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April 23

Pi and Even Zeta Constants

Why are all even zeta constants related to Pi ? — 79.113.230.93 (talk) 01:05, 23 April 2013 (UTC)[reply]

Have a look at Basel problem and how the squares could be split into two factors. It is just an indication but hopefully it will explain why for you. Dmcq (talk) 10:59, 23 April 2013 (UTC)[reply]

I did... before posting the question, of course... I just don't really see how it can be generalized... :-\ — 79.113.213.225 (talk) 04:02, 24 April 2013 (UTC)[reply]

Yeah, I guess you're right, it doesn't come out. Um I better extricate myself by showing I was right all along and it is obvious ;-) There's a couple of ways but going back to Euler's rather non-rigorous reasoning you can stick in

ix

for

x

to get

{\frac {\sin(ix)}{ix}}=\left(1+{\frac {x^{2}}{\pi ^{2}}}\right)\left(1+{\frac {x^{2}}{4\pi ^{2}}}\right)\left(1+{\frac {x^{2}}{9\pi ^{2}}}\right)\cdots

which can be multiplied by the original to get the fourth powers as the first term after 1, and in fact if you multiply out the

{\frac {\sin(x)\sin((ix)}{ix^{2}}}

you get

1-{\frac {x^{4}}{90}}+\cdots

and equating the term with the fourth power as in the original you get

\sum _{n=1}^{\infty }{\frac {1}{n^{4}}}={\frac {\pi ^{4}}{90}}

.

Putting in the nth roots of -1 and multiplying n terms like that should give the 2nth value of the zeta function and we'll end up with a rational fraction with the roots gone as they are used symmetrically, and a power of pi on the other. Hope that reasoning is actually all right. Dmcq (talk) 08:58, 24 April 2013 (UTC)[reply]

Euler's rather non-rigorous reasoning — Non-rigorous is fine by me... :-) Ramanujan himself was a total and complete disaster from this perspective, and they were the greatest! Sometimes, some things are just so simple, that no one can see them... — 79.113.213.225 (talk) 10:12, 24 April 2013 (UTC)[reply]

Resolved

Haar wavelet

In the article Haar wavelet it says "Any continuous real function can be approximated by linear combinations of and their shifted functions." Can someone give me a proof of this fact? Thanks. Money is tight (talk) 07:05, 23 April 2013 (UTC)[reply]

I doubt anybody can prove that. An integral of a single wavelet over all reals is zero, so any finite linear combination of wavelets will also have a zero integral. That implies no combination can approximate functions with nonzero (or infinite) integrals, like Normal distribution function, Heaviside step function or just

f(x) = 1

– at leas not over all reals.
However I suppose functions with a zero mean (meaning a₀ term of the Fourier series here) might have some good wavelet approximations... --CiaPan (talk) 14:17, 23 April 2013 (UTC)[reply]

I think most uses of the Haar wavelet and wavelets in general have a finite domain, eg a computer stream of data or an image. The article does mention using the unit interval in the Haar system section. The statement in the article is

Any continuous real function can be approximated by linear combinations of

\phi (t),\phi (2t),\phi (4t),\dots ,\phi (2^{k}t),\dots

and their shifted functions.

The

\phi (t)

are just Rectangular functions. Assuming we are working over the unit interval, you can just construct the approximation. A) Find the average value of f over [0-1], m₀ say. Let

f_{1}(t)=f(t)-m_{0}*\phi (t)

Then split the domain in two, rescale to [0-1] and repeat. You can continue for any desired degree of approximation.--Salix (talk): 16:05, 23 April 2013 (UTC)[reply]

You might also need to assume the function is bounded just to prevent any odd cases. Again a reasonable assumption for computer applications.--Salix (talk): 16:18, 23 April 2013 (UTC)[reply]

With regard to the integration point, wavelets are normally used for L^2 approximation and on infinite domains L^2 closeness does not imply L^1 closeness. Straightontillmorning (talk) 21:16, 23 April 2013 (UTC)[reply]