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\begin{document}
\chapter*{BSJ Bessel J}
\label{BSJ}
\section*{BSJ.1 Introduction}
\label{BSJ:intro}


Let $x$ be a complex variable of $\mathbb{C} \setminus \{0,\infty\}$ and let $\nu$ denote a parameter (independent of $x$).The function Bessel J (noted $\operatorname{J} _{\nu}$) is defined by the following second order differential equation
\begin{equation*}
\label{BSJ:diffeq}
\begin{split}
\bigl(x^{2} - \nu^{2}\bigr) y (x) + x \frac{\partial y (x)}{\partial x} + x^{2} \frac{\partial^{2} y (x)}{\partial x^{2}}& =0.
\end{split}\tag{BSJ.1.1}
\end{equation*}



Although $0$ is a singularity of BSJ.1.1, the initial conditions can be given by 
\begin{equation*}
\label{BSJ:inicond}
\begin{split}
\frac{\partial \frac{\operatorname{J} _{\nu} (x)}{x^{\nu}}}{\partial x}& =\frac{1}{\Gamma (\nu + 1) 2^{\nu}}.
\end{split}\tag{BSJ.1.2}
\end{equation*}



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


Related functions: \href{http://algo.inria.fr/esf/function/H1/H1.html#H1}{Hankel H1},\href{http://algo.inria.fr/esf/function/H2/H2.html#H2}{Hankel H2},\href{http://algo.inria.fr/esf/function/BSY/BSY.html#BSY}{Bessel Y}
\section*{BSJ.2 Series and asymptotic expansions}
\label{BSJ:asympt}
\subsection*{BSJ.2.1 Asymptotic expansion at $0$}
\label{743335624114640886}
\subsubsection*{BSJ.2.1.1 First terms}
\label{BSJ:asympt:0:termsec}



\begin{equation*}
\label{BSJ:asympt:0:terms}
\begin{split}
& \operatorname{J} _{\nu} (x)\approx x^{\nu} \Biggl(\frac{1}{\Gamma (\nu + 1) 2^{\nu}} - \frac{x^{2}}{4 \Gamma (\nu + 1) 2^{\nu} (\nu + 1)} +  \\
& \quad{}\quad{}\frac{x^{4}}{32 \Gamma (\nu + 1) 2^{\nu} (\nu + 1) (\nu + 2)} -  \\
& \quad{}\quad{}\frac{x^{6}}{384 \Gamma (\nu + 1) 2^{\nu} (\nu + 1) (\nu + 2) (\nu + 3)}\ldots\Biggr).
\end{split}\tag{BSJ.2.1.1.1}
\end{equation*}

\subsubsection*{BSJ.2.1.2 General form}
\label{BSJ:asympt:0:genf}



\begin{equation*}
\label{BSJ:asympt:0:genfsum}
\begin{split}
& \operatorname{J} _{\nu} (x)\approx x^{\nu} \sum_{n = 0}^{\infty} u (n) x^{n}.
\end{split}\tag{BSJ.2.1.2.1}
\end{equation*}
The coefficients $u (n)$ satisfy the recurrence
\begin{equation*}
\label{BSJ:asympt:0:genfrec}
\begin{split}
u (n) \bigl(-\nu^{2} + (\nu + n)^{2}\bigr) + u (n - 2)& =0.
\end{split}\tag{BSJ.2.1.2.2}
\end{equation*}
Initial conditions of BSJ.2.1.2.2 are given by
\begin{equation*}
\label{BSJ:asympt:0:genfic}
\begin{split}
u (1)& =0, \\
u (0)& =\frac{1}{\Gamma (\nu + 1) 2^{\nu}}.
\end{split}\tag{BSJ.2.1.2.3}
\end{equation*}
The recurrence BSJ.2.1.2.2 has the closed form solution
\begin{equation*}
\label{BSJ:asympt:0:RDINREFRDGENFROMRDCLOSED}
\begin{split}
u (2 n + 1)& =0, \\
u (2 n)& =\frac{(-1)^{n}}{2^{\nu} 4^{n} \Gamma (n + 1) \Gamma (n + \nu + 1)}.
\end{split}\tag{BSJ.2.1.2.4}
\end{equation*}

