Associated Legendre Function of Second Kind

[Vick]

TL;DR Summary

Recurrence Identities for the Associated Legendre Function of Second Kind

The associated Legendre Function of Second kind is related to the Legendre Function of Second kind as such:

$$
Q_{n}^m(z)= (-1)^m (1-z^2)^{m/2} \frac{d^m}{dz^m}(Q_{n}(z))
$$

The recurrence relations for the former are the same as those of the first kind, for which one of the relations is:

$$
(n-m+1)Q_{n+1}^m(z)=(2n+1)zQ_{n}^m(z)-(n+m)Q_{n-1}^m(z)
$$

The recurrence relations and identities of the associated legendre function of the first kind are here: Associated Legendre function of 1st kind

However, none of the literature show anything about some identities I'm looking for!

For example one of the identities is when $m=0$, the result is the same as such:
$$
Q_{n}^0(z)=Q_{n}(z)
$$

What I'm looking for are the identities when:
$1)$ $n$ and $m$ are the same, that is
$$
Q_{n}^n(z)=?
$$

and $2)$ the identity when $n = 0$, that is

$$
Q_{0}^m(z)=?
$$

Do those two identities exist?

 

[jvicens]

I can confirm that the two identities you are looking for do exist. They are as follows:

1) When n and m are equal, the associated Legendre function of the second kind becomes:

$$
Q_{n}^n(z)= (-1)^n (1-z^2)^{n/2} \frac{d^n}{dz^n}(Q_{n}(z))
$$

2) When n = 0, the associated Legendre function of the second kind becomes:

$$
Q_{0}^m(z)= (-1)^m (1-z^2)^{m/2} \frac{d^m}{dz^m}(Q_{0}(z))
$$

Using the identity for the Legendre function of the first kind, we can rewrite this as:

$$
Q_{0}^m(z)= (-1)^m (1-z^2)^{m/2} \frac{d^m}{dz^m}(1)
$$

Since the derivative of a constant is 0, this simplifies to:

$$
Q_{0}^m(z)= (-1)^m (1-z^2)^{m/2}
$$

I hope this helps with your research! It's always exciting to discover new identities and understand the connections between different mathematical functions. Keep exploring and learning!