similar to: integrate function fails! (PR#1718)

Displaying 20 results from an estimated 400 matches similar to: "integrate function fails! (PR#1718)"

2008 Aug 26
2
Problem with Integrate for NEF-HS distribution
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
2010 Dec 22
3
How to integrate a function with additional argument being a vector or matrix?
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
2013 Feb 12
2
integrate function
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) =
2011 Nov 10
2
performance of adaptIntegrate vs. integrate
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
2004 May 05
4
Discontinuities in a simple graph (machine precision?)
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),...){
2011 May 30
1
Error in minimizing an integrand using optim
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
2012 Oct 20
4
Error in integrate(integrand, 0, Inf) : non-finite function value
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
2011 Jun 25
1
integration function
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)$val } f2 <- function(x) {integrand <- function (x, mu){
2012 Jan 17
4
Problema para integrar una funcion ajustada a un conjunto de datos con la instruccion sm.density(x)
Estimada comunidad, nuevamente con algo que no se como hacer en R, pero aprendiendo .... El problema es el siguiente: 1. tengo un conjunto de 10 mil datos (n) cuyo valor va entre 0 y 10.000 a los que aplico una funcion sm.density() para obtener una especie de histograma pero con una curva que parece continua. 2. Una vez que obtengo esa curva necesito calcular el area bajo la curva en ciertos
2010 Oct 29
2
what´s wrong with this code?
Hello, I want to maximize a likelihood function expressed as an integral that can not be symbolically evaluated. I expose my problem in a reduced form. g<- function(x){ integrand<-function(y) {exp(-x^2)*y} g<-integrate(integrand,0,1) } h<-function(x) log((g(x))) g is an object of the class function, but g(2) is a integrate object, I can print(g(2))
2008 Sep 20
1
lower and upper limits in integrate as vectors
Dear R useRs, i try to integrate the following function for many values "integrand" <- function(z) { return(z * z) } i do this with a for-loop for(i in 2:4) { z <- integrate(integrand, i-1, i)$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
2011 May 03
2
adaptIntegrate - how to pass additional parameters to the integrand
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
2008 Aug 27
5
Integrate a 1-variable function with 1 parameter (Jose L. Romero)
Hey fellas: I would like to integrate the following function: integrand <- function (x,t) { exp(-2*t)*(2*t)^x/(10*factorial(x)) } with respect to the t variable, from 0 to 10. The variable x here works as a parameter: I would like to integrate the said function for each value of x in 0,1,..,44. I have tried Vectorize to no avail. Thanks in advance, jose romero
2009 Dec 18
1
Numerical Integration
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]))
2010 Feb 09
1
how to adjust the output
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     <-
2018 Mar 23
1
Integrate erros on certain functions
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
2008 Mar 07
3
Numerical Integration in 1D
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
2011 Jun 06
2
Taking Integral and Optimization using Integrate, Optim and maxNR
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:
2006 Nov 18
1
Questions regarding "integrate" function
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"
2010 Jan 02
2
ifelse and piecewise function
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)