Talk:Associated Legendre function: Difference between revisions

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imported>Paul Wormer
(→‎Problem: new section)
imported>Paul Wormer
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--[[User:Paul Wormer|Paul Wormer]] 11:15, 6 September 2009 (UTC)
--[[User:Paul Wormer|Paul Wormer]] 11:15, 6 September 2009 (UTC)
:I checked some and I believe that the equation must be
::<math>
\int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l}  dx = -\frac{2l}{2l+1} \int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l-1}  dx
</math>
:I changed the text accordingly, but there must be another sign error because the final result now obtains (&minus;1)<sup>''l''</sup>, which is not correct. --[[User:Paul Wormer|Paul Wormer]] 13:14, 6 September 2009 (UTC)

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 Definition Function defined by where P denotes a Legendre function. [d] [e]
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Started from scratch, because (as so often) the Wikipedia article contains many duplications.--Paul Wormer 10:08, 22 August 2007 (CDT)

Added link to proof of orthogonality and derivation of normalization constant

I added a link to a proof of the first integral equation in the Orthogonality relations section Dan Nessett 16:39, 11 July 2009 (UTC)

Formattting of proof

I formatted the proof somewhat to my taste (which is mainly determined by a careful reading of Knuth's the TeXbook). Most of my changes are self-explanatory. I added the integration by parts formula, because it was not clear at all where the u and the v' came from. Further

--Paul Wormer 12:43, 5 September 2009 (UTC)

Nice formating. Thanks for catching the problem. Also, I agree it is better to explicitly provide the integration by parts formula, rather than expecting the reader to already know it. Dan Nessett 00:49, 6 September 2009 (UTC)

Move of equation

I moved the dif. eq. from the proof page to the main article. I did this because the proof does not use the dif. eq., but uses the definition of the assleg's (which is given in the statement of the theorem). --Paul Wormer 09:24, 6 September 2009 (UTC)

Another (non-essential) error

I didn't like the reference to WP and tried my hand at the integral, which indeed is not difficult. Doing this I found another slip of the pen.--Paul Wormer 10:43, 6 September 2009 (UTC)

Problem

Let us apply the following equation (taken from Dan's proof and which holds for l ≥ 1) for l = 1:

Then is it true that

Integration of both sides gives

The problem is not trivial because the recursion ends with this integral. Dan, HELP!

--Paul Wormer 11:15, 6 September 2009 (UTC)

I checked some and I believe that the equation must be
I changed the text accordingly, but there must be another sign error because the final result now obtains (−1)l, which is not correct. --Paul Wormer 13:14, 6 September 2009 (UTC)