- #1

Taylor_1989

- 402

- 14

## Homework Statement

I am having a slight issue with generating function of legendre polynomials and shifting the sum of the genertaing function.So here is an example:

I need to derive the recurence relation ##lP_l(x)=(2l-1)xP_{l-1}(x)-(l-1)P_{l-2}##

so I start with the following equation:

$$(1-2xh+h^2)\frac{\partial\phi}{\partial h}=(x-h)\phi $$

now taking the differential of the series of the Legendre generating function

$$\frac{\partial}{\partial h}(\sum_{l=0}^{\infty} h^l P_l(x))$$

I make it equal to $$\sum_{l=1}^{\infty} lh^{l-1} P_l(x))$$

Some book I have read ignor the shift and keep the sum ##l=0## which I can see beacuse even at the ##l=0## the sums are the same.But when I expand the following is when I get in a bit of a mess.

So expanding both side I get the following:

$$\sum_{l=1}^{\infty} lh^{l-1} P_l(x))-2x\sum_{l=1}^{\infty} lh^{l} P_l(x))+\sum_{l=1}^{\infty} lh^{l+1} P_l(x)=x\sum_{l=0}^{\infty} h^l P_l(x)-\sum_{l=0}^{\infty} h^{l+1} P_l(x)[1]$$

So now if I make eq [1]=0 like so:

$$\sum_{l=1}^{\infty} lh^{l-1} P_l(x))-2x\sum_{l=1}^{\infty} lh^{l} P_l(x))+\sum_{l=1}^{\infty} lh^{l+1} P_l(x)-x\sum_{l=0}^{\infty} h^l P_l(x)+\sum_{l=0}^{\infty} h^{l+1} P_l(x)=0$$

So from here I know that I need to get all the powers of h to ##h^{l-1}##

So here what I do, which I believe is incorrect I just can't see why or to be honest understand why it wrong. My belief it it has some thing to do with how the generating function works, but every book I have read and youtube video I have watch completely ignores the step that I get confused with, and without any explanation just give the recurrence relation. Anyway here what I do.looking at the series I make all the powers of h equalt to ##l-1##

$$\sum_{l=1}^{\infty} lh^{l-1} P_l(x))-2x\sum_{l=2}^{\infty} (l-1)h^{l-1} P_{l-1}(x))+\sum_{l=3}^{\infty} (l-2)h^{l-1} P_{l-2}(x)-x\sum_{l=1}^{\infty} h^{l-1} P_{l-1}(x)+\sum_{l=2}^{\infty} h^{l-1} P_{l-2}(x)=0$$

Now youtube videos I have watched and book I have read just change the powers of h without make any shitf to the series like soo.

$$\sum_{l=0}^{\infty} lh^{l-1} P_l(x))-2x\sum_{l=0}^{\infty} (l-1)h^{l-1} P_{l-1}(x))+\sum_{l=0}^{\infty} (l-2)h^{l-1} P_{l-2}(x)-x\sum_{l=0}^{\infty} h^{l-1} P_{l-1}(x)+\sum_{l=0}^{\infty} h^{l-1} P_{l-2}(x)=0$$

Then just factor out the sum and the h and your left with the desired recurrence relation albeit with a little simplifying.

Where as when I do it my way I expand all the series unitl the are l=3 and then do the same thing. But as I said I believe this to be wrong, but I just can't see how that in the book I have read and video I have watched how they can just change the powers of the h without affecting the series itself.

I would much appreciate if someone could help me on understanding this.