# Find the Fourier series for $f(x) := |\sin(x)|$ and the sum of $\sum_{n=1}^{\infty} \frac {(-1)^{n+1}} {4n^2-1}$.

Find the Fourier series for $f(x) := |\sin(x)|$ and the sum of $\sum_{n=1}^{\infty} \frac {(-1)^{n+1}} {4n^2-1}$.

I have computed $$c_n = \frac 1 {2\pi} \int^{\pi}_{-\pi} |\sin(x)|e^{-inx} dx = \frac 1 {2\pi} \int^{\pi}_{0} |\sin(x)|e^{-inx} + |\sin(-x)|e^{-in(-x)} dx = \frac {1} {\pi} \frac {1} {n^2-1} ((-1)^{n+1} -1)$$

However, this doesn't correspond that well with the sum of the series I must find. What have I done wrong ?

• A good way for you and us to find your mistake (if there is one) is to completely write your calculation. – user37238 May 20 '14 at 8:09
• Did you think about the substitution $m=2n$? – Fabian May 20 '14 at 8:22
• try to find $x$ s.t all the terms beside those on the even spots will become 0. – Snufsan May 20 '14 at 8:35
• Similar question here math.stackexchange.com/questions/467787/… – Graham Hesketh May 20 '14 at 16:08

The exponential Fourier series for this particular $f(x)$ is \begin{align} |\sin(ax)| = \sum_{n= -\infty}^{\infty} c_{n} e^{-i nx}, \end{align} where \begin{align} c_{n} = \frac{1}{2\pi} \int_{-\pi}^{\pi} |\sin(ax)| \ e^{-i nx} dx. \end{align} This integral can be calculated as follows. \begin{align} c_{n} &= \frac{1}{2\pi} \int_{-\pi}^{\pi} |\sin(ax)| \ e^{-i nx} dx \\ &= \frac{1}{2\pi} \int_{-\pi}^{0} |\sin(ax)| \ e^{-i nx} dx + \frac{1}{2\pi} \int_{0}^{\pi} |\sin(ax)| \ e^{-i nx} dx \\ &= \frac{1}{2\pi} \int_{0}^{\pi} |\sin(-ax)| \ e^{i nx} dx + \frac{1}{2\pi} \int_{0}^{\pi} |\sin(ax)| \ e^{-i nx} dx \\ &= \frac{1}{\pi} \int_{0}^{\pi} \sin(ax) \cos(nx) dx \\ &= \frac{1}{2\pi} \int_{0}^{\pi} \left[ \sin(a+n)x + \sin(a-n)x \right] dx\\ &= \frac{1}{2\pi} \left[ \frac{2a}{a^{2} - n^{2}} - \frac{\cos(a\pi + n \pi)}{a+n} - \frac{\cos(a\pi - n \pi)}{a-n} \right] \\ c_{n} &= \frac{a}{\pi (a^{2} - n^{2})} \left[ 1 - (-1)^{n} \cos(a\pi) \right]. \end{align} From this the Fourier series is \begin{align} |\sin(ax)| = \frac{a}{\pi} \sum_{-\infty}^{\infty} \frac{(1 - (-1)^{n} \cos(a\pi))} {\pi (a^{2} - n^{2})} \ e^{-i nx}. \end{align} The summation can be changed and is seen in the following. \begin{align} \sum_{-\infty}^{\infty} b_{n} e^{-i nx} &= \sum_{-\infty}^{-1} b_{n} e^{-i nx} + b_{0} + \sum_{1}^{\infty} b_{n} e^{-i nx} \\ &= b_{0} + \sum_{n=0}^{\infty} \left( b_{-n} e^{i nx} + b_{n} e^{-i nx} \right). \end{align} From this it is seen that \begin{align} |\sin(ax)| = \frac{1 - \cos(a\pi)}{a \pi} - \frac{2a}{\pi} \sum_{n=0}^{\infty} \frac{(1-(-1)^{n} \cos(a\pi))}{n^{2}-a^{2}} \ \cos(nx) \end{align} or \begin{align} |\sin(ax)| = \frac{\sin^{2}\left(\frac{a\pi}{2}\right)}{2 a \pi} - \frac{2a}{\pi} \sum_{n=0}^{\infty} \frac{(1-(-1)^{n} \cos(a\pi))}{n^{2}-a^{2}} \ \cos(nx). \end{align}
Now, when $a=1$ this reduces to \begin{align} |\sin(x)| &= \frac{2}{\pi} - \frac{2}{\pi} \sum_{n=0}^{\infty} \frac{1+(-1)^{n}}{n^{2}-1} \ \cos(nx) \\ &= \frac{2}{\pi} - \frac{4}{\pi} \sum_{n=0}^{\infty} \frac{\cos(2nx)}{4n^{2}-1}. \end{align} In order to evaluate the series in question let $x = \pi/2$ for which the Fourier series is reduced to \begin{align} |\sin(\pi/2)| &= \frac{2}{\pi} - \frac{4}{\pi} \sum_{n=0}^{\infty} \frac{\cos(n \pi)}{4n^{2}-1} \\ &= \frac{2}{\pi} + \frac{4}{\pi} \sum_{n=0}^{\infty} \frac{(-1)^{n-1}}{4n^{2}-1}. \end{align} which leads to \begin{align} \sum_{n=0}^{\infty} \frac{(-1)^{n-1}}{4n^{2}-1} = \frac{\pi - 2}{4}. \end{align}