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ERFC Complementary Error Function

ERFC.1 Introduction

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Let $x$ be a complex variable of $\mathbb{C} \setminus \{\infty\}$ .The function Complementary Error Function (noted $\operatorname{erfc}$ ) is defined by the following second order differential equation


\begin{equation*} 
\begin{split} 
2 x \frac{\partial y (x)}{\partial x} + \frac{\partial^{2} y (x)}{\partial x^{2}}& =0. 
\end{split} 
\end{equation*}
ERFC.1.1

The initial conditions of ERFC.1.1 are given at $0$ by


\begin{equation*} 
\begin{split} 
\operatorname{erfc} (0)& =1, \\ 
\frac{\partial \operatorname{erfc} (x)}{\partial x} (0)& =\frac{-2}{\sqrt{\pi}}. 
\end{split} 
\end{equation*} 
 ERFC.1.2

Related function: Error Function

ERFC.2 Series and asymptotic expansions

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ERFC.2.1 Taylor expansion at $0$

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ERFC.2.1.2 General form

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\begin{equation*} 
\begin{split} 
\operatorname{erfc} (x)& =\sum_{n = 0}^{\infty} u (n) x^{n}. 
\end{split} 
\end{equation*} 
 ERFC.2.1.2.1
The coefficients $u (n)$ satisfy the recurrence

\begin{equation*} 
\begin{split} 
2 n u (n) + \bigl(n^{2} + 3 n + 2\bigr) u (n + 2)& =0. 
\end{split} 
\end{equation*}
ERFC.2.1.2.2
Initial conditions of ERFC.2.1.2.2 are given by

\begin{equation*} 
\begin{split} 
u (1)& =\frac{-2}{\sqrt{\pi}}, \\ 
u (0)& =1. 
\end{split} 
\end{equation*}
ERFC.2.1.2.3

ERFC.2.2 Asymptotic expansion at $\infty$

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ERFC.2.2.1 First terms

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\begin{equation*} 
\begin{split} 
& \operatorname{erfc} (x)\approx \operatorname{e} ^{\biggl(-\frac{1}{x^{2}}\biggr)} x y _{0} (x), 
\end{split} 
\end{equation*}
where

\begin{equation*} 
\begin{split} 
y _{0} (x)& =\pi^{\Bigl(-\frac{1}{2}\Bigr)} - \frac{x^{2}}{2 \sqrt{\pi}} + 2 \ldots 
\end{split} 
\end{equation*}

ERFC.2.2.2 General form

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ERFC.2.2.2.1 Auxiliary function $y _{0} (x)$

The coefficients $u (n)$ of $y _{0} (x)$ satisfy the following recurrence

\begin{equation*} 
\begin{split} 
2 n u (n) + u (n - 2) \bigl(-4 + 3 n + (n - 2)^{2}\bigr)& =0 
\end{split} 
\end{equation*}
whose initial conditions are given by

\begin{equation*} 
\begin{split} 
u (1)& =0 \\ 
u (0)& =\pi^{\Bigl(-\frac{1}{2}\Bigr)} 
\end{split} 
\end{equation*}
This recurrence has the closed form solution

\begin{equation*} 
\begin{split} 
u (2 n + 1)& =0, \\ 
u (2 n)& =\frac{(-1)^{n} \Gamma \Bigl(n + \frac{1}{2}\Bigr)}{\pi}. 
\end{split} 
\end{equation*}
 
 
 
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