Deriving Fourier Series for a Regular Sawtooth Wave

[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),&amp;\mbox{ if }<br /> =-\pi \leq x &lt; 0\\+\frac{1}{2}(\pi -x),&amp; \mbox{ if } 0 &lt; 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)

 

[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)

 

[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.

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)

 

[Galileo]

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

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.

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 { ??}

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.