1

The madness packages, as mentioned here, is nice for autodiff in R.

I would like to compute now a derivative wrt x of a derivative wrt y.

$\frac{\partial}{\partial x}\frac{\partial}{\partial y}xy$

how can this be done using madness?

update: actually here I guess it factors.. maybe this will be ok by just multiplying the two derivatives? Maybe this will only be difficult if x is a function of y.

apaderno
  • 28,547
  • 16
  • 75
  • 90
Sam Weisenthal
  • 2,791
  • 9
  • 28
  • 66

1 Answers1

1

Here's a way using the numderiv function in madness:

library(madness)

dxdy <- function(x, y, f) {
  dy <- function(x, y) {
    dvdx(f(x, y))
  }
  
  numderiv(dy, x, y = y)
}

x <- matrix(1:3, nrow = 1)
y <- matrix(3:1, ncol = 1)
# identity matrix, as expected
dvdx(dxdy(madness(x), madness(y), function(x, y) x%*%y))
#>      [,1] [,2] [,3]
#> [1,]    1    0    0
#> [2,]    0    1    0
#> [3,]    0    0    1
x <- matrix(2, ncol = 1)
y <- matrix(3, ncol = 1)
dvdx(dxdy(madness(x), madness(y), function(x, y) y^x))
#>          [,1]
#> [1,] 9.591674
# compare to analytical solution
y^(x-1)*(x*log(y) + 1)
#>          [,1]
#> [1,] 9.591674
x <- matrix(1:3, ncol = 1)
y <- matrix(3:1, ncol = 1)
dvdx(dxdy(madness(x), madness(y), function(x, y) sum(y^x)))
#>          [,1]     [,2] [,3]
#> [1,] 2.098612 0.000000    0
#> [2,] 0.000000 4.772589    0
#> [3,] 0.000000 0.000000    1
# compare to analytical solution
y^(x-1)*(x*log(y) + 1)
#>          [,1]
#> [1,] 2.098612
#> [2,] 4.772589
#> [3,] 1.000000
x <- matrix(1:3, ncol = 1)
y <- matrix(3:1, ncol = 1)
dvdx(dxdy(madness(x), madness(y), function(x, y) sum(sin(x*y))))
#>           [,1]     [,2]      [,3]
#> [1,] -1.413352 0.000000  0.000000
#> [2,]  0.000000 2.373566  0.000000
#> [3,]  0.000000 0.000000 -1.413353
# compare to analytical solution
cos(x*y) - x*y*sin(x*y)
#>           [,1]
#> [1,] -1.413353
#> [2,]  2.373566
#> [3,] -1.413353

Note that the madness derivatives are the partials of each x (columns) with respect to the y partials of each xy (rows).

jblood94
  • 10,340
  • 1
  • 10
  • 15
  • Thank you - this does answer the question, so I am accepting. I had been hoping to use however the autodiff functionality rather than the numerical derivative (I think madness provides both). – Sam Weisenthal Nov 14 '21 at 20:02
  • 1
    Understood. That's where I started, but I'm pretty convinced now that `madness` can't do that. See this question for working with Hessians symbolically in case it can help: https://stackoverflow.com/questions/20297711/r-hessian-matrix – jblood94 Nov 15 '21 at 01:53