Displaying 5 results from an estimated 5 matches for "richarson".
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richardson
2004 Apr 28
4
numericDeriv
Dear All,
I am trying to solve a Generalized Method of Moments problem which
necessitate the gradient of moments computation to get the
standard errors of estimates.
I know optim does not output the gradient, but I can use numericDeriv to
get that. My question is: is this the best function to do this?
Thank you
Jean,
2005 May 05
2
Numerical Derivative / Numerical Differentiation of unknown funct ion
...------------------------------------
for(m in 1:(r - 1)) {
for(i in 1:(r - m)) b.mtr[i,]<- (a.mtr[(i+1),]*(4^m)-a.mtr[i,])/(4^m-1)
# a.mtr<- b.mtr
# a.mtr<- (a.mtr[2:(r+1-m),]*(4^m)-a.mtr[1:(r-m),])/(4^m-1)
if(show & m!=(r-1) ) {
cat("\n","Richarson improvement group No. ", m, "\n")
print(a.mtr[1:(r-m),], 12)
}
}
a.mtr[length(a.mtr)]
}
## try it out
richardson.grad(function(x){x^3},2)
########################################################################################
Regards,
Tolga Uzuner
===========...
2002 Nov 06
1
tcltk package under Windows
...Tcl 8.4.1
is now stable, and I was wondering whether this newer release of Tcl can
be used with R under Windows? I've not used Tcl/Tk before but I am
considering using it for some applications.
Thank you for your time,
-Paul
Paul Schwarz
Department of Forest Science
Oregon State University
Richarson Hall
Corvallis, Oregon 97331
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2006 Mar 12
2
Numerical Derivatives in R
Hi,
Suppose I have an arbitrary function:
arbfun<-function(x) {...}
Is there a robust implementation of a numerical derivative routine in R
which I can use to take it's derivative ? Something a bit more than
simple division by delta of the difference of evaluating the function at
x and x+delta...
Perhaps there is a way to do this using D or deriv but I could not
figure it out.
2005 May 05
2
Numerical Derivative / Numerical Differentiation of unkno wn funct ion
...> for(m in 1:(r - 1)) {
> for(i in 1:(r - m)) b.mtr[i,]<-
> (a.mtr[(i+1),]*(4^m)-a.mtr[i,])/(4^m-1)
> # a.mtr<- b.mtr
> # a.mtr<- (a.mtr[2:(r+1-m),]*(4^m)-a.mtr[1:(r-m),])/(4^m-1)
> if(show & m!=(r-1) ) {
> cat("\n","Richarson improvement group No. ", m, "\n")
> print(a.mtr[1:(r-m),], 12)
> }
> }
> a.mtr[length(a.mtr)]
> }
>
> ## try it out
> richardson.grad(function(x){x^3},2)
> ##############################################################
> #############...