Re: Message using FindFit with LevenbergMarquardt Method

*To*: mathgroup at smc.vnet.net*Subject*: [mg125480] Re: Message using FindFit with LevenbergMarquardt Method*From*: Darren Glosemeyer <darreng at wolfram.com>*Date*: Thu, 15 Mar 2012 00:30:17 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201203140541.AAA25797@smc.vnet.net>

On 3/14/2012 12:41 AM, geraldine oliveux wrote: > Hello, > > I'm a new user of Mathematica. I'm trying to fit data to a model in > which I have one parameter. I try to determine this parameter by > Levenberg-Marquardt Method. > I wrote the following lines : > data = {{0., 0.425}, {60.6, 0.224}, {91.2, 0.1314}, {119.4, 0.0716}, > {150., 0.040}, {181.8, 0.0241}} > u = t*(a*sqrt[e0])/2 > x = Tanh[t*(a*sqrt[e0])/2] > Etac = e0*(1 - x^2) > fit = FindFit[data, Etac, a, t, Method -> "LevenbergMarquardt"] > > But it gives me the following message : > FindFit::nrlnum: "The function value {0.,-0.224+0.425\ (1. -1.\ > Tanh[Times[<<2>>]]^2),-0.1314+0.425\ (1. -1.\ > Tanh[Times[<<2>>]]^2),-0.0716+0.425\ (1. -1.\ > Tanh[Times[<<2>>]]^2),-0.04+0.425\ (1. -1.\ > Tanh[Times[<<2>>]]^2),-0.0241+0.425\ (1. -1.\ Tanh[Times[<<2>>]]^2)}\ > \n is not a list of real numbers with dimensions {6} at {a} = {1.}. " > > What does it mean ? I think there is something I don't do well, and > that it concerns the data declaration. > Thank you in advance for your help, > Best regards, > G=E9raldine > The message means that the model did not evaluate to a real number for at least one of the data points for a value of a that the algorithm tried. It the particular example sent, this happens at the starting value for a, which is 1 by default. There are a couple problems I see in the code. First sqrt should be Sqrt. With that fix, we can see there is still a problem by substituting the initial value of a and the t values from the data: In[1]:= data = {{0., 0.425}, {60.6, 0.224}, {91.2, 0.1314}, {119.4, 0.0716}, {150., 0.040}, {181.8, 0.0241}}; u = t*(a*Sqrt[e0])/2; x = Tanh[t*(a*Sqrt[e0])/2]; Etac = e0*(1 - x^2); In[5]:= Etac /. {a -> 1} /. t -> data[[All, 1]] Out[5]= {1. e0, e0 (1 - Tanh[30.3 Sqrt[e0]]^2), e0 (1 - Tanh[45.6 Sqrt[e0]]^2), e0 (1 - Tanh[59.7 Sqrt[e0]]^2), e0 (1 - Tanh[75. Sqrt[e0]]^2), e0 (1 - Tanh[90.9 Sqrt[e0]]^2)} The parameter e0 does not have a numeric value, so either it will need to be estimated by FindFit as well: In[6]:= fit = FindFit[data, Etac, {a, e0}, t, Method -> "LevenbergMarquardt"] Out[6]= {a -> -1.92292*10^23, e0 -> 0.425} Or a numeric value will need to be given for it: In[7]:= fit = FindFit[data, Etac /. e0 -> 3, a, t, Method -> "LevenbergMarquardt"] Out[7]= {a -> 0.21875} Darren Glosemeyer Wolfram Research

**References**:**Message using FindFit with LevenbergMarquardt Method***From:*geraldine oliveux <geraldine.oliveux@free.fr>