# Talk:Parseval's theorem

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Field: Analysis

## Proof

Can anybody put a proof of the theorem up?

## 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)

## 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)

## 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)

## "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 (talkcontribs) 06:50, 24 October 2006 (UTC)

The integeral variable used in this equation is f, not ω, where

$\omega = 2 \pi f$

so 1/(2*pi) could be omitted. — Preceding unsigned comment added by Dongliangseu (talkcontribs) 08:00, 23 December 2010 (UTC)

## "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)

I believe it simply depends on the definition of the transform that was used to go from x[n] to X[k]. The 1/N can be placed in either the forward or the inverse transform - or often 1/sqrt(N) is placed in both. It may be worth clarifying which transform is used. daviddoria (talk) 18:08, 13 September 2008 (UTC)

I agree with Gmoose1, the 1/N goes to the left side, as the total energy of a stationary random function is its _variance_ wich is defined like

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

See e.g.Uriel Frisch, Turbulence, Cambridge 1995 Page 53 —Preceding unsigned comment added by 130.183.84.206 (talk) 17:52, 18 February 2010 (UTC)

I think the comment of Daviddoria is resonable. If the fourier transform is defined by the following formula, which is the common case,

$X[k] = \sum_{n=0}^{N-1} x[n] exp( \frac{-j2 \pi kn}{N} )$

then

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

However, if

$X[k] = \frac{1}{N} \sum_{n=0}^{N-1} x[n] exp( \frac{-j2 \pi kn}{N} )$

then,

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

There is also another case, that is, if

$X[k] = \frac{1}{\sqrt N} \sum_{n=0}^{N-1} x[n] exp( \frac{-j2 \pi kn}{N} )$

then,

$\sum_{n=0}^{N-1} | x[n] |^2 = \sum_{k=0}^{N-1} | X[k] |^2$ — Preceding unsigned comment added by Dongliangseu (talkcontribs) 07:50, 23 December 2010 (UTC)

## Parseval's Power Theorem is not the same as Rayleigh's Energy Theorem

In the intro paragraph, it says "this is also known as Rayleigh's Energy Theorem". I don't believe this is the case, because Parseval's theorem is for periodic signals (power signals) and Rayleigh's theorem is for energy signals (usually finite duration - have 0 power). Can someone confirm this and then I will fix it? daviddoria (talk) 18:08, 13 September 2008 (UTC)

Parseval's theorem is not just for periodic functions, but it also applies to non-periodic functions. Thenub314 (talk) 19:37, 13 September 2008 (UTC)
I still think there is a difference though.. Isn't Parseval's Power Theorem a sum of the Fourier series coefficients = the power in the signal? And Rayleigh's Energy Theorem that the integral of the signal (energy) = the energy of the spectrum? —Preceding unsigned comment added by Daviddoria (talkcontribs) 16:17, 14 September 2008 (UTC)
I am not sure, other then this article I had not heard of "Rayleigh's Energy Theorem" but when I google the term everything I find seems to simply be Parseval's theorem. Thenub314 (talk) 11:33, 15 September 2008 (UTC)