Hello,
On 6/16/09, Stu @ AGS <s...@agstechnet.com> wrote:
> Thanks for your response!
> No, my basic equation does not use matrices at all. It takes scalar values
> and returns a scalar.
>
Not quite. Taking the example above, if you run the following:
> with(observs , {1*x1*x2^2*x3^3})
[1] 0.000e+00 4.267e+16 8.910e+15 2.293e+17 4.308e+16 1.207e+18
you get a vector 6x1. I may be wrong, but I would expect optim() or
any other optimiser (nlminb, etc.) to expect that the objective
function returns a 1x1 value. In my specific example, I arbitrarily
chose values for the parameters: c[1]=1,c[2]=2,c[3]=3.
> What I am trying to accomplish is to find the "best-fit" coefficients to the
> equation as follows:
> y ~ c1 * x1 * x2^c2 * x3^c3
> where y, x1, x2, and x3 are observed data and c1, c2, and c3 are regression
> coefficients.
>
Here I'm slightly confused. If it is a regression that you are trying
to do, and it seems non-linear, perhaps ?nls could help. I tried nls
with your data, but it ended up with an error:
> nls( y ~ c1 * x1 * x2^c2 * x3^c3, data=observs, start=list(c1=0.66, c2=0.999,
> c3 = 0.064))
Error in numericDeriv(form[[3]], names(ind), env) :
Missing value or an infinity produced when evaluating the model
Best,
Liviu
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