similar to: Solving quadratic equation in R

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]]

Hi there, I have a complex math equation which does not have a closed form solution. It is - y <- (p*exp(-a*d)*(1-exp((d-p)*(a-x[1]))))/((p-d)*(1-exp(-p*(a-x[1])))) For this equation, I have all the values except for x[1]. So I need to solve this problem numerically. Can anyone suggest an optimization package that I can use to estimate the value for x[1]? Thanks in advance, Diviya

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

Dear R Helpers, I need to find the root of following equationy=x^2+3*x-1 by substitution a random number from less to more in the certain segment.I had tried using this codef <- function (x) x^2+x-12str(xmin <- uniroot(f, c(2, 4), tol = 0.0001)) but $ f.root : num -7.85e-07not 0, and so I'm not shure it's right code and satisfactory answer. Please guide. sveta -- View this

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 would like to know if a R function have the same behaviour than the matlab solve function. I tried something with uniroot but I have some problems: The equation I need to solve is : exp(c0-r0)*(bb0+x)*(bb1-x)=(bb0+x+1)(bb1-x-1) So I tried this: STEP 1: my function test test <- function(x,bb0=-3,bb1=5,c0=2,r0=0) { + ((exp(c0-r0)*(bb0+x)*(bb1-x))/((bb0+x+1)(bb1-x-1))-1)} STEP 2: >

Hello all, I was going to solve (n-m)! * (n-k)! = 0.5 *n! * (n-m-k)! for m when values of n and k are provided n1<-c(10,13,18,30,60,100,500)         # values of n kx<-seq(1,7,1);                               # values of k slv2<-lapply(n1,function(n){    slv1<-lapply(kx,function(k){              lhs<-function(m)              {                

Dear R Helpers, I need to find the root of following equation. 0.0016^2 = (0.001*x)^2 + (0.002 * (1-x))^2 + 2 * 0.7 *0.001*0.002 * x * (1-x). I had tried using "animation " package as follows. # My Code library(animation) ani.options(nmax = 500) solu = newton.method(function(x) 0.0016^2 - 0.001^2*x^2 - 0.002^2*(1-x)^2 - 2*0.7*0.001*0.002*x*(1-x), 1, c(-1,1)) solu$root #

Hello there, Can anyone point me to the code for logLik of an nls object? I found the code for logLik of an lm but could not find exactly what function is used for calculating the logLik of nls function? I am using the nls to fit the following model to data - Model 1: y ~ Ae^(-mx) + Be^(-nx) +c and want to understand what is the likelihood function used by nls. Presumably it is using -

Hello there, I have a question regarding bar plots. I am trying to plot the data from the following matrix as a barplot - # input data mdat <- matrix(c(0.1,0.9,0.9,0.1,0.5,0.5,0.45,1-0.45,0.6,0.4,0.8,0.2), nrow = 6, ncol=2, byrow=TRUE, + dimnames = list(c("Mon", "Mon", "Tues", "Tues", "Thurs", "Friday"), +

Hello there, I have a matrix with some data and I want to split this matrix based on the values in one column. Is there a quick way of doing this? I have looked at cut but I am not sure how to exactly use it? for example: I would like to split the matrix "a" based on the spending such that the data is binned groups [0..99],[100..199]...and so on. a <- data.frame(patient=1:7,

Hello there, I am new to using regression in R. I wanted to solve a simple regression problem where I have 2 equations and 2 unknowns. So lets say - y1 = alpha1*A + beta1*B y2 = alpha2*A + beta2*B y1 <- runif(100000, 0,1) y2 <- runif(100000,0,1) alpha1 <- 0.6 alpha2 <- 0.75 beta1 <- 1-alpha1 beta2 <- 1-apha2 I now want this equation to estimate the values of A and B. Both A

Hi there, I was wondering if there is any R package that one can use for plotting that has more legend symbols - the standard pch has 18 symbols but I need ~30 for my application- and just using different colors is not an option. Thank you in advance, Diviya [[alternative HTML version deleted]]

Hello, I want to solve: x*(3^x)*log(4)-x*log(4/3)-(3^x)+1=0 for x. I used the following code, uniroot(function(x) x*(3^x)*log(4)-x*log(4/3)-(3^x)+1, lower = -2, upper = 2, tol = 0.001 ) While using this I am getting the following error. Can anyone please help me out. Error in uniroot(function(x) x * (3^x) * log(4) - x * log(4/3) - (3^x) + : f() values at end points not of opposite sign.

Hello there, I recently wrote some code to perform pairwise correlations between all samples in a large dataset. So we are talking about performing pairwise correlations between 400K vectors. Since R has a very rich library of functions, it was very easy to code this in R. However, R was probably not the best choice as it is super slow for this large job. So my plan is to recode it in C. I was

Hello R-list, How can I row wise solve a function for which the input variables are written in a matrix or a vector and have the calculated output added to the matrix or written in a vector? To specify my case in particular I continue: I would like R to calculate for me a value for ‘t’ which is function of input parameters A and B, which are comprised in a matrix called ‘ddt’ and some other

Hi R, I have a 100th order equation for which I need to solve the value for x. Is there a package to do this? For example my equation is: (x^100 )- (2*x^99) +(10*x^50)+.............. +(6*x ) = 4000 I have only one unknown value and that is x. How do I solve for this? BR, Shubha Shubha Karanth | Amba Research Ph +91 80 3980 8031 | Mob +91 94 4886 4510 Bangalore *

Hi all, I've a linear equation of the form: 0.95=2 ( [3+ln(x/3)]^-13  + 4 [3+2ln(x/3)]^-13  + [3+4ln(x/3)]^-13 ) and I want to solve it for x, can I do this using R? Thanks in advance. Maram. [[alternative HTML version deleted]]

Hi all, I have the following two function f1 and f2. f1 <- function(lambda,z,p1){ lambda*(p1*exp(-3*z-9/2)+(0.2-p1)*exp(4*z-8))-(1-lambda)*0.8} f2 <- function(p1,cl, cu){ 0.8*(pnorm(cl)+(1-pnorm(cu)))/(0.8*(pnorm(cl)+(1-pnorm(cu)))+p1*(pnorm(cl+3)+(1-pnorm(cu+3)))+(0.2-p1)*(pnorm(cl-4)+(1-pnorm(cu-4))))}-0.05 First fix p1 to be 0.15. (i) choose a lambda value, say lamda=0.6, (ii)

Hello there, I am using NLS for fitting a complex model to some data to estimate a couple of the missing parameters. The model is - y ~ (C+((log(1-r))*exp(-A*d)*(-1+r+exp(d*(A-B)))/(r*(-A*d+d*B+log(1-r))))) where A, B and C are unknown. In order to test the model, I generate data by setting values for all parameters and add some noise (C). A <- 20 B <- 500 r <- 0.6881 d <-