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
- References:
- Recursion
- From: JikaiRF@aol.com
- Recursion