Associated Legendre Special Properties

Question

Prove the relationship of the equation that : $$ P_n^{-m}(x)=(-1)^m\frac{(n-m)!}{(n+m)!}P_n^{m}(x) -------> equation (1)$$

This is what I know: $$ P_n^{m}(x)=\frac{1}{2^nn!}(1-x^2)^{\frac{m}{2}}\frac{d^{n+m}}{dx^{n+m}}(x^2-1)^n$$

Leibnitz's formula:

$$\frac{d^{n}}{dx^{n}}(A(x)B(x))=\sum_{s=0}^n \binom{n}{s}\frac{d^{n-s}}{dx^{n-s}}A(x)\frac{d^{s}}{dx^{s}}B(x)$$ and $$\binom{n}{s}=\frac{n!}{(n-s)!s!}$$

then

$$\frac{d^{n+m}}{dx^{n+m}}(x^2-1)^n=\frac{d^{n+m}}{dx^{n+m}}(x-1)^n(x+1)^n=\sum_{s=0}^{n+m} \binom{n+m}{s}\frac{d^{n+m-s}}{dx^{n+m-s}}(x+1)^n\frac{d^{s}}{dx^{s}}(x-1)^n$$

$$\binom{n+m}{s}=\frac{(n+m)!}{(n+m-s)!s!}$$

$$ P_n^{-m}(x)=\frac{1}{2^nn!}(1-x^2)^{\frac{-m}{2}}\frac{d^{n-m}}{dx^{n-m}}(x^2-1)^n$$

$$\frac{d^{n-m}}{dx^{n-m}}(x^2-1)^n=\frac{d^{n-m}}{dx^{n-m}}(x-1)^n(x+1)^n=\sum_{s=0}^{n-m} \binom{n-m}{s}\frac{d^{n-m-s}}{dx^{n-m-s}}(x+1)^n\frac{d^{s}}{dx^{s}}(x-1)^n$$

$$\binom{n-m}{s}=\frac{(n-m)!}{(n-m-s)!s!}$$

$$ P_n^{m}(x)=\frac{1}{2^nn!}(1-x^2)^{\frac{m}{2}}\frac{d^{n+m}}{dx^{n+m}}(x^2-1)^n$$

$$\frac{1}{2^nn!}=P_n^{m}(x)(1-x^2)^{\frac{-m}{2}}\frac{1}{\sum_{s=0}^{n+m} \frac{(n-m)!}{(n-m-s)!s!}\frac{d^{n+m-s}}{dx^{n+m-s}}(x+1)^n\frac{d^{s}}{dx^{s}}(x-1)^n}$$

$$ P_n^{-m}(x)=P_n^{m}(x)(1-x^2)^{\frac{-m}{2}}\frac{\sum_{s=0}^{n-m} \frac{(n-m)!}{(n-m-s)!s!}\frac{d^{n-m-s}}{dx^{n-m-s}}(x+1)^n\frac{d^{s}}{dx^{s}}(x-1)^n}{\sum_{s=0}^{n+m} \frac{(n+m)!}{(n+m-s)!s!}\frac{d^{n+m-s}}{dx^{n+m-s}}(x+1)^n\frac{d^{s}}{dx^{s}}(x-1)^n}(1-x^2)^{\frac{-m}{2}}$$

$$ P_n^{-m}(x)=P_n^{m}(x)(1-x)^{-m}(1+x)^{-m}\frac{\sum_{s=0}^{n-m} \frac{(n-m)!}{(n-m-s)!s!}\frac{d^{n-m-s}}{dx^{n-m-s}}(x+1)^n\frac{d^{s}}{dx^{s}}(x-1)^n}{\sum_{s=0}^{n+m} \frac{(n+m)!}{(n+m-s)!s!}\frac{d^{n+m-s}}{dx^{n+m-s}}(x+1)^n\frac{d^{s}}{dx^{s}}(x-1)^n}$$

$$ P_n^{-m}(x)=P_n^{m}(x)\frac{(n-m)!}{(n+m)!}(1-x)^{-m}(1+x)^{-m}\frac{\sum_{s=0}^{n-m} \frac{1}{(n-m-s)!s!}\frac{d^{n-m-s}}{dx^{n-m-s}}(x+1)^n\frac{d^{s}}{dx^{s}}(x-1)^n}{\sum_{s=0}^{n+m} \frac{1}{(n+m-s)!s!}\frac{d^{n+m-s}}{dx^{n+m-s}}(x+1)^n\frac{d^{s}}{dx^{s}}(x-1)^n}$$

$$ P_n^{-m}(x)=P_n^{m}(x)\frac{(n-m)!}{(n+m)!}(-1)^{-m}(x-1)^{-m}(x+1)^{-m}\frac{\sum_{s=0}^{n-m} \frac{1}{(n-m-s)!s!}\frac{d^{n-m-s}}{dx^{n-m-s}}(x+1)^n\frac{d^{s}}{dx^{s}}(x-1)^n}{\sum_{s=0}^{n+m} \frac{1}{(n+m-s)!s!}\frac{d^{n+m-s}}{dx^{n+m-s}}(x+1)^n\frac{d^{s}}{dx^{s}}(x-1)^n}$$

