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From this

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

I should derive that

$$P^{-m}_n(x)=(-1)^n\displaystyle\frac{(n-m)!}{(n+m)!}P_m^n(x)$$

Using the Leibniz's formula to $(x^2-1)^n=(x-1)^n(x+1)^n$

But I get lost adding terms in the Leibniz expansion...I appreciate any help to get this result.

Thanks a lot!!!

EQJ
  • 4,369

1 Answers1

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Equate the coefficients of equal powers on the left and right hand side of $$ \frac{d^{n-m}}{dx^{n-m}} (x^2-1)^{n} = c_{nm} (1-x^2)^m \frac{d^{n+m}}{dx^{n+m}}(x^2-1)^{n}, $$ then it follows that the proportionality constant is $$ c_{nm} = (-1)^m \frac{(n-m)!}{(n+m)!} , $$ so that $$ P^{-m}_n(x) = (-1)^m \frac{(n-m)!}{(n+m)!} P^{m}_n(x). $$

alexjo
  • 14,976
  • Why this proportionality holds? – EQJ Apr 21 '14 at 00:16
  • Since the associated Legendre equation is invariant under the substitution $m → −m$, the equations for $P_n^{±m}$ are proportional.

    To obtain the proportionality constant we consider $$ (1-x^2)^{-m/2} \frac{d^{n-m}}{dx^{n-m}} (x^2-1)^{\ell} = c_{nm} (1-x^2)^{m/2} \frac{d^{n+m}}{dx^{n+m}}(x^2-1)^{n},\qquad 0 \le m \le n, $$ and we bring the factor $(1−x²)^{−m/2}$ to the other side; and then, equate...

     – alexjo Apr 21 '14 at 00:23
  • Amazing way to get the result! But I don't understand yet why the invariance implies the proporcionality. But assuming it holds, I get the result. Thanks! – EQJ Apr 21 '14 at 00:58