Where did I go wrong in computing Fourier Series for f(x)=x^2 on [-\pi,\pi]?

[δοτ]

Homework Statement

Let f(x) = x^2 on [-\pi,\pi]. Computer the Fourier Coefficients of the 2π-periodic extension of f. Use Dirichlet's Theorem to determine where the Fourier Series of f converges. Use the previous two conclusions to show that \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}

Homework Equations

My book (baby Rudin) uses
c_n = \frac{1}{2\pi} \int_{-\pi}^\pi f(x) e^{-inx} dx
as the equation for the coefficients of the series.

The Attempt at a Solution

My main problem is the last part, but I'm sure it's rooted in another piece of the puzzle. I've found that
c_n = \frac{2(-1)^n}{n^2}
for n ≠ 0 via Maple and
c_0 = \frac{\pi^2}{3}
by computation.

My book doesn't actually have anything I can see called "Dirichlet's Theorem", but I found a paper that said if f(x) was continuous then the series would converge to that point, otherwise it would converge to the midpoint of the jump discontinuity. Is this right?
If so, then the series ought to converge at every point, even the endpoints, because x^2 is continuous and even. So then we'd have, because \exp(i n \pi) = (-1)^n, that'd cancel the (-1)^n in the series and thus given the equation
\frac{\pi^2}{3} + 2\sum_{n=1}^\infty \frac{1}{n^2} = \pi^2 = f(\pi)
but that doesn't work out too well.Where'd I go wrong?

This is the first of four almost identical problems, but if I can do one I can do them all. Thanks for any input.

 

[Dick]

Homework Statement

Let f(x) = x^2 on [-\pi,\pi]. Computer the Fourier Coefficients of the 2π-periodic extension of f. Use Dirichlet's Theorem to determine where the Fourier Series of f converges. Use the previous two conclusions to show that \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}

Homework Equations

My book (baby Rudin) uses
c_n = \frac{1}{2\pi} \int_{-\pi}^\pi f(x) e^{-inx} dx
as the equation for the coefficients of the series.

The Attempt at a Solution

My main problem is the last part, but I'm sure it's rooted in another piece of the puzzle. I've found that
c_n = \frac{2(-1)^n}{n^2}
for n ≠ 0 via Maple and
c_0 = \frac{\pi^2}{3}
by computation.

My book doesn't actually have anything I can see called "Dirichlet's Theorem", but I found a paper that said if f(x) was continuous then the series would converge to that point, otherwise it would converge to the midpoint of the jump discontinuity. Is this right?
If so, then the series ought to converge at every point, even the endpoints, because x^2 is continuous and even. So then we'd have, because \exp(i n \pi) = (-1)^n, that'd cancel the (-1)^n in the series and thus given the equation
\frac{\pi^2}{3} + 2\sum_{n=1}^\infty \frac{1}{n^2} = \pi^2 = f(\pi)
but that doesn't work out too well.


Where'd I go wrong?

This is the first of four almost identical problems, but if I can do one I can do them all. Thanks for any input.

Check your value for c0. I think it's wrong.

 

[δοτ]

I'm positive it's correct.
\frac{1}{2\pi} \left(\frac{\pi^3}{3} - \frac{(-\pi)^3}{3}\right) = \frac{1}{2\pi} * \frac{2\pi^3}{3} = \frac{\pi^2}{3}

 

[Dick]

I'm positive it's correct.
\frac{1}{2\pi} \left(\frac{\pi^3}{3} - \frac{(-\pi)^3}{3}\right) = \frac{1}{2\pi} * \frac{2\pi^3}{3} = \frac{\pi^2}{3}

Oh yeah, you are right. Now I'm thinking you forgot to sum over the negative values of n as well. How about that?

 

[δοτ]

Yea, I think that's totally the problem. In the book they kind of glossed over that the negatives disappeared at some point and a coefficient of 2 popped out front. But the negative sums would be the same as the positive sums here because c_n is even.

Thanks for guiding me to my error.