Skip to main content

Assigning an analytical approximation to the error function erf(x)


Working with some iterative integral equations, I have Gaussian density functions involved therein. Integrating the Gaussian function I obtain the error function. When I take the second integration, Mathematica is not able to provide an analytical expression. Because apparently there is no closed-form expression for the integral of the error function.


Question: Is it possible the tell Mathematica to consider the error function as some given exact expression which is analytically integrable?


For example


$$\mathrm{erf}(x)=\frac{1}{(1+a_1x+a_2x^2+a_3x^3+a_4x^4)^4}$$


for some appropriate constants $a_1$, $a_2$, $a_3$, $a_4$?




Answer



You could unprotect Erf and redefine it as you like it.


Unprotect[Erf];
(* Here you define the constants and your erf approximation *)
a1=...; etc.
Erf[x_] := 1/(1 + a1 x + a2 x^2 + a3 x^3 + a4 x^4)^4;
(* Here you perform your calculations with the approximate value of erf *)
myComplexCalculation = Erf[x]/2
(* When you are done, clear your definition and reprotect Erf *)
Clear[Erf];

Protect[Erf];

This can be dangerous but, as long as you know what you are doing, it will work.


A more local workaround is to enclose your 'approximate' calculation within a Block that blocks the definition of Erf:


Block[ {Erf},
(* Here you define the constants and your erf approximation *)
a1=...; etc.
Erf[x_] := 1/(1 + a1 x + a2 x^2 + a3 x^3 + a4 x^4)^4;

(* Here you perform your calculations with the approximate value of erf *)

myComplexCalculation = Erf[x]/2
]

Is this what you were looking for or was it the actual expression for the approximation?


As a more general approach to temporarily modifying built-in functions within the walls of a scoping construct, you might also want to have a look at the undocumented function InheritedBlock, described in this post on Stack Overflow. It appears that it will copy all definitions associated with the symbol you want to block, so that you can change, within the block only the part you need to. Basically, while


Block[ {builtinFun},
code redefining builtinFun from scratch
and using the new builtinFun
]


would give you a completely clean symbol for builtinFun to code from scratch, using


Internal`InheritedBlock[ {builtinFun},
code redefining parts of the inherited builtinFun
and using the modified builtinFun
]

would give you a completely modifiable copy of builtinFun, so that you can modify only the parts you need to modify.


Comments

Popular posts from this blog

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.