Derivative of Associated Legendre polynomials at $x = \pm 1$

Question

I'm creating meshes for spherical harmonics, and I need a normal at a given point. Whenever I'm at the poles, $\cos{\theta} = \pm 1$, and I do not know how to find the derivative there. All the formulas I have found to describe the derivative have an $1 - x^2$ in the denominator, and I get an indeterminate form. For reference, the one I'm using is: $$(P_\ell^m)^\prime(x) = \frac{\sqrt{1-x^2} P_\ell^{m+1}(x) + mx P_\ell^m (x)}{x^2 - 1}$$

I found the derivatives for some cases, and it seems that $m = \pm 1$ results in $\pm \infty$, $m = 0$ yields triangular numbers, and $|m| \ge 3$ makes the derivative $0$. But I can't find an overarching pattern or algorithm I can use to produce these. Is there a nice way?

Answer 1

The singularity at the denominator can be eliminated using L'Hospital's theorem, once you notice that the associated Legendre function has value of $0$ at $\pm 1$.

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Maybe this is not a right solution, because I found another formula about the derivative of the associated Legendre function here,

• Spherical Harmonics

and it gives a difference solution when I apply the same method.

Answer 2

Could you try using the formula from Wikipedia ?

\[ P_m^l (x) = \frac{(-1)^m}{2^\ell \ell!} (1-x^2)^{m/2} \frac{d^{\ell + m}}{dx^{\ell + m}}(1 - x^2)^{\ell} \]

Let $y = 1-x \approx 0$,

\[ P_m^l (y) = \frac{(-1)^m}{2^\ell \ell!} y^{m/2}(2-y)^{m/2} \frac{d^{\ell + m}}{dy^{\ell + m}} y^{\ell}(2-y)^\ell \propto y^{m/2} \approx 0\]

I am guessing the derivative is always zero unless $m = 0$, the only interesting case.

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Or plug in $x = \cos \theta$ and take the limit $\theta \to 0$:

\[ (P_\ell^m)^\prime(\cos \theta) = \frac{\sin \theta \, P_\ell^{m+1}(\cos \theta) + m \cos \theta \, P_\ell^m (\cos \theta )}{\sin^2 \theta} \]

Not sure how to evaluate this... spherical harmonics are rotationally symmetric, so maybe they shouldn't have non-zero derivatives at the poles?

Answer 3

As suggested by user116561, L'Hopital and induction is the way to go. Gotta break it into cases though. And I have to apply it to a different recurrence relation, one that fixes $m$:

$$( P_\ell^m)'(x) = \frac{(\ell + 1 - m) P_{\ell+1}^m(x) - (\ell + 1)xP_\ell^m(x)}{x^2 - 1}$$

which gives:

$$\lim_{x \to c}~( P_\ell^m)'(x) = \lim_{x \to c} \frac{(\ell + 1 - m) (P_{\ell+1}^m)'(x) - (\ell + 1)(P_\ell^m(x) + x(P_\ell^m)'(x))}{2x}$$

when L'Hopital is applicable, of course. But it is applicable for all $\ell,m$, when $c = 1$.

When $m = 0$, one can check $P_\ell^0(1) = 1$ for all $\ell$. So the numerator is $(\ell + 1) 1 - (\ell + 1) 1$, which is $0$. When $m \ne 0$, then $P_\ell^0(1) = 0$ for all $\ell$. So then the numerator obviously goes to $0$.

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$m = 0 : \frac{\ell(\ell + 1)}{2}$

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Base: $P_0^0(x) = 1$, so its derivative is $0$ everywhere. In particular it is $\frac{0\cdot 1}{2} = 0$ at $1$.

Assume $\lim_{x \to 1}( P_\ell^0 )'(x) = \frac{\ell(\ell + 1)}{2}$ for some $\ell$:

$$ \begin{align*} \lim_{x \to 1}~( P_\ell^0)'(x) &= \lim_{x \to 1} \frac{(\ell + 1) (P_{\ell+1}^0)'(x) - (\ell + 1)(P_\ell^0(x) + x(P_\ell^0)'(x))}{2x} \\ \frac{\ell(\ell + 1)}{2} &= \lim_{x \to 1} \frac{\ell + 1}{2} \left( (P_{\ell+1}^0)'(x) - P_\ell^0(x) - x(P_\ell^0)'(x) \right) \\ \ell &= \left[ \lim_{x \to 1} (P_{\ell+1}^0)'(x) \right] - 1 - \frac{\ell(\ell + 1)}{2} \\ \lim_{x \to 1} (P_{\ell+1}^0)'(x) &= \ell + 1 + \frac{\ell(\ell + 1)}{2} = \frac{(\ell + 1)(\ell + 2)}{2} \\ \end{align*} $$

So $\lim_{x \to 1}( P_\ell^0 )'(x) = \frac{\ell(\ell + 1)}{2}$ for all $\ell$.

