Re: Using FindFit with constraints ??

*To*: mathgroup at smc.vnet.net*Subject*: [mg75710] Re: [mg75655] Using FindFit with constraints ??*From*: DrMajorBob <drmajorbob at bigfoot.com>*Date*: Wed, 9 May 2007 04:25:23 -0400 (EDT)*References*: <12357150.1178621269861.JavaMail.root@m35>*Reply-to*: drmajorbob at bigfoot.com

I can't read your code -- what the heck is =CF=89? -- but this code works just fine in version 6: model = a Cos[c t]; data = Table[{t, Cos[2.1 t]}, {t, 0, 10, .25}] + RandomReal[.1, 41]; fit = FindFit[data, {model, {a > 0, 1 < c < 2}}, {a, c}, t] {a -> 0.867956, c -> 2.} It doesn't work in version 5.2, because (1) RandomReal isn't defined and = (2) constraints (like {a > 0, 1 < c < 2}) are not supported in the older = version. Bobby On Tue, 08 May 2007 04:45:28 -0500, Alex <axel.kowald at rub.de> wrote: > Hello everybody, > > I saw at http://reference.wolfram.com/mathematica/ref/FindFit.html > that FindFit can be used with constraints for the parameters. However > when I try the example with Mathematica 5.2 I get an error. > > model = a Cos[=CF=89 t]; > data = Table[{t, Cos[2.1 t]}, {t, 0, 10, .25}] + RandomReal[.1, 41]; > fit = FindFit[data, {model, {a > 0, 1 < =CF=89 < 2}}, {a, =CF== 89}, t] > > => > FindFit::nlnum: The function value {-1. + Cos[RandomReal[0.1, 41.]] - > \ > 1=2E\RandomReal[0.1, 41.], \[LeftSkeleton]9\[RightSkeleton], \ > [LeftSkeleton]113\ > \[RightSkeleton]} is not a list of numbers with dimensions {123} at > {a, =CF=89} = \ > {1, 1}. > > > What am I doing wrong ?? > > Alex > > > -- = DrMajorBob at bigfoot.com