similar to: Solution to differential equation

Hello, I am trying to find the analytical solution to this differential equation dR/dt = k1*R (1-(R/Rmax)^k2); R(0) = Ro k1, k2, and Rmax are parameters that need to fitted, while Ro is the baseline value (which can be fitted or fixed). The response (R) increases initially at an exponential rate governed by the rate constant k1. Response has a S-shaped curve as a function of time and it

Hi, I would like to plot the following equation: dF(x)/dx=(k1+k2F(x))(1-F(x)) where k1 and k2 are parameters that I have estimated already. How can I plot the curve in R? Thanks! Montserrat Rue Universitat de Lleida (Spain) [[alternative HTML version deleted]]

Hi, I would like to solve a system of coupled ordinary differential equations, where there is a delay (time lag) term. I would like to use the "lsoda" function "odesolve" package. However, I am not sure how to specify the delay term using the syntax allowed by odesolve. Here is an example of the kind of problem that I am trying to solve: > library(odesolve)

Hello, I am trying to compute the density of a variable k that is either (1) Normally distributed; (2) Log-Normally distributed; or (3) follows proportional error distribution. I tried to search R-help and the answer for normal distribution was easy to find (please see 1c). I am not sure if my formula for dlnorm is correct (please see 2c below)? I really don't know what function to use for the

Hello, I am trying to add a less than equal to symbol in a ggplot2 legend text. See sample code below. I have tried using the expression function and \u2264. I also tried adding labels to legend.text under theme. Neither of these 3 options work. Please help, Mahesh ++++++++++++++ Extra.column=ifelse(data[,covariate]>cutpoint,1,0) Grp1 <- "\u2264 1.5" Grp2 <-

Hello, I am trying to use ≥ symbol in the Trellis custom strip. The ascii ALT Key for this symbol is 242. However, when I try "\242 90 kg" I get the ¢ symbol in the trellis custom strip. Can some one kindly help me? Thanking you in advance, MNS [[alternative HTML version deleted]]

To whom it may concern, I am a student at Macaleste College, and next semester Macalester is going to offer a course for CellBio that is mainly statistically based. For the most part the students will be using R for analysis. The problem is there will be some simple differential equations for the students to solve. The committee that in charge of the classes corriculam would like only to

Hi all, I try to assess the parameters (K1,K2) of a model that describes the adsorption of a molecule onto on adsorbent. equation: dq/dt = K1*C*(qm-q)-K2*q I know the value of 'qm' and I experimentally measure the variables 'q', 'C', and the time 't'. t C q 1 0 144.05047 0.0000000 2 565 99.71492 0.1105625 3 988 74.99426

Hi.. i have an expression of the form: model1<-nls(y~beta1*(x1+(k1*x2)+(k1*k1*x3)+(k2*x4)+(k2*k1*x5)+(k2*k2*x6)+(k3*x7)+(k3*k4*x8)+(k3*k2*x9)+(k3*k3*x10)+ (k4*x11)+(k4*k1*x12)+(k4*k2*x13)+(k4*k3*x14)+(k4*k4*x15)+(k5*x16)+(k5*k1*x17)+(k5*k2*x18)+(k5*k3*x19)+

Hi everyone! I'm matching two samples to create one sample that have pairs of observations equal for the k1 variable. Merge() doesn't work because I dont't want to recycle the values. x <- data.frame(k1=c(1,1,2,3,3,5), k2=c(20,21,22,23,24,25)) x y <- data.frame(k1=c(1,1,2,2,3,4,5,5), k2=c(10,11,12,13,14,15,16,17)) y merge(x,y,by="k1") k1 k2.x k2.y 1 1 20

Hello, I am writing a package with a collection of several models. In order to allow users to play interactively with the models (in contrast to hacking lengthy scripts), I want to put all what is needed to run a particular model into a single list object for each model. Then there will be a collection of functions to run the model or to modify parameters, time steps, integration method ...,

Dear Colleagues, Our group is also working on implementing the use of R for pharmacokinetic compartmental analysis. Perhaps I have missed something, but > fit <- nls(noisy ~ lsoda(xstart, time, one.compartment.model, c(K1=0.5, k2=0.5)), + data=C1.lsoda, + start=list(K1=0.3, k2=0.7), + trace=T + ) Error in eval(as.name(varName), data) : Object

Hey all, I am trying to work under a State Space form, but I didn't get the help exactly. Have anyone eles used this functions? I was used to work with S-PLUS, but I have some codes I need to adpt. Thanks alot, Bernardo [[alternative HTML version deleted]]

Dear R Gang, I'm interested in using R and the nls package for fitting kinetic models. I'm having some difficulty getting a model specified for nls though. The math for the model that I want to fit is dg(t)/dt = K1 f(t) - k2 g(t) where g(t) and f(t) are measured data at a sequence of times t. K1 and k2 are the parameters of the model. If I solve this, the solution is g(t) = K1

Dear list, If I know the standard error for k1 and k2, is there anything I can call in R to calculate the standard error of k1/k2? Thanks. Jun [[alternative HTML version deleted]]

Hi, The code below is giving me this error message: Error in while (err > eps) { : missing value where TRUE/FALSE needed In addition: Warning messages: 1: In dnbinom(x, size, prob, log) : NaNs produced 2: In dnbinom(x, size, prob, log) : NaNs produced I know from the help files that for dnbinom "Invalid size or prob will result in return value NaN, with a warning", but I am not able

--- v2v/convert_linux.ml | 18 +++++++++++++++--- 1 file changed, 15 insertions(+), 3 deletions(-) diff --git a/v2v/convert_linux.ml b/v2v/convert_linux.ml index f670812..39a520c 100644 --- a/v2v/convert_linux.ml +++ b/v2v/convert_linux.ml @@ -49,13 +49,14 @@ type kernel_info = { ki_modules : string list; (* The list of module names. *) ki_supports_virtio : bool; (* Kernel has

Thank you Jim, I read the data as you suggested but I could not find K1 in col1. rbind(preval,mydat) Col1 Col2 col3 1 <NA> <NA> <NA> 2 X1 <NA> <NA> 3 Y1 <NA> <NA> 4 K2 <NA> <NA> 5 W1 <NA> <NA> 6 Z1 K1 K2 7 Z2 <NA> <NA> 8 Z3 X1 <NA> 9 Z4 Y1 W1 On Sat, Feb 24, 2018 at 6:18 PM, Jim

HI Jim and all, I want to put one more condition. Include col2 and col3 if they are not in col1. Here is the data mydat <- read.table(textConnection("Col1 Col2 col3 K2 X1 NA Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) The desired out put would be Col1 Col2 col3 1 X1 0 0 2 K1 0 0 3 Y1 0 0 4 W1 0 0 6 K2 X1

Hi, I'm new to R and have a problem with a little test program (see below). Why doesn't <<- in function rk4 assign the new value to y so that it is seen in rktest. I thought that <<- does exactly this. But it seems that I didn't get it right. I would be very appreciative for an explanation of that behaviour of <<-. I know how to write the whole thing so that it