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]
what I am talking about.
Ravi.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of sammyny
Sent: Monday, August 02, 2010 4:49 AM
To: r-help@r-project.org
Subject: Re: [R] newton.method
I have this function:
function(x)
-0.3*x*exp
sammyny wrote:
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)))
..
The newton method is not generating good result.
for( x in
Thanks everyone for their help. I am able to see things more clearly now.
cheers,
Samit
--
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Sent from the R help mailing list archive at Nabble.com.
__
Sometimes it is easier to just write it. See below.
On 10-07-30 06:00 AM, r-help-requ...@r-project.org wrote:
Date: Thu, 29 Jul 2010 11:15:05 -0700 (PDT)
From: sammynysj...@caa.columbia.edu
To:r-help@r-project.org
Subject: Re: [R] newton.method
Message-ID:1280427305687-2306895.p...@n4
/SendEmail.jtp?type=nodenode=2308224i=2
Subject: Re: [R] newton.method
Message-ID:[hidden
email]http://user/SendEmail.jtp?type=nodenode=2308224i=3
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newton.method is in package 'animation'.
Thanks Ravi.
BBSolve/BBOptim seems to work very
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
On Jul 29, 2010, at 4:32 AM, sammyny wrote:
Hi,
Is this method broken in R?
?newton.method
No documentation for 'newton.method' in specified packages and
libraries:
you could try '??newton.method'
If you are asking about a non-base function, you are asked by the
Posting Guide
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
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
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
sammyny wrote:
If someone could point me a correct working version of newton method for
finding roots and its usage, that would be helpful.
You mentioned in your original post that you had no idea how to use nleqslv.
nleqslv provides a Broyden and Newton method with several different
The function newton.method() in the package 'animation' is merely for
demonstration purpose instead of serious computation -- it illustrates
how Newton's method works step by step. But I'm unable to reproduce
your bogus results:
library(animation)
par(pch = 20)
ani.options(nmax = 50, interval =
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