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\begin{document}

\begin{itemize}

\item \( \frac{\partial ^{2}R}{\partial x^{2}}+\frac{1}{x}\frac{\partial R}{\partial x}+(1-\frac{\nu ^{2}}{x^{2}})R=0 \) is the bessel equation. 

\item Try power series solution with \( x^{\alpha }\sum _{j=0}^{\infty }a_{j}x^{j} \) with \( a_{0}\neq 0 \). Lots of algebra gives \( \nu ^{2}=\alpha ^{2} \) and
a recurrence relation, \( a_{j}=-\frac{1}{j(j+2\alpha )} \). Also \( a_{1}=0 \). This means \( a_{2j}=\frac{(-1)^{j}a_{0}\Gamma (\alpha +1)}{2^{2j}j!\Gamma (j+\alpha +1)} \) where \( \Gamma (z) \) is the gamma
function. Defining \( a_{0}=(2^{\gamma }\Gamma (\alpha +1))^{-1} \)

\item \( J_{\nu }(x)=(\frac{x}{2})^{\nu }\sum _{j=0}^{\infty }(-1)^{j}\frac{1}{\Gamma (j+1)\Gamma (\nu +j+1)}(\frac{x}{2})^{2j} \) and \( J_{\nu }(x)=(\frac{x}{2})^{-\nu }\sum _{j=0}^{\infty }(-1)^{j}\frac{1}{\Gamma (j+1)\Gamma (j-\nu +1)}(\frac{x}{2})^{2j} \) 

\item For \( \nu =n \), you get: \( J_{-n}(x)=(-1)^{n}(\frac{x}{2})^{n}\sum _{j=0}^{\infty }(-1)^{j}\frac{1}{\Gamma (j+1)\Gamma (j+n+1)}(\frac{x}{2})^{2j} \) \( =(-1)^{n}J_{n}(x) \). To get the second linearly independent solution
when \( \nu =n \), you take the derivative with respect of \( \nu  \) of the Bessel equation,
and get the solution: \( N_{\nu }(x)=\frac{1}{\pi }[\frac{\partial J_{\nu }}{\partial \nu }-(-1)^{\nu }\frac{\partial J_{-\nu }}{\partial \nu }] \) for \( \nu =n \). To get a solution always, redefine
\( N_{\nu }=\frac{cos(\nu \pi )J_{\nu }(x)-J_{-\nu }(x)}{sin(\nu \pi )} \)for any \( \nu  \). Typically, use, \( J_{v},N_{\nu } \).

\item \( J_{\nu }(x)\rightarrow \frac{1}{\Gamma (\nu +1)}(\frac{x}{2})^{\nu } \) as x-\(>\)0

\item \( N_{\nu }(x)\rightarrow \{\begin{array}{cc}
\frac{2}{\pi }(ln(\frac{x}{2})+\gamma ) & \nu =0\\
\frac{-\Gamma (\nu )}{\pi }(\frac{2}{x})^{\nu } & \nu >0
\end{array}
 \). these are singular.

\item \( J_{\nu }(x)\sim \sqrt{\frac{2}{\pi x}}cos(x-\frac{\nu \pi }{2}-\frac{\pi }{4}) \) for x \(>\)\(>\) \( \nu  \)

\item \( N_{\nu }(x)\sim \sqrt{\frac{2}{\pi x}}sin(x-\frac{\nu \pi }{2}-\frac{\pi }{4}) \) for x \(>\)\(>\)\( \nu  \)

\item \( J_{\nu -1}+J_{\nu +1}=\frac{2\nu }{x}J_{\nu } \)

\item The generating function is: \( e^{\frac{x}{2}(t-\frac{1}{t})}=\sum _{n=-\infty }^{\infty }J_{n}(x)t^{n} \). This lets you get an expansion of \( e^{xsin(\theta )} \).

\end{itemize}

Orthogonality relations

\begin{itemize}

\item \( \int _{0}^{a}d\rho *\rho J_{\nu }(k_{1}\rho )J_{\nu }(k_{2}\rho )=\frac{a^{2}}{2}J^{2}_{\nu +1}(x_{\nu n})\delta _{k_{1}k_{2}} \) where \( x_{\nu n} \) is the nth root of \( J_{\nu } \) if \( k_{i}a \) = 0 of \( J_{\nu } \). or \( k_{i}a= \)0 of \( J_{\nu }' \)

\item \( f(\rho )=\sum _{n=1}^{\infty }c_{\nu n}J_{\nu }(x_{\nu n}\frac{\rho }{a}) \) with \( c_{\nu n}=\frac{2}{a^{2}J_{\nu +1}^{2}(x_{\nu n})}\int _{0}^{a}d\rho *\rho f(\rho )J_{\nu }(x_{\nu n}\frac{\rho }{a}) \)

\end{itemize}

\end{document}
