What is $\lim\limits_{n\to\infty} (\text j_{x,x}-\text y_{x,x})$ with the BesselYZero and BesselJZero function?

Solution 1:

According to §10.21(viii) in the DLMF, $$ j_{n,n} - y_{n,n} = n(z(n^{ - 2/3} a_n ) - z(n^{ - 2/3} b_n )) + \mathcal{O}\!\left( {\frac{1}{n}} \right) $$ where the function $z(\zeta)$ is defined implicitly by $$ \tfrac{2}{3}( - \zeta )^{3/2} = \sqrt {z^2 (\zeta ) - 1} - \operatorname{arcsec}z(\zeta ) $$ so that $z>1$ and $\zeta<0$. The $a_n$ and the $b_n$ are the $n$th negative real zeros of the Airy functions $\mathrm{Ai}$ and $\mathrm{Bi}$. By §9.9(iv) in the DLMF, $$ n^{ - 2/3} a_n = - \left( {\frac{3}{8}\pi \frac{{4n - 1}}{n}} \right)^{2/3} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right) = - \left( {\frac{{3\pi }}{2}} \right)^{2/3} + \frac{1}{6}\left( {\frac{{3\pi }}{2}} \right)^{2/3} \frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right) $$ and $$ n^{ - 2/3} b_n = - \left( {\frac{3}{8}\pi \frac{{4n - 3}}{n}} \right)^{2/3} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right) = - \left( {\frac{{3\pi }}{2}} \right)^{2/3} + \frac{1}{2}\left( {\frac{{3\pi }}{2}} \right)^{2/3} \frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right). $$ Hence, by a Taylor approximation, $$ j_{n,n} - y_{n,n} = - \frac{1}{3}\left( {\frac{{3\pi }}{2}} \right)^{2/3} z'\!\left( { - \left( {\tfrac{{3\pi }}{2}} \right)^{2/3} } \right) + \mathcal{O}\!\left( {\frac{1}{n}} \right). $$ Therefore, your limit is $$ - \frac{1}{3}\left( {\frac{{3\pi }}{2}} \right)^{2/3} z'\!\left( { - \left( {\tfrac{{3\pi }}{2}} \right)^{2/3} } \right). $$ Using implicit differentiation, $$ - z'(\zeta ) = z(\zeta )\sqrt { - \frac{\zeta }{{z^2 (\zeta ) - 1}}} . $$ The $z\!\left( { - \left( {\tfrac{{3\pi }}{2}} \right)^{2/3} } \right)$ is the unique positive solution of $\pi =\sqrt{z^2-1}-\operatorname{arcsec}z$. Hence, in summary, $$ \lim_{n\to +\infty}(j_{n,n} - y_{n,n})=\frac{\pi }{2}\sqrt {\frac{{z^2 }}{{z^2 - 1}}} $$ where $$ z=4.603338848751700352556582029103016513067397134\ldots $$ is the unique positive solution of $\pi =\sqrt{z^2-1}-\operatorname{arcsec}z$. In particular, $$\lim_{n\to +\infty}(j_{n,n} - y_{n,n})= 1.609225204679256120709524517222109882356759015\ldots\, .$$ Using identities between inverse trigonometric functions, you can re-phrase the result as $$ \lim_{n\to +\infty}(j_{n,n} - y_{n,n})=\frac{\pi }{2}\frac{{\sqrt {w^2 + 1} }}{w} =-\frac{\pi}{2} \csc w $$ where $w$ is the unique positive solution of $\pi =w-\arctan w$, or the smallest positive root of $\tan w=w$.