I want Sinc'[0] to return 0, but instead it returns Indeterminate.
I've tried
Unprotect[Sinc]
Unprotect[Derivative]
Derivative[1][Sinc][0] ^= 0
But it doesn't work.
Maybe this needs to be similar to this (from the help files for Derivative)
f'[x_] := If[PossibleZeroQ[x], N[0, Precision[x]], (x Cos[x] - Sin[x])/x^2];
Here is a slightly more robust way to modify the default behavior. It works for all positive derivatives of Sinc:
Unprotect[Sinc];
Sinc /: Derivative[n_Integer?Positive][Sinc] :=
Piecewise[{
{(-1)^(n/2)/(n + 1), # == 0 && EvenQ[n]},
{0, # == 0 && OddQ[n]},
{Derivative[n - 1][Cos[#]/# - Sin[#]/#^2 &][#], True}
}] &;
Protect[Sinc];
Usage:
Sinc'[0]
(* 0 *)
Sinc''[0]
(* -1/3 *)
Derivative supports direct assignment without the need to unprotect. So Derivative[n_Integer?Positive][Sinc] := ... is sufficient (without TagSet or Unprotect).
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TagSet. See e.g. <<GeneralUtilities`; PrintDefinitions[PolyLog]; PrintDefinitions[MittagLefflerE], etc... So, I follow the WRI way for Sinc.
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Commented
Oct 12, 2016 at 15:05
One solution is making your own function:
MySinc[x_] := Sinc[x]
Derivative[1][MySinc] ^=
If[# == 0, 0, Derivative[1][Sinc] // Evaluate] &;
MySinc[0]
(* 1 *)
MySinc'[0]
(* 0 *)
And then in expressions which use Sinc use expr/.Sinc->MySinc. To me this seems like the cleanest solution. However, this can be done with Sinc, too. But it is difficult to undo!
Unprotect[Sinc];
tmp = Derivative[1][Sinc] // Evaluate;
Derivative[1][Sinc] ^= If[# == 0, 0, tmp] &;
Protect[Sinc];
Sinc[0]
(* 1 *)
Sinc'[0]
(* 0 *)
The function is:
f[x_] = D[Sinc[x], x];
f[x]
Cos[x]/x - Sin[x]/x^2
In its current form, the value at x=0 is indeterminate. It is only when taking the limit as x->0 that a value emerges. Hence:
Limit[D[Sinc[x], x], x -> 0]
0
Or, more succinctly:
Limit[f[x], x -> 0]
0
Sinc alone. So I agree with @Rojo above that the reason to have a Sinc function should be that you don't have to evaluate Limit[Sin[x]/x,x->0] every time. It would be nice if the same was true for Sinc'.
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Sincfunction in the first place so this itches me a little $\endgroup$sinc[x_] := Piecewise[{{Sin[x]/x, x != 0}}, 1]. The mathematically proper derivative ofSincshould be whatsinc'[x]returns. But now let's take the derivative of this once more:D[sinc'[x],x]. This will be a piecewise that's still 0 in the point $x=0$. However the actual second derivative sinc''(0) should be -1/3. This illustrates how Piecewise itself is unable to handle derivatives ... $\endgroup$Piecewisewon't give us proper derivative, it doesn't make a lot of sense to return a piecewise forSinc'[x]. It would just delay the problem until the next derivative. OK, this is somewhat subjective, but after thinking this through I start to accept why they might have consciously made the decision not to both with Sinc'[x] in x=0. My point is that while they could have implementedDerivative[n][Sinc][x]to return a correct Piecewise for anyn, taking another derivative of that result would fail anyway. $\endgroup$