Hh Function

Author

Eric W. Weisstein
December 2, 2019

This notebook downloaded from http://mathworld.wolfram.com/notebooks/SpecialFunctions/HhFunction.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/HhFunction.html.

©2019 Wolfram Research, Inc. except for portions noted otherwise

Definition

Q[x_]:=(1/2) Erfc[(x/(Sqrt[2]))]
Z[x_]:=1/Sqrt[2Pi]Exp[-x^2/2]

Hh[n_Integer,x_]:=If[n≥0,
(-1)^n/n! Hh[-1,x]D[Q[x]/Z[x],{x,n}],
(-1)^(-n-1)Sqrt[2Pi]D[Z[x],{x,-n-1}]]

D[Q[x]/Z[x],{x,n}]

Integrals

(∫_0^∞)(((t^n) exp(-(1/2) (((t+x))^2)))/(n!))dt

i1cond=Integrate[((t^n)/(n!))Exp[-(((t+x))^2)/2],{t,0,∞}]

i1=Assuming[n∈Integers&&n>=0,FullSimplify[i1cond]]

TraditionalForm[i1]

Table[i1,{n,0,5}]//FullSimplify//Column

(∫_x^∞)((exp(((u^2)/2)) (((u-x))^n))/(n!))du

i2cond=Integrate[((((u-x))^n)/(n!))Exp[-(u^2)/2],{u,x,∞}]

i2=Assuming[n∈Integers&&n>=0,FullSimplify[i2cond]]

TraditionalForm[i2]

Table[i2,{n,0,5}]//FullSimplify//Column

Table

TextGrid[Table[{n,Hh[n,x]//FullSimplify},{n,-4,4}],Dividers->All]

Plot

V5

Needs["Graphics`Colors`"]//Quiet

Plot[Evaluate[Table[Hh[n,x],{n,-3,3}]],{x,-4,4},PlotRange->{Automatic,{-1,2}},PlotStyle->{Red,Yellow,Green,Black,Blue,Indigo,Violet},AxesLabel->{x,HoldForm[(Subscript[Hh, n])[x]]}]

V6

Plot[Evaluate[Table[Hh[n,x],{n,-3,3}]],{x,-4,4},PlotRange->{Automatic,{-1,2}},PlotStyle->{Red,Yellow,Green,Black,Blue,Darker[Blue],Purple},AxesLabel->{x,Subscript[Hh,n][x]},
LabelStyle->Black,
BaseStyle->FontFamily->Times,PlotLegends->(Style[Subscript[Hh,#][x],FontFamily->Times,Black]&/@Range[-4,4])]

Identities

(i^n) erfc(z)==((Hh(n,(Sqrt[2]) z))/(Sqrt[π (2^(n-1))]))

Abramowitz and Stegun 1972, p. 300, 7.2.10

Seems wrong

Subtract@@@Table[{I^n Erfc[(z/(Sqrt[2]))],(1/(Sqrt[(2^(n-1))π]))Hh[n,z]},{n,-3,3}]//FullSimplify