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MathGroup Archive 2005

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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



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