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Math Help - Absolute value of Legendre polynomials ??

  1. #1
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    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
     \left| P_{n}(x) \right | \le \sqrt{\pi/(1-x^{2})}

    Thank you in advance!
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    Re: Absolute value of Legendre polynomials ??

    What have you tried so far? Have you tried induction?
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    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.
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    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 n, |P_n(x)| \leq f(x) for some function f(x). We have the Bonnet recursion formula

    (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

    (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 x. So,

    |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 n. I'll take a look at this complex integrals approach and let you know.
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    Re: Absolute value of Legendre polynomials ??

    Rearrange the formula for P_n(x) to get

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

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

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

    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
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    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 epsilon, however I don't understand why should it. Could you please give me some more help?
    Many thanks
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    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.
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