Talk:Parseval's theorem

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Mathematics rating: B Class Mid Priority  Field: Analysis

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[edit] Proof

Can anybody put a proof of the theorem up?

[edit] problem solved in Good Will Hunting?

What was solved by the main character in Good Will Hunting? No problems are mentioned up to this point in the text. Cgibbard 16:40, 26 February 2006 (UTC)

[edit] Added "1/2*pi" in section "Modern interpretation"

Hi,

I'm not a Mathematician, but am involved in control projects, where I often make use of the z-transform formalism. I think there was a "half pi" factor missing following the "For discrete time signals, ..." paragraph, so I added it.

Regards,

Biscay.

[edit] reason for edit

Lest the reader be misled, it should be noted that the article is not a math article. Clearly mathematics is not the context here. the discussion is non-mathematical. there are no real mathematical reference listed. article should be expanded or title should be changed to "...in physics and engineering." Mct mht 14:17, 17 May 2006 (UTC)

[edit] sum of inverse squares

There's a very simple proof using this theorem, that

\sum_{k=1}^\infty {1\over k^2} = {\pi^2\over 6}

Anyone know it? It might be cool to add it to the article. Phr (talk) 09:20, 2 August 2006 (UTC)


This actually just showed up on my homework - it consists of finding the Fourier series for the period 2pi extension of the function defined as f(x) = (pi - x)^2 on [0,2pi]. The value of this series for x = 0 gives the summation of that series; Parseval's Theorem gives the sum of 1/k^4. It's a bit computational, though; might be worth mentioning briefly how to do it but without details. Moocowpong1 (talk) 03:57, 11 March 2008 (UTC)

[edit] "1/2*pi" in section "Applications"

It looks to me like there is a 1/2*pi missing from the first equation after the heading "Applications," also. Can someone with a bit of knowledge check that, please? —The preceding unsigned comment was added by 4.242.147.1 (talk • contribs) 06:50, 24 October 2006 (UTC)

[edit] "Discrete time Parseval's Theorem for a Periodic Function"

The discrete time Parseval's Theorem for a periodic function should have the 1/N term in the discrete time domain instead of in the frequency domain. I don't have a textbook to confirm this so I did not edit it in the article.

Here is the original (current) equation:

\sum_{n=0}^{N-1} | x[n] |^2 = \frac{1}{N} \sum_{k=0}^{N-1} | X[k] |^2

Here is what I beleive is correct:

\frac{1}{N} \sum_{n=0}^{N-1} | x[n] |^2 = \sum_{k=0}^{N-1} | X[k] |^2

Any agreements/oppositions? --Gmoose1 (talk) 02:53, 26 November 2007 (UTC)