Re: Re: problem with nonlinearfit

*To*: mathgroup at smc.vnet.net*Subject*: [mg25589] Re: [mg25578] Re: [mg25489] problem with nonlinearfit*From*: "Dragan Grgic" <Dragan.Grgic at ensg.inpl-nancy.fr>*Date*: Mon, 9 Oct 2000 21:43:29 -0400 (EDT)*References*: <001901c031c0$5a879250$f8dcc180@als2127a.als.orst.edu>*Sender*: owner-wri-mathgroup at wolfram.com

Mark, Thanks for your help, but I have not converted some of my parameters to logs (e). In fact, I have first use reduced y variable in order to reduce the scale of its values: y=e'=e*100000; then K'=(100000^(1/N*a))/K. But the scale of the y variable is too large again (0 to ~ 200) And I have convert the y' variable into Log(e); then the scale of the values is lineary and compressed. Dragan GRGIC. > Dragan, > A rough answer to your question: > > >Why the y variable of a function y=f(x) must have a reduced range of values > to perform a nonlinear fit? > > Because the part of NonlinearRegress that calculates step sizes needs to > invert the curvature matrix of the problem, and that matrix depends on the > scales of the different parameters in the parameter vector. The routine > doesn't "know" anything about the units in which those parameters are > expressed, it only "sees" a matirx of elements that are, maybe, different in > magnitude by 10^10, or something, so the problem of inverting the matrix is > unstable, and it either catches this problem and generates an error msg., or > it goes ahead and calculates an unreasonable step size, sending your search > into the nether reaches of parameter space, where, maybe, your model > function could take on even imaginary values, or whatever. > This is commonly a problem when fitting chemical models with chemical > data, where one is trying to determine equilibrium constants that could be > very small, or very much larger than one. In this case, it is better to > express the model in terms of standard free energies, which are proportional > to Ln[equilibrium constant], which compresses those parameters. > I suspect that since converting some of your parameters to logs has > helped you, you might be having the same problem. Either use logs, or scale > linearly (see a reference on nonlinear regression, or optimization, for a > discussion). > -mark harder > > > > -----Original Message----- > From: Dragan Grgic <Dragan.Grgic at ensg.inpl-nancy.fr> To: mathgroup at smc.vnet.net > To: mathgroup at smc.vnet.net <mathgroup at smc.vnet.net> > Date: Sunday, October 08, 2000 10:30 PM > Subject: [mg25589] [mg25578] Re: [mg25489] problem with nonlinearfit > > > >Thank you for your answer Timothy, > >but I have already realized that it is not possible to specify a range to > >use for parameter values in NonlinearFit (or NonlinearRegress). In fact, > the > >problem seems to be the difference between values of the two variables (e > >and t): > >e is the deformation and its range is: 0 to 10^-5 (without unit) > >t is the time and its range is: 0 to 10^6 seconds > >a and N are exposants: 0,2 to 1 for a and ~5 for N. > >For the time it is possible to use an other unit (days for example); then > >the range of the unit is: 0 to 10 days. > >For the deformation, the only solution is to use a reduced variable (i.e. > >e'=e*100000). Then the parameter K is reduced too. But mathematica reports > >always the same error messages: > >"Objetive function or gradient is not real at {K,N,A}={in general starting > >values}" > >The final solution that I have found is to use the Log function (logarithm > >to base e) for the variable e', then the function is: Log[e'] = > >Log[(s/K')^(Na)*(t/a)^a] > >The Logarithm makes it possible to reduce the range of the e' variable (~0 > >to ~5). Then, all the variables and all the parameters are of the same > >order; and the program work well. > >Why the y variable of a function y=f(x) must have a reduced range of values > >to perform a nonlinear fit? > > > >Thanks, Dragan. > > > > > > > > > >----- Original Message ----- > >From: "Timothy Stiles" <tastiles at students.wisc.edu> To: mathgroup at smc.vnet.net > >To: mathgroup at smc.vnet.net > ><mathgroup at smc.vnet.net> > >Subject: [mg25589] [mg25578] Re: [mg25489] problem with nonlinearfit > > > > > >> I'm not sure if it's possible to specify a range to use for parameter > >> values in NonlinearFit, but it is possible to specify starting values by > >> giving a list of {parameter, starting value} instead of just a list of > >> parameters. It may be an "undocumented" feature, but it seems to work > >well. > >> For instance if your data is contained in the variable data and the > >> function is as you describe, you could use > >> > >> NonlinearFit[data, (s/K)^(N a)*(t/a)^a, t, {{K, 5}, {N, 7}, {a, 9}}] > >> > >> if you wanted to start the fit with K=5, N=7, a=9. You could use any > other > >> starting values for the parameters, but if you want to specify a starting > >> value for one parameter, you have to specify a starting value for all > >> parameters, you can's use the following > >> > >> NonlinearFit[data, (s/K)^(N a)*(t/a)^a, t, {{K, 5}, N, a}] > >> > >> to specify a starting value for K but not for N or a. > >> > >> Since Mathematica does not have a constrained minimization function, it > >may > >> not be possible to specify a "range" for the parameters. > >> > >> -- Tim Stiles > >> > >> Dragan Grgic wrote: > >> > >> > Hi everybody, > >> > > >> > I have to fit the function e = (s/K)^(Na)*(t/a)^a (s = constant; > >> > t = variable; K,N,a = parameters) to a given set of datas. > >> > Then, I have used the NonlinearFit function (Statistics'NonlinearFit' > >> > package), but I have two problems: > >> > > >> > 1) I can't specify any start values or range using. I tried to give > them > >> > as any forms, but I always got error messages concerning these > >> > parameters. > >> > > >> > 2) mathematica shows kind of a strange behavior when repeating the fit > >> > operation, sometimes it keeps old values and sometimes it reports > >> > different error messages. > >> > > >> > Any help welcome, > >> > Thanks in advance, Dragan. > >> > >> > > > > > > > >