Re: Differences between recursions and limit

*To*: mathgroup at smc.vnet.net*Subject*: [mg60536] Re: [mg60509] Differences between recursions and limit*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Tue, 20 Sep 2005 05:18:51 -0400 (EDT)*Reply-to*: hanlonr at cox.net*Sender*: owner-wri-mathgroup at wolfram.com

The recursion is faster if you give it memory so that it doesn't have to redo calculations. Clear[x]; x[0]=0; x[k_]:=x[k]=0.25`30*(1+x[k-1]+x[k-1]^2+x[k-1]^3); Table[{k,x[k]},{k,0,150,10}]//TableForm The limit can be found exactly Solve[z==(1+z+z^2+z^3)/4,z][[3]] {z -> -1 + Sqrt[2]} N[%,30] {z -> 0.414213562373095048801688724209698078569671875377`30.} Bob Hanlon > > From: "anbra1" <xyxanbra1 at tiscalixxxyxxx.it> To: mathgroup at smc.vnet.net > Date: 2005/09/19 Mon AM 04:45:30 EDT > Subject: [mg60536] [mg60509] Differences between recursions and limit > > Please, > I want to know > why these two recursions,doing the same thing, > are the first one much more slow than the second > > ------------------------- > > x[0]=0; > x[k_] := .25 *(1 + x[k-1] + (x[k-1])^2 + (x[k-1])^3); > > For[k=1,k<12,Print[k," ",SetPrecision[x[k],30]];k++] > > ------------------------- > > a=0; > > For[k=1,k<12,k++, > > b= .25 (1 + a + a^2 + a^3);a:=b;Print[k," ",SetPrecision[b,30]]] > > ------------------------- > Another question > How can I calculate > the limit of this recursion for k->Infinite ? > Thank you all > > >