similar to: what´s wrong with this code?

I need to calcuate the cumulative probability for the Natural Exponential Family - Hyperbolic secant distribution with a parameter theta between -pi/2 and pi/2. The integration should be between 0 and 1 as it is a probability. The function "integrate" works fine when the absolute value of theta is not too large. That is, the NEF-HS distribution is not too skewed. However, once the

Dear list, [cross-posting from Stack Overflow where this question has remained unanswered for two weeks] I'd like to perform a numerical integration in one dimension, I = int_a^b f(x) dx where the integrand f: x in IR -> f(x) in IR^p is vector-valued. integrate() only allows scalar integrands, thus I would need to call it many (p=200 typically) times, which sounds suboptimal. The

Dear expeRts, I somehow don't see why the following does not work: integrand <- function(x, vec, mat, val) 1 # dummy return value A <- matrix(runif(16), ncol = 4) u <- c(0.4, 0.1, 0.2, 0.3) integrand(0.3, u, A, 4) integrate(integrand, lower = 0, upper = 1, vec = u, mat = A, val = 4) I would like to integrate a function ("integrand") which gets an "x" value (the

Hi All, Can any one help to explain why min and max function couldn't work in the integrate function directly. For example, if issue following into R: integrand <- function(x) {min(1-x, x^2)} integrate(integrand, lower = 0, upper = 1) it will return this: Error in integrate(integrand, lower = 0, upper = 1) : evaluation of function gave a result of wrong length However, as min(U,V) =

Hi, I am trying to update a CentOS 4.4 but the yum update gives me error messages Setting up Update Process Setting up repositories http://mirror.centos.org/centos/4/updates/i386/repodata/repomd.xml: [Errno 4] IOError: <urlopen error nonnumeric port: '8080?'> Trying other mirror. Cannot open/read repomd.xml file for repository: update failure: repodata/repomd.xml from update:

Hi all, Can anyone please take a look at the following two functions. The answer does not seem to be right. Thank you very much! f1 <- function(x) {integrand <- function (x, mu){ dnorm(x, mean=mu, sd=1)*dnorm(mu, mean=2, sd=1) } integrate(integrand, -Inf, Inf,x)$val } f2 <- function(x) {integrand <- function (x, mu){

Hi, Am not sure if my code itself is correct. Here's what am trying to do: Minimize integration of a function of gaussian distributed variable 'x' over the interval qnorm(0.999) to Inf by changing value of parameter 'mu'. mu is the shift in mean of 'x'. Code: # x follows gaussian distribution # fx2 to be minimized by changing values of mu # integration to be done over

Full_Name: José Enrique Chacón Version: 1.5.0 and 1.3.1 OS: Windows Millenium Submission from: (NULL) (158.49.28.155) Dear reader: I was trying to evaluate the L2 error produced when estimating the density function N(0,1) from a sample of size 100 using a kernel density estimate. It produced a strange value. You can reproduce the process by typing samp<-rnorm(100)

Hi R-users,   I have this code below and I understand the error message but do not know how to correct it.  My question is how do I get rid of “with absolute error < 7.5e-06” attach to value of cdf so that I can carry out the calculation.   integrand <- function(z) { alp  <- 2.0165   rho  <- 0.868   # simplified expressions   a      <- alp-0.5   c1     <-

Dear R users, When I run the code below, I get the error "Error in integrate(integrand, 0, Inf) : non-finite function value". The code works if the function returns only "sum(integ)". However, I want to add "cmh" to it. When I add "cmh" I get that error. I can't figure out why this is happening because my integrate function has nothing to do with

Dear R useRs, i try to integrate the following function for many values "integrand" <- function(z) { return(z * z) } i do this with a for-loop for(i in 2:4) { z <- integrate(integrand, i-1, i)$value cat("z", z, "\n") } to speed up the computation for many values i tried vectors in integrate to do this. vec1<-1:3 vec2<-2:4

Dear @ll. I have to calculate numerical integrals for triangular and trapezoidal figures. I know you can calculate the exactly, but I want to do it this way to learn how to proceed with more complicated shapes. The code I'm using is the following: integrand<-function(x) { print(x) if(x<fx[1]) return(0) if(x>=fx[1] && x<fx[2]) return((x-fx[1])/(fx[2]-fx[1]))

Hi there. Thanks for your time in advance. I am using R 2.2.0 and OS: Windows XP. My final goal is to calculate 1/2*integral of (f1(x)^1/2-f2(x)^(1/2))^2dx (Latex codes: $\frac{1}{2}\int^{{\infty}}_{\infty} (\sqrt{f_1(x)}-\sqrt{f_2(x)})^2dx $.) where f1(x) and f2(x) are two marginal densities. My problem: I have the following R codes using "adapt" package. Although "adapt"

In the help for ?integrate: >When integrating over infinite intervals do so explicitly, rather than just using a large number as the endpoint. This increases the chance of a correct answer ? any function whose integral over an infinite interval is finite must be near zero for most of that interval. I understand that and there are examples such as: ## a slowly-convergent integral integrand

Hi! I am a beginner of R. I am trying to calculate integrate and draw a graph of the output, but just kept on getting error messages. I list my program and error message below. Please help. Many Thanks! ======================= + > a<--11 > b<-0.1 > c<-0.012 > x<-0:110 > t<-0:15 > integrand<-function(x) {exp(-exp(a-c*t)*(exp(b*x)-exp(c*x))/(b-c))} >

Dear All, Hello! I have some questoins in R programming as follows: Question 1- How to take the integral of this function with respect to y, such that x would appear in the output after taking integral. f(x,y)=(0.1766*exp(-exp(y+lnx))*-exp(y+lnx))/(1-exp(-exp(y+lnx))) y in (-6.907,-1.246) It is doable in maple but not in R. At least I could not find the way. p.s: result from maple is:

Hi, I've got an ugly but fairly simple function: mdevstdev <- function(a){ l <- dnorm(a)/(1-pnorm(a)) integrand <- function(z)(abs(z-l)*dnorm(z)) inted <- integrate(integrand, a, Inf) inted[[1]]/((1- pnorm(a))*sqrt((1 + a*l - l^2))) } I wanted to quickly produce a graph of this over the range [-3,3] so I used: plotit <-function(x=seq(-3,3,0.01),...){

I am a novice user of "R" and I'm learning with R version 2.8.1, using WinEdt_1.8.1, under Widows Vista Home Version. ## The test function below, from a vector input, returns vector values: # and it contains an "ifelse"statement TEST<- function(x) { low<- -x^2 up<- x^4 ifelse(x>=0,up,low ) } u<- seq(-1,1,0.5) TEST(u)

Dear UseRs, I'm curious about the derivative of n!. We know that Gamma(n+1)=n! So when on takes the derivative of Gamma(n+1) we get Int(ln(x)*exp(-x)*x^n,x=0..Inf). I've tried code like > integrand<-function(x) {log(x)*exp(x)*x^n} > integrate(integrand,lower=0,upper=Inf) It seems that R doesn't like to integrate for any n, and I was wondering if anyone knew a way around

Hello, I am trying to use adaptIntegrate function but I need to pass on a few additional parameters to the integrand. However, this function seems not to have the flexibility of passing on such additional parameters. Am I missing something or this is a known limitation. Is there a good alternative to such restrictions, if there at all are? Many thanks for your time. HC -- View this message in