On Apr 4, 2011, at 6:35 AM, kitty wrote:
> 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
>
> #
y~'A'/(1+exp((4*'mu'/'A')*('lambda'-'time'))+2))
# from the
> grofit (
> package I am using to fit the model) documentation.
>
> I have attempted to use the R function 'deriv' to get the
> first derivative from which I can then reuse the deriv function to
> get the
> second derivative.... unfortunately this does not seem to work
>
> ###
>>
express<-expression(y~'A'/(1+exp((4*'mu'/'A')*('lambda'-'time'))+2))
>> express
> expression(y ~ "A"/(1 + exp((4 * "mu"/"A") *
("lambda" - "time")) +
> 2))
>>
>> d1<-deriv(express)
> Error in deriv.default(express) : element 2 is empty;
> the part of the args list of '.Internal' being evaluated was:
> (expr, namevec, function.arg, tag, hessian)
> ####
For one thing you are not specifying what variable you want to
differentiate with-respect-to: Assuming this to be `A` then:
express <-
expression( A/(1+exp((4*mu/A)*(lambda-time))+ 2))
# The quotes looked "wrong" inside an expression, so I removed them
d1<-deriv(express, "A") # but the diff w.r.t variable needs to
be
quoted.
d1
expression({
.expr1 <- 4 * mu
.expr3 <- lambda - time
.expr5 <- exp(.expr1/A * .expr3)
.expr7 <- 1 + .expr5 + 2
.value <- A/.expr7
.grad <- array(0, c(length(.value), 1L), list(NULL, c("A")))
.grad[, "A"] <- 1/.expr7 + A * (.expr5 * (.expr1/A^2
* .expr3))/.expr7^2
attr(.value, "gradient") <- .grad
.value
})
All this should have been clear if you had looked at the examples in
help(deriv).
>
> Why is this not working and how do I get the second derivative?
That , too, is clearly exemplified in the help page.
>
> Thank you for reading my post, all help is appreciated,
> Kitty
--
David Winsemius, MD
West Hartford, CT