bugs - Terrible accuracy of DawsonF

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.

Answer

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.

Comments

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.

Blog

Search This Blog