similar to: how to plot a differential equation?

I’m trying to obtain a plot of given function, say the density of a normal, in which I could indicate the area under the curve between two points (a,b). I.e. the P(a<X<b). Could anyone indicate me which is the instruction for obtain such a shadow area under a curve? I guess the solution is not difficult, but I can not find the procedure. Thanks Albert Sorribas Departament de Ciències

This is a generic request concerning were to look for finding appropriate information on a precise procedure in R. I’m using R for teaching introductory statistics and my students are learning how to deal with it. However, I find it difficult to locate some of the procedures. For instance, for basic crosstabulation, it is obvious that basic functions as table, ftable, and prop.table can be used.

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

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 compare the hazard functions of two samples using the Cox proportional hazards model. For sample 1 I have individual time-to- event data. For sample 2 I don't have individual data, but grouped data that allows to obtain a hazard function. I am wondering if there is an R function that allows to obtain a hazard ratio of the two hazard funtions (under the

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)

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

Library gmodels include a function CrossTable that is useful for crosstabulation. In the help, it is indicated that one can call this function as CrossTable(data), were data is a matrix. However, when I try to use this option, it doesn't help. Any idea? Is there a bug? Thanks for your help. Prof. Albert Sorribas Grup de Biomatem?tica i Bioestad?stica Departament de Ci?ncies M?diques B?siques

Hi r-users, I would like to solve system of nonlinear equation using BBsolve function and below is my code.  I have 4 parameters and I have 4 eqns. mgf_gammasum <- function(p) { t  <- rep(NA, length(p)) mn <- 142.36 vr <- 9335.69 sk <- 0.8139635 kur <- 3.252591 rh  <- 0.896 # cumulants k1 <- p[1]*(p[2]+p[3]) k2 <- p[1]*(2*p[2]*p[3]*p[4] +p[2]^2+p[3]^2) k3 <-

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

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]]

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

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

I'm having trouble reshaping a data.frame from long to wide. (I think that's the right terminology; feel free to educate me.) I've looked at the reshape function and package and plyr package, but I can't quite figure out how to do this after a dozen variations. I have a data.frame with more levels than this, but similar to: > tst K1 K2 K3 V1 V2 V3 1 10 D a 0.08 99