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Old 2013-03-20, 21:15   #5
Basketry That Evening!
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"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:
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{<f, \quad sin(kx)>}{<sin(kx), \quad sin(kx)>} (where sin(kx) = \sqrt{\pi}e_k), or re-writing to put it in the form of Parseval's identity, <f, \quad e_k> = \frac{<f, \quad sin(kx)>}{\sqrt{\pi}} = \frac{<f, \quad sin(kx)>}{\sqrt{<sin(kx), \quad sin(kx)>}} = b_k \sqrt{<sin(kx), \quad sin(kx)>} = \sqrt\pi b_k. Thus <f, \quad f> = \sum_{k=1}^{\infty}{|<f, \quad e_k>|^2} = \sum_{k=1}^{\infty}{|\sqrt\pi b_k|^2} = \sum_{k=1}^{\infty}{\pi \frac{4}{k^2}}. Meanwhile, <f, \quad f> = \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)
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