Displaying 3 results from an estimated 3 matches for "expr9".
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2007 May 26
1
bug from nlm function (PR#9711)
...;-formula(~log(a)-log(b)-(a-1)*(log(x)-log(b))- (x^a)/(b^a))
> fllwfuncH <-deriv(fllwform,c("a","b"),function(a,b,x){})
> fllwfuncH
function (a, b, x)
{
.expr2 <- log(b)
.expr4 <- a - 1
.expr5 <- log(x)
.expr6 <- .expr5 - .expr2
.expr9 <- x^a
.expr10 <- b^a
.expr19 <- .expr10^2
.expr23 <- 1/b
.value <- log(a) - .expr2 - .expr4 * .expr6 - .expr9/.expr10
.grad <- array(0, c(length(.value), 2), list(NULL, c("a",
"b")))
.grad[, "a"] <- 1/a - .e...
2007 Mar 20
2
Problems about Derivaties
...example: Suppose that I have a function of this form:
f(x,y)=x^3+y^3+(x^2)*(y^2). With "deriv3" I can evaluate the first derivative
and the hessian matrix, as follows:
> d<-deriv3(~x^3+y^3+(x^2)*(y^2),c("x","y"))
> d[[1]]
{
.expr4 <- x^2
.expr5 <- y^2
.expr9 <- 2 * x
.expr15 <- 2 * y
.value <- x^3 + y^3 + .expr4 * .expr5
.grad <- array(0, c(length(.value), 2), list(NULL, c("x",
"y")))
.hessian <- array(0, c(length(.value), 2, 2), list(NULL,
c("x", "y"), c("x", "y")))
.grad[, &qu...
2007 Jan 12
1
incorrect result of deriv (PR#9449)
...0, c(length(.value), 1), list(NULL, c("z")))
.grad[, "z"] <- -(z * dnorm(z))
attr(.value, "gradient") <- .grad
.value
})
deriv(~exp(-z^2/(2*s^2))/s/sqrt(2*pi),"z")
expression({
.expr4 <- 2 * s^2
.expr6 <- exp(-z^2/.expr4)
.expr9 <- sqrt(2 * pi)
.value <- .expr6/s/.expr9
.grad <- array(0, c(length(.value), 1), list(NULL, c("z")))
.grad[, "z"] <- -(.expr6 * (2 * z/.expr4)/s/.expr9)
attr(.value, "gradient") <- .grad
.value
})
should provide the same expression...