Re: Recursion

• To: mathgroup at smc.vnet.net
• Subject: [mg62305] Re: [mg62274] Recursion
• From: Pratik Desai <pdesai1 at umbc.edu>
• Date: Sat, 19 Nov 2005 23:18:32 -0500 (EST)
• References: <200511191053.FAA16405@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```JikaiRF at aol.com wrote:

>Dear Sirs,
>
>I would like to obtain trajectories with respect to x, from the following
>relationship:
>
>
>
>                    x(t+1)=a x(t)(1-x(t))
>
>
>
>Here, a means a constant with a constraint, 1 < a < 4, t   a positive integer
>and x a variable.
>I know this difference equation cannot be solved by mathematics directly,
>because it is not linear.
>So, I would like to follow a numerical series in terms of x, and I need to
>know how to program Mathematica for this.
>
>I look forward to your response.
>
>
>
>     Sincerely,
>
>     Fujio Takata.
>
>
>
The correct syntax for entering a reccurence equation in Mathematica is

eqn=x[1 + t] == (1 - x[t])*x[t]

I tried to solve it using RSolve but it was not able to solve it.
Mathematica Version 5.1. Perhaps in the newer version this is possible.

How about taking Z-transform? You still have to deal with the quadriatic
term

The other approach that I can think of is to use perturbation methods.
Programming that in mathematica is probably more trouble than it is
worth-- I have tried it previously. Again this is just my opinion and my
expertise in using pattern matching and such is quite limited.

Hope this helps

Pratik

--
Pratik Desai