Re: iteration question

*To*: mathgroup at smc.vnet.net*Subject*: [mg122601] Re: iteration question*From*: "Dr. Wolfgang Hintze" <weh at snafu.de>*Date*: Thu, 3 Nov 2011 03:44:20 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <j8r9b8$3ie$1@smc.vnet.net>

Why not use the Reap, Sow mechanism described in the Help file? (* 1 *) Reap[FindRoot[Cos[x] == x, {x, 0}, EvaluationMonitor :> Sow[x]]] {{x -> 0.7390851332151607}, {{0., 1., 0.7503638678402439, 0.7391128909113617, 0.739085133385284, 0.7390851332151607}}} You can also combine your idea with this (* 2 *) c = 0; Reap[FindRoot[Cos[x] == x, {x, 0}, StepMonitor :> Sow[x, c++]]] c {{x -> 0.7390851332151607}, {{1.}, {0.7503638678402439}, {0.7391128909113617}, {0.739085133385284}, {0.7390851332151607}}} Out[35]= 5 Or, using only c (* 3 *) c = 0; Reap[FindRoot[Cos[x] == x, {x, 0}, StepMonitor :> Sow[c++]]] c Out[36]= {{x -> 0.7390851332151607}, {{0, 1, 2, 3, 4}}} Out[37]= 5 Regards, Wolfgang "Francisco Gutierrez" <fgutiers2002 at yahoo.com> schrieb im Newsbeitrag news:j8r9b8$3ie$1 at smc.vnet.net... > Dear Group: > > I have a function, and I iterate it using fixedpoint. Something of > this sort: > > FixedPointList[ > function[arg1,arg2,arg3, #] &, arg4, > SameTest -> (Max[Abs[Flatten[#1] - Flatten[#2]]] < 0.01 &)] > > > I would like to know how many steps it takes this function to > converge. I have tried with EvaluationMonitor to no avail: > > Block[{veamos, c = 0}, > veamos = FixedPoint[ > function[arg1,arg2,arg3, #] &, arg4, > SameTest -> (Max[Abs[Flatten[#1] - Flatten[#2]]]< 0.01 &)]; > EvaluationMonitor :> c++; {veamos, c}] > > > The iterator c simply does not move, and the previous function works > ok but returns the value of c as 0. > > I tried if this was true with Length[FixedPointList[...]] and the > answer was 20 (which should be the value of c in the immediately > previous function). In principle, this would solve my problem, but it > seems > rather inefficient, especially when the convergence criterion is > severe (not 0.01, but say 10^-4). > > Is there an efficient and nice way to solve this? > > Thanks, > > Francisco