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