[R] newton.method

[Ravi Varadhan]

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: rvaradhan at jhmi.edu


----- Original Message -----
From: Dennis Murphy <djmuser at gmail.com>
Date: Thursday, July 29, 2010 8:28 am
Subject: Re: [R] newton.method
To: sammyny <sjain at caa.columbia.edu>
Cc: r-help at 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 <sjain at 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 at 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 at r-project.org mailing list
>  
>  PLEASE do read the posting guide 
>  and provide commented, minimal, self-contained, reproducible code.