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DawsonF[30.] returns 0. The correct value is 0.016676... At least it prints a warning message,

General::munfl: Exp[-900.] is too small to represent as a normalized
machine number; precision may be lost.

I am on Mathematica 11.3.

The problem is that DawsonF[x] is being computed as Exp(-x^2) * Erfi[x] (times constant factors), which in this case is a product of a very small quantity times a very large one, resulting in under/overflow. This is a VERY bad algorithm. The point of having a DawsonF in the first place is to bypass this multiplication and return the result without under/overflow (see the section on Numerical Recipes book, for example).

I know I can use N[DawsonF[30], 20] to obtain an accurate result, but this can be slower, and there is no reason why DawsonF could not work in machine precision.

I will submit a bug report to Wolfram, but I wanted to post this here to get some feedback before. If the community agrees please tag this as a bug.

Are there other examples like this in Mathematica of special functions that just don't work in machine precision?

Update: I submitted a bug report and they replied (see my answer). See J.M.'s answer for a workaround.

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    $\begingroup$ DawsonF[30.] returns 0.01667594140106 (MMA 11.0.1 Windows) $\endgroup$ – Ulrich Neumann Sep 20 '18 at 14:48
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    $\begingroup$ See this question Numerical underflow for a scaled error function. It might be regarded as a duplicate as well. $\endgroup$ – Artes Sep 20 '18 at 15:01
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    $\begingroup$ I obtain also an underflow in version 11.3 for macOS. $\endgroup$ – Henrik Schumacher Sep 20 '18 at 19:38
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    $\begingroup$ @J.M.issomewhatokay. I got a reply from Wolfram Support. The relevant line is this: Starting with Version 11.3 underflow in no longer trapped in machine arithmetic and Mathematica doe s not switch automatically to arbitrary precision. This provides a more efficient way to handle numerical calculations and brings Mathematica much more in line with the IEEE 754 standard for how floating point numbers are to be handled ( https://en.wikipedia.org/wiki/IEEE_754 ) Apparently the bad machine precision algorithm was there all along, but in previous versions it fell back to arbitrary precision. $\endgroup$ – becko Sep 24 '18 at 16:30
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    $\begingroup$ DawsonF[30`10] $\endgroup$ – Αλέξανδρος Ζεγγ Sep 26 '18 at 4:36
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Before DawsonF[] became built-in in Mathematica, I used the following method for (small to moderately-sized) real arguments:

dawson = With[{eps = $MachineEpsilon, e2 = $MachineEpsilon^2},
              Compile[{{z, _Real}},
                      Module[{a, b, c, d, f, h, w},
                             a = 2. z^2;
                             f = c = b = a + 1.;
                             a = w = -2. a; d = 0.;
                             If[c == 0., c = e2];
                             While[b += 2.;
                                   d = b + a d; If[d == 0., d = e2]; d = 1/d;
                                   c = b + a/c; If[c == 0., c = e2];
                                   a += w; f *= (h = c d);
                                   Abs[h - 1] > eps];
                             z/f], RuntimeAttributes -> {Listable}]];

This is based on using the Lentz-Thompson-Barnett algorithm to evaluate this CF representation for Dawson's integral. It is not the most efficient method, since power series will outperform the CF near the origin, and asymptotic series will do best for really large arguments. It is quite compact and respectably performant in its intended argument range, however.

Here is a plot of it compared against the built-in:

Plot[dawson[x] - DawsonF[x], {x, -20, 20}, Frame -> True, PlotRange -> All]

absolute error

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  • $\begingroup$ Very nice! And dawson[x] remains accurate even for large x (in machine precision). Thanks for this. $\endgroup$ – becko Sep 26 '18 at 12:39
  • $\begingroup$ @becko, it does work, but at a point it gets less efficient than evaluating an asymptotic series; I just didn't want to post something long ;) $\endgroup$ – J. M. is away Sep 26 '18 at 13:02
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I submitted a bug report and this is the relevant line from what they replied:

Starting with Version 11.3 underflow in no longer trapped in machine arithmetic and Mathematica does not switch automatically to arbitrary precision. This provides a more efficient way to handle numerical calculations and brings Mathematica much more in line with the IEEE 754 standard for how floating point numbers are to be handled ( https://en.wikipedia.org/wiki/IEEE_754 )

This explains the origin of the bug. Apparently the bad machine precision algorithm was there all along, but in previous versions it fell back to arbitrary precision and thus went unnoticed (though it probably impacted performance).

Hope it gets fixed soon.

