similar to: how to adjust the output

Hi R-users,   I have this code here: library(numDeriv)   fprime <- function(z) { alp  <- 2.0165;   rho  <- 0.868;   # simplified expressions   a      <- alp-0.5   c1     <- sqrt(pi)/(gamma(alp)*(1-rho)^alp)   c2     <- sqrt(rho)/(1-rho)   t1     <- exp(-z/(1-rho))   t2     <- (z/(2*c2))^a   bes1   <- besselI(z*c2,a)   t1bes1 <- t1*bes1   c1*t1bes1*t2 }   ## Newton

Hi r-users,   I would like to solve for z values using newton iteration method.  I 'm not sure which part of the code is wrong since I'm not very good at programming but would like to learn.  There seem to be some output but what I expected is a vector of z values.  Thank you so much for any help given.   newton.inputsingle <- function(pars,n) {  runi    <- runif(974, min=0, max=1)

I would like to solve a nonlinear eqn using newton method and here is the code:   library(numDeriv)   fprime <- function(z) { alp  <- 2.0165;   rho  <- 0.868;   # simplified expressions   a      <- alp-0.5   c1     <- sqrt(pi)/(gamma(alp)*(1-rho)^alp)   c2     <- sqrt(rho)/(1-rho)   t1     <- exp(-z/(1-rho))   t2     <- (z/(2*c2))^a    bes1   <- besselI(z*c2,a)  

Hi r-users,   I hope somebody can help me with this code. I would like to solve for z values using newton iteration method.  I 'm not sure which part of the code is wrong since I'm not very good at programming but would like to learn.  There seem to be some output but what I expected is a vector of z values.  Thank you so much for any help given.   newton.inputsingle <-

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 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 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)

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"

Hi r-users,   I hope somebody can help me to understand the error message.  Here is my code; ## Newton iteration newton_gam <- function(z) { n   <- length(z)   r   <- runif(n)   tol <- 1E-6   cdf <- vector(length=n, mode="numeric")   fprime <- vector(length=n, mode="numeric")   f   <- vector(length=n, mode="numeric")     for (i in 1:1000)   {

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

Hello All, I have to create a variable that is a function of another one (already created), its cumulative distribution function and the integral of this cumulative distribution, with limits: 0 and the value of the variable. To be clear, I have the variable called “cip”. And its cdf called “cdfcip” I need to create the variable: bip = cip + ((1 – cdfcip)^4)*integral((1-cdf(u))^4*du, 0, value

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))} >