but $$(-1)^{-m}=(-1)^m $$ therefore,

$$ P_n^{-m}(x)=P_n^{m}(x)\frac{(n-m)!}{(n+m)!}(-1)^{m}(x-1)^{-m}(x+1)^{-m}\frac{\sum_{s=0}^{n-m} \frac{1}{(n-m-s)!s!}\frac{d^{n-m-s}}{dx^{n-m-s}}(x+1)^n\frac{d^{s}}{dx^{s}}(x-1)^n}{\sum_{s=0}^{n+m} \frac{1}{(n+m-s)!s!}\frac{d^{n+m-s}}{dx^{n+m-s}}(x+1)^n\frac{d^{s}}{dx^{s}}(x-1)^n}$$

$$ P_n^{-m}(x)=P_n^{m}(x)\frac{(n-m)!}{(n+m)!}(-1)^{m}\frac{\sum_{s=0}^{n-m} \frac{1}{(n-m-s)!s!}\frac{d^{n-m-s}}{dx^{n-m-s}}(x+1)^{n-m}\frac{d^{s}}{dx^{s}}(x-1)^{n-m}}{\sum_{s=0}^{n+m} \frac{1}{(n+m-s)!s!}\frac{d^{n+m-s}}{dx^{n+m-s}}(x+1)^n\frac{d^{s}}{dx^{s}}(x-1)^n}$$

and that leads to $$ P_n^{-m}(x)=(-1)^m\frac{(n-m)!}{(n+m)!}P_n^{m}(x)A$$

where

$$A=\frac{\sum_{s=0}^{n-m} \frac{1}{(n-m-s)!s!}\frac{d^{n-m-s}}{dx^{n-m-s}}(x+1)^{n-m}\frac{d^{s}}{dx^{s}}(x-1)^{n-m}}{\sum_{s=0}^{n+m} \frac{1}{(n+m-s)!s!}\frac{d^{n+m-s}}{dx^{n+m-s}}(x+1)^n\frac{d^{s}}{dx^{s}}(x-1)^n} $$

This is almost the equation (1) derived except for variable A. Is A equal to 1? My problem is that I do not know how to evaluate multiderivative and go further to prove the relationship. Thank you in advance.

Answer 1

You're trying to compare constant multiples of $$ (1-x^{2})^{m/2}\frac{d^{n+m}}{dx^{n+m}}(x^{2}-1)^{n},\;\;\; (1-x^{2})^{-m/2}\frac{d^{n-m}}{dx^{n-m}}(x^{2}-1)^{n}. $$ Multiplying by $(1-x^{2})^{m/2}$ allows comparison of $$ (x^{2}-1)^{m}\frac{d^{n+m}}{dx^{n+m}}(x^{2}-1)^{n},\;\;\; \frac{d^{n-m}}{dx^{n-m}}(x^{2}-1)^{n}. $$ As you noted, it is convenient to write $(x^{2}-1)^{n}=(x+1)^{n}(x-1)^{n}$. Then, $$ \frac{d^{n-m}}{dx^{n-m}}(x+1)^{n}(x-1)^{n}=\sum_{j=0}^{n-m}{n-m\choose j}\frac{d^{j}}{dx^{j}}(x+1)^{n}\frac{d^{n-m-j}}{dx^{n-m-j}}(x-1)^{n} \\ = \sum_{j=0}^{n-m}{n-m\choose j}\frac{n!}{(n-j)!}(x+1)^{n-j}\frac{n!}{(m+j)!}(x-1)^{m+j}. $$ And $$ (x^{2}-1)^{m}\frac{d^{n+m}}{dx^{n+m}}(x^{2}-1)^{n}=(x^{2}-1)^{m}\sum_{j=0}^{n+m}{n+m\choose j} \frac{d^{j}}{dx^{j}}(x+1)^{n}\frac{d^{n+m-j}}{dx^{n-j}}(x-1)^{n} \\ =(x^{2}-1)^{m}\sum_{j=m}^{n}{n+m\choose j}\frac{n!}{(n-j)!}(x+1)^{n-j}\frac{n!}{(j-m)!}(x-1)^{j-m} \\ = (x^{2}-1)^{m}\sum_{j=0}^{n-m}{n+m\choose j+m}\frac{n!}{(n-m-j)!}(x+1)^{n-m-j}\frac{n!}{j!}(x-1)^{j} \\ = \sum_{j=0}^{n-m}{n+m\choose j+m}\frac{n!}{(n-m-j)!}(x+1)^{n-j}\frac{n!}{j!}(x-1)^{m+j} $$ Compare the coefficients of like powers: $$ {n-m\choose j}\frac{n!}{(n-j)!}\frac{n!}{(m+j)!}=\frac{(n-m)!}{j!(n-m-j)!}\frac{n!}{(n-j)!}\frac{n!}{(m+j)!} \\ {n+m\choose j+m}\frac{n!}{(n-m-j)!}\frac{n!}{j!}=\frac{(n+m)!}{(n-j)!(m+j)!}\frac{n!}{(n-m-j)!}\frac{n!}{j!} $$ All ratios of corresponding coefficients are independent of $j$--dividing the second into the first gives a common ratio of $$ \frac{(n-m)!}{(n+m)!} $$ Therefore, $$ \frac{d^{n-m}}{dx^{n-m}}(x^{2}-1)^{n}=\frac{(n-m)!}{(n+m)!}(x^{2}-1)^{m}\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n+m}. $$ Finally, $$ (1-x^{2})^{-m/2}\frac{d^{n-m}}{dx^{n-m}}(x^{2}-1)^{n}= \frac{(-1)^{m}(n-m)!}{(n+m)!}(1-x^{2})^{m/2}\frac{d^{n+m}}{dx^{n+m}}(1-x^{2})^{n}\\ P_{n}^{-m}(x)=\frac{(-1)^{m}(n-m)!}{(n+m)!}P_{n}^{m}(x). $$