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WM Whittaker M

WM.1 Introduction

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Let $x$ be a complex variable of $\mathbb{C} \setminus \{0,\infty\}$ and let $\mu,\nu$ denote a set of parameters (independent of $x$ ).The function Whittaker M (noted $\operatorname{WM} _{\mu , \nu}$ ) is defined by the following second order differential equation


\begin{equation*} 
\begin{split} 
-x^{2} - 4 \mu x - 1 + 4 \nu^{2}y (x) + 4 x^{2} \frac{\partial^{2} y (x)}{\partial x^{2}}& =0. 
\end{split} 
\end{equation*}
WM.1.1

Although $0$ is a singularity of WM.1.1, the initial conditions can be given by


\begin{equation*} 
\begin{split} 
\frac{\partial \frac{\operatorname{WM} _{\mu , \nu} (x)}{x^{\bigl(\nu + \frac{1}{2}\bigr)}}}{\partial x}& =1. 
\end{split} 
\end{equation*} 
 WM.1.2

The formulae of this document are valid for $2 \nu \not\in \mathbb{Z} .$

Related function: Whittaker W

WM.2 Series and asymptotic expansions

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WM.2.1 Asymptotic expansion at $0$

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WM.2.1.2 General form

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\begin{equation*} 
\begin{split} 
& \operatorname{WM} _{\mu , \nu} (x)\approx x^{\Bigl(\nu + \frac{1}{2}\Bigr)} \sum_{n = 0}^{\infty} u (n) x^{n}. 
\end{split} 
\end{equation*}
WM.2.1.2.1
The coefficients $u (n)$ satisfy the recurrence

\begin{equation*} 
\begin{split} 
& u (n) \Biggl(4 \biggl(\nu + \frac{1}{2} + n\biggr)^{2} - 4 \nu - 1 - 4 n - 4 \nu^{2}\Biggr) + 4 u (n - 1) \mu - u (n - 2)=0. 
\end{split} 
\end{equation*}
WM.2.1.2.2
Initial conditions of WM.2.1.2.2 are given by

\begin{equation*} 
\begin{split} 
u (1)& =\frac{-4\mu}{8 \nu + 4}, \\ 
u (0)& =1. 
\end{split} 
\end{equation*}
WM.2.1.2.3
The recurrence WM.2.1.2.2 has the closed form solution

\begin{equation*} 
\begin{split} 
u (n)& =0. 
\end{split} 
\end{equation*}
WM.2.1.2.4

WM.2.2 Asymptotic expansion at $\infty$

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WM.2.2.1 First terms

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\begin{equation*} 
\begin{split} 
& \operatorname{WM} _{\mu , \nu} (x)\approx \operatorname{e} ^{\frac{1}{2 x}} x^{\mu} y _{0} (x), 
\end{split} 
\end{equation*}
where

\begin{equation*} 
\begin{split} 
y _{0} (x)& =\frac{\Gamma (2 \nu + 1)}{\Gamma \Bigl(\frac{1}{2} + \nu - \mu\Bigr)} + \frac{-\bigl(4 \nu^{2} - 4 \mu - 4 \mu^{2} - 1\bigr) \Gamma (2 \nu + 1) x}{4 \Gamma \Bigl(\frac{1}{2} + \nu - \mu\Bigr)} +  \\ 
& \quad{}\quad{}\frac{\bigl(4 \nu^{2} - 12 \mu - 4 \mu^{2} - 9\bigr) \bigl(4 \nu^{2} - 4 \mu - 4 \mu^{2} - 1\bigr) \Gamma (2 \nu + 1) x^{2}}{32 \Gamma \Bigl(\frac{1}{2} + \nu - \mu\Bigr)}  \\ 
& \quad{}\quad{}+ \Bigl(-\bigl(4 \nu^{2} - 20 \mu - 4 \mu^{2} - 25\bigr) \bigl(4 \nu^{2} - 12 \mu - 4 \mu^{2} - 9\bigr)  \\ 
& \quad{}\quad{}\bigl(4 \nu^{2} - 4 \mu - 4 \mu^{2} - 1\bigr) \Gamma (2 \nu + 1) x^{3}\Bigr)\Bigg/\Biggl(384 \Gamma \biggl(\frac{1}{2} + \nu - \mu\biggr)\Biggr) + 2 \ldots 
\end{split} 
\end{equation*}

WM.2.2.2 General form

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WM.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} 
& -4u (n) n +  \\ 
& u (n - 1) \bigl(-4\nu^{2} + 4 \mu + 4 \mu^{2} - 3 + 4 n + 8 (n - 1) \mu + 4 (n - 1)^{2}\bigr)=0 
\end{split} 
\end{equation*}
whose initial conditions are given by

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

\begin{equation*} 
\begin{split} 
u (n)& =\Biggl(\operatorname{sin} \Biggl(\frac{\pi (1 + 2 \nu - 2 \mu)}{2}\Biggr) (-2)^{n} 2^{n} (-1)^{n} \Gamma \biggl(n - \nu + \frac{1}{2} + \mu\biggr)  \\ 
& \quad{}\quad{}\Gamma \biggl(n + \nu + \frac{1}{2} + \mu\biggr) \Gamma (2 \nu + 1)\Biggr)\Bigg/\Biggl(4^{n} \pi \Gamma \biggl(\nu + \frac{1}{2} + \mu\biggr) \Gamma (n + 1)\Biggr). 
\end{split} 
\end{equation*}
 
 
 
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