From: Radka Pancheva <radica_at_abv.bg>

Date: Fri, 25 Apr 2008 15:12:45 +0300 (EEST)

}

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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Fri 25 Apr 2008 - 12:18:08 GMT

Date: Fri, 25 Apr 2008 15:12:45 +0300 (EEST)

ex <- 0.3914877

ex2 <- 0.2671597

p <- x[1] q <- x[2] p <- .Machine$double.eps q <- .Machine$double.eps F <- rep(NA,2) F[1] <- p/(p + q) F[2]<- (p*q + (p + q + 1)*p^2)/((p + q + 1)*(p + q)^2) return(F)

}

p0 <- c(ex,ex2)

dfsane(par=p0, fn=my.mm,control=list(maxit=50000))

and I became the following output:

…

iteration: 3640 ||F(xn)|| = 0.7071068
iteration: 3641 ||F(xn)|| = 0. 7071068
…

iteration: 49990 ||F(xn)|| = 0. 7071068
iteration: 50000 ||F(xn)|| = 0. 7071068
$par

[1] -446.2791 -446.4034

$residual

[1] 0.5

$fn.reduction

[1] 0

$feval

[1] 828495

$iter

[1] 50001

$convergence

[1] 1

$message

[1] "Maximum limit for iterations exceeded"

I have tried maxiter=100000 but the output is the same. I know that ex and ex2 are bringing the problems, but I am stuck with them. How can I make it convergent?

Thank you,

Evgeniq

>2008/4/25 Radka Pancheva <radica@abv.bg>:

* >> I am trying to estimate the parameters of a bimodal normal distribution using moments matching, so I have to solve a non-linear system of equations. How can I solve the following simple example?
** >>
** >> x^2 - y^2 = 6
** >> x – y = 3
** >>
** >> I heard about nlsystemfit, but I don't know how to run it exactly. I have tried the following code, but it doesn't really work:
** >>
** >>
** >> f1 <-y~ x[1]^2-x[2]^2-6
** >> f2 <-z~ x[1]-x[2]-3
** >> f <- list(f1=0,f2=0)
** >> nlsystemfit("OLS",f,startvals=c(0,0))
** >
** >You could try the recent package BB by Ravi Varadhan. The code could
** >be the following:
** >
** >library(BB)
** >
** >f <- function(x) {
** > x1 <- x[1]
** > x2 <- x[2]
** >
** > F <- rep(NA, 2)
** >
** > F[1] <- x1^2 - x2^2 - 6
** > F[2] <- x1 - x2 - 3
** >
** > return(F)
** >}
** >
** >p0 <- c(1,2)
** >dfsane(par=p0, fn=f,control=list(maxit=3000))
** >
** >I got the solution:
** >
** >x1 = 2.5
** >x2 = -0.5
** >
** >Paul
** >
** >______________________________________________
** >R-help_at_r-project.org mailing list
** >https://stat.ethz.ch/mailman/listinfo/r-help
** >PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
** >and provide commented, minimal, self-contained, reproducible code.
** >
*

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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Fri 25 Apr 2008 - 12:18:08 GMT

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