[R] linear constrained optimization in R

I am trying to use http://rss.acs.unt.edu/Rdoc/library/stats/html/constrOptim.html in R to do optimization in R with some given linear constraints but not able to figure out how to set up the problem. For example, I need to maximize $f(x,y) = log(x) + \frac{x^2}{y^2}$ subject to constraints $g_1(x

[R] calculate phase/amplitude of fourier transform function in R

I did a fourier transform on a function in time domain to get the following functions in frequency domain (in latex): $Y_1[\omega] = \frac{1}{1-\phi_1 e^{-jw}}$ $Y_2[\omega] = \frac{1}{1-(\phi_1 + \phi_2)e^{-jw} +\phi_1\phi_2e^{-2jw}}$ How do I find the spectrum of this function for given $\phi

Re: [R] newton.method

Thanks everyone for their help. I am able to see things more clearly now. cheers, Samit -- View this message in context: http://r.789695.n4.nabble.com/newton-method-tp2306111p2310426.html Sent from the R help mailing list archive at Nabble.com. __ R

Re: [R] newton.method

I have this function: function(x) -0.3*x*exp(-(((log(x)+(0.03+0.3*0.3/2)*0.5)/(0.3*sqrt(0.5)))^2)/2)/(2*sqrt(2*pi*0.5)) + 0.03*exp(-0.03*0.5)*pnorm(-(log(x)+(0.03-0.3*0.3/2)*0.5)/(0.3*sqrt(0.5))) uniroot is giving the correct results. > uniroot(f,c(0,10)) $root [1] 0.7347249 $f.root [1] -1.955740

Re: [R] newton.method

> > On 10-07-30 06:00 AM, [hidden > email]<http://user/SendEmail.jtp?type=node&node=2308224&i=0>wrote: > > > Date: Thu, 29 Jul 2010 11:15:05 -0700 (PDT) > > From: sammyny<[hidden > > email]<http://user/SendEmail.jtp?type=node&node=2308224&

Re: [R] newton.method

newton.method is in package 'animation'. Thanks Ravi. BBSolve/BBOptim seems to work very well although I am not familiar with the optimization methods being used there. Is there a way to specify a tolerance in the function to get the required precision. I did something like this to use newton m

[R] newton.method

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 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.