similar to: uniroot function question

I am calling the uniroot function from inside another function using these lines (last two lines of the function) : d <- uniroot(k, c(.001, 250), tol=.05) return(d$root) The problem is that on occasion there's a problem with the values I'm passing to uniroot. In those instances uniroot stops and sends a message that it can't calculate the root because f.upper * f.lower is greater

Hi, I'm using the uniroot function, and would like to detect an error which occurs, for instance, when the values at endpoints are not of opposite signs. For example: uniroot( function(x) x^2+1, lower=1, upper=2 ). I want to say something like: if "error in uniroot(...)" return NA else return uniroot$root Thanks a lot! Asaf -- View this message in context:

On 06/10/2020 11:00 a.m., Sorkin, John wrote: > Colleagues, > I am trying to learn to use uniroot to solve a simple linear equation. I define the function, prove the function and a call to the function works. When I try to use uniroot to solve the equation I get an error message, > Error in yfu n(x,10,20) : object 'x' not found. > > I hope someone can tell we how I can fix

curiosity---given that vector operations are so much faster than scalar operations, would it make sense to make uniroot vectorized? if I read the uniroot docs correctly, uniroot() calls an external C routine which seems to be a scalar function. that must be slow. I am thinking a vectorized version would be useful for an example such as of <- function(x,a) ( log(x)+x+a ) uniroot( of, c(

With main R releases only happening yearly in spring, now is good time to consider *and* discuss new features for what we often call "R-devel" and more officially is R Under development (unstable) (.....) -- "Unsuffered Consequences" Here is one such example I hereby expose to public scrutiny: A few minutes ago, I've committed the following to R-devel (the

Dear all I am trying to calculate the value of n using uniroot. Here is the message I am having: <<< Error in uniroot(integ, lower = 0, upper = 1000, n) : 'interval' must be a vector of length 2 >>> Please would you be able to give me an indication on why I am having this error message. Many thanks EXAMPLE BELOW: ## t = statistics test from t -distribution

Dear All I am trying to find a uniroot of a function within another function (see example) but I am getting an error message (f()values at end points not of opposite sign). I was wondering if you would be able to advise how redefine my function so that I can find the solution. In short my first function calculates the intergrale which is function of "t" , I need to find the uniroot of

I have a strange problem with uniroot() function. Here is the result : > uniroot(th, c(-20, 20)) $root [1] 4.216521e-05 $f.root [1] 16.66423 $iter [1] 27 $estim.prec [1] 6.103516e-05 Pls forgive for not reproducing whole code, here my question is how "f.root" can be 16.66423? As it is finding root of a function, it must be near Zero. Am I missing something? -- View this message

Colleagues, I am trying to learn to use uniroot to solve a simple linear equation. I define the function, prove the function and a call to the function works. When I try to use uniroot to solve the equation I get an error message, Error in yfu n(x,10,20) : object 'x' not found. I hope someone can tell we how I can fix the problem

Hi, I have tried to use uniroot to solve a value (value a in my function) that gives f=0, and I repeat this process for 10000 times(stimulations). However error occures from the 4625th stimulation - Error in uniroot(f, c(0, 2), maxiter = 1000, tol = 0.001) : f() values at end points not of opposite sign I have also tried interval of (lower=min(U), upper=max(U)) and it won't work as well.

Hello, I am struggling to find the root of a exponent function. "uniroot" is complaining about a values at end points not of opposite sign? s<- sapply(1:length(w),function(i) + { + + + + + uniroot(saeqn,lower=-5000,upper=0.01036597923,l=list(t=w[i],gp=gp))$root + }) Error in uniroot(saeqn, lower = -5000, upper = 0.01036597923, l = list(t = w[i], : f() values at

A recent message on ansari.test() prompted me to play with the examples. This doesn't work for me in R version 2.4.1 R> ansari.test(rnorm(100), rnorm(100, 0, 2), conf.int = TRUE) Error in uniroot(ab, srange, tol = 1e-04, zq = qnorm(alpha/2, lower = FALSE)) : object "ab" not found It looks like there's a small typo in ccia() inside ansari.test.default() in which

In looking at rootfinding for the histoRicalg project (see gitlab.com/nashjc/histoRicalg), I thought I would check how uniroot() solves some problems. The following short example ff <- function(x){ exp(0.5*x) - 2 } ff(2) ff(1) uniroot(ff, 0, 10) uniroot(ff, c(0, 10), trace=1) uniroot(ff, c(0, 10), trace=TRUE) shows that the trace parameter, as described in the Rd file, does not seem to be

I wrote first cut at unirootR for Martin M and he revised and put in Rmpfr. The following extends Ben's example, but adds the unirootR with trace output. c1 <- 4469.822 c2 <- 572.3413 f <- function(x) { c1/x - c2/(1-x) }; uniroot(f, c(1e-6, 1)) uniroot(f, c(1e-6, 1)) library(Rmpfr) unirootR(f, c(1e-6, 1), extendInt="no", trace=1) This gives more detail on the iterations,

Hi all, This is probably a blindingly obvious question: Why does it matter in the uniroot function whether the f() values at the end points that you supply are of the same sign? For example: f <- function(x,y) {y-x^2+1} #this gives a warning uniroot(f,interval=c(-5,5),y=0) Error in uniroot(f, interval=c(-5, 5), y = 0) : f() values at end points not of opposite sign #this doesn't give a

Despite my years with R, I didn't know about trace(). Thanks. However, my decades in the minimization and root finding game make me like having a trace that gives some info on the operation, the argument and the current function value. I've usually found glitches are a result of things like >= rather than > in tests etc., and knowing what was done is the quickest way to get there.

Le 18/02/2023 ? 21:44, J C Nash a ?crit?: > I wrote first cut at unirootR for Martin M and he revised and put in > Rmpfr. > > The following extends Ben's example, but adds the unirootR with trace > output. > > c1 <- 4469.822 > c2 <- 572.3413 > f <- function(x) { c1/x - c2/(1-x) }; uniroot(f, c(1e-6, 1)) > uniroot(f, c(1e-6, 1)) > library(Rmpfr) >

Hallo, I am trying to define expectation as an integral and use uniroot to find the distribution parameter for a given expectation. However I fail to understand how to define properly the functions involved and pass the parameters correctly. Can anyone help me out? Thanks, Gerrit Draisma. This what I tried: ======= > # exponential density > g <- function(x,lambda){ lambda

Dear all, I have the following problem with the uniroot function. I want to find roots for the fucntion "Fp2" which is defined as below. Fz <- function(z){0.8*pnorm(z)+p1*pnorm(z-u1)+(0.2-p1)*pnorm(z-u2)} Fp <- function(t){(1-Fz(abs(qnorm(1-(t/2)))))+(Fz(-abs(qnorm(1-(t/2)))))} Fp2 <- function(t) {Fp(t)-0.8*t/alpha} th <- uniroot(Fp2, lower =0, upper =1,

Hi, I am using the uniroot function in order to carry out a bivariate Monte Carlo simulation using the logistics model. I have defined the function as: BV.FV <- function(x,y,a,A) (((x^(-a^-1)+y^(-a^-1))^(a-1))*(y^(a-1/a))*(exp(-((1^(-a^-1)+y^(-a^-1))^a)+y^-1)))-A and the procedure is as follows: Randomly generate values of A~(0,1), y0 = -(lnA)^-1 Where: A=Pr{X<xi|Y=yi-1} and a is the