Re: Differences between recursions and limit

*To*: mathgroup at smc.vnet.net*Subject*: [mg60552] Re: [mg60509] Differences between recursions and limit*From*: <bsyehuda at gmail.com>*Date*: Tue, 20 Sep 2005 05:19:08 -0400 (EDT)*References*: <200509190845.EAA23449@smc.vnet.net> <76e8f8180509190525636016ea@mail.gmail.com>*Reply-to*: bsyehuda at gmail.com*Sender*: owner-wri-mathgroup at wolfram.com

p.s. RSolve is not able to solve you recursive equation analytically. However, you can get a numerical estimate by using FixedPoint FixedPoint[0.25(1+#+#^2+#^3)&,0] Returns 0.414214 You can also use FixedPointList to look at intermediate values returned by the evaluation FixedPointList[0.25(1+#+#^2+#^3)&,0] this returns {0, 0.25, 0.332031, 0.36972, 0.389238, 0.399929, 0.40596, 0.409417, 0.411416, \ 0.412579, 0.413258, 0.413654, 0.413886, 0.414022, 0.414101, 0.414148, \ 0.414175, 0.414191, 0.4142, 0.414206, 0.414209, 0.414211, 0.414212, 0.414213, \ 0.414213, 0.414213, 0.414213, 0.414213, 0.414213, 0.414214, 0.414214, \ 0.414214, 0.414214, 0.414214, 0.414214, 0.414214, 0.414214, 0.414214, \ 0.414214, 0.414214, 0.414214, 0.414214, 0.414214, 0.414214, 0.414214, \ 0.414214, 0.414214, 0.414214, 0.414214, 0.414214, 0.414214, 0.414214, \ 0.414214, 0.414214, 0.414214, 0.414214, 0.414214, 0.414214, 0.414214, \ 0.414214, 0.414214, 0.414214, 0.414214, 0.414214, 0.414214, 0.414214, \ 0.414214} yehuda On 9/19/05, bsyehuda at gmail.com <bsyehuda at gmail.com> wrote: > > Hi, > With the first programming style you recalculate each value right from the > begining, so to calculate x[3] in terms of x[2], x[2] is calculated in terms > of x[1] which is calculated from x[0], etc. > The same happens for x[4], x[5], etc. > The second programming style use a temporaty storage (your recursion is > only one step back, so for this case it is sufficient), so calculation are > replaced by looking at the storage, instead of re-computing as in the first > exaample (one note, replace a:=b with a=b) > > You can achieve that automatically by using > x[0]=0; > x[k_]:=x[k]= .25 *(1 + x[k-1] + (x[k-1])^2 + (x[k-1])^3); > This works for any degree of recursion. However you need to be aware that > it consums memory. > yehuda > > On 9/19/05, anbra1 <xyxanbra1 at tiscalixxxyxxx.it> wrote: > > > > 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 > > > > > > >

**References**:**Differences between recursions and limit***From:*"anbra1" <xyxanbra1@tiscalixxxyxxx.it>