\subsection*{BSJ.2.2 Asymptotic expansion at $\infty$}
\label{743337269292822488}
\subsubsection*{BSJ.2.2.1 First terms}
\label{743337314658131821}



\begin{equation*}
\begin{split}
& \operatorname{J} _{\nu} (x)\approx \operatorname{e} ^{\biggl(-\frac{\operatorname{RootOf} _{\xi,2} (1 + \xi^{2})}{x}\biggr)} \sqrt{x} y _{0} (x) +  \\
& \quad{}\quad{}\operatorname{e} ^{\biggl(-\frac{\operatorname{RootOf} _{\xi,1} (1 + \xi^{2})}{x}\biggr)} \sqrt{x} y _{1} (x),
\end{split}
\end{equation*}
where
\begin{equation*}
\begin{split}
y _{0} (x)& =\frac{\sqrt{2} \operatorname{e} ^{\bigl(-\frac{i}{4}\pi (2 \nu + 1)\bigr)}}{2 \sqrt{\pi}} - \frac{-\bigl(4 \nu^{2} - 1\bigr) \sqrt{2} \operatorname{e} ^{\bigl(-\frac{i}{4}\pi (2 \nu + 1)\bigr)} x}{16 \sqrt{\pi} \operatorname{RootOf} _{\xi,2} \bigl(1 + \xi^{2}\bigr)} +  \\
& \quad{}\quad{}\frac{\bigl(4 \nu^{2} - 9\bigr) \bigl(4 \nu^{2} - 1\bigr) \sqrt{2} \operatorname{e} ^{\bigl(-\frac{i}{4}\pi (2 \nu + 1)\bigr)} x^{2}}{256 \sqrt{\pi} \operatorname{RootOf} _{\xi,2} \bigl(1 + \xi^{2}\bigr)^{2}} -  \\
& \quad{}\quad{}\frac{-\bigl(4 \nu^{2} - 25\bigr) \bigl(4 \nu^{2} - 9\bigr) \bigl(4 \nu^{2} - 1\bigr) \sqrt{2} \operatorname{e} ^{\bigl(-\frac{i}{4}\pi (2 \nu + 1)\bigr)} x^{3}}{6144 \sqrt{\pi} \operatorname{RootOf} _{\xi,2} \bigl(1 + \xi^{2}\bigr)^{3}} +  \\
& \quad{}\quad{}2 \ldots \\
y _{1} (x)& =\frac{\sqrt{2} \operatorname{e} ^{\frac{i}{4} \pi (2 \nu + 1)}}{2 \sqrt{\pi}} - \frac{-\bigl(4 \nu^{2} - 1\bigr) \sqrt{2} \operatorname{e} ^{\frac{i}{4} \pi (2 \nu + 1)} x}{16 \sqrt{\pi} \operatorname{RootOf} _{\xi,1} \bigl(1 + \xi^{2}\bigr)} +  \\
& \quad{}\quad{}\frac{\bigl(4 \nu^{2} - 9\bigr) \bigl(4 \nu^{2} - 1\bigr) \sqrt{2} \operatorname{e} ^{\frac{i}{4} \pi (2 \nu + 1)} x^{2}}{256 \sqrt{\pi} \operatorname{RootOf} _{\xi,1} \bigl(1 + \xi^{2}\bigr)^{2}} -  \\
& \quad{}\quad{}\frac{-\bigl(4 \nu^{2} - 25\bigr) \bigl(4 \nu^{2} - 9\bigr) \bigl(4 \nu^{2} - 1\bigr) \sqrt{2} \operatorname{e} ^{\frac{i}{4} \pi (2 \nu + 1)} x^{3}}{6144 \sqrt{\pi} \operatorname{RootOf} _{\xi,1} \bigl(1 + \xi^{2}\bigr)^{3}} + 2 \ldots
\end{split}
\end{equation*}

