If $P_n(1)=1$ calculate $P'_n(1)$ in Legendre polynomials

Question

$P_n(x)$ is in $[-1,1]$ and $P_n(1)=1$ .The problem is getting $P'_n(1)$. On Wikipedia it says that it is $\frac{n(n+1)}2$.

I derive the problem showed here How could I prove that $P_n (1)=1=-1$ for the Legendre polynomials? in order to get P'(n) but it didn't helped so much.

Answer 1

It follows by induction from the recurrence $$ (n+1) P_{n+1}(x) = (2n+1) x P_n(x) - n P_{n-1}(x) $$ Just differentiate both sides and use that $P_n(1)=1$.

Answer 2

The generating function for Legendre polynomials is \begin{eqnarray*} \sum_{n=0}^{\infty} P_n(z) h^n =\frac{1}{\sqrt{1-2hz+h^2}}. \end{eqnarray*} Differentiate this w.r.t $z$ \begin{eqnarray*} \sum_{n=0}^{\infty} P^{'}_n(z) h^n =\frac{(-1/2)(-2h)}{(1-2hz+h^2)^{3/2}}. \end{eqnarray*} Set $z=1$ \begin{eqnarray*} \sum_{n=0}^{\infty} \color{red}{P^{'}_n(1)} h^n =\frac{h}{(1-h)^{3}} =\sum_{n=0}^{\infty} \color{red}{\binom{n+1}{2}} h^n. \end{eqnarray*}

Answer 3

Alternatively, it follows directly from Legendre's differential equation of $$(1 - x^2) y'' - 2xy' + n(n + 1) y = 0.\tag1$$

Since $y = P_n(x)$ is a solution to (1) we have $$(1 - x^2) P''_n(x) - 2x P'_n (x) + n(n + 1) P_n (x) = 0.$$ Setting $x = 1$ in the above equation leads to $$P'_n (1) = \frac{n(n + 1)}{2} P_n (1).$$ But since you already known that $P_n (1) = 1$, then the desired result immediately follows.