Fourier series of t on (-π, π): where does the n² come from?

[thomas49th]

Homework Statement

A 2pi peroidic function f is defined in the interval (-pi, pi) by f = t. Sketch the graph of the function and show that it's Fourier series is given by

$$\frac{\pi^{2}}{3} + 4\sum^{\infty}_{n=1}\frac{(-1)^{n} \cos(nt)}{n^{2}}$$

Homework Equations

The Attempt at a Solution

Well if you draw the function you can see that it's odd therefore $$a_{n} = 0 a_{0} = 0$$

$$b_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} t\sin(nt)dt$$

This has to be done by parts

giving $$\frac{1}{\pi}[\frac{t-\cos(nt)}{n} + \frac{1}{n}\int \cos(nt) dt$$

We can ignore the latter term as our limits are pi and the integral of cosine is sine and sine of any multiple of pi is 0. This means

$$b_{n} = \frac{1}{n\pi}[t.-\cos(nt)]^{\pi}_{-\pi}$$

After plugging in the limits I find this to eqal -2pi cos(npi) which is -2pi(-1)^n

Not what it's meant to equal :(

I don't see where the n^2 comes from in the original question nor the DC value. Actually I don't see where there answer comes from altogether.

Any suggestions

Thanks
Thomas

 

[thomas49th]

apparently the question set was incorrect... After being given the correct question I solved it (f(t) = t^2)