|
| Author | Message / Information |
| jim28 Quote | Reply | | This is driving me completely NUTS! posted on: 11/17/2005 2:24:41 PM Hi everyone I have the following pattern I need to try and make since of and I really need help soooo bad! First of all given any period for n I want to be able to find P n+1 Pn * x – P0*y ^ n - P0 * y^n+1=Pn+1 Where P0 is the first in the Pn sequence. Does this make since? There is a definite pattern but I just don't seem able to crack it! Many thanks in advance! |
|
Phoenix1177
Quote | Reply | |
This is driving me completely NUTS!
replied on: 8/17/2006 12:43:05 AM We define the generating function of s sequence f(n) as G[f](X)G(X)=sum(i=0 to infinity, f(n)X^n). Let G(X) be the gen function of your sequence Pn, let p=P0, a=x , b=y; I am replacing your notation from the orginal equation. We have that: aG(X)-pG[b^n](X)-pG[b^(n+1)](X)=G[P(n+1)](X)=(G(X)-p)/X. We simplify: G(X)=-p(XG[b^n](X)+bXG[b^n](X)-1)/(1-aX). Using a little manipulation of formal power series will quickly give that G[b^(n)](X)=1/(1-bX). Thus, G(X)=-p(X+2bX-1)/(1-ax)(1-bX). As power series are term by term equal, the coefficents of G(X) will be equal to the coefficents of the power series expansion of the right hand side. It would be tedious to perform that calculation here; so I leave it to you. If you are unsure of how to proceed; for a differentiable function f(x) the nth coefficent of its power series expansion about 0 is [D^n]f(0)/n!, where [D^n]f is the nth derrivative of f. If you have any questions about anything in my reply, feel free to ask:) |
|
LinkBot
|
Gamers Wanted is looking for people to write game reviews and post news, |
|
|
| Tired of seeing ads? Click here to upgrade to Elite Membership! |
ChatArea.com Help & News Forums | Terms of Use | Contact ChatArea.com | Advertising
Powered By ChatArea.com - Get your free Society today! © Copyright 2003 Wewp!