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HATN Inverse Hyperbolic Tangent

HATN.1 Introduction

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Let $x$ be a complex variable of $\mathbb{C} \setminus \{-1,1\}$ .The function Inverse Hyperbolic Tangent (noted $\operatorname{arctanh}$ ) is defined by the following second order differential equation


\begin{equation*} 
\begin{split} 
2 x \frac{\partial y (x)}{\partial x} + \bigl(x^{2} - 1\bigr) \frac{\partial^{2} y (x)}{\partial x^{2}}& =0. 
\end{split} 
\end{equation*}
HATN.1.1

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


\begin{equation*} 
\begin{split} 
\operatorname{arctanh} (0)& =0, \\ 
\frac{\partial \operatorname{arctanh} (x)}{\partial x} (0)& =1. 
\end{split} 
\end{equation*} 
 HATN.1.2

Related function: Inverse Hyperbolic Cotangent

HATN.2 Series and asymptotic expansions

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HATN.2.1 Taylor expansion at $0$

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HATN.2.1.2 General form

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

\begin{equation*} 
\begin{split} 
n u (n) - -(-n - 2) u (n + 2)& =0. 
\end{split} 
\end{equation*}
HATN.2.1.2.2
Initial conditions of HATN.2.1.2.2 are given by

\begin{equation*} 
\begin{split} 
u (1)& =1, \\ 
u (0)& =0. 
\end{split} 
\end{equation*}
HATN.2.1.2.3

HATN.2.2 Asymptotic expansion at $1$

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HATN.2.2.1 First terms

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\begin{equation*} 
\begin{split} 
& \operatorname{arctanh} (x)\approx \Biggl(\frac{i}{2} \pi + \frac{\operatorname{ln} (2)}{2} - \frac{x}{4} + \frac{1}{4} + \frac{(x - 1)^{2}}{16} - \frac{(x - 1)^{3}}{48} + \frac{(x - 1)^{4}}{128} -  \\ 
& \quad{}\quad{}\frac{(x - 1)^{5}}{320} + \frac{(x - 1)^{6}}{768} - \frac{(x - 1)^{7}}{1792} + \frac{(x - 1)^{8}}{4096} + \frac{\operatorname{ln} (x - 1)}{2}\ldots\Biggr). 
\end{split} 
\end{equation*} 
 HATN.2.2.1.1

HATN.2.2.2 General form

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The general form of is not easy to state and requires to exhibit the basis of formal solutions of ?? (coming soon).

HATN.2.3 Asymptotic expansion at $-1$

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HATN.2.3.1 First terms

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\begin{equation*} 
\begin{split} 
& \operatorname{arctanh} (x)\approx \Biggl(\frac{-\operatorname{ln} (2)}{2} - \frac{x}{4} - \frac{1}{4} - \frac{(x + 1)^{2}}{16} - \frac{(x + 1)^{3}}{48} - \frac{(x + 1)^{4}}{128} - \frac{(x + 1)^{5}}{320}  \\ 
& \quad{}\quad{}- \frac{(x + 1)^{6}}{768} - \frac{(x + 1)^{7}}{1792} - \frac{(x + 1)^{8}}{4096} - \frac{\operatorname{ln} (x + 1)}{2}\ldots\Biggr). 
\end{split} 
\end{equation*} 
 HATN.2.3.1.1

HATN.2.3.2 General form

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The general form of is not easy to state and requires to exhibit the basis of formal solutions of ?? (coming soon).
 
 
 
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