# Fourier Series Problem

1. Aug 26, 2012

### Ratpigeon

1. The problem statement, all variables and given/known data
Use the Fourier series of
f(x) = { 1 |x|<a
{ 0 a<|x|<$\pi$
for 0<a<$\pi$
extended as a 2-Pi periodic function for x $\in$R
to find
$\sum$ Sin2(na)/n2
2. Relevant equations[/b

I got that the fourier series of f(x) was
a/$\pi$+$\sum$ (2/(m$\pi$) sin(ma) sin(mx)

3. The attempt at a solution
I'm not sure what to do. I'm guessing that since its meant to use the Fourier series of f; it should be related to f(a), but f(a) is undefined, and the denominator is wrong...
Could anyone please point me in the right direction?

2. Aug 26, 2012

### vela

Staff Emeritus
Your f(x) is an even function, so you should have cosines instead of sines:
$$f(x) = \frac{a}{\pi} + \sum_{m=1}^\infty \frac{2}{m\pi} \sin ma \cos mx.$$ Now if you look at the sum you're trying to evaluate, you need another power of m on the bottom and you want to turn the cosine into a sine. Any idea of how to get that?

3. Aug 26, 2012

### gabbagabbahey

The formulas you are likely using for the Fourier Series of a function $f(x)$ require the function to be piecewise smooth and periodic over the Reals, as well as being defined at each discontinuity as equal to the mean value of the one-sided limits, so you require

$$f(x=\pm a) = \frac{1}{2} \left( \lim_{x \to \pm a^+}f(x) + \lim_{x \to \pm a^-}f(x) \right) = \frac{1}{2}$$

4. Aug 28, 2012

### Ratpigeon

I integrate it from zero to a to get the extra sine term?

5. Aug 28, 2012

### vela

Staff Emeritus
Sounds good. What do you get?

6. Aug 28, 2012

### Ratpigeon

a Pi/2. Thanks for the help :)

7. Aug 28, 2012

### uart

Ok, you're on the right track but not quite there yet. I think there's another term.

8. Aug 28, 2012

### Ratpigeon

I don't think there's another term - the 1/2 A0 term goes to zero when you integrate it from
-a to a.

9. Aug 28, 2012

### uart

No actually it doesn't.

BTW. Previously you said you would integrate from zero to pi? Either way will work, but zero to pi is easier.

10. Aug 28, 2012

### Ratpigeon

Integrating from zero to a got that it was Pi/2 (a+a^2/Pi); is that right?
Thanks for pointing it out

11. Aug 28, 2012

### uart

12. Aug 28, 2012

### Ratpigeon

Minus, sorry, thanks - I need sleep...

13. Aug 28, 2012

### uart

Yep, minus.

And as a nice little "sanity" check, notice that the sum is now zero when $a = \pi$.

14. Aug 28, 2012

Thank you :)