Generating function for Legendre polynomials

[NanakiXIII]

Homework Statement

Using binomial expansion, prove that

$$\frac{1}{\sqrt{1 - 2 x u + u^2}} = \sum_{k} P_k(x) u^k.$$

Homework Equations

$$\frac{1}{\sqrt{1 + v}} = \sum_{k} (-1)^k \frac{(2k)!}{2^{2k} (k!)^2} v^k$$

The Attempt at a Solution

I simply inserted $v = u^2 - 2 x u$, then expanded the [itex]v^k[/tex] to obtain the double sum<br /> <br /> $$\sum_{k} (-1)^k \frac{(2k)!}{2^{2k} (k!)^2} \sum_{n \leq k} \left( \begin{array}{c} k \\ n \end{array} \right) (-2 x)^n u^{2 k - n}.$$<br /> <br /> Now I need to turn this into a single sum by collecting like powers of $u$, which is what I'm stuck at. I don't see how to go about that.[/itex]

 

[vela]

You might find this relationship helpful. It's just a rearrangement of terms in a series.

$$\sum_{p=0}^\infty\sum_{q=0}^p\,a_{q,p-q} = \sum_{r=0}^\infty\sum_{s=0}^{[r/2]}\,a_{s,r-2s}$$

where [r/2] = r/2 if r is even or (r-1)/2 if r is odd.

 

[NanakiXIII]

I suppose you mean to apply this using $p = 2 k$, so that my powers of $u$ become $2 k - 2 n$ instead of $2 k - n$? I'm not sure how that simplifies things. I'm still not sure what kind of expressions I'm going to get in front of my $u$'s.

 

[vela]

I think you have to have p=k and q=n from comparing your expression to the LHS of the relationship.

What definition do you have for the Legendre polynomials? I'm wondering what you should be shooting for.

 

[vela]

You can factor out u^k from the second summation and swap the exponents on the (-2x) and u to get

$$\sum_{k=0}^\infty (-1)^k \frac{(2k)!}{2^{2k} (k!)^2} u^k \sum_{n=0}^k \begin{pmatrix}k \\ n \end{pmatrix} (-1)^{k-n} (2x)^{k-n} u^n = \sum_{k=0}^\infty \sum_{n=0}^k (-1)^{n} \frac{(2k)!}{2^{2k} (k!)^2} u^k \begin{pmatrix}k \\ n \end{pmatrix} (2x)^{k-n} u^n$$

After you switch to the new indices, u^s will be multiplied by a polynomial with just even or odd powers of x, just like the Legendre polynomials.

 

[NanakiXIII]

Though I hadn't given it a lot of thought yet, I expected to have to use the Rodriguez formula. I now see where you are going, though, since I just discovered the expression

$$P_k(x) = \frac{1}{2^n} \sum_{n=0}^{[k/2]} (-1)^n \frac{(2k-2n)!}{n!(k-n)!(k-2n)!} x^{k-2n}.$$

So, after using the relation you suggested, I get the sum

$$\sum_k \sum_{n=0}^{[k/2]} (-1)^k \frac{(2k-2n)!}{2^{2k-n} (k-n)! n! (k-2n)!} x^n u^{2k-3n}.$$

This is starting to look like what I need, but not quite yet.

As for your second post, I'm not sure what you're doing. First of all, your expression isn't equal to mine, since you have $u^{k+n}$, while I have $u^{2k-n}$. Is my expression incorrect? Second, what do you mean by 'swap the exponents'?

 

[vela]

As for your second post, I'm not sure what you're doing. First of all, your expression isn't equal to mine, since you have $u^{k+n}$, while I have $u^{2k-n}$. Is my expression incorrect? Second, what do you mean by 'swap the exponents'?

Both expressions are correct. It's just how you applied the binomial theorem. You wrote

$$(x+y)^n = \sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} x^k y^{n-k}$$

while I'm suggesting you instead write

$$(x+y)^n = \sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} x^{n-k} y^k$$

That's what I meant by "swap the exponents." It's just to make the exponent of u look more like what you're shooting for.

 

[NanakiXIII]

Ah, I see what you mean now. I hadn't actually used this form of the binomial expansion, I was using the special case $(1+x)^k$ and was factoring out to get that form, so the symmetry got a little lost. Using the second form of the expansion indeed made things a lot neater and I've got my answer now. Thanks a lot!