Decompositoin of f(x) in Legendre polynomials

[Apteronotus]

Hi,

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

<br /> \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 &lt;u_k, v&gt;= &lt;u_k, \sum a_iu_i&gt;= \sum a_i&lt;u_k, u_i&gt;= 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
&lt;f, P_k&gt;= \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 \frac{1}{2n+1}in front of the integral come from?&lt;br /&gt; 2. why do we integrate from 0-1?&lt;br /&gt; &lt;br /&gt; and lastly,&lt;br /&gt; 3. would the value of /etain our function \frac{1}{\sqrt{1 + \eta^{2} - 2\eta x}} play any role in our calculation of Pn&amp;amp;amp;amp;amp;#039;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 <br /> \frac{1}{2n+1}<br />
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\leqx\leq1.
Having said that, the term that is multiplied by the integral would be <br /> \frac{2}{2n+1}<br />
rather than <br /> \frac{1}{2n+1}<br />.

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.