# Fourier series question

I got this question out of a book, but I can't get the book's answer. Since I can't draw, I'll just describe the graph given. Express q(t) as a Fourier series expansion.

The charge q(t) on the plates of a capacitor at time t is shown as a saw-tooth wave with period $2\pi$ and its peak is at $t = \pi$, where q(t) = Q.

At first, I tried to use the expressions for open boundary conditions (as q(t) is 0 at t = 0 and t = 2 pi). I got stuck, so I started again, with

$$q(t) = \frac{a_{0}}{\sqrt{L}} + \sum_{n=1}^\infty a_{n} \left(\frac{2}{L}\right)^\frac{1}{2} \cos \frac{2\pi nt}{L} + \sum_{n=1}^\infty b_{n} \left(\frac{2}{L}\right)^\frac{1}{2} \sin \frac{2\pi nt}{L}$$

where

$$a_{0} = \int_{0}^{L} \left(\frac{1}{L}\right)^\frac{1}{2} q(t) dt$$

$$a_{n} = \int_{0}^{L} \left(\frac{2}{L}\right)^\frac{1}{2} \cos \frac{2\pi nt}{L} q(t) dt$$

$$b_{n} = \int_{0}^{L} \left(\frac{2}{L}\right)^\frac{1}{2} \sin \frac{2\pi nt}{L} q(t) dt$$

Now, since the saw-tooth is symmetric, there was no need to calculate $b_{n}$, since they'd have been 0 anyway.

I represented q(t) using top hat functions:

$$q(t) = \left[\theta(t) - \theta(t - \frac{L}{2})\right]\frac{2tQ}{L} + \left[\theta(t - \frac{L}{2}) - \theta(t - L)\right]\frac{2(L - t)Q}{L}$$

Then I used that to calculate $a_{0}[/tex] and [itex]a_{n}$, using $L = 2\pi$

After pages of working out, I got:

$$a_{0} = Q\pi \left(\frac{1}{2\pi}\right)^\frac{1}{2}$$

$$a_{n} = \left(\frac{1}{\pi}\right)^\frac{1}{2} \frac{Q}{\pi} \frac{2}{n^2} (\cos n\pi - 1)$$

So,

$$q(t) = Q\left[\frac{1}{2} + \frac{2}{\pi^2} \sum_{n=1}^\infty \frac{\cos nt \cos (n\pi - 1)}{n^2}\right]$$

The book's answer is $$q(t) = Q\left[\frac{1}{2} - \frac{4}{\pi^2} \sum_{n=1}^\infty \frac{\cos (2n -1)t}{(2n - 1)^2}\right]$$

I don't know where I've gone wrong. I can't post all my working, as it's a lot and would take forever to LaTeX.

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Nylex said:
So,

$$q(t) = Q\left[\frac{1}{2} + \frac{2}{\pi^2} \sum_{n=1}^\infty \frac{\cos nt \cos (n\pi-1)}{n^2}\right]$$

The book's answer is $$q(t) = Q\left[\frac{1}{2} - \frac{4}{\pi^2} \sum_{n=1}^\infty \frac{\cos (2n -1)t}{(2n - 1)^2}\right]$$

I don't know where I've gone wrong. I can't post all my working, as it's a lot and would take forever to LaTeX.
Well, your results are equivalent I'm poor at LaTeX, but I'll try to explain.

First, there is a typo in a final result (I suspect it is a typo), there should be

$$q(t) = Q\left[\frac{1}{2} + \frac{2}{\pi^2} \sum_{n=1}^\infty \frac{\cos nt (\cos (n\pi)-1)}{n^2}\right]$$

(notice the braces in second cos), but you got it right in a_n expression.

OK, now whenever n is even, expression under sumation is equal to 0 because $$(\cos (n\pi)-1) = 0$$ for even values of n.

So, we have to sum only if n is odd, otherwise written as $$n = 2k-1$$, for k going from 1 to inf.

For odd numbers $$(\cos (n\pi)-1) = -2$$

If you put it all together you can write
$$\sum_{k=1}^\infty \frac{-2\cos (2k-1)t}{(2k-1)^2}\right]$$

you can take the "-2" out of summation, and then you get the book's answer.

Hope my explanation was clear,
Cheers

Atrus said:
Well, your results are equivalent I'm poor at LaTeX, but I'll try to explain.

First, there is a typo in a final result (I suspect it is a typo), there should be

$$q(t) = Q\left[\frac{1}{2} + \frac{2}{\pi^2} \sum_{n=1}^\infty \frac{\cos nt (\cos (n\pi)-1)}{n^2}\right]$$

(notice the braces in second cos), but you got it right in a_n expression.

OK, now whenever n is even, expression under sumation is equal to 0 because $$(\cos (n\pi)-1) = 0$$ for even values of n.

So, we have to sum only if n is odd, otherwise written as $$n = 2k-1$$, for k going from 1 to inf.

For odd numbers $$(\cos (n\pi)-1) = -2$$

If you put it all together you can write
$$\sum_{k=1}^\infty \frac{-2\cos (2k-1)t}{(2k-1)^2}\right]$$

you can take the "-2" out of summation, and then you get the book's answer.

Hope my explanation was clear,
Cheers
Yay, thanks very much. Yes, I did make a typo with the brackets in my post.

Thanks again :D.