HI, Thank you all for the reply to my nls Q. after i put a variable at the left hand side of "~", and feed initial values of coefficients, it works now:) I have a related Q: I apply arima() and nls() on the same time series. and try to compare which one fits better. the result from nls() provides "residual sum-of-squares" the result from arima() gives "sigma2, var.coef" and residuals at each points. My Q is: what is the meaningful way to compare these two fittings from the returned value from nls() and arima()? should I also compute the sum-of-squares residuals for arima() from the big individual residual array? Is there any easy way to do this?
thanks a lot, have a nice weekend, yan On 17 Jan 2003, Douglas Bates wrote: > "kjetil brinchmann halvorsen" <[EMAIL PROTECTED]> writes: > > > On 16 Jan 2003 at 22:15, Yan Yu wrote: > > > > > HI, > > > i have some prob when i try to use nls(). > > > my data is 1D vector, I tried to use a polynomial function(order is 3) to > > > fit it. > > > the data series is stored in x. > > > the a0, a1, a2, a3 below is coefficient, which i hope i can get from > > > calls "nls" > > > > > > > z <- nls( ~ a0 + a1 * x + a2 * x * x + a3 * x * x * x, data = x ) > > > > You haven't given a response in the formula? nls could possibly give > > a more informative error message. > > and you didn't give starting estimates for the parameters and the > model should not be fit with nls in the first place. A polynomial in > x is a linear model in the coefficients a0, a1, a2, and a3 and should > be fit with lm, not nls. > ______________________________________________ [EMAIL PROTECTED] mailing list http://www.stat.math.ethz.ch/mailman/listinfo/r-help
