Fourier series

[hb9112]

(a) Compute the Fourier coefficient associated with
[MATH]f(t) =\sum_{n\in \mathbb{Z}} g(t-2n\pi)[/MATH], where [MATH]g(x) = x^2 \mathbb{1}_{[-\pi,\pi]}(x)[/MATH].

b) Use the convergence theorem and show that
[MATH]\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{8}[/MATH]
I did a) and got that
[MATH]a_0 = \frac{\pi^2}{3}[/MATH] and [MATH]a_n = \frac{4\cdot (-1)^n}{n^2}[/MATH].

I am not quite sure about b).

We have that [MATH]x^2 =\frac{\pi^2}{3} +\sum_{n=1}^{\infty} \frac{1}{n^2}cos(nx)[/MATH]. I started with setting [MATH]x = \pi[/MATH], which gives:
[MATH]\frac{\pi^2}{6} = \sum_{n=1}^{\infty} \frac{1}{n^2}[/MATH].
I don't really know where to go from here. I can see that the lower limit of the sum needs to be changed, but then I get [MATH]\sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{n=0}^{\infty} \frac{1}{(n+1)^2}[/MATH]Thanks.

 

[ksdhart2]

I don't understand your notation. What does \(\displaystyle \displaystyle g(x) = x^2 \mathbb{1}_{[-\pi,\pi]}(x)\) mean? Can you explain it in words, perhaps?

 

[LCKurtz]

I think he is using [MATH]1_{[-\pi,\pi]}[/MATH] for the characteristic function. His sum is just a complicated way of saying the periodic extension of [MATH]x^2[/MATH] on [MATH][-\pi,\pi][/MATH].

 

[ksdhart2]

I think he is using [MATH]1_{[-\pi,\pi]}[/MATH] for the characteristic function. His sum is just a complicated way of saying the periodic extension of [MATH]x^2[/MATH] on [MATH][-\pi,\pi][/MATH].

Ah, I see. That does make sense, but I was simply unfamiliar with that particular way of writing it.

(a) Compute the Fourier coefficient associated with
[MATH]f(t) =\sum_{n\in \mathbb{Z}} g(t-2n\pi)[/MATH], where [MATH]g(x) = x^2 \mathbb{1}_{[-\pi,\pi]}(x)[/MATH].

[...]

I did a) and got that
[MATH]a_0 = \frac{\pi^2}{3}[/MATH] and [MATH]a_n = \frac{4\cdot (-1)^n}{n^2}[/MATH].

Assuming that LCKurtz is correct about your notation, then your \(a_0\) is wrong. It should be:

\(\displaystyle a_0 = \frac{1}{\pi} \int\limits_{-\pi}^{\pi} g(x) \: dx = \frac{1}{\pi} \int\limits_{-\pi}^{\pi} x^2 \: dx = \frac{2\pi^2}{3}\)

Your \(a_n\) is correct though. You also didn't specify what you found for \(b_n\) but I can only presume you correctly deduced that \(b_n = 0\) for all integer \(n\).

We have that [MATH]x^2 =\frac{\pi^2}{3} +\sum_{n=1}^{\infty} \frac{1}{n^2}cos(nx)[/MATH]. I started with setting [MATH]x = \pi[/MATH], which gives:
[MATH]\frac{\pi^2}{6} = \sum_{n=1}^{\infty} \frac{1}{n^2}[/MATH].

This part is also not correct. You tried to follow the usual formula of:

\(\displaystyle g(x) = \frac{a_0}{2} + \sum\limits_{n=1}^{\infty} a_n \cos(nx) + \sum\limits_{n=1}^{\infty} b_n \sin(nx)\)

Plugging in the requisite values would give:

\(\displaystyle g(x) = \frac{\pi^2}{3} + \sum\limits_{n=1}^{\infty} \frac{4 \cdot (-1)^n}{n^2} \cos(nx)\)

But where did the 4 and the \((-1)^n\) disappear to in your workings? If you account for those, you should notice that

\(\displaystyle g(x) = \frac{\pi^2}{3} + 4 \left( \sum\limits_{\text{n is even}} \frac{1}{n^2} \cos(nx) - \sum\limits_{\text{n is odd}} \frac{1}{n^2} \cos(nx) \right)\)

Where does that lead you?