0

One is asked to determine the Fourier series of the function

$$ f(x)= \left\{\matrix{ 0 & \hbox{(for $-\pi\le x<0$)} \cr x & \hbox{(for $0\le x<\pi $)} }\right. $$ where $f(x+2\pi)$ = $f(x)$. Hence calculate the value of the infinite sum

$$\sum_{n=1}^\infty\frac{1}{(2n-1)^2}$$

For the coefficients, $A_0$ = $\frac{1}{2\pi}$$\int_{-\pi}^{+\pi}f(x)dx$ = $\frac{1}{2\pi}$[$\int_{-\pi}^0$$f(x)dx$ + $\int_0^{\pi}$$f(x)dx$]

$A_n$ = $\frac{1}{\pi}$$\int_{-\pi}^{+\pi}$$f(x)cos\frac{nx\pi}{\pi}dx$

I know that need to plug in x=0 to f(x) but can not get the fourier series f,from this kinda stuck how to proceed any help will be appriciated

Gevi
  • 15
  • What is your expression for the Fourier series of $f$? –  Apr 14 '16 at 23:50
  • Consider that $$ f(x) = \int\limits_{ - ,\pi }^x {\left( {0: - ,\pi \le t < 0,;1:0 \le t < \pi } \right)dt} $$ – G Cab Apr 14 '16 at 23:58
  • (I couldn't resist converting/cleaning-up the question to a form that I see as quasi-iconic, so, both cliched-boring and important. :) If there's a problem with my choices, please do revert...) – paul garrett Apr 15 '16 at 00:15
  • 1
    Apologies, I am new to using Mathjax formatting tool, forgot to write up f,I am currently writing up Fourier series of f with Mathjax tool, taking some time to actually write it up I will upload f as soon as I can, apologies. – Gevi Apr 15 '16 at 00:17

1 Answers1

0

We presume the following form for the Fourier series of $f$: $$\frac{a_0}{2} + \sum_{n=1}^{\infty}a_n \cos(nx) + \sum_{n=1}^{\infty}b_n \sin(nx)$$ where $$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x) \cos(nx) dx$$ We intend to evaluate the Fourier series only at $x=0$, so we do not care about the $b_n$ coefficients because $\sin(nx) = 0$ when $x=0$. As $$f(x) = \begin{cases} 0 & \text{if } -\pi \leq x < 0 \\ x & \text{if } 0 \leq x < \pi \\ \end{cases}$$ we have $$a_n = \frac{1}{\pi}\int_0^{\pi}x\cos(nx) dx$$ For $n = 0$, this becomes $$a_0 = \frac{1}{\pi}\int_0^{\pi}x dx = \frac{1}{2\pi}(\pi^2 - 0) = \frac{\pi}{2}$$ For $n \neq 0$, we can use integration by parts, with $u = x$ and $dv = \cos(nx) dx$: $$\begin{aligned} \pi a_n &= \left.\frac{1}{n}x\sin(nx)\right|_{0}^{\pi} - \frac{1}{n}\int_0^{\pi}\sin(nx) dx \\ &= \frac{\pi}{n} \sin(n\pi) + \frac{1}{n^2}(\cos(n\pi) - 1) \\ &= \frac{1}{n^2}(\cos(n\pi) - 1) \end{aligned}$$ where we have used that $\sin(n\pi) = 0$ when $n$ is an integer.

For even $n$, we have $\cos(n\pi) = 1$, so $$a_n = \frac{1}{\pi n^2}(1 - 1) = 0$$ For odd $n$, we have $\cos(n\pi) = -1$, so $$a_n = \frac{1}{\pi n^2}(-1 - 1) = -\frac{2}{\pi n^2}$$ This means that the Fourier series for $f$ is (ignoring the sine terms) $$\begin{aligned} \frac{a_0}{2} + \sum_{n=1}^{\infty}a_n \cos(nx) &= \frac{\pi}{4} + \sum_{m=1}^{\infty}a_{2m-1} \cos((2m-1)x) \\ &= \frac{\pi}{4} - \sum_{m=1}^{\infty}\frac{2}{\pi(2m-1)^2}\cos((2m-1)x) \\ \end{aligned}$$ Plugging in $x=0$, we have $\cos((2m-1)0) = 1$ for all $m$, so we end up with $$0 = f(0) = \frac{\pi}{4} - \sum_{m=1}^{\infty}\frac{2}{\pi(2m-1)^2}$$ assuming we can justify that the Fourier series converges to the original function at $x=0$ (e.g. this is true because $f$ is continuous and has bounded variation). This allows us to conclude that $$\sum_{m=1}^{\infty}\frac{2}{(2m-1)^2} = \frac{\pi^2}{4}$$