Re: Recursion question

*To*: mathgroup at smc.vnet.net*Subject*: [mg55527] Re: Recursion question*From*: "Dana DeLouis" <delouis at bellsouth.net>*Date*: Mon, 28 Mar 2005 02:42:05 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

In Mathematica 5.1.1, it seems to work better if you take out the definition of c[n] for now. Here, A1 would equal 1/10, and B1 = 8/10. v = { a[n] == 4*a[n - 1]*(1 - a[n - 1]), b[n] == 4*b[n - 1]*(1 - b[n - 1]), a[1] == A1, b[1] == B1}; Off[Solve::ifun] sol = Flatten[FullSimplify[ {a[n], b[n]} /. RSolve[v, {a[n], b[n]}, n]]] So, here's a[n], and b[n]. { Sin[2^(n - 2)*ArcCos[1 - 2*A1]]^2, Sin[2^(n - 2)*ArcCos[1 - 2*B1]]^2 } (* Definition for c[n] - > c[n]==(a[n]+b[n])/2 *) FullSimplify[Total[sol]/2] (1/2)* (Sin[2^(n - 2)*ArcCos[1 - 2*A1]]^2 + Sin[2^(n - 2)*ArcCos[1 - 2*B1]]^2) I note that when n gets to only about 3 or 4, both a[n] & b[n] tend to be equal, and the average of the two (c[n]) is the same as either a[n] or b[n]. HTH -- Dana DeLouis <rbedient at hamilton.edu> wrote in message news:d2347h$lhc$1 at smc.vnet.net... >I have a set of single step recursion equations that I want to simplify > into a single multi-step equation. Here's what it looks like: > > a[n]=4*a[n-1]*(1-a[n-1]) > b[n]=4*b[n-1]*(1-b[n-1]) > c[n]=(a[n]+b[n])/2 > a[1]=.1 <-arbitrary starting value > b[1]=.8 <-arbitrary starting value > > What I'm hoping for is something like: > > c[n]=some function of c[n-1], c[n-2]... > > I've tried various combinations of Solve, RSolve, Simplify etc. to no > avail. Any help would be appreciated. > > Fairly Newbie > > Dick