For illustration purposes, consider the function $$ f(x, y) = (x - 1)^2 + (y - 1)^2 + \sin[25(x+y)] . $$ The mixed partial derivatives at the origin are $f^{(x,x)}(0,0) = f^{(y,y)}(0,0) = 2$ and $f^{(x,y)}(0,0) = 0$ (the $\sin$-term does not give any contribution).

I want to determine these derivatives numerically. That is, let's imagine I have only access to the function

f[(x_)?NumericQ, (y_)?NumericQ] := (x - 1)^2 + (y - 1)^2 + Sin[25*(x + y)]

as a numerical black box. I would first try to solve the problem as follows:

N[D[f[x, y], {x}, {x}] /. x -> 0 /. y -> 0]
(* 2. *)

N[D[f[x, y], {y}, {y}] /. x -> 0 /. y -> 0]
(* 2. *)

N[D[f[x, y], {x}, {y}] /. x -> 0 /. y -> 0]
(* Derivative[1, 1][f][0., 0.] *)

In the first two cases, it just works. However, for the mixed derivative, it is well known that the simple approach fails and one must use nested calls to ND instead. (To keep it short, I will do that the simple way, not using the trick described here to reduce the number of function calls.) Let's try it:

ND[f[x, 0], {x, 2}, 0]
(* 54.6838 *)

ND[f[0, y], {y, 2}, 0]
(* 54.6838 *)

ND[ND[f[x, y], x, 0], y, 0]
(* -8.40318 *)


The problem, of course, lies in the "scale" used for the differentiation. We can explore how the results change as a function of the employed scale:

   ND[f[x, 0], {x, 2}, 0, Scale -> 10^-s],
   ND[f[0, y], {y, 2}, 0, Scale -> 10^-s],
   ND[ND[f[x, y], x, 0, Scale -> 10^-s], y, 0, Scale -> 10^-s]
   }, {s, 0, 6}

{{54.6838, 54.6838, -8.40318},
 {2.00013, 2.00013, 7.46312*10^-6},
 {2., 2., -1.89464*10^-7},
 {2.00001, 2.00001, 9.40594*10^-6},
 {2.00063, 2.00063, 0.00106619},
 {2.08502, 2.08502, -0.0788629},
 {-5.68049, -5.68049, -20.9135}}

As we can see, there is only a very narrow window for Scale where ND yields great results. Yet, using N[D[...] /. ...] in the first example, the results for the non-mixed derivatives were perfect. My question is: how does Mathematica manage to choose the scale perfectly using that approach? Is there a way to make Mathematica automatically choose the correct scale also for the mixed derivatives?

Bonus question: could someone explain why the accuracy of the results starts to get worse at a scale of $10^{-3}$ already? I would have expected the accuracy to increase until we get close to the machine precision. This works, for example:

h = 10^-8;
(f[h, 0] + f[-h, 0] - 2*f[0, 0])/h^2
(* 2 *)
  • 1
    $\begingroup$ For the bonus, try h = 10.^-8; (f[h, 0] + f[-h, 0] - 2*f[0, 0])/h^2. Your calculation with an exact h does not suffer from catastrophic round-off error in subtractive cancellation. $\endgroup$ – Michael E2 Apr 4 at 20:23
  • 1
    $\begingroup$ The error $\Delta^n f/h^n - f^{(n)}(c)$ consists of truncation and round-off errors. The truncation error has the form $O(h^p)$ or roughly $\alpha \,h^p$, where $p$ is the order of approximation. The round-off error in $\Delta^n f$ is scaled by $1/h^n$. So the round-off error has the form $O(h^{-n})$ or $\beta \,\epsilon/h^n$, where we explicitly factor out machine epsilon. As $h$ approaches zero, the truncation error decreases and the round of error increases. Depending on $\alpha/\beta$, the turning point when round-off error will start to dominate will occur around $\epsilon^{1/(n+p)}$. $\endgroup$ – Michael E2 Apr 4 at 21:35
  • $\begingroup$ @MichaelE2 Thank you, that is quite enlightening already! I did absolutely not realize I was doing an exact calculation in that last code snippet. (Especially since f was supposed to only accept numerical parameters?) In your second comment, what exactly does "order of approximation" refer to? Is it (related to) the number of function evaluations that enter the approximation of the derivative? $\endgroup$ – Noiralef Apr 4 at 21:46
  • 1
    $\begingroup$ A finite difference formula of order $p$ for $f^{(n)}(c)$ is equal to $f^{(n)}(c) + O(h^p) = f^{(n)}(c) + \alpha h^p + O(h^{p+1})$. For instance the order of approximation of your second derivative formula is two, which can be seen from the output of Series: Clear[h]; Series[(g[h] + g[-h] - 2*g[0])/h^2, {h, 0, 2}]. Finite difference formulas are derived from the Taylor series of $f(x)$ so that all the leading terms cancel up to $f^{(n)}(c)$ and a certain number of terms cancel after it up to the order of approximation. $\endgroup$ – Michael E2 Apr 4 at 22:03
  • $\begingroup$ As the result is calculated in an expansion, increasing the number of terms will help. Of course, too many terms will lead to cancellation errors, so to not over do it. In your case Terms-> 10..12 will do the job. $\endgroup$ – Daniel Huber Apr 10 at 15:31

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