What does this error message mean?

I've got two vectors:

times =

           0    0.0005    0.0050    0.0500    0.5000    5.0000

mittel =

           0    2.0505    5.7940    8.8363   14.1563   35.6821

I would like to fit two functions to these vectors:

func1 = 'y ~ b0*(x + b1)^(1/2)';
fitresult = NonLinearModel.fit(times,mittel,func1,[ 2.775e+06 0.008033 ]);

and

func2 = 'y ~ b0*(x + b1)^(1/3)';
fitresult2 = NonLinearModel.fit(times,mittel,func2,[2.1e+01 0 ]);

The first works absolutly fine, but the second gives me an error I don't understand:

Error using internal.stats.getscheffeparam>ValidateParameters (line 182)
If non-empty, JW must be a numeric, real matrix.

Error in internal.stats.getscheffeparam (line 110)
    [J,VF,VP,JW,Intopt,TolSVD,TolE,VQ,usingJ] =
    ValidateParameters(J,VF,VP,JW,Intopt,TolSVD,TolE,VQ,allowedIntopt);

Error in nlinfit (line 340)
        sch =
        internal.stats.getscheffeparam('WeightedJacobian',J(~nans,:),'Intopt','observation','VQ',VQ);

Error in NonLinearModel/fitter (line 1121)
                [model.Coefs,~,J_r,model.CoefficientCovariance,model.MSE,model.ErrorModelInfo,~]
                = ...

Error in classreg.regr.FitObject/doFit (line 219)
            model = fitter(model);

Error in NonLinearModel.fit (line 1484)
            model = doFit(model);

Answer 1

[Sean de Wolski]

For some reason the model is returning an imaginary component. I would contact tech support, that error message is useless.

[Matt J]

Probably because (x+b1).^(1/3) is being used instead of nthroot(x+b1,3). E.g.,

    >> (-1)^(1/3)

    ans =

       0.5000 + 0.8660i

    >> nthroot(-1,3)

    ans =

        -1

[Sean de Wolski]

Learn something new every day :)

Answer 2

[Matt J]

In addition to the need for nthroot(q,3) instead of q^(1/3), your model equations are non-differentiable w.r.t. b1. This makes me wonder if the Jacobian calculations are creating trouble.

I recommend FMINSPLEAS ( Download ) for your problem, since it does not rely on differentiability. Also, it can take advantage of the fact your model is linear w.r.t. b0.

[Matt J]

As an example

z=[times(:),ones(numel(times),1)]\mittel(:).^3;
b1start=z(2)./z(1); %starting guess for b1

[b1,b0]=fminspleas( {@(b1,x) nthroot(x + b1, 3)}, b1start, times, mittel)

Answer 3

[Star Strider]

For whatever reason, nlinfit does not have a problem with either one:

func1 = @(b,x) b(1).*(x + b(2)).^(1/2);
func2 = @(b,x) b(1).*(x + b(2)).^(1/3);

B1 = nlinfit(times, mittel, func1, [ 2.775e+06 0.008033 ])
B2 = nlinfit(times, mittel, func2, [2.1e+01 0 ])

producing:

B1 =

      16.2925e+000    47.9075e-003

B2 =

      20.5786e+000 -850.3513e-021i     6.1977e-012 +156.4066e-024i

but it returns complex parameter estimates for ‘B2’, although with negligible imaginary components.