Taylor series approximation of Gaussian $Q$ function , $Q(x) = \frac{1}{{\sqrt {2\pi } }}\int_x^\infty {{e^{ - \frac{{{v^2}}}{2}}}dv}$

Question

$ \require{newcommand}\newcommand{\Erfc}{\operatorname{Erfc}}$

I am trying to find a Taylor series expansion for the Gaussian $Q$ function. I have seen that error function $\Erfc(x)$ is an approximation of $Q(x)$ (Is my assumption correct?).

$\Erfc(x)$ has a Taylor approximation. Is it possible to construct such for $Q(x)$ function? Is there any condition for that? Thank you.

Answer 1

You can obtain a Taylor series as follows: \begin{align*} Q(x) & = \frac{1}{{\sqrt {2\pi } }}\int_0^{ + \infty } {e^{ - t^2 /2} dt} - \frac{1}{{\sqrt {2\pi } }}\int_0^x {e^{ - t^2 /2} dt} = \frac{1}{2} - \frac{1}{{\sqrt {2\pi } }}\int_0^x {e^{ - t^2 /2} dt} \\ & = \frac{1}{2} - \frac{1}{{\sqrt {2\pi } }}\int_0^x {\sum\limits_{n = 0}^\infty {\frac{1}{{n!}}\left( { - \frac{{t^2 }}{2}} \right)^n } dt} = \frac{1}{2} - \frac{1}{{\sqrt {2\pi } }}\sum\limits_{n = 0}^\infty {\frac{1}{{n!}}\left( { - \frac{1}{2}} \right)^n \frac{{x^{2n + 1} }}{{2n + 1}}} . \end{align*} Addendum: A different series expansion is $$ Q(x) = \frac{1}{2} - \frac{1}{{\sqrt {2\pi } }}e^{ - x^2 /2} \sum\limits_{n = 0}^\infty {\frac{{x^{2n + 1} }}{{1 \cdot 3 \cdots (2n + 1)}}} . $$