similar to: roots of a function

Hello, I have a problem finding a root of a function, which I define like this: tuki <- function(u, x, a, lambda){ if((lambda%%1>0) & u<0) {u<-(u+0i)} f <- Re(x-(a*(u)^lambda-(1-(u))^lambda)/lambda) f } What I want to do is to find the root, but without specifying the interval within which to search for it. I can do it easily in MATLAB with fsolve() or fzero() functions. I

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

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

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) >

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 everyone. I have a problem that I have been unable to determine either the best way to proceed and why the methods I'm trying to use sometimes fail. I'm using the pf() function in an optimization function to find a noncentrality parameter that leads to a specific value at a specified quantile. My goal is to have a general function that returns the noncentrality parameter that

Hi! R is an amazing piece of software and with so many libraries it can do almost anything... but I was very surprised that a standard "binary search" function seems not to exist. I can find other much more highly-complex search routines (optimize, uniroot, nlm) in the standard no-extra-packages-loaded version of R, but not this simple alternative. I searched and found an

I would like to solve the equation is is the sum from k = i to N of choose(N,k) * MR ^ k * (1 - MR) ^ (N - k) - 0.50 = 0 I want to solve for MR. This seems like a non-linear equation to me. But I am having a hard time writing the function that implements the above. I could use 'for(...) as a brute force appoarch but I would like a more "elegant" solution. The variables 'N'

Hi! I´m need a function that solves the equation f(x) = 0 (i.e. the root of the function) when f is a nonlinear function. Is there any? I´ve tried nlm and optim on the square of the function but the solution is very unstable. Thanks before hand. / Fredrik Thuring ------------------------------------------------------------------------------ This e-mail and any attachment may be confidential

Dear r-help, I would like to use the 'ppm' function of the 'spatstat' package to fit a Strauss inhibition model. I understand that I can specify a parametric model for the "background" trend, but how would I specify a trend which is estimated using a Kernel density smoother? In particular, I would like to use the 'kde' function of the 'ks' package to

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,

hello, I have got some trouble with R functions nlm(), nls() or optim() : I would like to fit 3 parameters which must stay in a precise interval. For exemple with nlm() : fn<-function(p) sum((dN-estdata(p[1],p[2],p[3]))^2) out<-nlm(fn, p=c(4, 17, 5), hessian=TRUE,print.level=2) with estdata() a function which returns value to fit with dN (observed data vactor) My problem is that only

Hi, Below I have a function mlogl_k, later it's called with "nlm" . __BEGIN__ vsamples<- c(14.7, 18.8, 14, 15.9, 9.7, 12.8) mlogl_k <- function( k_func, x_func, theta_func, samp) { tot_mll <- 0 for (comp in 1:k_func) { curr_mll <- (- sum(dgamma(samp, shape = x_func, scale=theta_func, log = TRUE))) tot_mll <- tot_mll + curr_mll }

Hi R-users,I'm trying to solve a non linear equation, to find the degrees of freedom of a mixture of t student. I'm sure i wrote the minimization equation in the right way, but when i try to run the EM algorithm to estimate the parameters of the mixture, the following error will appear: Error in uniroot(function(z) log(z/2) - digamma(z/2) + 1 - log((z + d)/2) + : f() values at end

dear R experts--- I have (many) unidimensional root problems. think loc.of.root <- uniroot( f= function(x,a) log( exp(a) + a) + a, c(.,9e10), a=rnorm(1) ) $root (for some coefficients a, there won't be a solution; for others, it may exceed the domain. implied volatilities in various Black-Scholes formulas and variant formulas are like this, too.) except I don't need 1 root, but a

I tend to avoid the the trace/verbose arguments for the various root finders and optimizers and instead use the trace function or otherwise modify the function handed to the operator. You can print or plot the arguments or save them. E.g., > trace(ff, print=FALSE, quote(cat("x=", deparse(x), "\n", sep=""))) [1] "ff" > ff0 <- uniroot(ff, c(0, 10))

Dear list, I've had this problem for a while and I'm looking for a more general and robust technique than I've been able to imagine myself. I need to find N (typically N= 3 to 5) zeros in a function that is not a polynomial in a specified interval. The code below illustrates this, by creating a noisy curve with three peaks of different position, magnitude, width and

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

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