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Re: sequence of functions

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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...@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

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Re: sequence of functions Dana DeLouis <dana01@me.com> - 2011-07-10 09:06 +0000

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