similar to: Simultaneous equations

In this email to the R-help list: https://stat.ethz.ch/pipermail/r-help/2017-June/447474.html and in this question on Stackoverflow: https://stackoverflow.com/questions/44486643/nleqslv-memory-use-in-r Andrew Leach has raised a question about the memory usage of my package nleqslv. In a model with a loop within a function he has experienced continuously increasing memory usage by package nleqslv

Dear R-users, I use the "nleqslv" function to get parameter estimates by solving a system of non-linear equations. But I also need standard error for each of estimates. I checked the nleqslv manual but it didn't mention about SE. Is there any way to get the SE for each estimate? Thank you very much. [[alternative HTML version deleted]]

Dear all, I initially ran into this problem while rebuilding a package dependent on nleqslv. I got the following error: Error in match.arg(global) : 'arg' must be of length 1 This didn't occur in previous versions of nleqslv, but did in the current one (2.4). I think I pinned the problem down to the following example: Take two functions: test <-

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

Hello all, I am relatively new to R, but enjoying it very much. I am hoping that someone on this list can help me with an issue I am having. I am having issues with iterations over nleqslv, in that the solver does not appear to clean up memory used in previous iterations. I believe I've isolated the/my issue in a small sample of code: library(nleqslv) cons_ext_test <- function(x){

Hi, I am trying to find a simple way to numerically solve a system of two equations equal to zero with two unknowns (x_loc and y_loc). Here is a mock data set and below it, the equations I need to solve. theta<-c(180,135,90)/(2*pi) x<-c(0,0,15) y<-c(20,0,0) 0 = -sum((y_loc-y)*(sin(theta)*(x_loc-x)-cos(theta)*(y_loc-y))/(((x_loc-x)^2+(y_loc-y)^2)^0.5)^3) 0 =

Hi there, I have following equations to be solved for a and b: a/(a+b) = x1 ab/((a+b)^2 (a+b+1)) = x2 Is there any direct function available to solve them without disentangling them manually? Thanks for your help.

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]]

> > On 9 Feb 2017, at 16:00, G?ran Brostr?m <goran.brostrom at umu.se> wrote: > > > > In my package 'glmmML' I'm using old C code and linpack in the optimizing procedure. Specifically, one part of the code looks like this: > > > > F77_CALL(dpoco)(*hessian, &bdim, &bdim, &rcond, work, info); > > if (*info == 0){ > >

I have two matrix s1 and s2, each of them is 1000*1. and I have two equations: digamma(p)-digamma(p+q)=s1, digamma(q)-digamma(p+q)=s2, and I want to sovle these two equations to get the value of x and y, which are also two 1000*1 matrices. I write a program like this: f <- function(x) { p<- x[1]; q <- x[2]; ((digamma(p)-digamma(p+q)-s1[2,]) )^2 +((digamma(q)-digamma(p+q)-s2[2,]) )^2

Dear R users, I have a problem. I would like to solve the following: I have pL = 1/(1+e^(-b0+b1)) pM = 1/(1+e^(-b0)) pH = 1/(1+e^(-b0-b1)) My target function is TF= mean(pL,pM,pH) which must equal 0.5% My non-linear constraint is nl.Const = 1-(pM/pH), which must equal 20%, and would like the values of both b0 and b1 where these conditions are met. I have searched widely for an answer,

In my package 'glmmML' I'm using old C code and linpack in the optimizing procedure. Specifically, one part of the code looks like this: F77_CALL(dpoco)(*hessian, &bdim, &bdim, &rcond, work, info); if (*info == 0){ F77_CALL(dpodi)(*hessian, &bdim, &bdim, det, &job); ........ This usually works OK, but with an ill-conditioned data

I'm solving 4 complex equations simultaneously. Code is below. The code returns only zero's for the solution though there should also be a non-zero result. I'm pretty confident that the equations are correct because they are straight from a published paper and I checked them pretty thoroughly. The parameter values I used are from the published paper as well. Any suggestions for how

Dear all, I am relatively new to R and have had some difficulty in understanding an error i get when running a code to solve a system of non-linear equations, with four equations and two variables. This is my code: ALPHA <- c(-0.0985168033402, 0.1, 0.15, 0.2, 0.25, 0.3, 0.35, 0.4) BETA <- c(-0.0985168033402, 0.1, 0.15, 0.2, 0.25, 0.3, 0.35, 0.4) GAMMA <- c(0.3940672148378, 0.1, 0.15,

People, I'm researching some Bayesian statistic topics and in the midle of my study i found a very simple problem and i'm trying to find a simple package to solve this type of equations: Lets say that i need to compute beta values for the beta distribution and i now for example: E(teta)=a/(a+b) = 0,5 Var(teta)=ab/((a+b)^2(a+b+1))=0.05 So if i want to solve this to non-linear system to

> On 10 Feb 2017, at 14:53, G?ran Brostr?m <goran.brostrom at umu.se> wrote: > > Thanks to all who answered my third question. I learned something, but: > > On 2017-02-09 17:44, Martin Maechler wrote: >> >>>> On 9 Feb 2017, at 16:00, G?ran Brostr?m <goran.brostrom at umu.se> wrote: >>>> >>>> In my package 'glmmML'

I have a question regarding the most efficient way to select a substring of a vector: I have a vector of value v, and I want to select a subspace of this vector called w such that: w=v[1:n] where sum(w) = x I am interested in what you thing would be the most efficient way to do this - I would like to avoid slowing down my simulations as much as possible. Thank you very much for any help that

Santi, In the second line of your function you have the following: f <- numeric(length(x)) This sets the length of this numeric vector (i.e., "f") to the length of the vector "x". Later, inside the function you assign to values to 4 elements of the vector "f". This assumes that "f" is at least 4 element in length. However, you define

Hi, I would appreciate if someone could help me on track with this problem. I want to compute some parameters from a system of equations given a number of sample observations. The system looks like this: sum_i( A+b_i>0 & A+b_i>C+d_i) = x sum_i( C+d_i>0 & C+d_i>A+b_i) = y sum_i( exp(E+f_i) * ( A+b_i>0 & A+b_i>C+d_i) = z A, C, E are free variables while the other

Hello there, I wish to solve the following nonlinear System of equations: + u1 - Vmax11*S1/(S1 + Km11 *(1 + S2/Km21)) - Vmax12*S1/( S1 + Km12 *(1+S2/Km22)) == 0 + u2 - Vmax22*S2/(S2 + Km22 *(1 + S1/Km12)) - Vmax21*S2/( S2 + Km21 *(1+S1/Km11)) == 0 + Vmax11*S1/(S1 + Km11 *(1 + S2/Km21)) + Vmax12*S1/( S1 + Km12 *(1+S2/Km22)) - d1*P1 == 0 + Vmax22*S2/(S2 + Km22 *(1 + S1/Km12)) + Vmax21*S2/( S2 +