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

*To*: mathgroup at smc.vnet.net*Subject*: [mg80389] Re: Iteratively determing successive values from previous values....*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Tue, 21 Aug 2007 05:00:04 -0400 (EDT)*Organization*: Uni Leipzig*References*: <fabnc6$419$1@smc.vnet.net>*Reply-to*: kuska at informatik.uni-leipzig.de

Hi, > Example: > > x(i+1) = x(i) + 1; would be Nest[#+1 &,x1,n] or f[x1_,0]:=x1 f[x1_,n_Integer]:=1+f[x1,n-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)) > With[{n=2}, Fold[ # + {-1, 1}*#*#2/2 &, {x, y}, Table[s[i], {i, n}]] ] > 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. hmm -- it seem s that it ork fine with Nest[] and Fold[] so what you have done ?? Regards Jens