Re: Problem solving a difference equation
- To: mathgroup at smc.vnet.net
- Subject: [mg68332] Re: Problem solving a difference equation
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Tue, 1 Aug 2006 07:00:04 -0400 (EDT)
- Organization: The University of Western Australia
- References: <eaer8e$5a4$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <eaer8e$5a4$1 at smc.vnet.net>,
"aTn" <ayottes at dms.umontreal.ca> wrote:
> I am trying to solve the following difference equation:
>
> s[i] = s[i-1] ( 1 + d ( a/ (b + c s[i-1]) - 1 ))
>
> , where a,b,c and d are fixed real numbers.
>
> Here is what happens when I try RSolve:
>
> In[1] := RSolve[s[i] == s[i-1] ( 1+ d ( a/ (b + c s[i-1]) - 1)),
> s[i],i]
> Out[1] := RSolve[s[i] == s[i-1] (1+d ( a/ (b + c s[i-1]) - 1)),
> s[i],i]
>
> ,that is, nothing much happens :).
>
> I have a few questions:
>
> 1) Why are the input and output of RSolve the same ?
Because RSolve cannot solve the difference equation, at least as posed.
> 2) Can one solve the difference equation using RSolve ?
I do not know if RSolve can solve rational difference equations.
Note that as i -> Infinity, the limiting behaviour is
Solve[s == s ((a/(b + c s) - 1) d + 1), s]
{{s -> 0}, {s -> (a - b)/c}}
> 3) If the answer to question (2) is no, then do you have any
> suggestions on how to solve the equation.
It may be interesting to know how the recurrence arises.
Have a look at the literature on (first order) rational difference
equations. For example, Yanagihara [1] shows that for any rational
function R(y), the difference equation
y[z + 1] == R(y[z])
has a non-trivial meromorphic solution.
Also, perhaps "generatingfunctionology" by Herbert Wilf, at
http://www.math.upenn.edu/~wilf/DownldGF.html
will help.
[1] Yanagihara N 1980 Meromorphic solutions of some difference equations
Funkcial. Ekvac. 23 30926
Cheers,
Paul
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