       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 == A1,

b == 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  <-arbitrary starting value
> b=.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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