# Absolute value of Legendre polynomials ??

• Oct 15th 2012, 07:52 AM
Johan13
Absolute value of Legendre polynomials ??
Hi guys,

I would like to ask for some help regarding a problem with Legendre polynomials:

P_n(x) is an arbitrary Legendre polynomial (0<=n and -1<x<1). Prove the following
$\displaystyle \left| P_{n}(x) \right | \le \sqrt{\pi/(1-x^{2})}$

• Oct 17th 2012, 01:08 AM
Vlasev
Re: Absolute value of Legendre polynomials ??
What have you tried so far? Have you tried induction?
• Oct 18th 2012, 12:58 PM
Johan13
Re: Absolute value of Legendre polynomials ??
Well, I don't think that induction is the right choice, because my professor said that I should start from the Rodrigues-formula (Rodrigues' formula - Wikipedia, the free encyclopedia) and than use some complex Cauchy-integrals. However, my efforts weren't successful yet.
So, I hope someone can give me some ideas how to approach this problem.
• Oct 18th 2012, 11:33 PM
Vlasev
Re: Absolute value of Legendre polynomials ??
Alright, let's see. Here is what would have happened if you tried induction specifically. Suppose that up to and including some $\displaystyle n$, $\displaystyle |P_n(x)| \leq f(x)$ for some function $\displaystyle f(x)$. We have the Bonnet recursion formula

$\displaystyle (n+1)P_{n+1}(x) = (2n+1)x P_n(x) - n P_{n-1}(x)$

which you can derive from the other formula. Now

$\displaystyle (n+1)|P_n(x)| \leq (2n+1)|x||P_n(x)| + n |P_{n-1}(x)| \leq (3n+1)f(x)$

all obtained by the triangle inequality and the bounds on $\displaystyle x$. So,

$\displaystyle |P_{n+1}(x)| \leq \frac{3n+1}{n+1}f(x) = \left(3-\frac{2}{n+1}\right)f(x) \leq 3f(x)$

So, technically, you can do an induction but the function you bound with will depend on $\displaystyle n$. I'll take a look at this complex integrals approach and let you know.
• Oct 18th 2012, 11:57 PM
Vlasev
Re: Absolute value of Legendre polynomials ??
Rearrange the formula for $\displaystyle P_n(x)$ to get

$\displaystyle 2^n n! P_n(x) = \frac{d^n}{dx^n}(x^2-1)^n$

The last one we can obtain by the $\displaystyle n$-th Cauchy integral formula

$\displaystyle \frac{d^n}{dx^n}(x^2-1)^n)= \frac{n!}{2\pi i} \oint_{\gamma}\frac{(w^2-1)^n}{(w-x)^{n+1}}dw$

where $\displaystyle \gamma$ is a circle with radius, say $\displaystyle 1+\epsilon$, centered at the origin. So, we have obtained an integral formula for the Legendre polynomials

$\displaystyle P_n(x) = \frac{2^{-n}}{2\pi i}\oint_{\gamma}\frac{(w^2-1)^n}{(w-x)^{n+1}}dw$

Now it might just be an exercise in estimating this integral
• Oct 22nd 2012, 01:18 PM
Johan13
Re: Absolute value of Legendre polynomials ??
I really appreciate your help Vlasev.
Unfortunately, I think that I am stuck again. I tried to estimate a absolute value of the integral, but what I get that isn't close to the original upper limit. I tried using the fact the absolute value of an integral is smaller than the integral of an absolute value.
The results somehow seems to depend on the value of $\displaystyle epsilon$, however I don't understand why should it. Could you please give me some more help?
Many thanks
• Oct 23rd 2012, 12:31 AM
Vlasev
Re: Absolute value of Legendre polynomials ??
Are you sure that your bound is correct, because in actuality, you can even divide by n or something inside the square root and it'll still be good.