# Determining the Fourier Series and its convergence

Let $ƒ$ denote the function with period 2π, in the range $0≤x≤2π$. The function is given as,

$$f(x)=\begin{cases} x,\quad \mbox{for}\quad 0<x<π \\ 0\quad \mbox{for}\quad π<x<2π \end{cases}$$

I) Determine the fourier series for the given function.

II) Does fourier series converges for all $x\in\mathbb{R}$? And if so, what does it converges towards?

My Approach So I have almost solved the first part. I used the following complex method formula to determine the fourierseries.

$$\sum _{ n=-\infty }^{ \infty }{ c_{ n }\cdot e^{ inx } }$$

Where I found the fourier coefficient using the following equation, $$c_{ n }=\frac { 1 }{ 2π } \cdot \int _{ -π }^{ π }{ f(x)·e^{ -inx }dx }$$

And I have come to the following expression for the fourier series,

$$f(x)=\sum _{ n=-\infty }^{ \infty }{ \left( \frac { \left( -1 \right) ^{ n }i }{ n } \right) } ·e^{inx}$$

But I cannot go any further to get the correct real fourier series. Because I have tried using the following method.

$$f(x)=\sum _{ n=-\infty }^{ \infty }{ \left( \frac { \left( -1 \right) ^{ n }i }{ n } \right) } ·(\cos(nx)+i\sin(nx))$$

$$f(x)=\sum _{ n=-\infty }^{ \infty }{ \left( \frac { \left( -1 \right) ^{ n }i }{ n } \right) \cdot \cos(nx) + \left( \frac { \left( -1 \right) ^{ n }i }{ n } \right) \cdot i\sin(nx)}$$

$$f(x)=\sum _{ n=-\infty }^{ \infty }{ \left( \frac { \left( -1 \right) ^{ n }i }{ n } \right) \cdot \cos(nx) - \left( \frac { \left( -1 \right) ^{ n } }{ n } \right) \cdot \sin(nx)}$$

Because I want the real so I removed the $\cos(nx)$ and ended up with the following, $$f(x)=\sum _{ n=-\infty }^{ \infty }{ - \left( \frac { \left( -1 \right) ^{ n } }{ n } \right) \cdot \sin(nx)}$$

I have no clue what I am doing wrong and I have no clue how I can bring a 2 into the equation. Any help on this will be great.

For II, I can not come up with a method to find the convergence. Any hint on that will be great. Thank you.

• 1) Did you mean $x \texttt{for} 0 \leq x \leq \pi$? 2) If you are looking for a factor of two couldn't you look for it with the even symmetry in $n$ for your expression? 3) Could you find the convergence of the series as the real part of the sum of exponentials? You could also use an integration/derivative trick to get rid of the $n$ in the denominator. – Kitter Catter Nov 10 '16 at 22:49
• 1) Yes. 2) Yes I can try to do that. Thanks 3) ah ok thanks. I will give that a try as well. – MathCurious314 Nov 10 '16 at 23:09
• @MathCurious314 Wolfram alpha gives a different expression for your coefficients $c_n$. Note that you have to integrate from $0$ to $\pi$. – A.Γ. Nov 10 '16 at 23:34

Suppose $f$ is a piecewise function in the interval $(-L, L)$. The Fourier series of $f$ will converge to:

• The periodic extension of $f$ if the periodic extension is continuous

• The average of the two one-sided limits if the periodic extension has a jump discontinuity at $x_0$ $$\frac{1}{2}\left[f(x_0^+) + f(x_0^-)\right] = \frac{1}{2}\lim_{x\to x_0^+}f(x) + \frac{1}{2}\lim_{x\to x_0^-}f(x)$$

Now the first question. There are basically two ways you can periodically extend $f$, you can make it either even or odd.

ODD PERIODIC EXTENSION

Let us define

$$f_{\rm odd}(x)=\begin{cases} 0\quad \mbox{for}\quad -2\pi < x < -\pi \\ x\quad \mbox{for}\quad -\pi<x<\pi \\ 0\quad \mbox{for}\quad \pi< x <2\pi \end{cases}$$

Note that $f(x) = f_{\rm odd}(x)$ for $0<x<2\pi$ which is actually the region you care about. Now we calculate the Fourier coefficients, it is important to realize that the period of the function $f_{\rm odd}$ is $4\pi$

$$b_n = \frac{1}{2\pi}\int_{-2\pi}^{2\pi}dx\; f_{\rm odd}(x) \sin nx/2 = \frac{4 \sin n\pi/2-2 \pi n \cos n\pi/2}{\pi n^2}$$

so that

$$f_{\rm odd}(x) = \sum_{n=1}^{+\infty}b_n\sin nx/2$$

EVEN PERIODIC EXTENSION

For this case $$f_{\rm even}(x)=\begin{cases} 0\quad \mbox{for}\quad -2\pi < x < -\pi \\ |x|\quad \mbox{for}\quad -\pi<x<\pi \\ 0\quad \mbox{for}\quad \pi< x <2\pi \end{cases}$$

and the coefficients are

$$a_n = \frac{1}{2\pi}\int_{-2\pi}^{2\pi}dx\; f_{\rm even}(x) \cos nx/2 = \begin{cases} 2-4/\pi \quad \mbox{for}\quad n = 0 \\ 2 (\pi n \sin n\pi/2+2 \cos n\pi/2-2)/(\pi n^2) \quad \mbox{for}\quad n > 0\end{cases}$$

And then

$$f_{\rm even}(x) = a_0 + \sum_{n=1}^{+\infty}a_n\cos nx/2$$