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)

In the equation for $K_{kl}^{m}$ lacked a factor (−1)^l. I added it. --Paul Wormer 13:25, 6 September 2009 (UTC)

Paul,

I had family responsibilities yesterday, so I only read your talk comments this morning. I see you have made some changes to the style of the proof, e.g., converted some factorial expressions to the binomial coefficient, moved the Kroneker delta around and moved some factorials out of parentheses. That's fine. These sorts of things are a matter of taste and I have no problem with your choices.

Good catch on the fraction in front of the integral for the $\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx\$ recursion relation. However, I think the problem with the negative sign doesn't exist.

First,

$K_{kl}^{m}=\delta _{kl}\;{\frac {(-1)^{l+m}}{2^{2l}\,(l!)^{2}}}{\binom {l+m}{2m}}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}{\frac {d^{2m}}{dx^{2m}}}\left[\left(1-x^{2}\right)^{m}\right]{\frac {d^{2l}}{dx^{2l}}}\left[\left(1-x^{2}\right)^{l}\right]dx,$

and

${\frac {d^{2k}}{dx^{2k}}}\left[\left(1-x^{2}\right)^{k}\right]=(-1)^{k}\,(2k)!\,.$

So,

$K_{kl}^{m}=\delta _{kl}\;{\frac {(-1)^{l+m}}{2^{2l}\,(l!)^{2}}}{\binom {l+m}{2m}}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}(-1)^{l}\ (2l)!\ (-1)^{m}\ (2m)!\ dx\ =\ \delta _{kl}\;{\frac {\ (2l)!\ (2m)!}{2^{2l}\,(l!)^{2}}}{\binom {l+m}{2m}}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}$

In other words, the powers of -1 multiply so the exponent on (-1) is a power of 2.

But, there isn't really a problem anyway. Applying:

$\int \limits _{0}^{\pi }\sin ^{n}\theta d\theta ={\frac {\left(n-1\right)}{n}}\int \limits _{0}^{\pi }\sin ^{n-2}\theta d\theta$

to:

$\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx=\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l+1}d\theta$

yields:

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

That is, there is no negative sign in front of the integral.

However, the equation immediately prior to the final result changes because of the new fractional component and I am in the process of checking that. Dan Nessett 17:18, 7 September 2009 (UTC)

Update: I fixed the logic on the computation of the fraction within the equation immediately prior to the final result. Dan Nessett 20:52, 7 September 2009 (UTC)

Note: Paul refers above to "Dan's" proof. Just to ensure clarity, the proof originally placed on the proof page was a 50/50 collaboration between a friend and former colleague and me.

[Unindent]

Is it true that (for l = 1),

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

Integration of both sides gives

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

--Paul Wormer 05:33, 8 September 2009 (UTC)

Whew! Calculus gymnastics is a sport that requires constant training. Leave it alone for a while and doing the requisite back-flips and cartwheels becomes very difficult.

You correctly identified the problem Paul. A factor of (-1)^l is required both in the recursion and in K(m,k,l). These cancel out at the end. I have made the necessary changes to the proof. Interestingly, I came to the same conclusion as you by noticing an error in the change of variable equation you inserted on 5 Sept. I have corrected this also.

I think it is useful to get on record the following derivation. It doesn't belong in the proof itself, but putting it on this talk page gets it into the history of the article:

$x=\cos \theta \;\Longrightarrow \;dx=-\sin \theta d\theta \quad {\hbox{and}}\quad 1-x^{2}=(\sin \theta )^{2}\Longrightarrow \;x^{2}-1=\ -(\sin \theta )^{2}.$

$\left(x^{2}-1\right)^{l}\ =\ (-1)^{l}\left(\sin \theta \right)^{2l}$

$\left(x^{2}-1\right)^{l-1}\ =\ (-1)^{l-1}\left(\sin \theta \right)^{2l-2}$

$\int \limits _{-1}^{1}(x^{2}-1)^{l}dx\ =\ (-1)^{l+1}\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l+1}d\theta$

$\int \limits _{-1}^{1}(x^{2}-1)^{l-1}dx\ =\ (-1)^{l}\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l-1}d\theta$

$\int \limits _{0}^{\pi }\sin ^{n}\theta d\theta ={\frac {\left(n-1\right)}{n}}\int \limits _{0}^{\pi }\sin ^{n-2}\theta d\theta$

$\int \limits _{-1}^{1}(x^{2}-1)^{l}dx\ =\ (-1)^{l+1}\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l+1}d\theta \ =\ (-1)^{l+1}{\frac {2l}{2l+1}}\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l-1}d\theta \ =\ -\ {\frac {2l}{2l+1}}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l-1}dx$

Dan Nessett 17:20, 8 September 2009 (UTC)

Reinserted comment about evaluating overlap integral whether or not k=l

I have reintegrated the comment that one can evaluate $K_{kl}^{m}$ whether or not k=l. This is pertinent, since the eventual expression for $K_{kl}^{m}$ contains $\delta _{kl}$. So, $K_{kl}^{m}$ not only provides the normalization constant, but also shows orthogonality when k=l.