Finding the Sum of Sin2(na)/n2 Using Fourier Series for f(x)

[Ratpigeon]

Homework Statement

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. Homework Equations [/b

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

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?

 

[vela]

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?

 

[gabbagabbahey]

it should be related to f(a), but f(a) is undefined

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

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

 

[Ratpigeon]

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

 

[vela]

Sounds good. What do you get?

 

[Ratpigeon]

a Pi/2. Thanks for the help :)

 

[uart]

a Pi/2. Thanks for the help :)

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

 

[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.

 

[uart]

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

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.

 

[Ratpigeon]

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

 

[uart]

Pi/2 (a+a^2/Pi); is that right?

Almost. Check your "signs".

 

[Ratpigeon]

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

 

[uart]

Yep, minus.

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

 

[Ratpigeon]

Thank you :)