similar to: Derivative

I'd like to plot a point at the intersection of these two curves. Thanks x <- seq(.2, .3, by = .01) f <- function(x){ x*cos(x)-2*x**2+3*x-1 } plot(x,f(x), type = "l") abline(h = 0)

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

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.

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

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

I think the function you need is 'help.search'; try: help.search("binomial") and look for something obvious in the 'stats' package. A good deal quicker and easier than posting to an internet forum! Cheers, Mike. cathelf wrote: > > Dear all, > > I am trying a find the value "p" in binomial. > > X ~ Bin(n,p) > > I want to find the

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

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.

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,

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

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

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 have the following problem. inf p= Int [dnorm(x-d) * pnorm(x) ^ (m-1)] dx -inf Given p and m, I want to find d. For example, p m d .9 2 1.81 Is there a way to solve this problem in R? Ideally I would have a function with arguments p and m, and output d. Thanks very much for any help. Bill -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing

Hi all, I want to solve the following equation for x with rho <- 0.5 pnorm(-x)*pnorm((rho*dnorm(x)/pnorm(x)-x)/sqrt(1-rho^2))==0.05 Is there a function in R to do this? Thank you very much! Hannah [[alternative HTML version deleted]]

Hi r-users,   I would like to solve for z values using newton iteration method.  I 'm not sure which part of the code is wrong since I'm not very good at programming but would like to learn.  There seem to be some output but what I expected is a vector of z values.  Thank you so much for any help given.   newton.inputsingle <- function(pars,n) {  runi    <- runif(974, min=0, max=1)

Is there a function to compute the derivative of the probit (qnorm) function in R, or in any of the packages? Thanks, -Andrew [[alternative HTML version deleted]]

hi: could someone please point me to a function that allows me to solve general non-linear functions? > irr.in <- function(r, c1, c2, c3 ) { return(c1+c2/(1+r) + c3/(1+r)^2); } > solve.nonlinear( irr.in, -100, 60, 70 ); 0.189 If someone has written an irr function, this would be helpful, too---though not difficult to write, either. thanks for any pointers. Regards, /iaw

Hi, Is this method broken in R? I am using it to find roots of the following function: f(x) = 2.5*exp(-0.5*(2*0.045 - x)) + 2.5*exp(-0.045) + 2.5*exp(-1.5*x) - 100 It is giving an answer of -38.4762403 which is not even close (f(x) = 2.903809e+25 for x=-38.4762403). The answer should be around 0.01-0.1. This function should converge.. Even for a simple function like f(x) = exp(-x) * x, it gives

Hi there, I would like to solve the following equation in R to estimate 'a'. I have the amp, d, x and y. amp*y^2 = 2*a*(1-a)*(-a*d+(1-a)*x)^2 test data: amp = 0.2370 y= 0.0233 d= 0.002 x= 0.091 Can anyone suggest how I can set this up? Thanks, Diviya [[alternative HTML version deleted]]