What is the Fourier Series for (sin(x))^2 on the interval [-π, π]?

[Hertz]

Hey there!
I'm trying to calculate the Fourier Series for sin²x on [-π, π]

For a₀ I found 1/2. (By determining the average value of the function on the interval)

Since sin²x is even, I know that b_n = 0.
Now, for a_n.. The following link shows the integral I used to try to evaluate a_n.
http://www.wolframalpha.com/...

Since a_n and b_n are both coming out to be zero, doesn't this imply that sin²x = a₀ = 1/2? Will someone please show me where I am going wrong? Can't find anything online about this. Everyone just always uses the sin² identity and calls it good.

 

[lurflurf]

hint: Recall that
(sin(x))^2=(1/2) (1-cos(2 x))

as for the integral we have

$$a_2=\frac{1}{\pi} \int _{- \pi}^\pi \! \sin^2(x) \cos(2x) \, \mathrm{dx}=\frac{1}{2\pi} \int_{-\pi}^\pi \! (1-\cos(2 x)) \cos(2x) \, \mathrm{dx}$$

 

[Hertz]

What can I say... My mind is blown...

When you take the integral of sin²(x)cos(nx) you get -\frac{4sin(πn)}{n^3-4n}

It is easy to assume that this is equal to zero for all values of n, but in hindsight it is clear that this does not apply for n = 2. :S!

I'm utterly confused now. Just when I thought I was beginning to understand how to compute Fourier Series...

 

[mathman]

What can I say... My mind is blown...

When you take the integral of sin²(x)cos(nx) you get -\frac{4sin(πn)}{n^3-4n}

It is easy to assume that this is equal to zero for all values of n, but in hindsight it is clear that this does not apply for n = 2. :S!

I'm utterly confused now. Just when I thought I was beginning to understand how to compute Fourier Series...

For n = 2, use L'Hopital's rule. I got -2π.

One could also go to basics, using cos2x = cos²x-sin²x
Therefore sin²x = 1/2(1 - cos2x)

 

[vanhees71]

It's clear that a_0=1 and a_1=-1/2, because as said in the previous posting from the double-angle theorem you get
\sin^2 x=\frac{1}{2}[1-\cos(2 x)].
For the Fourier coefficients you indeed have
a_n=\frac{1}{\pi} \int_{-\pi}^{\pi} \mathrm{d} x \sin^2 x \cos(n x)=\frac{4}{\pi(4n-n^3)} \sin (n \pi).
This is indeed 0 for all n \in \mathbb{N} \setminus \{2 \}, and for n=2 you get by direct evaluation of the integral a_2=-1/2.

Of course you can also take the limit n \rightarrow 2 by using de L'Hospital's rule, because it's a limit of the indefinite form 0/0:
\lim_{n \rightarrow 2} a_n=\lim_{n \rightarrow 2} \frac{4}{\pi} \frac{n \pi \cos(n \pi)}{4-3n^2}=-\frac{1}{2}.