Skip to main content

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

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

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}]
Post a Comment
Read more