Yes, there are two roots. Try this to get the 2 roots: require(BB)
p0.mat <- matrix(rnorm(10), 10, 1) # 10 random starting values ans <- multiStart(par=p0, fn=f) ans$par You can see that the 2 roots are -2.438285 and 7.419378. Hope this helps, Ravi. ____________________________________________________________________ Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: rvarad...@jhmi.edu ----- Original Message ----- From: Dennis Murphy <djmu...@gmail.com> Date: Thursday, July 29, 2010 8:28 am Subject: Re: [R] newton.method To: sammyny <sj...@caa.columbia.edu> Cc: r-help@r-project.org > Hi: > > Interesting. Try the following; f is copied and pasted directly from > your > e-mail: > > f <- function(x) 2.5*exp(-0.5*(2*0.045 - x)) + 2.5*exp(-0.045) + > 2.5*exp(-1.5*x) - 100 > curve(f, -3, 8) # plot it in a localized region > abline(0, 0, lty = 2) > > There are two places where the function crosses the x-axis, but not where > you were expecting. The graph suggests one root between -3 and -2 and > another between 7 and 8: Note that uniroot() takes an object of class > function as its first argument and a search interval as its second. > It is > the R function that finds a zero of a univariate function. See > ?uniroot for > details. (And f is a function object; try class(f).) > > On Thu, Jul 29, 2010 at 1:32 AM, sammyny <sj...@caa.columbia.edu> wrote: > > > > > Hi, > > Is this method broken in R? I am using it to find roots of the following > > function: > > f(x) = 2.5*exp(-0.5*(2*0.045 - x)) + 2.5*exp(-0.045) + > 2.5*exp(-1.5*x) - > > 100 > > > > Which method? It is never (conveniently) mentioned... Function? Package? > > > > > It is giving an answer of -38.4762403 which is not even close (f(x) > = > > 2.903809e+25 for x=-38.4762403). The answer should be around > 0.01-0.1. This > > function should converge.. > > > > > Even for a simple function like f(x) = exp(-x) * x, it gives > answer as > > 8.89210984 for which f(x) = 0.001222392 and I set tolerance to 10^-12.. > > > > I see handwaving and fulminating, but no code... > > > > > Also, is there a non graphical version of newton method? I looked at > > nleqslv > > but have no idea how to use it.. > > > > > Thanks for your help. > > > > HTH, > Dennis > > > -- > > View this message in context: > > > > Sent from the R help mailing list archive at Nabble.com. > > > > ______________________________________________ > > R-help@r-project.org mailing list > > > > PLEASE do read the posting guide > > > > and provide commented, minimal, self-contained, reproducible code. > > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help@r-project.org mailing list > > PLEASE do read the posting guide > and provide commented, minimal, self-contained, reproducible code. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
