Talk:Associated Legendre function

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

$1-x^{2}=\sin ^{2}\theta \quad {\hbox{not}}\quad \sin \theta$

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

Nice formating. Thanks for catching the $\sin \theta$ 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:

$\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx={\frac {2\left(l+1\right)}{2l+1}}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l-1}dx$

Then is it true that

$\int \limits _{-1}^{1}\left(x^{2}-1\right)dx={\frac {2\left(1+1\right)}{2+1}}\int \limits _{-1}^{1}dx\quad ?$

Integration of both sides gives

${\frac {1}{3}}\left[x^{3}\right]_{-1}^{1}\;-\;\left[x\right]_{-1}^{1}\;{\stackrel {?}{=}}\;{\frac {4}{3}}\;\left[x\right]_{-1}^{1}\;\Rightarrow \;{\frac {2}{3}}-2\;{\stackrel {?}{=}}\;{\frac {8}{3}}\;\Rightarrow \;-4\;{\stackrel {?}{=}}\;8$

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

$\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx=-{\frac {2l}{2l+1}}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l-1}dx$

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)