Decompositoin of f(x) in Legendre polynomials

[Apteronotus]

Hi,

In Wikipedia it's stated that
"...
Legendre polynomials are useful in expanding functions like

$$\frac{1}{\sqrt{1 + \eta^{2} - 2\eta x}} = \sum_{k=0}^{\infty} \eta^{k} P_{k}(x)$$
..."

Unfortunately, I am failing to see how this can be true. Is there a way of showing this?

I know that Legendre polynomials form an orthonormal set, and so given any function, we should be able to decompose it into a 'linear combination' of these polynomials. But what form does this decomposition take?

 

[HallsofIvy]

If u_i is an orthonormal basis for a vector space and v is any vector in that space, then v can be written as $v= \sum a_i u_i$. Taking the inner product of v with any member of the basis, say u_k gives $<u_k, v>= <u_k, \sum a_iu_i>= \sum a_i<u_k, u_i>= a_k$ because <u_k, u_k>= 1 while <u_k, u_i>=0 for any value of i other than k. That is, you can find the coefficient of P_k by taking the innerproduct (defined as an integral) of the function with P_k. According to Wikipedia (yes, I had to look it up!) the coefficients of the function f(x) for the Legendre polynomials is
$$<f, P_k>= \frac{1}{2n+1}\int_0^1 f(x)P_k(x)dx$$

 

[Apteronotus]

HallsofIvy
Thank you very much for your explanation. I understand now, the reasoning behind the equation.
On a technical note,
1. where does the term [tex] \frac{1}{2n+1}[tex]in front of the integral come from?
2. why do we integrate from 0-1?

and lastly,
3. would the value of [tex]/eta[tex] in our function [tex] \frac{1}{\sqrt{1 + \eta^{2} - 2\eta x}} [tex] play any role in our calculation of Pn's coefficients?

 

[Apteronotus]

Ok, so I think I have the answer to my first two questions and for anyone reading this thread, I'm going to try to answer them.
1.
The term $$\frac{1}{2n+1}$$
in front of the integral comes from the fact that the Lagendre polynomials are only orthogonal and not orthonormal.
By normalizing the polynomials, we can follow HallsofIvy's reasoning.
2.
The integral I believe should be taken from -1 to 1, since the polynomials Pn are orthogonal on -1$$\leq$$x$$\leq$$1.
Having said that, the term that is multiplied by the integral would be $$\frac{2}{2n+1}$$
rather than $$\frac{1}{2n+1}$$.

What HallsofIvy has done is that he's taken the parity (even/odd) of the functions into account. Since for any function f, the product of f and Pn is even.

I'm still working on 3. Hope this helps.

 

[Apteronotus]

Actually I think the terms
$$\frac{1}{2n+1}$$
and
$$\frac{2}{2n+1}$$
should in fact be
$$\frac{2n+1}{1}$$
and
$$\frac{2n+1}{2}$$
respectively.