See @J.M. answer for a an algorithm that works in MachinePrecision.

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Based on @EddieXiao's answer to Numerical underflow for a scaled error function, you could define your own DawsonF with:

dawsonF[x_] := -(I/2) E^-x^2 Sqrt[π]+I HermiteH[-1,I x]

For example:

Chop @ dawsonF[30.] //InputForm

General::munfl: Exp[-900.] is too small to represent as a normalized machine number; precision may be lost.

0.01667594140105917

vs.

DawsonF[30`19]

0.01667594140105918

This is faster than the built-in machine precision code for DawsonF, e.g.:

dawsonF[N @ Range[-10, 10]]; //AbsoluteTiming
DawsonF[N @ Range[-10, 10]]; //AbsoluteTiming

{0.005917, Null}

{0.008466, Null}

On the other hand, @JM's compiled code is about 4 times faster:

dawson[N @ Range[30, 40]] //AbsoluteTiming
Chop @ dawsonF[N @ Range[30, 40]] //AbsoluteTiming

{0.000078, {0.0166759, 0.0161374, 0.0156326, 0.0151585, 0.0147123, 0.0142916, 0.0138943, 0.0135185, 0.0131625, 0.0128247, 0.0125039}}

{0.000323, {0.0166759, 0.0161374, 0.0156326, 0.0151585, 0.0147123, 0.0142916, 0.0138943, 0.0135185, 0.0131625, 0.0128247, 0.0125039}}

Of course, @JM's compiled code is expecting real numbers, and so complex input doesn't use the compiled code:

dawson[3. + 2. I] //AbsoluteTiming
dawsonF[3. + 2. I] //AbsoluteTiming

CompiledFunction::cfsa: Argument 3. +2. I at position 1 should be a machine-size real number.

{0.001332, 0.110514 - 0.0771238 I}

{0.001298, 0.110514 - 0.0771238 I}

The non-compiled code is about the same speed as dawsonF. Finally, as @JM mentions, his compiled code is not meant to work for all possible arguments, e.g.:

dawson[3 + 5 I]
dawsonF[3. + 5. I]
N[DawsonF[3 + 5 I], 20]

CompiledFunction::cfsa: Argument 3+5 I at position 1 should be a machine-size real number.

14.7334 + 6.88742 I

-7.78086*10^6 + 1.21475*10^6 I

-7.7808580812920342136*10^6 + 1.2147471245770455984*10^6 I

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    $\begingroup$ If one is evaluating Dawson's integral for general complex arguments, the conventional approach is to use its relation with the Faddeeva function $w(z)$: $$F(z)=\frac {i\sqrt\pi}{2}\left(\exp(-z^2)-w(z)\right)$$ for which there are methods available. $\endgroup$ – J. M. is away Sep 27 '18 at 7:16
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How about these

DawsonF[30`20]
DawsonF[30.0000000000000000000]
0.016675941401059176

Then I think that it might not be a bug.


Update

One can use SetPrecision

DawsonF[SetPrecision[30, #]] & /@ {5, 10, 20}
{0.017, 0.01667594, 0.016675941401059176}
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    $\begingroup$ That works, but in cases where one wishes to stick to machine precision, we have the OP's problem. $\endgroup$ – J. M. is away Sep 26 '18 at 4:48
  • $\begingroup$ @J.M.issomewhatokay. Above 20 can be changed to 10, or even 5. However it can not be MachinePrecision $ (\approx 15.95) $, which I cannot explain. $\endgroup$ – Αλέξανδρος Ζεγγ Sep 26 '18 at 4:49
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    $\begingroup$ As the OP mentioned, that is because the internal implementation is using the naive method to evaluate the Dawson integral (that is, computing Erfi[] and then multiplying with a Gaussian factor), which can be confirmed through spelunking. $\endgroup$ – J. M. is away Sep 26 '18 at 4:57
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    $\begingroup$ However it can not be MachinePrecision ... Because SetPrecision[..] means that the computation uses arbitrary precision, which relies on different algorithms (even if the precision is quite low). $\endgroup$ – becko Sep 26 '18 at 12:47
  • $\begingroup$ @becko: Exactly. Calculations done with machine precision do not track errors, so you never really have any guarantee about the outcome. Arbitrary precision arithmetic, on the other hand, does track errors and ensures correctness of the requested digits. That's why asking for just 5 digits of precision can still give you a more accurate result than using machine numbers. $\endgroup$ – Sjoerd Smit Sep 26 '18 at 20:58

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