Re: Iteratively determing successive values from previous values....

*To*: mathgroup at smc.vnet.net*Subject*: [mg80388] Re: [mg80369] Iteratively determing successive values from previous values....*From*: DrMajorBob <drmajorbob at bigfoot.com>*Date*: Tue, 21 Aug 2007 04:59:33 -0400 (EDT)*References*: <28579907.1187597940895.JavaMail.root@m35>*Reply-to*: drmajorbob at bigfoot.com

Clear[x] x[i_] /; i > 1 := x[i - 1] + 1 x /@ Range[10] {x[1], 1 + x[1], 2 + x[1], 3 + x[1], 4 + x[1], 5 + x[1], 6 + x[1], 7 + x[1], 8 + x[1], 9 + x[1]} Clear[x, y] x[i_] /; i > 1 := Simplify[x[i - 1] + s[i - 1] y[i - 1]/2^(i - 1)] y[i_] /; i > 1 := Simplify[y[i - 1] + s[i - 1] x[i - 1]/2^(i - 1)] x[5] (1/1024)((8 (128 + s[3] s[4] + 2 s[2] (2 s[3] + s[4])) + s[1] (32 (2 s[3] + s[4]) + s[2] (128 + s[3] s[4]))) x[1] + 2 (32 (2 s[3] + s[4]) + s[2] (128 + s[3] s[4]) + 2 s[1] (128 + s[3] s[4] + 2 s[2] (2 s[3] + s[4]))) y[1]) You should also look at Help for RSolve. Bobby On Mon, 20 Aug 2007 02:37:48 -0500, ashesh <ashesh.cb at gmail.com> wrote: > Hi, > > Need one help in determining the next value of a function from the > previous values. > > Example: > > x(i+1) = x(i) + 1; > > If one iterates to find x(5), we should get it as (x(1) + 4), as the > initial value is x(1), which is obtained > > x2 = x1+1; > x3 = x2 + 1 = (x1+1) + 2 = x1 + 2 > x4 = x3 + 1 = (x2+1) + 1 = x2 + 2 = (x1+1) + 2 = x1 + 3 > x5 = x4 + 1 = (x3+1) + 1 = x3 + 2 = (x2+1) + 2 = x2 + 3 = x1 + 4 > > Similary, I would like to know how to deal if there are two variable > expressions, say > > x(i+1) = x(i) - (s_i) * (2^-i) * y(i) > y(i+1) = y(i) + (s_i) *(2^-i) * x(i) > > for i =1, we have > > x_2 = x1 - s_1 (2^-1) y_1 > y_2 = y1 + s_1 (2^-1) x_1 > > and for i = 2 it becomes > x_3 = x1* (1 - (s_1) * (s_2) * (2^-3)) - y1* ((s_1) * (2^-1) + (s_2) > * (2^-2)) > y_3 = y1* (1 - (s_1) * (s_2) * (2^-3)) + x1* ((s_1) * (2^-1) + (s_2) > * (2^-2)) > > and so on ..... > > I have tried to use Table, Nest, and few other functional iteration > procedures. But, they seem to be giving the product of the terms > rather than substituting to give the SUM of the terms in iterative > manner. > > Hope some one can help me on this. > > > -- DrMajorBob at bigfoot.com