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 \inR
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 :)