Mathematica's definition of the connection of associated Legendre polynomials with $m$ and $-m$ is:

$P_l^{-m}=(-1)^m \frac{(l-m)!}{(l+m)!} P_l^m$.

We also now that $|m|>l \Rightarrow P_l^m=0$. Try this in Mathematica and tell me if it is a bug or can be explained by something else:




The first one will give 0, which to my knowledge is correct. The second one yields $\sqrt{1-x}/\sqrt{1+x}$ although it should also yield 0! Why?


1 Answer 1


The identity $P_{\ell}^{-m}(z) =(-1)^m \frac{(\ell-m)!}{(\ell+m)!} P_{\ell}^m (z)$ only holds for $$(\ell, m) \in \left\{ \mathbb{Z}_{\geqslant 0} \times \mathbb{Z} \colon -\ell \leqslant m \leqslant \ell\right\}.$$

Mathematica's definition for associated Legendre's polynomial is given by formula $$ P_{\ell}^{\mu}(z) = \frac{\left(1+z\right)^{\mu/2}}{\left(1-z\right)^{\mu/2}} \sum_{k}^{\ell} \frac{(-\ell)_k \cdot \left(\ell+1\right)_k}{\Gamma\left(k + 1-\mu\right)} \cdot \frac{\left(1-z\right)^2}{2^k \cdot k!} $$ This formula is also consistent with NIST's DLMF/14.3.E1.

In particular, for $\ell = 0$: $$ P_0^{\mu}\left(z\right) = \frac{\left(1+z\right)^{\mu/2}}{\left(1-z\right)^{\mu/2}} \cdot \frac{1}{\Gamma(1-\mu)} $$ which makes it zero for positive integer values of $\mu$, but not for non-positive integer values.

  • 1
    $\begingroup$ Actually, I think there is something wrong here. The answer is correct if taken by itself, but it is inconsistent with the definition of SphericalHarmonicY because there you do get 0 for {l,m} = {1,-1}. $\endgroup$
    – Jens
    Jan 22, 2015 at 20:48

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