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$m = 1 : \infty$

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What do? I can't throw infinities about willy-nilly, because limit laws don't work as nicely and I can't "subtract infinity from both sides".

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$m = 2 : -\frac{(\ell - 1) \ell (\ell + 1)(\ell + 2)}{4}$

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Base: $P_2^2(x) = 3(1 - x^2)$. The derivative is continuous, and at $1$ it is $-6$, which is $-\frac{1 \cdot 2 \cdot 3 \cdot 4}{4}$.

Assume: $\lim_{x \to 1}( P_\ell^2 )'(x) = -\frac{(\ell - 1) \ell (\ell + 1)(\ell + 2)}{4}$ for some $\ell$:

$$ \begin{align*} \lim_{x \to 1}~( P_\ell^2)'(x) &= \lim_{x \to 1} \frac{(\ell - 1) (P_{\ell+1}^2)'(x) - (\ell + 1)(P_\ell^2(x) + x(P_\ell^2)'(x))}{2x} \\ -\frac{(\ell - 1) \ell (\ell + 1)(\ell + 2)}{4} &= \frac{(\ell - 1) \left[ \lim_{x \to 1} (P_{\ell+1}^2)'(x) \right] - (\ell + 1) \left(0 + -\frac{(\ell - 1) \ell (\ell + 1)(\ell + 2)}{4} \right)}{2} \\ 2 \left( - \frac{(\ell - 1) \ell (\ell + 1)(\ell + 2)}{4} \right) &= (\ell - 1) \left[ \lim_{x \to 1} (P_{\ell+1}^2)'(x) \right] - (\ell + 1) \left(-\frac{(\ell - 1) \ell (\ell + 1)(\ell + 2)}{4} \right) \\ -\frac{(\ell - 1) \ell (\ell + 1)(\ell + 2)}{4} (2 + (\ell + 1)) &= (\ell - 1) \left[ \lim_{x \to 1} (P_{\ell+1}^2)'(x) \right] \\ -\frac{\ell (\ell + 1)(\ell + 2)(\ell + 3)}{4} &= \left[ \lim_{x \to 1} (P_{\ell+1}^2)'(x) \right] \end{align*} $$

So $\lim_{x \to 1}( P_\ell^2 )'(x) = -\frac{(\ell - 1) \ell (\ell + 1)(\ell + 2)}{4}$ for all $\ell$.

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$m \ge 3 : 0$

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Base: can't prove it, but $P_\ell^m(1) = 0$ for all $\ell \ge m \ge 3$.

Assume: $\lim_{x \to 1}( P_\ell^m )'(x) = 0$ for all $m \ge 3$ and some $\ell \ge m$:

$$ \begin{align*} \lim_{x \to 1}~( P_\ell^m)'(x) &= \lim_{x \to 1} \frac{(\ell + 1 - m) (P_{\ell+1}^m)'(x) - (\ell + 1)(P_\ell^m(x) + x(P_\ell^m)'(x))}{2x} \\ 0 &= \frac{(\ell + 1 - m) \left[ \lim_{x \to 1} (P_{\ell+1}^m)'(x) \right] - (\ell + 1)(0 + 0)}{2} \\ 0 &= (\ell + 1 - m) \left[ \lim_{x \to 1} (P_{\ell+1}^m)'(x) \right] \\ \ell \ge m &\implies (\ell + 1 - m) \ne 0 \\ 0 &= \lim_{x \to 1} (P_{\ell+1}^m)'(x) \end{align*} $$

So for all $\ell \ge m \ge 3$: $\lim_{x \to 1}( P_\ell^m )'(x) = 0$.

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What I'm still stuck on:

• The recursive step of $m = 1$.
• The base case of $m \ge 3$.

Everything else can be found using $P_\ell^m(-x) = (-1)^{\ell + m} P_\ell^m(x)$ and $P_\ell^{-m}(x) = (-1)^m \frac{(\ell - m)!}{(\ell + m)!} P_\ell^m(x)$.