1
$\begingroup$

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

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

$\endgroup$
2
2
$\begingroup$

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 $P_n$.

$\endgroup$
2
  • $\begingroup$ What does "when t is large" mean here? The setting is L_2[-1, 1], so isn't |t| <= 1? $\endgroup$ – Xindaris Oct 17 '14 at 20:27
  • 1
    $\begingroup$ @Xindaris Yes. But $x_n,y_n$ are two polynomials, if they agrees on the some interval, they must agree on $\mathbb{R}$. $\endgroup$ – John Oct 18 '14 at 3:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.