# Fouiers series Calculate sum

by dikmikkel
Tags: fouier series, fouiers law, infinite series
 P: 169 1. The problem statement, all variables and given/known data Consider the 2$\pi$-periodic function f(t) = t t in [-Pi;Pi] a) show that the real fouier series for f(t) is: $f(t) ~ \sum\limits_{n=1}^{\infty}\frac{2}{n}(-1)^{n+1}\sin nt$ b) Use the answer to evaluate the following : $\sum\limits_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{2n-1}$ Hint: Use Fouier's law with t = $\pi$/2 2. Relevant equations Fouiers Law? I'm danish, and therefore i'm not really sure what it's called. 3. The attempt at a solution Part a i have done by finding the coefficients. Part b) I can't see where the problem in part b and the answer to a relates. I've tried with Maple 15 to calculate the value and i'm getting Pi/4, but i keep getting something different for the series from a) Please Help me, i would really like to understand this as i'm studying physics.
 P: 6 The answer should be pi/4 ... from fourier series we got ... t = $\sum\frac{2}{n}(-1)n+1sinnt$ for t = $\frac{\pi}{2}$ , $\frac{\pi}{2}$ = 2 - 2/3 + 2/5 - 2/9 + ...... which followed , $\frac{\pi}{2}$ = 2 * $\sum\frac{(-1)n+1}{2n - 1}$ hence, $\sum\frac{(-1)n+1}{2n - 1}$ = $\frac{\pi}{4}$
 P: 169 Hmm, isn't that a backwards proof? Can't i use something like the even terms of sin(n*Pi/2) = 0 and the odd ones alternates between -1 and 1 And how is t = ... isn't it f(t)?

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