Hh Function

Let

$\displaystyle Z(x)$	$\textstyle \equiv$	$\displaystyle {1\over\sqrt{2\pi}} e^{-x^2/2}$	(1)
$\displaystyle Q(x)$	$\textstyle \equiv$	$\displaystyle {1\over\sqrt{2\pi}} \int_x^\infty e^{-t^2/2}\,dt,$	(2)

where

and

are closely related to the Normal Distribution Function, then

$\displaystyle {\rm Hh}_{-n}(x)$	$\textstyle =$	$\displaystyle (-1)^{n-1}\sqrt{2\pi}\,Z^{(n-1)}(x)$	(3)
$\displaystyle {\rm Hh}_n(x)$	$\textstyle =$	$\displaystyle {(-1)^n\over n!} {\rm Hh}_{-1}(x){d^n\over dx^n} \left[{Q(x)\over Z(x)}\right].$	(4)