similar to: first derivative of a time series

I upgraded from Fedora Core 4 to Fedora Core 5 and I find a lot of previously installed packages won't run because shared libraries or other system things have changed "out from under" the installed R libraries. I do not know for sure if the R version now from Fedora-Extras (2.2.1) is exactly the same one I was using in FC4. I see problems in many packages. Example, Hmisc: unable

Dear all: Does anyone has harmonic regresssion analysis package written in R (to be used in Windows platform) ? Thanks _______________________________________ YM - Â÷½u°T®§ ´Nºâ§A¨S¦³¤Wºô¡A§AªºªB¤Í¤´¥i¥H¯d¤U°T®§µ¹§A¡A·í§A¤Wºô®É´N¯à¥ß§Y¬Ý¨ì¡A¥ô¦ó»¡¸Ü³£ÉN¨«¥¢¡C http://messenger.yahoo.com.hk [[alternative HTML version deleted]]

Is there some reason why > (1:2)+(1:3) [1] 2 4 4 Warning message: In (1:2) + (1:3) : longer object length is not a multiple of shorter object length can't be made into an error? I realise it was there in S-PLUS, but since it produces a warning there can't be many examples on CRAN or Bioconductor using it, and I can't think of any situation where it would be used deliberately.

Hi All Please see the Rprofile file which i have modified as follows and after that when I start R then I see that R says to me "TRUE" for all the packages implying that all loaded at once. But when i try to use commands as simple as help("lm"), it doesnt work nor any of the menu "Packages" is not working. Although the regression using lm ( Y ~ X ) is working

Hello,        Dr. Simon Wood told me how to force a cubic spline passing through a point. The code is as following. Anyone  who knows how I can change the code to force the first derivative to be certain value. For example, the first derivative of the constrained cubic spline equals 2 at point (0, 0.6).        I really appreciate your help!        Thanks!                 Best             Victor   

Hello, I've noticed that SPLUS has a function for evaluating the derivative matrices of splines. I've also noticed that there's a function in R for evaluating matrices from smooth.spline; maybe someone knows if something has been written to evaluate matrices from 'smooth.basis'? regards, s

I would like to calculate a first and second derivative and am having problems finding a simple solution. My syntax may be off as I am not a mathematician, so pardon ahead of time. data: t<-seq(0,4, by=.1) The function is: H(t) = t^3-6*t^2+5*t + 30 from here I plot the curve: plot(x,y ,xlab="x-values", ylab="f(x)", type="l") But would like to similarly plot the

Hello, I've noticed that SPLUS seems to have a function for evaluating derivative matrices of splines. I've found the R function that evaluates matrices from 'smooth.spline'; maybe someone has written something to do the same with smooth.basis? regards, s

How would I numerically find the x value where the derivative of the function below is zero? x <- seq(1,2, by = .01) y <- 5*cos(2*x)-2*x*sin(2*x) plot(x,abs(y), type = "l", ylab = "|y|")

Hi, Everyboday, Does anybody know how to get the first derivative form the cubic natural spline matrix? Thanks !!! Jassy ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Nuoo-Ting (Jassy) Molitor, MS, Junior Statistician Division of Biostatistics Department of Preventive Medicine University of Southern California Tel: (323) 442-2584 1540 Alcazar St., CHP-210F Fax:

Good evening! I have a multi-dimensional minimization problem whose gradient is pretty hard to code. I tried Nelder and Mead method implemented in function optim and it does not work well. I also tried the quasi Newton method in optim using difference as approximate derivative. It does not work well either. I just went through Numerical Recipes book. The book discusses another method without

Dear All, I'd like to ask fro any pointers to code in any package out there that can (even partially) handle the following situation: say we have the linear model # toy data y <- rnorm(100) time <- runif(100, 0, 5) treat <- gl(2, 50, labels = c("placebo", "active")) sex <- gl(2, 1, 100, labels = c("male", "female")) bmi <- runif(100,

If I'm not mistaken, that's sort of local polynomial with even degree and fixed bandwidth (based on my own interpretation of description in Numerical Recipes). You can do that with functions in the KernSmooth package. HTH, Andy > -----Original Message----- > From: wolf at micro-biolytics.com [mailto:wolf at micro-biolytics.com] > Sent: Friday, March 21, 2003 1:43 PM > To:

Dear All I have a problem of calculating the derivative of dxm matrix A with respect to another dxm matrix B, where A= [a1 a2 ... am] and B =[b1 b2 ... bm] with ai and bj are vectors. Actually the matrix A itself is the first order derivative of a scalar J with respect to B, i.e., A = dJ/dB, where a1 = dJ/db1, a2 = dJ/db1, and so on. Now dA/dB is the second order derivative dA/dB =

Suppose I have a simple function that returns a matrix, such as: test <- function(x){ return(matrix(c(x,x^2,x^3,x^4),2,2)) } so that test returns: [ x x^3 ] [ x^2 x^4 ] Is it possible for me to get the derivative of an expression such as: c(1,0) %*% test() %*% c(0,1) The vectors are used just to "index" the matrix. I don't want a value, but the expression to work with

Hello, I am trying to find a hessian matrix that involves log(gamma(1/p)) second derivative, p being one of the parameters of the function. I am using a function "deriv" with the hessian=TRUE option, but psigamma is not on the list of derivative functions. I know that it is possible to use 'psigamma(p,deriv)', but it doesn't work with 1/p. Does anybody can help with this?

Dear list, This is quite a specific question requiring the package orthopolynom. This package provides a nice implementation of the Legendre polynomials, however I need the associated Legendre polynomial which can be readily expressed in terms of the mth order derivative of the corresponding Legendre polynomial. (For the curious, I'm trying to calculate spherical harmonics [*]).

Dear All, I'm looking for?a way on computing the derivative of first and second order?of a smoothing curve produced by a nonprametric regression. For instance, if we run the R script below, a smooth nonparametric regression curve is produced. provide.data(trawl) Zone92?? <- (Year == 0 & Zone == 1) Position <- cbind(Longitude - 143, Latitude) dimnames(Position)[[2]][1] <-

Is there a function to compute the derivative of the probit (qnorm) function in R, or in any of the packages? Thanks, -Andrew [[alternative HTML version deleted]]

Hi! I'm faced the next problem: suppose I have a vector of some coefficient and for some purpose from these coefficients I need dynamically build a functional expression (because, the number of elements in the vector of coefficients might differ each time). So, I'm trying to do like this (let the number of coefficients be 3): Building expression > foo <-