# Fourier series of a function

I'm trying to find Fourier series for the following function:
$$f(x) = \begin{cases}1, & \mbox{if x \in (-\frac{\pi}{2}+2\pi n,\frac{\pi}{2}+2\pi n) } \\ -1, & \mbox{if x \in [\frac{\pi}{2}+2\pi n,\frac{3\pi}{2} + 2\pi n]} \end{cases}$$

This is how I calculated a_n and b_n:

So I got the following series: $$\sum_{n=1}^{\infty} \cos{nx}\frac{4(-1)^{n+1}}{\pi(2n-1)}$$

But when I checked if it converges to f(x) at point \pi I get that it diverges, however all requirements of Fourier theorem are met.

What am I doing wrong?

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LCKurtz
Homework Helper
Gold Member
I'm trying to find Fourier series for the following function:
$$f(x) = \begin{cases}1, & \mbox{if x \in (-\frac{\pi}{2}+2\pi n,\frac{\pi}{2}+2\pi n) } \\ -1, & \mbox{if x \in [\frac{\pi}{2}+2\pi n,\frac{3\pi}{2} + 2\pi n]} \end{cases}$$

This is how I calculated a_n and b_n:

So I got the following series: $$\sum_{n=1}^{\infty} \cos{nx}\frac{4(-1)^{n+1}}{\pi(2n-1)}$$

But when I checked if it converges to f(x) at point \pi I get that it diverges, however all requirements of Fourier theorem are met.

What am I doing wrong?

The cos(nx) term in the sum should be cos((2n-1)x) since you only pick up the odd subscripts. Otherwise it is correct. You could have saved half the work by noting that since the function is period 2pi, you could have used the formula on (-pi,pi) and observed the function is even. That's why the bn are 0. And using the half range integral for an would have made less work there too.

Thanks, indeed it saved me a lot of work!