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MathGroup Archive 2005

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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


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