A133842 Decimal expansion of value of the maximum of Dawson's integral D_+(x).

5, 4, 1, 0, 4, 4, 2, 2, 4, 6, 3, 5, 1, 8, 1, 6, 9, 8, 4, 7, 2, 7, 5, 9, 3, 3, 0, 2, 4, 1, 4, 7, 7, 1, 8, 6, 3, 9, 0, 6, 0, 4, 6, 7, 6, 8, 2, 6, 8, 2, 5, 8, 8, 7, 4, 5, 6, 3, 5, 4, 2, 1, 6, 9, 0, 8, 4, 2, 0, 5, 5, 8, 7, 9, 8, 1, 4, 6, 9, 7, 5, 3, 1, 4, 3, 9, 1, 9, 1, 0, 2, 1, 3, 2, 2, 7, 7, 7, 1, 0, 2, 4, 4, 9, 6

Offset

0,1

Links

Stanislav Sykora, Table of n, a(n) for n = 0..2000

Eric Weisstein's World of Mathematics, Dawson's Integral

Wikipedia, Dawson function

Formula

Equals 1/(2*A133841), since F'(x)=1-2*x*F(x). - Stanislav Sykora, Sep 17 2014

Example

0.54104422463518169847...

Mathematica

DawsonF[x_] := Sqrt[Pi]*Erfi[x]/(2*Exp[x^2]); y0 = DawsonF[x] /. FindRoot[ DawsonF'[x], {x, 1}, WorkingPrecision -> 110]; RealDigits[y0][[1]][[1 ;; 105]] (* Jean-François Alcover, Oct 26 2012, after Eric W. Weisstein *)

Prog

(PARI) Erfi(z) = -I*(1.0-erfc(I*z));
Dawson(z) = 0.5*sqrt(Pi)*exp(-z*z)*Erfi(z); \\ same as F(x)=D_+(x)
DDawson(z) = 1.0 - 2*z*Dawson(z); \\ Derivative of the above
x = 0.5/solve(z=0.1, 2.0, real(DDawson(z))) \\ Stanislav Sykora, Sep 17 2014

Crossrefs

Cf. A133841, A133843, A243433.

Sequence in context: A361050 A325221 A129522 * A199453 A245699 A115637

Adjacent sequences: A133839 A133840 A133841 * A133843 A133844 A133845

Keyword

nonn,cons

Author

Eric W. Weisstein, Sep 26 2007

Status

approved