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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


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