On 6/21/05, Gabor Grothendieck <[EMAIL PROTECTED]> wrote: > On 6/21/05, Christfried Kunath <[EMAIL PROTECTED]> 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. >
One other comment. If you do want to use your model try fitting 1/y first to get your starting value since that model has a unique solution: > res1 <- nls(1/y ~ a*x^2 + b*x + c, start = list(a=0,b=0,c=0)) > res2 <- nls(y ~ 1/(a*x^2 + b*x + c), start = as.list(coef(res1))) > res2 Nonlinear regression model model: y ~ 1/(a * x^2 + b * x + c) data: parent.frame() a b c 9.690187e-06 6.885577e-03 2.729825e+00 residual sum-of-squares: 0.000547369 ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html