My Math Forum Legendre reccurence relation

 Real Analysis Real Analysis Math Forum

 November 6th, 2017, 12:23 AM #1 Senior Member   Joined: Sep 2012 Posts: 201 Thanks: 1 Legendre reccurence relation I am having a slight issue with generating function of Legendre polynomials and shifting the sum of the generating function. So here is an example: I need to derive the recurrence 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 books I have read ignore the shift and keep the sum $l=0$ which I can see because 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 sides 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 jus't cant see why or to be honest understand why it wrong. My belief it has some thing to do with how the generating function works, but every book I have read and youtube video I have watched completely ignores the step that I get confused with, and without any explanation just gives the recurrence relation. Anyway, here's what I do. Looking at the series, I make all the powers of h equal 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 books I have read just change the powers of h without make any shift 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 you're left with the desired recurrence relation, albeit with a little simplifying. Whereas when I do it my way, I expand all the series until the are l=3 and then do the same thing. 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 it if someone could help me on understanding this. Last edited by skipjack; November 6th, 2017 at 07:31 AM.

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post esthetic Calculus 1 April 27th, 2015 10:10 AM uniquegel Algebra 4 September 8th, 2014 05:18 PM Bromster Math Events 2 May 25th, 2013 01:00 PM tinynerdi Number Theory 8 September 14th, 2010 05:58 PM randarrthebarbarian Number Theory 3 October 8th, 2009 06:51 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top