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Re: Recursion

  • To: mathgroup at
  • Subject: [mg62305] Re: [mg62274] Recursion
  • From: Pratik Desai <pdesai1 at>
  • Date: Sat, 19 Nov 2005 23:18:32 -0500 (EST)
  • References: <>
  • Sender: owner-wri-mathgroup at

JikaiRF at wrote:

>Dear Sirs,
>I would like to obtain trajectories with respect to x, from the following
>                    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 

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

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