Displaying 4 results from an estimated 4 matches for "mfun1".
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2004 Oct 11
1
Puzzled on nlm
Dear R People:
Here is a function to minimized:
>mfun1
function(x,a) {
x[1] <- a[1]*x[2] + a[3] - a[2]*(a[1]-a[2])*a[3]
x[2] <- a[1]*x[1] - a[2]*a[3]
return(x)
}
Here is my first try:
>nlm(mfun1,c(1,1))
Error in f(x, ...) : Argument "a" is missing, with no default
>
>nlm(mfun1,c(1,1),a=c(0.8,0.5,1))
Error in nlm(mfun1,...
2004 Oct 11
1
nlm question
...ar R People:
I am trying to duplicate the example from Dennis and Schnabel's "Numerical
Methods for Unconstrained Optimization and Nonlinear Equations", which
starts on page 149.
My reason for doing so: to try to understand the "nlm" function.
Here is the function:
>mfun1
function(x) {
z <- matrix(0,nrow=2,ncol=1)
z[1,1] <- x[1]^2 + x[2]^2 - 2
z[2,1] <- exp(x[1]-1) + x[2]^3 - 2
res <- 0.5*t(z)%*%z
res
}
This function has a root at c(1,1).
When I use the following:
>nlm(mfun1,c(2,0.5))
$minimum
[1] 0.09168083
$...
2011 Sep 01
0
UNSOLVED: Fwd: generate correlated qualitative data
The problem remain unsolved. If you have any idea please do suggest
....................
thank you;
Ram H
On Wed, Aug 31, 2011 at 12:06 PM, Ram H. Sharma <sharma.ram.h@gmail.com>wrote:
> Dear R experts:
>
> I have following problem:
>
> # myfunction
> mfun1 <- function(x) {
> if ( x == 2){
> xv <- sample(c(2,1,0),100, replace = T, prob = c(0.6, 0.2, 0.2))
> }
> if ( x == 1){
> xv <- sample(c(1,0),100, replace = T, prob = c(0.6, 0.4))
> }
> if ( x == 0) {
> xv <- sample(c(0,0),100, replace = T, prob = c(0.5,...
2011 Aug 31
0
generate correlated qualitative data
Dear R experts:
I have following problem:
# myfunction
mfun1 <- function(x) {
if ( x == 2){
xv <- sample(c(2,1,0),100, replace = T, prob = c(0.6, 0.2, 0.2))
}
if ( x == 1){
xv <- sample(c(1,0),100, replace = T, prob = c(0.6, 0.4))
}
if ( x == 0) {
xv <- sample(c(0,0),100, replace = T, prob = c(0.5, 0.5))
}
return(xv)
}
# applying the...