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Groups > comp.soft-sys.math.mathematica > #3624

Re: sequence of functions

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What about

f[0] = Function[x, x (1 - x)]
f[n_] := f[n] = 
   Function[x, Evaluate[Simplify[f[n - 1][2 x] + f[n - 1][2 x - 1]]]];

Then

f[2][x];
?f

returns:

Global`f
f[0]=Function[x,x (1-x)]
f[1]=Function[x$,-2 (1-2 x$)^2]
f[2]=Function[x$,-20+64 x$-64 x$^2]
f[n_]:=f[n]=Function[x,Evaluate[Simplify[f[n-1][2 x]+f[n-1][2 x-1]]]]


Heike



On 8 Jul 2011, at 09:55, rych 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 Heike Gramberg <heike.gramberg@gmail.com> - 2011-07-09 11:42 +0000
  Re: sequence of functions rych <rychphd@gmail.com> - 2011-07-11 10:56 +0000
  Re: sequence of functions DrMajorBob <btreat1@austin.rr.com> - 2011-07-12 11:07 +0000

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