Re: Recursion question

*To*: mathgroup at smc.vnet.net*Subject*: [mg55519] Re: [mg55498] Recursion question*From*: DrBob <drbob at bigfoot.com>*Date*: Sun, 27 Mar 2005 02:43:02 -0500 (EST)*References*: <200503260739.CAA21875@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

Solving for a, b, and c is easy: Off[Solve::"ifun"] Clear[a, b, c] a[n] = a[n] /. First@RSolve[{a[n] == 4*a[n - 1]*(1 - a[n - 1]), a[1] == k1}, a[n], n] b[n] = b[n] /. First@RSolve[{b[n] == 4*b[n - 1]*(1 - b[n - 1]), b[1] == k2}, b[n], n] c[n] = (a[n] + b[n])/2 // Simplify (1/2)*(1 - Cos[2^(-1 + n)*ArcCos[1 - 2*k1]]) (1/2)*(1 - Cos[2^(-1 + n)*ArcCos[1 - 2*k2]]) (1/4)*(2 - Cos[2^(-1 + n)*ArcCos[1 - 2*k1]] - Cos[2^(-1 + n)*ArcCos[1 - 2*k2]]) Set k1 and k2 equal to 1/10 and 8/10. > What I'm hoping for is something like: > > c[n]=some function of c[n-1], c[n-2]... I doubt that's possible, or... if it is, it won't be anything you'll like. Bobby On Sat, 26 Mar 2005 02:39:30 -0500 (EST), <rbedient at hamilton.edu> wrote: > 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 > > > > -- DrBob at bigfoot.com

**References**:**Recursion question***From:*rbedient@hamilton.edu