Re: Functions That Remember Values They Have Found

*To*: mathgroup at smc.vnet.net*Subject*: [mg127655] Re: Functions That Remember Values They Have Found*From*: Dana DeLouis <dana01 at me.com>*Date*: Sun, 12 Aug 2012 01:48:57 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net

> Hgeom[n_] := > H[n] = Sum[(1 - 0.5)^i/(1 - 0.5^i) Hgeom[n - i], {i, 1, n}]/n > Hgeom[0] = 1; Hi. I see you have a solution. If you later want to eliminate the recursive part... fx[n_]:=((-1)^-n 2^(1/2 (-1+n) n))/QPochhammer[2,2,n] Hgeom3[18]//Timing {0.00296, = 11417981541647679048466287755595961091061972992/864390930149924278258644465301081415783295060766875} fx[18]//Timing {0.000073, = 11417981541647679048466287755595961091061972992/864390930149924278258644465301081415783295060766875} %%[[-1]]==%[[-1]] True fx[4]//N 0.203175 = = = = = = = = = = HTH :>) =E2=80=A8Dana DeLouis Mac & Mathematica 8 To understand recursion, one must first understand recursion. =E2=80=A8= = = = = = = = = = On Aug 6, 4:53 am, Esteban Gonz=C3=A1lez Morales <yo8... at gmail.com> wrote: > Hi, I was trying to create a recursive function and I read the help about it, and wrote this code > > Hgeom[n_] := > H[n] = Sum[(1 - 0.5)^i/(1 - 0.5^i) Hgeom[n - i], {i, 1, n}]/n > Hgeom[0] = 1; > > However, when I calculate Hgeom[10] it gives me the right value, and then I ask for the information about Hgeom and get that it has values calculated. > > Have you any idea what could have gone wrong?