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