4
$\begingroup$

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:

LegendreP[0,1,x]

and

LegendreP[0,-1,x]

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?

$\endgroup$
9
$\begingroup$

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 05.07.02.0001.01: $$ P_{\ell}^{\mu}(z) = \frac{\left(1+z\right)^{\mu/2}}{\left(1-z\right)^{\mu/2}} \sum_{k}^{\ell} \frac{(-\ell)_k \cdot \left(\ell+1\right)_k}{\Gamma\left(k + 1-\mu\right)} \cdot \frac{\left(1-z\right)^2}{2^k \cdot k!} $$ 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.

| improve this answer | |
$\endgroup$
  • 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 '15 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.