Re: sequence of functions
- To: mathgroup at smc.vnet.net
- Subject: [mg120130] Re: sequence of functions
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Sat, 9 Jul 2011 07:32:15 -0400 (EDT)
- References: <201107080855.EAA28789@smc.vnet.net>
- Reply-to: drmajorbob at yahoo.com
ClearAll[f, x] f[0][x_] = x (1 - x); f[n_Integer?Positive][x_] := f[n][x] = Simplify[f[n - 1][2 x] + f[n - 1][2 x - 1]] f[0][x] (1 - x) x f[1][x] -2 (1 - 2 x)^2 f[2][x] -20 + 64 x - 64 x^2 f[3][x] -8 (21 - 64 x + 64 x^2) ?f (output omitted) Bobby On Fri, 08 Jul 2011 03:55:33 -0500, rych <rychphd 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 > > -- DrMajorBob at yahoo.com
- References:
- sequence of functions
- From: rych <rychphd@gmail.com>
- sequence of functions