Discovering the Fourier Series of a 2pi Periodic Function | Homework Help

[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)