Re: iterative method?

*To*: mathgroup at smc.vnet.net*Subject*: [mg95742] Re: iterative method?*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>*Date*: Sun, 25 Jan 2009 21:51:24 -0500 (EST)*Organization*: The Open University, Milton Keynes, UK*References*: <glhjnu$is$1@smc.vnet.net>

In article <glhjnu$is$1 at smc.vnet.net>, matt <trex1704 at yahoo.com> wrote: > How to find x_(n+1)=(a*x_n+b*y_n)/c*x_n , y_(n+1)=d*x_n*y_n using > mathematica? > I want to find a list of x and y when n=1,2,...,50 > so I defined the function as f[{x_,y_}]:={a*x_n+b*y_n)/c*x_n,d*x_n*y_... } > and then use the function NestList[f,{2,5},50] with initial value of x is 2 > and y is 5 > with this, I got the list of 50 values of (x,y). > The problem is I want to plot graphs of x against n and y against n, n from 1 > to 50 because I want to see what happened if n goes to infinity. But if use > the method I just mentioned, I can only plot x against y. > So do anyone have a clever method?? > Btw...a,b,c and d are constants and given and x_n is supposed to be x > (subscript n)...same case with x_(n+1) and the others You could use recursive definitions for x_n and y_n. For instance, ClearAll[x, y]; x[1] = 2; y[1] = 5; x[n_Integer /; n > 1] := x[n] = (a*x[n - 1] + b*y[n - 1])/(c*x[n - 1]) y[n_Integer /; n > 1] := y[n] = d*x[n - 1]*y[n - 1] Block[{a = 2, b = 3, c = 5, d = 7}, Print[Table[{x[n], y[n]}, {n, 1, 5}]]; Print[ListLogPlot[Table[{n, x[n]}, {n, 1, 30}], Filling -> Bottom]]; Print[ListLogPlot[Table[{n, y[n]}, {n, 1, 30}], Filling -> Bottom]]; ] Regards, --Jean-Marc