Talk:Parseval's theorem

message: 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)
thread_title:  problem solved in Good Will Hunting? 

message: 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)
thread_title:  reason for edit 

message: 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)
thread_title:  sum of inverse squares 

message: 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)
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 (talk • contribs) 08:00, 23 December 2010 (UTC)
thread_title:  "1/2*pi" in section "Applications" 

message: 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)
replies: text: 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)
replies: text:  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 (talk • contribs) 16:17, 14 September 2008 (UTC)
replies: text:  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)
thread_title:  Parseval's Power Theorem is not the same as Rayleigh's Energy Theorem 

message: What is the usual procedure on this?
If someone searches "Parsevals Theorem" (without an apostrophe) I'd say this page should come up right away, without having to look through the search results to find "Parseval's Theorem".  Can I create the article "Parsevals Theorem" and redirect it to here? What is usually done about this?
daviddoria (talk) 18:08, 13 September 2008 (UTC)
replies: text: Yes; it's pretty obvious that this will only help people without confusing any. I've just done exactly this BTW. C xong (talk) 01:39, 10 August 2010 (UTC)
thread_title:  Link from "Parsevals Theorem 

message: In the statement of the theorem in the article, it's written
:where i is the imaginary unit and horizontal bars indicate complex conjugation.
Aren't both these notations extremely well established, rendering the clarification a bit superfluous? ✎ HannesP · talk 22:30, 8 April 2012 (UTC)
thread_title:  Unnecessary clarification 

message: This article makes Parseval's theorem needlessly difficult. It badly needs clarification for people who are not mathematicians.
In almost any book on Fourier series that one looks at, Parseval's theorem is as follows:
Suppose f(x) is a square-integrable function over (-\pi, \pi) (i.e., f(x) and f^2(x) are integrable on that interval), with the Fourier series
:f(x) \simeq \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx)).
Then
:\frac{1}{\pi} \int_{-\pi}^{\pi} f^2(x) dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty} (a_n^2 + b_n^2).
Is there any objection to adding the above statement to the article?
MathPerson (talk) 20:54, 17 July 2017 (UTC)
replies: text: Seeing no objection, I added it, with references. MathPerson (talk) 00:22, 29 July 2017 (UTC)
thread_title:  This article is way too obscure