Use optimize() to find the minimum and feed that value into uniroot().
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Apr 16, 2015 at 7:47 AM, li li <hannah.hlx at gmail.com> wrote:
Hi Jeff,
Thanks for the reply. I am aware that the sign needs to be different at
the ends of the starting interval.
Another question:
Is there a way to set the right end point ( (the "upper" argument in the
uniroot function below) as the point where the function takes on its
minimun, for example my function f1 below?
Thanks very much!
u1 <- -3
u2 <- 4
pi0 <- 0.8
f1 <- function(lambda,z,p1){
lambda*(p1*exp(u1*z-u1^2/2)+(0.2-p1)*exp(u2*z-u2^2/2))-(1-lambda)*pi0}
x <- seq(-20,20, by=0.1)
y <- numeric(length(x))
for (i in 1:length(x)){y[i] <- f1(x[i],p1=0.15,lambda=0.998)}
plot(y ~ x, ylim=c(-1,1))
abline(h=0)
a <- uniroot(f1, lower =-10, upper = 0,
tol = 1e-20,p1=0.15,lambda=0.998)$root
2015-04-15 22:57 GMT-04:00 Jeff Newmiller <jdnewmil at dcn.davis.ca.us>:
You really need to read the help page for uniroot. The sign needs to be
different at the ends of the starting interval. This is a typical
limitation of numerical root finders.
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On April 15, 2015 7:20:04 PM PDT, li li <hannah.hlx at gmail.com> wrote:
Hi all,
In the following code, I am trying to use uniroot function to solve for
the root (a and b in code below) for function f1.
I am not sure why uniroot function does not give the answer since when
we
look the graph, the function does cross 0 twice.
Any suggestion?
Thanks.
Hanna
u1 <- -3
u2 <- 4
pi0 <- 0.8
f1 <- function(lambda,z,p1){
lambda*(p1*exp(u1*z-u1^2/2)+(0.2-p1)*exp(u2*z-u2^2/2))-(1-lambda)*pi0}
a <- uniroot(f1, lower =-10, upper = 0,
tol = 1e-20,p1=0.15,lambda=0.998)$root
b <- uniroot(f1, lower =0, upper = 10,
tol = 1e-20,p1=0.15,lambda=0.998)$root
x <- seq(-20,20, by=0.1)
y <- numeric(length(x))
for (i in 1:length(x)){y[i] <- f1(x[i],p1=0.15,lambda=0.998)}
plot(y ~ x, ylim=c(-1,1))
abline(h=0)
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