Partial Derivatives in Haskell

A while back a friend wanted help with a program that could solve for the roots of functions using Newton's method, and naturally for that I needed some way to calculate the derivative of a function numerically, and this is what I came up with:

deriv f x = (f (x+h) - f x) / h where h = 0.00001

Newton's method was a fairly easy thing to implement, and it works rather well. But now I've started to wonder - Is there some way I could use this function to solve partial derivatives in a numerical manner, or is that something that would require a full-on CAS? I would post my attempts but I have absolutely no clue what to do yet.

Please keep in mind that I am new to Haskell. Thank you!

2 answers

  • answered 2018-01-11 21:21 luqui

    This is called automatic differentiation and there is a lot of really neat work in this area in Haskell, though I don't know how accessible it is.

    From the wiki page:

  • answered 2018-01-11 21:34 leftaroundabout

    You can certainly do much the same thing as you already implemented, just with multivariate pertubations. But first (as you should always do with top-level functions) add a type signature:

    deriv :: (Double -> Double) -> Double -> Double

    (That's not the most general possible signature, but probably sufficiently general for everything you'll need.) I'll call

    type ℝ = Double

    in the following for brevity, i.e.

    deriv :: (ℝ -> ℝ) -> ℝ -> ℝ

    Now what you want is, for example in ℝ²

    grad :: ((ℝ,ℝ) -> ℝ) -> (ℝ,ℝ) -> (ℝ,ℝ)
    grad f (x,y) = ((f (x+h,y) - f (x,y)) / h, (f (x,y+h) - f (x,y)) / h)
     where h = 0.00001

    It's awkward having to write out the components individually and making the definition specific to a particular-dimensional vector space. A generic way of doing it:

    import Data.VectorSpace
    import Data.Basis
    grad :: (HasBasis v, Scalar v ~ ℝ) => (v -> ℝ) -> v -> v
    grad f x = recompose [ (e, (f (x ^+^ h*^basisValue b) - f x) ^/ h)
                         | (e,_) <- decompose x ]
     where h = 0.00001

    Note that this pre-chosen-step–finite-differentiation is always a tradeoff between inaccuracy from higher-order terms and from floating-point errors, so definitely check out automatic differentiation.