Skip to main content

numerics - Numerical underflow for a scaled error function


I calculate a scaled error function defined as


f[x_] := Erfc[x]*Exp[x^2]

but it can not calculate f[30000.]. f[20000.] is not very small (0.0000282). I think Mathematica is supposed to switch to high precision instead of machine precision, but it does not. It says:




General::unfl: Underflow occurred in computation. >>
General::ovfl: Overflow occurred in computation. >>

How can I calculate f for large values of x? Even with N[f[30000], 50], it does not use high precision and fails.



Answer



If you have an analytic formula for f[x_] := Erfc[x]*Exp[x^2] not using Erfc[x] you could do what you expect. However it is somewhat problematic to do in this form because Erfc[x] < $MinNumber for x == 27300.


$MinNumber


1.887662394852454*10^-323228468


N[Erfc[27280.], 20]


5.680044213569341*10^-323201264

Edit


A very good approximation of your function f[x] for values x > 27280 you can get making use of these bounds ( see e.g. Erfc on Mathworld) :


enter image description here


which hold for x > 0.



Here we find values of the lower and upper bounds with relative errors for various x:


T = Table[ 
          N[#, 15]& @ {2 /(Sqrt[Pi] (x + Sqrt[2 + x^2])), 
                       2 /(Sqrt[Pi] ( x + Sqrt[x^2 + 4/Pi])), 
                       1 - ( x + Sqrt[x^2 + 4/Pi])/(x + Sqrt[2 + x^2]),
          {x, 30000 Table[10^k, {k, 0, 5}]}];

Grid[ Array[ InputField[ Dynamic[T[[#1, #2]]], FieldSize -> 13] &, {6, 3}]]

enter image description here



Therefore we propose this definition of the function f (namely the arithetic mean of its bounds for x > 27280 ) :


f[x_]/; x >= 0 := Piecewise[ { { Erfc[x]*Exp[x^2],                      x < 27280 },

                               { 1 /( Sqrt[Pi] ( x + Sqrt[2 + x^2])) 
                               + 1 /( Sqrt[Pi] ( x + Sqrt[x^2 + 4/Pi])), x >= 27280}}
                           ]
f[x_] /; x < 0 := 2 - f[-x]

I.e. we use the original definition of the function f for 0 < x < 27280, the approximation for x > 27280 and for x < 0 we use the known symmetry of the Erfc function, which is relevant when we'd like to calculate f[x] for x < - 27280. Now we can safely use this new definition for a much larger domain :


{f[300], f[300.], f[30000.], f[-30000.]}



{E^90000 Erfc[300], 0.0018806214974, 0.0000188063, 1.99998}

and now we can make plots of f around of the gluing point ( x = 27280.)


GraphicsRow[{ Plot[ f[x], {x, 2000, 55000}, 
                      Epilog -> {PointSize[0.02], Red, Point[{27280., f[27280.]}]}, 
                      PlotStyle -> Thick, AxesOrigin -> {0, 0}], 
              Plot[ f[x], {x, 27270, 27290}, 
                      Epilog -> {PointSize[0.02], Red, Point[{27280., f[27280.]}]}, 

                      PlotStyle -> Thick]}]

enter image description here


Comments

Post a Comment

Popular posts from this blog

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

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

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...
Post a Comment
Read more