Legendre Polynomial orthogonality

Question

It is given that $P_n (x) = \frac{d^n}{dx^n}(x^2-1)^n $

I have to show that these polynomials are a system of orthogonal polynomials.

I started like this :

$ \int_{-1}^{1} \frac{d^n}{dx^n}(x^2-1)^n * \frac{d^m}{dx^m}(x^2-1)^m $

with partial integration I came to

$ \left [ \frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n* \frac{d^m}{dx^m}(x^2-1)^m \right ]_{-1}^1 - \int_{-1}^1 \frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n * \frac{d^{m+1}}{dx^{m+1}}(x^2-1)^m dx $

So the first part is $0$ because of $|x|=1$ and the integral equals $0$ because the $(m+1)$ Derivation of a polynomial of order $m$ is $0$ ?

I am not sure at all if this is right, can someone tell me if it's okay or if I cant do it this way?

Your argument for the integral to vanish is not quite correct. Note that $(x^2-1)^m$ has degree $2m$ and thus $m+1$ derivative will in general not vanish — Severin Schraven, Jun 12 '17 at 18:01
i am not allowed to use the recursion formual unfortunatly :/. — wondering1123, Jun 12 '17 at 18:44
so it means i have to integrate the hole Thing one more time. Because then it will be m+2 ..and that would vanish? — wondering1123, Jun 12 '17 at 18:45
Did you read the lead from @StephenG ? ... Possible duplicate https://math.stackexchange.com/questions/2319494/how-do-i-calculate-the-right-value-of-this-integral — Donald Splutterwit, Jun 12 '17 at 18:46
hint: see my post on https://math.stackexchange.com/questions/725519/legendre-polynomial-orthogonality-and-size/765495#765495 — Maestro13, Jun 14 '17 at 16:29

Gabriel Romon · Answer 1 · 2022-12-02T16:33:14.650

The following derivation is self-contained, in the sense that it does not leverage properties of Legendre polynomials.

To facilitate notations, let $Q_n(X) = (X^2-1)^n$ and let $Q_n^{(n)}$ denote the $n$-th derivative of $Q_n$, so that $P_n = Q_n^{(n)}$.

Let $1\leq n< m$ and note that $$\langle P_n , P_m \rangle = \int_{-1}^1 Q_n^{(n)}(x) Q_m^{(m)}(x) dx = \Big[Q_n^{(n)}(x) Q_m^{(m-1)}(x)\Big]_{-1}^1 - \int_{-1}^1 Q_n^{(n+1)}(x) Q_m^{(m-1)}(x) dx.$$

Since $m\geq 1$, $1$ and $-1$ are roots of $Q_m$, each with multiplicity $m$ hence $$Q_m^{(m-1)}(-1) = Q_m^{(m-1)}(1)=0.$$

If $n=1$ we stop here, otherwise we integrate by parts repeatedly until $Q_m^{(m-n)}$ appears and we find $$\langle P_n , P_m \rangle = (-1)^n \int_{-1}^1 Q_n^{(2n)}(x) Q_m^{(m-n)}(x) dx,$$ since for every $k\in \{2,\ldots,n\}$, $0\leq m-k\leq m-1 \implies Q_m^{(m-k)}(-1) = Q_m^{(m-k)}(1)=0$.

The degree of $Q_n$ is $2n$, thus $Q_n^{(2n)}$ is a constant polynomial: $Q_n^{(2n)} = (2n)!$, hence $$\langle P_n , P_m \rangle = (-1)^n (2n)! \int_{-1}^1 Q_m^{(m-n)}(x) dx = (-1)^n (2n)! \Big[Q_m^{(m-n-1)}(x)\Big]_{-1}^1. $$ Since $0\leq m-n-1\leq m-1$, this last bracket is $0$, and finally $\langle P_n , P_m \rangle = 0$.

The same technique yields the value of $\|P_n\|$.

Legendre Polynomial orthogonality

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