Re: [R] nls(): Levenberg-Marquardt, Gauss-Newton, plinear - PI curve fitting

From: Gabor Grothendieck <ggrothendieck_at_gmail.com>
Date: Tue 21 Jun 2005 - 20:57:19 EST

On 6/21/05, Christfried Kunath <mailpuls@gmx.net> wrote:
> Hello,
>
> i have a problem with the function nls().
>
> This are my data in "k":
> V1 V2
> [1,] 0 0.367
> [2,] 85 0.296
> [3,] 122 0.260
> [4,] 192 0.244
> [5,] 275 0.175
> [6,] 421 0.140
> [7,] 603 0.093
> [8,] 831 0.068
> [9,] 1140 0.043
>
> With the nls()-function i want to fit following formula whereas a,b, and c
> are variables: y~1/(a*x^2+b*x+c)
>
> With the standardalgorithm "Newton-Gauss" the fitted curve contain an peak
> near the second x,y-point.
> This peak is not correct for my purpose. The fitted curve should descend
> from the maximum y to the minimum y given in my data.
>
> The algorithm "plinear" give me following error:
>
>
> phi function(x,y) {
> k.nls<-nls(y~1/(a*(x^2)+b*x+c),start=c(a=0.0005,b=0.02,c=1.5),alg="plinear")
> coef(k.nls)
> }
>
> phi(k[,1],k[,2])
>
> Error in qr.solve(QR.B, cc) : singular matrix `a' in solve
>
>
> I have found in the mailinglist
> "https://stat.ethz.ch/pipermail/r-help/2001-July/012196.html" that is if t
> he data are artificial. But the data are from my measurment.
>
> The commercial software "Origin V.6.1" solved this problem with the
> Levenberg-Marquardt algorithm how i want.
> The reference results are: a = 9.6899E-6, b = 0.00689, c = 2.72982
>
> What are the right way or algorithm for me to solve this problem and what
> means this error with alg="plinear"?
>
> Thanks in advance.

This is not a direct answer to your question but log(y) looks nearly linear in x when plotting them together and log(y) ~ a + b*x or y ~ a*exp(b*x) will always be monotonic. Also, this model uses only 2 rather than 3 parameters.



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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Tue Jun 21 21:00:32 2005

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