similar to: bug from nlm function (PR#9711)

Full_Name: Joerg Polzehl Version: 2.3.1 OS: x86_64, linux-gnu Submission from: (NULL) (62.141.176.22) I observed an incorrect behavior of function deriv when evaluating arguments of dnorm deriv(~dnorm(z,0,s),"z") expression({ .value <- dnorm(z, 0, s) .grad <- array(0, c(length(.value), 1), list(NULL, c("z"))) .grad[, "z"] <- -(z * dnorm(z))

Dear participants to the list, this is my problem: I want to obtain an expression that represents the second derivative of one function. With "deriv3" (package "stats") it is possible to evaluate the second derivative, but I do not know how I can get the (analytical) expression of this derivative. For example: Suppose that I have a function of this form:

Also, this should have gone in R-bugs quite a while ago : ------- start of forwarded message ------- From: Martin Maechler <maechler@stat.math.ethz.ch> To: R-core@stat.math.ethz.ch Subject: PROTECT() bugs in deriv(*, *, function.arg = ) Date: Mon, 16 Apr 2001 21:02:10 +0200 In R versions 0.50 and 0.64.2 , the following worked > deriv(expression(sin(cos(x) * y)),

Dear all i want to fit the self start model in nls. i have two question. i have a function, (asfr ~ I(((a*b)/c))+ ((c/age)^3/2)+ exp((-b^2)*(c/age)+(age/c)-2) i am wondering how to build the selfstart model. there is lost of example, (i.e. SSgompertz, SSmicman, SSweibull, etc). my question is, how to derive the function of parameters. and also which model to use for get the initials values. In the

------- start of forwarded message ------- From: Martin Maechler <maechler@stat.math.ethz.ch> To: R-core@stat.math.ethz.ch Subject: PROTECT() bugs in deriv(*, *, function.arg = ) Date: Mon, 16 Apr 2001 21:02:10 +0200 In R versions 0.50 and 0.64.2 , the following worked > deriv(expression(sin(cos(x) * y)), c("x","y"), function(x,y){}) function (x, y)

Hello, nls library provides 6 self-starting models, among them: SSfp, a four parameters logistic function. Its self-starting procedure involves several steps. One of these steps is: pars <- as.vector(coef(nls(y ~ cbind(1, 1/(1 + exp((xmid - x)/exp(lscal)))), data = xydata, start = list(lscal = 0), algorithm = "plinear"))) which assumes an initial value of lscal equal to 0. If lscal

deriv seems to have problems with a minus-sign before a bracket. Below are four examples of the same function, the top one is wrong, all others are correct (hopefully). Rest of expression not shown, it is the same for all versions. _ platform i386-pc-mingw32 arch x86 os Win32 system x86, Win32 status major 1 minor 3.0 year 2001 month 06 day 22 language R

This example is pretty compelling, but I still think I want to limit the scope of this change to not include this feature. It's probably reasonable to hold onto the idea as future work, though. In the meantime, frontends can decide for themselves whether they want code size or stronger optimization of cleanups by doing early outlining or not. The only way we can hit the quadratic growth from

Hi, I was hoping to get some advice on how to derive estimates of slopes from four parameter logistic models fit with SSfpl. I fit the model using: model<-nls(temp~SSfpl(time,a,b,c,d)) summary(model) I am interested in the values of the lower and upper asymptotes (parameters a and b), but also in the gradient of the line at the inflection point (c) which I assume tells me my rate of

Hello, I am wondering if someone can help me. I have the following function that I derived using nls() SSlogis. I would like to find its derivative. I thought I had done this using deriv(), but for some reason this isn't working out for me. Here is the function: asym <- 84.951 xmid <- 66.90742 scal <- -6.3 x.seq <- seq(1, 153,, 153) nls.fn <- asym/((1+exp((xmid-x.seq)/scal)))

I think adding transitions to cleanupblocks on the normal execution path would be an optimization barrier. Most passes would see the cleanupblock instruction and run the other way. It's definitely appealing from the perspective of getting the smallest possible code, but I'm OK with having no more than two copies of everything in the finally block. I think with the addition of the