Re: sequence of functions

*To*: mathgroup at smc.vnet.net*Subject*: [mg120191] Re: sequence of functions*From*: Dana DeLouis <dana01 at me.com>*Date*: Tue, 12 Jul 2011 06:59:36 -0400 (EDT)

Hi. Just to be different... What we have... Function[x,a+b x + c x^2][2x]+Function[x,a+b x + c x^2][2x-1]; Collect[%,x] 2 a-b+c+(4 b-4 c) x+8 c x^2 We start with: x (1-x)//Expand x-x^2 Our data: equ={ a[0]==0, b[0]==1, c[0]==-1, a[n]==2 a[n-1]-b[n-1]+c[n-1], b[n]==(4 b[n-1]-4 c[n-1]), c[n]==8 c[n-1]}; RSolve[equ,{a[n],b[n],c[n]},n]//FullSimplify {{ a[n]->-(1/3) 2^n (4^n-1), b[n]->8^n, c[n]->-8^n }} So, our 5th sequence: -(1/3) 2^n (-1+4^n) /.n->5 -10912 8^n /.n->5 32768 -8^n /.n->5 -32768 Which checks with the answer given by others: -10912 + 32768*x - 32768*x^2 = = = = = = = = = = HTH : >) Dana DeLouis $Version 8.0 for Mac OS X x86 (64-bit) (November 6, 2010) To understand recursion, one must first understand recursion. = = = = = = = = = = On Jul 8, 5:07 am, rych <rych... at gmail.com> wrote: > I would like to build a function sequence inductively, for example > f[n][x]=f[n-1][2x]+f[n-1][2x-1], f[0][x] = x(1-x) > > This is what I tried in Mathematica, > > ClearAll["Global`*"] > f[n_] = Function[x, Simplify[f[n - 1][2 x] + f[n - 1][2 x - 1]]]; > f[0] = Function[x, x (1 - x)]; > > f[2][x] > Out: > -20 + 64 x - 64 x^2 > > ?f > Out: > Global`f > f[0]=Function[x,x (1-x)] > f[n_]=Function[x,Simplify[f[n-1][2 x]+f[n-1][2 x-1]]] > > My questions: it doesn't seem to cache the previous f[n-1] etc? If I > do it this way instead, > f[n_] := f[n] = Function[... > > Then it does cache the f[n-1] etc. but still in the unevaluated form, > Global`f > f[0]=Function[x,x (1-x)] > f[1]=Function[x$,Simplify[f[1-1][2 x$]+f[1-1][2 x$-1]]] > f[2]=Function[x$,Simplify[f[2-1][2 x$]+f[2-1][2 x$-1]]] > f[n_]:=f[n]=Function[x,Simplify[f[n-1][2 x]+f[n-1][2 x-1]]] > > How do I force the reduction so that I see f[1] = -2 (1 - 2 x)^2, etc. > in the definitions? > > Thanks, > Igor