View Single Post
2013-03-20, 21:15   #5
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

722110 Posts

I can't say much on this particular solution (I haven't had analysis yet), but I can say that another method I've seen to calculate this series (and which generalizes well to larger even n, I believe) is a combination of Fourier analysis/linear algebra/Parseval's identity (the latter is essentially the Pythagorean theorem for an arbitrary inner product space), as well as being quite accessible to undergrads with little analysis.

The following were questions on a lin. alg. quiz I took in high school:
Quote:
 Originally Posted by Dr. Fogel 2) Compute the Fourier coefficients (for the interval [-$\pi$, $\pi$]) of sines and cosines for the function f(x) = x. .... 4) Use Parseval's identity on the result from problem 2 to obtain an interesting result.
The answer to 2) is easily calculated to be 0 cosine coeffiecients, and the sine coefficients are $b_k = \frac{-2(-1)^k}{k}$.

For problem 4), we apply Parseval's identity with our orthonormal basis $\{e_k\}$ being $e_k = \frac{sin(kx)}{\sqrt \pi}$. We also note that $b_k = \frac{}{}$ (where $sin(kx) = \sqrt{\pi}e_k$), or re-writing to put it in the form of Parseval's identity, $ = \frac{}{\sqrt{\pi}} = \frac{}{\sqrt{}} = b_k \sqrt{} = \sqrt\pi b_k$. Thus $ = \sum_{k=1}^{\infty}{||^2} = \sum_{k=1}^{\infty}{|\sqrt\pi b_k|^2} = \sum_{k=1}^{\infty}{\pi \frac{4}{k^2}}$. Meanwhile, $ = \int_{-\pi}^{\pi}{x^2dx} = \frac{2\pi^3}{3}$, so we have $\frac{2\pi^3}{3} = 4\pi \sum_{k=1}^{\infty}{\frac{1}{k^2}}$, or $\sum_{k=1}^{\infty}{\frac{1}{k^2}} = \frac{\pi^2}{6}$.

The only squishy part (AFAICT) is proving Parseval's identity (esp. in the infinite dimensional case). We can also see why we get only even powers -- because of the squaring of the fourier coefficients. It ought to be easy (though I haven't really thought about it) to use f(x)=x^2, x^3, ... to get $\zeta(4), \quad \zeta(6)...$. (The only part that I haven't confirmed is that the Fourier coefficients take the form I think they take for the higher order f's.)

Last fiddled with by Dubslow on 2013-03-20 at 21:20 Reason: (esp. in the infinite dimensional case)