similar to: uniroot() problem

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,

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

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) >

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.

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

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

This below is not solvable with uniroot to find "a": fn=function(a){ b=(0.7/a)-a (1/(a+b+1))-0.0025 } uniroot(fn,c(-500,500)) gives "Error in uniroot(fn, c(-500, 500)) : f() values at end points not of opposite sign" I read R-help posts and someone wrote a function: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/92407.html but it is not very precise. Is there any

Hi, I wonder if it would make sense to make uniroot detect zeros at the endpoints, eg if f(lower)==0, return lower as the root, else if f(upper)==0, return upper as the root, else stop if f(upper)*f(lower) > 0 (currently it stops if >=), else proceed to the algorithm proper. Currently I am using a wrapper function to implement this, and I found it useful. But I didn't want to send a

Dear friends - occurring in Windows R2.6.2 I am modeling physical chemistry in collaboration with a friend who has preferred working in Excel. I used uniroot, and find a solution to a two buffer problem in acid-base chemistry which I believe is physiologically sensible. Using "goal seek" in Excel my friend found another plausible root, quite close to zero, and a plot of the function

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

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))

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)) provides a root at -6.00e-05, which is outside of the specified bounds. The default value of the "extendInt" argument to uniroot() is "no", as far as I can see ... $root [1] -6.003516e-05 $f.root [1] -74453981 $iter [1] 1 $init.it [1] NA

Hi all. I am an R newbie and trying to grasp how the simple optimization routines in R work. Specifically, I would like some guidance on how to set up a code to calculate the internal rate of return on an investment project (http://en.wikipedia.org/wiki/Internal_rate_of_return). My main problem (I think) is that I want a generic code where N (number of periods) can be easily changed and set

I need to solve a equation like this : a = b/(1+x) + c/(1+x)^2 + d/(1+x)^3 where a,b,c,d are known constant. Is there any R-way to do that? Thanks in advance -- View this message in context: http://www.nabble.com/How-to-solve-following-equation--tp20785063p20785063.html Sent from the R help mailing list archive at Nabble.com.

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

I have one equation, two unknowns, so I am trying to build the solution set by running through possible values for one unknown, and then using uniroot to solve for the accompanying second solution, then graphing the two vectors. p0 = .36 f = function(x) 0.29 * exp(5.66*(x - p0)) f.integral = integrate(f, p0, 1) p1 = p0 + .01 i = 1 n = (1 - p0)/.01 p1.vector = rep(0,n) p2.vector = rep(0,n) for (i