanorequalto x lessthanorequalto pi Then the fourier series expansion of f(x) is f(x) = pi^2/3 + Sigma^infinity_n = 1 ((- 1)^n 4/n^2 cos (nx)) ---(1) pi^2 = f(pi) = pi^2/3 + Sigma^infinity_n = 1 ((- 1)^n 4/n^2 (-1)^n) = > 2pi^2/3 = Sigma^infinity_n = 1 4/n^2 = > pi^2/6 = Sigma^infinity_n = 1 1/n^2 = Sigma^infinity_k = 0 1/(2k)^2 + Sigma^infinity_k = 0 1/(2k + 1)^2 = > pi^2/6 = 1/4(Sigma^infinity_k = 0 1/k^2) + Sigma^infinity_k = 0 1/(2k + 1)^2 = > pi^2/6 - pi^2/24 = Sigma^infinity_k = 0 (1/(2k + 1)^2) = > pi^2/8 = 1 + 1/9 + 1/25 + --- \r\n (b) f(x) = pi - x/2 0 lessthanorequalto x lessthanorequalto pi The Fourier series expression of f(x) is f(x) = pi - x/2 = Sigma^infinity_n = 1 1/n sin(nx) 0 lessthanorequalto x lessthanorequalto pi x = pi/2 pi/4 = f(pi/2) = Sigma^infinity_n = 1 /1/n(sin(n . pi/2)) Therefore pi/4 = 1 - 1/3 + 1/5 - 1/7 --- = Sigma^infinity_k = 1 (- 1)^k - 1/2k - 1.