Question

2. Prove Leibniz' formula: 0 2k + 1-4 A Road Map to Glory a. Explain why k = 0, 1, 2, .. .. sin(kπ + π/2) = (-1)" (a) Explain why the trigonometric Fourier series of the function f (x)- be expressed solely as a sine series, specifically: ,sin(nz) sin(n) c. Compute(f,sinn). Simplify your work by explaining why 〈f,sin(nz)) = sin(nz) dr d. Does the Fourier series converge at x = π/2? Evaluate the Fourier series and f at π/2 to derive Leibniz formula.

Answer 1

Sin (n pi + pi/2) = { sin (pi/2) if k is even -sin pi/2 if is odd = (-1)^k given f(x) = { -pi/4: x epsilon [- pi, 0] pi/4: x epsilon [0, pi] this f(x) is an odd function i.e, f(-x) = -f(x) and thus can be represented as f(x) = a_0 + sum_(n = 1)^infinity a_n sin nx where, a_0 = 1/2pi integral_-pi^pi f(x) dx = 0 and a_n = 1/pi integral_(-pi)^pi f(x) sin nx dx = 1/pi (f sin (nx)) therefore f(x) = sum_(n = 1)^infinity (f sin nx)/pi sin nx \r\n (f sin nx) = integral_(-pi)^pi f(x) sin n x dx = integral_(- pi)^0 f(x) sin nx dx + integral_0^pi f(x) sin nx dx = -pi/4 integral_-pi^0 sin nx dx + pi/4 integral_0^pi sin n x dx = pi/2 integral_0^pi sin n x dx = -pi/2 n [cos nx]_0^pi = {pi/n: n is odd 0: n is even