# Fourier Sawtooth

1. Nov 8, 2004

### cj

I can derive the Fourier series for a regular
sawtooth wave.

A different kind of sawtooth is represented by:

$$f(x)=\left\{\begin{array}{cc}-\frac{1}{2}(\pi +x),&\mbox{ if } =-\pi \leq x < 0\\+\frac{1}{2}(\pi -x),& \mbox{ if } 0 < x \leq \pi\end{array}\right.$$

For the life of me I can't figure out how
to derive the series for this, which is:

$$f(x)=\sum_{n=1}^{\infty} sin (nx/n)$$

2. Nov 8, 2004

### Galileo

Probably a typo, but:
$$f(x)=\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}$$

It's obtained the usual way. The function is odd, so all the cosine coefficients are zero.
Now just get:

$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx$$

where $a_n$ is the coefficient of sin(nx)

3. Nov 8, 2004

### cj

Yes, you're right -- there was a typo.

Does the integration breaks down
into

$$a_n=\frac{1}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx + \frac{1}{\pi}\int_{-\pi}^{0}f(x)\sin(nx)dx \text { ??}$$

Also, does an $a_0$ term need to be
determined? I'm not sure when,
or when not, to include an $a_n$.

Thanks a lot.

4. Nov 8, 2004

### Galileo

No. There's an easy way to remember/see it. If n=0, then sin(nx)=0.
So the zeroth coeff. of the sine is always zero.
For the cosine: cos(nx)=1 if n=0.

That's correct, so that's all there's to it.
Both integrals are equal though, since f(x) and sin(nx) are odd, f(x)sin(nx) is even.