Behavior of Legendre polynomials $P_\ell(\cos\theta)$ under $\theta\to\pi-\theta$

Question

I'm studying some notes on the hydrogen atom which define the Legendre polynomials via Rodrigues's formula, $$P_\ell(z)=\frac{1}{2^\ell \ell!}\frac{d^\ell}{dz^\ell}(z^2-1)^\ell,\quad z=\cos\theta.$$ I'm trying to figure out what happens when $\theta\to\pi-\theta$ or equivalently, $z\to-z$. By computing some of the lowest order Legendre polynomials I believe I've convinced myself that $P_\ell(-z)=(-1)^\ell P_\ell(z)$, but I don't see how I can prove this given the above formula, i.e. I don't see what exactly will happen to the $\ell$-th order derivative. Is it mathematically meaningful to write $$P_\ell(-z)=\frac{1}{2^\ell \ell!}\frac{d^\ell}{d(-z)^\ell}[(-z)^2-1]^\ell?$$

Answer 1

If $w = -z$, then by the chain rule $$\frac{d}{dw} = \frac{dz}{dw}\frac{d}{dz} = -\frac{d}{dz}$$ Now $\ell$-application of the above yields $$\frac{d^\ell}{dw^\ell} = (-1)^\ell \frac{d^\ell}{dz^{\ell}}$$ Since also $(w^2 - 1)^\ell = (z^2 - 1)^\ell$, then $$P_\ell(w) = \frac{1}{2^\ell \ell!}\frac{d^\ell}{dw^{\ell}} (w^2 - 1)^\ell = \frac{1}{2^\ell \ell!} (-1)^\ell \frac{d^\ell}{dz^\ell}(z^2 - 1)^\ell = (-1)^\ell P_\ell(z)$$