Re: [R] nls in r

Thank you very much Prof.JC Nash I am still don't understand about that Jacobian. why the model not approprite. Really appreciate King regards -- View this message in context: http://r.789695.n4.nabble.com/nls-in-r-tp4711012p4711044.html Sent from the R help mailing list archive at Nabble.co

Re: [R] nls in r

This subject is introduced in multivariable calculus, and in more detail in some numerical analysis courses. Graphing is one common technique for identifying promising search ranges if the number of variables can be reduced to one or two. Analytical identification of asymptotes, extrema, and zer

Re: [R] nls in r

How can you find that starting value ? is there a trick for that ?. Really appreciate. Thank you very much. King Regards -- View this message in context: http://r.789695.n4.nabble.com/nls-in-r-tp4711012p4711043.html Sent from the R help mailing list archive at Nabble.com.

Re: [R] nls in r

I believe that if your try these starting values the sum of squares is considerably smaller a=1.0851e-06 b=1.4596e-01 delta=9.1375e-01 something like SS= 0.005236471 vs SS= 0.01597071 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and mor

Re: [R] nls in r

With package nlmrt, I get a solution, but the Jacobian is essentially singular, so the model may not be appropriate. You'll need to read the documentation to learn how to interpret the Jacobian singular values. Or Chapter 6 of my book "Nonlinear parameter optimization with R tools." Here's the scr

[R] nls in r

I get this error Error in numericDeriv(form[[3L]], names(ind), env) : Missing value or an infinity produced when evaluating the model I was replace the starting value but still get error. Here is my code: library(stats) x=c(30:110) y=c(0.000760289, 0.000800320, 0.000830345, 0.000840353, 0.0008