User:Dnessett/Sandboxes/First Sandbox

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.