Re: sequence of functions

*To*: mathgroup at smc.vnet.net*Subject*: [mg120159] Re: sequence of functions*From*: Dana DeLouis <dana01 at me.com>*Date*: Sun, 10 Jul 2011 05:02:24 -0400 (EDT)

f[n_]:= f[n]= Function[x,.... Hi. You have solutions. FindSequenceFunction suggests the following if interested: Your first two solutions: {x(1-x),-2 (1-2 x)^2}//Expand {x-x^2, -2+8 x-8 x^2} f[n_]:=1/3 2^(-3+n) (4-4^n+3 4^n x-3 4^n x^2) Table[f[k],{k,5}]//Expand {x-x^2, -2+8 x-8 x^2, -20+64 x-64 x^2, -168+512 x-512 x^2, -1360+4096 x-4096 x^2} = = = = = = = = = = HTH : >) Dana DeLouis $Version 8.0 for Mac OS X x86 (64-bit) (November 6, 2010) On Jul 8, 5:07 am, rych <rych... at gmail.com> wrote: > I would like to build a function sequence inductively, for example > f[n][x]=f[n-1][2x]+f[n-1][2x-1], f[0][x] = x(1-x) > > This is what I tried in Mathematica, > > ClearAll["Global`*"] > f[n_] = Function[x, Simplify[f[n - 1][2 x] + f[n - 1][2 x - 1]]]; > f[0] = Function[x, x (1 - x)]; > > f[2][x] > Out: > -20 + 64 x - 64 x^2 > > ?f > Out: > Global`f > f[0]=Function[x,x (1-x)] > f[n_]=Function[x,Simplify[f[n-1][2 x]+f[n-1][2 x-1]]] > > My questions: it doesn't seem to cache the previous f[n-1] etc? If I > do it this way instead, > f[n_] := f[n] = Function[... > > Then it does cache the f[n-1] etc. but still in the unevaluated form, > Global`f > f[0]=Function[x,x (1-x)] > f[1]=Function[x$,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