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MathGroup Archive 2005

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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
Graduate Student
UMBC
Department of Mechanical Engineering
Phone: 410 455 8134



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