search for: derivatives

Hi All, I copied code from an Splus manual to take symbolic derivatives: my.deriv <- function( mathfunc, var ) { tmp <- substitute(mathfunc) name <- deparse(substitute(var)) D(tmp, name) } (The code also w...

...SE, ...) This doesn't seem to include an "order of derivative" argument, in fact, in the examples section, it is outlined, how one can build a higher deriv. function oneself... Any hints are much appreciated! alex -- View this message in context: http://r.789695.n4.nabble.com/higher-derivatives-using-deriv-tp4256118p4256118.html Sent from the R help mailing list archive at Nabble.com.

Trying to resolve some general tablegen questions. Consider the test case for Tablegen called eq.td class Base<int V> { int Value = V; } class Derived<string Truth> : Base<!if(!eq(Truth, "true"), 1, 0)>; def TRUE : Derived<"true">; def FALSE : Derived<"false">; If I process this through tablegen I get: ------------- Classes

Hello, I am trying to construct two matrices, F and V, composed of partial derivatives and then find the eigenvalues of F*Inverse(V). I have the following equations in ryacas notation: > library(Ryacas) > FIh <- Expr("betah*Sh*Iv") > FIv <- Expr("betav*Sv*Ih") > VIh <- Expr("(muh + gamma)*Ih") > VIv <- Expr("muv*Iv...

Quick question: Which function do you use to calculate partial derivatives from a model equation? I've looked at deriv(), but think it gives derivatives, not partial derivatives. Of course my equation isn't this simple, but as an example, I'm looking for something that let's you control whether it's a partial or not, such as: somefunction(y~a+bx, wi...

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)))

Dear list, Hi, I am trying to get the second derivative of a logistic formula, in R summary the model is given as : ### >$nls >Nonlinear regression model >model: data ~ logistic(time, A, mu, lambda, addpar) >data: parent.frame() > A mu lambda >0.53243 0.03741 6.94296 ### but I know the formula used is #

Dear All, I tried following for getting derivative of a polynomial in R i<- -10:10 x<-i*i*i+3*i*i+2 fun_spline<-splinefun(i,x) plot(x,type="l") lines(x,fx_spline(x, deriv=1), col='green') lines(x,fx_spline(x, deriv=2), col='green') Now when I plot 3*i*i + 6*i and 6*i + 6 the plot was not same for first deivative. where as the 2nd derivative was same Is this a

>From my understanding, NAME is a special builtin entry and dedicated for things related multiclass, So, is the following rewrite what you want? class Base<int V> { int Value = V; } class Derived<string Truth> : Base<!if(!eq(Truth, "true"), 1, 0)>; multiclass Derived_m<string T> { def #NAME# : Derived<T>; } defm TRUE :

I don't understand why I get the following results. I define two classes 'Base' and 'Derived', the latter of which 'contains' the first. I then define a generic method 'test' and overload it for each of these classes. I call 'callNextMethod()' in the overload for Derived. From the output, it appears that the overload for Base gets called twice. Why is

> Hi! i'm tring to figure out how classes and dependencies > can be mapped to llvm. > [snip] > how do i encode a function that takes "base" type > so that it also takes "derived" ? You'll notice that your function doesn't take an object of type base, but a pointer to it. This is important - in a language such as LLVM where actual structures can be

...stg <- outer(xg, yg, function(x,y) 1.0) Error in outer(xg, yg, function(x, y) 1) : dims [product 16] do not match the length of object [1] Of course there are simpler ways to construct a constant matrix, that is not my point. It happens for me in the context of generating matrices of partial derivatives, and if on of these partial derivatives happens to be constant it fails. So e.g this works: library(Deriv) f <- function(x,y) (x-1.5)*(y-1)*(x-1.8)+(y-1.9)^2*(x-1.1)^3 fx <- Deriv(f,"x") fy <- Deriv(f,"y") fxy <- Deriv(Deriv(f,"y"),"x") fxx <...

Hi! i'm tring to figure out how classes and dependencies can be mapped to llvm. I've read on docs that nesting can be used: class base { int Y; }; class derived : base { short Z; }; becomes: %base = type { int } %derived = type { %base, short } That's ok, but now the question is: how do i encode a function that takes "base" type so that it also takes "derived"

Hi, Trying this: deriv(expression(sum(x)), "x") Gives the error message: Function 'sum' is not in the derivatives table I'd like to add it, is this difficult? If not, where is the derivatives table? However, give how basic "sum" is, I suspect it would have been added if it were straightforward. Do functions in the derivatives table need to be vectorized? i.e. given a vector of length n, t...

...rouble implementing a function which computes the k-th derivative of a specified function f and returns it as a function. I tried to adapt what I found under ?deriv but could not get it to work. Here is how it should look like: ## specify the function f <- function (x,alpha) x^alpha ## higher derivatives DD <- function(expr, variable, order = 1) { if(order < 1) stop("'order' must be >= 1") if(order == 1) deriv(expr, variable) else DD(deriv(expr, variable), variable, order - 1) } ## compute the second derivative of f f.prime.prime <- DD(f,"x",2) ##...

> In any case when you > want to obtain a base pointer to a "derived" object (where derivation > is implemented as you showed - with nesting) it's simply a matter of > using getelementptr to obtain a pointer to the nested base object and > using that. Umm ok, but i've the strange feeling that i'm missing something.. First a simple question: getelementptr with

Hi I am currently (for pedagogical purposes) writing a simple numerical analysis library in R. I have come unstuck when writing a simple Newton-Raphson implementation, that looks like this: f <- function(x) { 2*cos(x)^2 + 3*sin(x) + 0.5 } root <- newton(f, tol=0.0001, N=20, a=1) My issue is calculating the symbolic derivative of f() inside the newton() function. I cant seem to get R to

...mates. The reason I don't want to use smooth.spline is that there is no way to impose constraints (e.g. >=0, <=1, and monotonicity) as there is with COBS. However, since COBS doesn't have the 'deriv =' option, the only way I can think of doing it with COBS is to evaluate the derivatives numerically. Regards, Jim McDermott

Okay, I deleted most of this thread, so I was waiting for another message to respond to, so unfortunately this will be out of place in the thread. This is in response to Miroslav's idea about variable block sizes. I may be a bit out of my league here as I'm just starting to look at how the actual encoding gets done. But it seems to me that you could make a decent guess about when

UseRs, I used the optim function valor.optim <- optim(c(1,1,1),logexp1,method ="BFGS",control=list(fnscale=-1),hessian=T); and I want to calculate the derivates, psi1<-valor.optim$par[1] psi2<-valor.optim$par[2] psi3<-valor.optim$par[3] a0=exp(psi1); a1=exp(psi2)/(20+exp(psi2)+exp(psi3)); a2=exp(psi3)/(20+exp(psi2)+exp(psi3))