# Proving Rodrigues's formula

Recall that the Legendre polynomials $\{P_n(t)\}, n = 0, 1,\dotsc$ are defined by applying the Gram–Schmidt process to the monomials $\{1, t, t^2,\dotsc \}$ in $L^2[−1, 1]$ and by rescaling the resulting orthonormal vectors ${x_n(t)}$ as follows:

$$P_n(t) = \sqrt{\frac{2}{2n+1}} \cdot x_n(t).$$

Show that

$$P_n(t) = \frac{1}{2^n n!} \frac{d^n}{dt^n}(t^2 - 1)^n.$$

Let $Q_n(t)=\frac{1}{2^n n!}\frac{d^n}{dt^n}(t^2-1)^n$,$y_n=\frac{\sqrt{2n+1}}{2}Q_n(t)$. We aim to show $y_n=x_n \forall n\in \mathbb{N}$

First we show $y_n$ is orthonormal, which can be check by definition after some calculations. I skip the details here.

Let $Y_n=\text{span} \{x_0, x_1, \cdots, x_n\}$. Then $Y_n=\text{span} \{y_0, y_1, \cdots, y_n\}$, this is because $\{y_i\}_{i=0}^{n}$ are orthogonal, hence $\text{dim} \,\text{span} \{y_0, y_1, \cdots, y_n\}=n+1$ and each $y_i \in Y_n$. Thus $y_n=\sum_{k=0}^{n} \langle y_n,x_k\rangle x_k$. Since $y_n$ is orthogonal to $Y_{n-1}$, $\langle y_n,x_k\rangle>=0,\, \forall k=0,\cdots , n-1$. So $y_n=\langle y_n,x_n\rangle x_n$.

Take norm on both sides, $|y_n|=|\langle y_n,x_n\rangle||x_n|$, by normality, $\langle y_n,x_n\rangle=1$ or $-1$.

Now observe $y_n$ has positive leading coefficients by definition, so $y_n(t)>0$ when $t$ is large.

$x_n$ has positive leading coefficient 1 by Grad-Schmidt Process, then $x_n(t)>0$ when $t$ is large.

Hence $\langle y_n,x_n\rangle=1$, we can conclude $Q_n=P_n$ by definition of $P_n$.

• What does "when t is large" mean here? The setting is L_2[-1, 1], so isn't |t| <= 1? – Xindaris Oct 17 '14 at 20:27
• @Xindaris Yes. But $x_n,y_n$ are two polynomials, if they agrees on the some interval, they must agree on $\mathbb{R}$. – John Oct 18 '14 at 3:12