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I need to prove that, over the monic polynomials $f$ of degree $n$, the integral $$\int_{-1}^1\bigl(f(x)\bigr)^2\,dx$$ takes its minimum value when $$f(x)=\frac{2^n}{\binom{2n}n}L_n(x),$$ where $L_n(x)$ is the $n$th Legendre polynomial.

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What do you mean by "an is the reciprocal of the leading coefficient of Ln(x)"?, in the proof it self? Your result is not what I want to prove, it's upside down, it should be the inverse value right? –  user68315 Mar 24 '13 at 19:44
    
In my answer, I am using $a_n$ to denote the coefficient in front of $L_n(x)$ in the expansion of $f(x)$ as I write in the first bullet point in my answer. In order for $a_nL_n(x)$ to be monic, $a_n$ must be the reciprocal of the leading coefficient of $L_n(x)$. –  anon Mar 24 '13 at 23:18

1 Answer 1

Make sure you are either granted use of or can independently verify these facts:

  • The Legendre polynomials form a basis for the space of polynomials, so we may write $$f(x)=\sum_{k=0}^n a_kL_k(x)$$ for some real numbers $a_k\in\bf R$, for any $f(x)$ of degree $n$.
  • If $f(x)$ is monic, then $a_n$ is the reciprocal of the leading coefficient of $L_n(x)$. Looks like that means you have to verify that the leading coefficient of $L_n(x)$ is $2^{-n}{2n\choose n}$.
  • The Legendre polynomials are orthogonal, in the sense that $$\langle L_r,L_s\rangle=\int_{-1}^{+1}L_r(x)L_s(x)dx=0$$ when $r\ne s$. Therefore (with $\|\cdot\|$ induced from $\langle\cdot,\cdot\rangle$ defined directly above) $$\|f(x)\|^2=\sum_{k=0}^na_k^2\|L_k(x)\|^2\ge a_n^2\|L_n(x)\|^2$$ with equality if and only if $a_0=a_1=a_2=\cdots=a_{n-1}=0$.
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I really understood the idea of this proof but I've been working around with fourier coificients and all that's to get to the final result, verify that the leading coeficient of Ln(x) is that number. Does anyone help me out on that a little bit more?? Thanks –  deiota Mar 23 '13 at 23:52
    
@deiota: Do you have Rodrigues formula available to you? –  anon Mar 24 '13 at 1:09
    
Yes, it's[url]s2.postimg.org/4nnf7cwqx/image.jpg[/URL] –  deiota Mar 24 '13 at 10:32
    
@deiota Indeed. If you expand $(x^2-1)^n$ with the binomial expansion the leading term is $x^{2n}$; if you differentiate that $n$ times the leading coefficient is ... ? –  anon Mar 24 '13 at 10:50
    
HI got to the expansion expression with the leading term x^2n and th constant (-1)^n, I suppose the leading coeficient you asked me about is n. I'm I correct? Or (-1)^n.. –  deiota Mar 24 '13 at 11:39

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