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 November 5th, 2017, 11:23 PM #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: $$\sum_{l=1}^{\infty} lh^{l-1} P_l(x))-2x\sum_{l=1}^{\infty} lh^{l} P_l(x))+\sum_{l=1}^{\infty} lh^{l+1} P_l(x)=x\sum_{l=0}^{\infty} h^l P_l(x)-\sum_{l=0}^{\infty} h^{l+1} P_l(x)$$ So now if I make eq =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 06:31 AM. Tags legendre, realtion, reccurence, relation, series or generating Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post esthetic Calculus 1 April 27th, 2015 09:10 AM uniquegel Algebra 4 September 8th, 2014 04:18 PM Bromster Math Events 2 May 25th, 2013 12:00 PM tinynerdi Number Theory 8 September 14th, 2010 04:58 PM randarrthebarbarian Number Theory 3 October 8th, 2009 05:51 AM

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