\subsubsection*{BSJ.2.2.2 General form}
\label{74333787672314883}
\paragraph*{BSJ.2.2.2.1 Auxiliary function $y _{0} (x)$}
\label{74333769961484650}
The coefficients $u (n)$ of $y _{0} (x)$ satisfy the following recurrence
\begin{equation*}
\begin{split}
& 8 u (n) n \operatorname{RootOf} _{\xi,2} \bigl(1 + \xi^{2}\bigr) + u (n - 1) \bigl(-4\nu^{2} - 3 + 4 n + 4 (n - 1)^{2}\bigr)=0
\end{split}
\end{equation*}
whose initial conditions are given by
\begin{equation*}
\begin{split}
u (0)& =\frac{\sqrt{2} \operatorname{e} ^{\bigl(-\frac{i}{4}\pi (2 \nu + 1)\bigr)}}{2 \sqrt{\pi}}
\end{split}
\end{equation*}
This recurrence has the closed form solution
\begin{equation*}
\begin{split}
u (n)& =\Biggl(2^{\Bigl(n + \frac{1}{2}\Bigr)} \operatorname{RootOf} _{\xi,2} \bigl(1 + \xi^{2}\bigr)^{n} \Gamma \biggl(n - \nu + \frac{1}{2}\biggr) \Gamma \biggl(\nu + \frac{1}{2} + n\biggr)  \\
& \quad{}\quad{}\operatorname{sin} \Biggl(\frac{\pi (2 \nu + 1)}{2}\Biggr) (-2)^{n} (-1)^{n} \operatorname{e} ^{\Bigl(-\frac{i}{4}\pi (2 \nu + 1)\Bigr)}\Biggr)\Bigg/ \\
& \quad{}\quad{}\Biggl(2 \Gamma (n + 1) 8^{n} \pi^{\frac{3}{2}}\Biggr).
\end{split}
\end{equation*}
\paragraph*{BSJ.2.2.2.2 Auxiliary function $y _{1} (x)$}
\label{743337727864330445}
The coefficients $u (n)$ of $y _{1} (x)$ satisfy the following recurrence
\begin{equation*}
\begin{split}
& 8 u (n) n \operatorname{RootOf} _{\xi,1} \bigl(1 + \xi^{2}\bigr) + u (n - 1) \bigl(-4\nu^{2} - 3 + 4 n + 4 (n - 1)^{2}\bigr)=0
\end{split}
\end{equation*}
whose initial conditions are given by
\begin{equation*}
\begin{split}
u (0)& =\frac{\sqrt{2} \operatorname{e} ^{\frac{i}{4} \pi (2 \nu + 1)}}{2 \sqrt{\pi}}
\end{split}
\end{equation*}
This recurrence has the closed form solution
\begin{equation*}
\begin{split}
u (n)& =\Biggl(2^{\Bigl(n + \frac{1}{2}\Bigr)} \operatorname{RootOf} _{\xi,1} \bigl(1 + \xi^{2}\bigr)^{n} \Gamma \biggl(n - \nu + \frac{1}{2}\biggr) \Gamma \biggl(\nu + \frac{1}{2} + n\biggr)  \\
& \quad{}\quad{}\operatorname{sin} \Biggl(\frac{\pi (2 \nu + 1)}{2}\Biggr) (-2)^{n} (-1)^{n} \operatorname{e} ^{\frac{i}{4} \pi (2 \nu + 1)}\Biggr)\Bigg/ \\
& \quad{}\quad{}\Biggl(2 \Gamma (n + 1) 8^{n} \pi^{\frac{3}{2}}\Biggr).
\end{split}
\end{equation*}
\end{document}