[R] NLSR package newDeriv function

[Debangan Dey]

Hi Soren,

Thanks for the reply. The main reason I want to compute the derivative is to add it to the derivative table in R so that I can use it later in my optimization. And, my complicated function doesn’t have an analytical form of the derivative as it involves distribution function of a bivariate normal distribution.

Thanks,
Debangan

--
Debangan Dey,
PhD Student in Biostatistics,
Johns Hopkins Bloomberg School of Public Health,
Baltimore, MD

From: Søren Højsgaard <sorenh using math.aau.dk>
Date: Wednesday, April 14, 2021 at 6:56 PM
To: Debangan Dey <ddey1 using jhu.edu>, r-help using r-project.org <r-help using r-project.org>
Cc: Mikkel Meyer Andersen <mikl using math.aau.dk>
Subject: Re: [R] NLSR package newDeriv function

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One approach is to compute things exact using the caracas package; see
below.

Best regards
Søren

----

> library(caracas)
> f <- function(x,y){x+y}
> def_sym(x,y,z)
> f <- x+y^2+y*z^2
> f
[caracas]:      2      2
           x + y  + y⋅z
>
> d1  <- der(f, c(x,y,z))
> d2 <- der2(f, c(x,y,z))
>
> d1
[caracas]: ⎡          2       ⎤
           ⎣1  2⋅y + z   2⋅y⋅z⎦
> d2
[caracas]: ⎡0   0    0 ⎤
           ⎢           ⎥
           ⎢0   2   2⋅z⎥
           ⎢           ⎥
           ⎣0  2⋅z  2⋅y⎦
>
> tex(d1)
[1] "\\left[\\begin{matrix}1 & 2 y + z^{2} & 2 y
z\\end{matrix}\\right]"
> tex(d2)
[1] "\\left[\\begin{matrix}0 & 0 & 0\\\\0 & 2 & 2 z\\\\0 & 2 z & 2
y\\end{matrix}\\right]"
>
> as_expr(d1)
expression(cbind(1,  2*y + z^2,  2*y*z))
> as_expr(d2)
expression(rbind(cbind(0, 0, 0), cbind(0, 2, 2*z), cbind(0, 2*z, 2*y)))
>



On Wed, 2021-04-14 at 19:35 +0000, Debangan Dey wrote:
> Hi,
>
> I am trying to solve a non-linear least square which has a function
> from R^3 -> R. Is it possible to define gradient using newDeriv for a
> 3-variate scalar function?
>
> I am trying to use the genD function in numDeriv package to define
> numerical gradient and treat them as a function. So far, I have
> failed to do it for simple function as follows �
>
> f <- function(x,y){x+y}
> grad.f <- function(x,y){z <- c(x,y)
>   f2 <- function(z){f(z[1],z[2])}
>   gd <- genD(f2,z)$D[,1:length(z)]
>   return(gd)
> }
> newDeriv(f(x,y), grad.f(x,y))
>
> This derivative definition is not working as the following error pops
> up �
>
> fd <- fnDeriv(~ f(x,y), c("x",�y�))
> fd(c(1,2))
> Error in .grad[, "x"] <- grad.f(x) :
>   number of items to replace is not a multiple of replacement length
>
>
> I know the above example would work if I just give expressions
> instead of using grad function in numDeriv but I am trying to use
> numDeriv::grad here to see if my toy example can be translated to the
> bigger function I would later use it on.
>
> Thanks,
> Debanga
>
> --
> Debangan Dey,
> PhD Student in Biostatistics,
> Johns Hopkins Bloomberg School of Public Health,
> Baltimore, MD
>
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