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MathGroup Archive 2005

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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
> 


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