Re: Recursion question

*To*: mathgroup at smc.vnet.net*Subject*: [mg55513] Re: Recursion question*From*: dh <dh at metrohm.ch>*Date*: Sun, 27 Mar 2005 02:42:47 -0500 (EST)*References*: <d2347h$lhc$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi Dick, Consider 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, c[n - 1] == (a[n - 1] + b[n - 1])/2 We have the variables a[n],a[n-1],b[n],b[n-1] that we would like to eliminate. Obviously we have not enough equations because we need 4 equationes for the elimination and we need an additional one for the recursion. Therefore, we need one more step in the recursion. This will add 3 equations and 2 variables to eliminate: eq = { a[n] == 4*a[n - 1]*(1 - a[n - 1]), b[n] == 4*b[n - 1]*(1 - b[n - 1]), a[n - 1] == 4*a[n - 2]*(1 - a[n - 2]), b[n - 1] == 4*b[n - 2]*(1 - b[n - 2]), c[n] == (a[n] + b[n])/2, c[n - 1] == (a[n - 1] + b[n - 1])/2 c[n - 2] == (a[n - 2] + b[n - 2])/2 } Now we eliminate the unwanted variables: Eliminate[eq, {a[n], b[n], a[n - 1], b[n - 1], a[n - 2], b[n - 2]}] giving: c[-1 + n]*c[n] == c[-1 + n]*(16 - 128*c[-1 + n] + 384*c[-1 + n]^2 - 512*c[-1 + n]^3 + 256*c[-1 + n]^4) Dividing by c[-1 + n] gives: c[n] == 16 - 128*c[-1 + n] + 384*c[-1 + n]^2 - 512*c[-1 + n]^3 + 256*c[-1 + n]^4 Sincerely, Daniel 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 >