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
- References:
- Recursion question
- From: rbedient@hamilton.edu
- Recursion question