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

Question

This question is similar to:

Conjecture: $$\lim\limits_{x\to\infty}\operatorname{Re}\text W_x(x)\mathop=\limits^?-\ln(2\pi)$$

I have come across BesselJZero and BesselYZero function as a form of “Inverse Bessel function” with the following definition as zeros of the real and imaginary parts of the First Hankel function with the Real Bessel J part and Imaginary Bessel Y function part:

$$\text J_v(\text j_{v,k})=0\text{ and }\text Y_v(\text y_{v,k})=0,0\ne k\in\Bbb N$$

with $k$ being the $kth$ real zero. After some experimentation, the following was found:

$$\lim_{x\to\infty} \text j_{x,x}=\lim_{x\to\infty}\text y_{x,x}=\infty$$

but equating both $v,k$ makes a less trivial limit and subtracting both limits is not zero:

$$\lim_{x\to\infty}(\text j_{x,x}-\text y_{x,x })= \lim_{x\to\infty}(\text j_{x,x}-\text y_{-x,x })= \lim_{x\to\infty}(\text j_{-x,x}-\text y_{x,x })= \lim_{x\to\infty}(\text j_{-x,x}-\text y_{-x,x }) ≈1.6092344923020101524314251158845736064326608605905438683176022909...$$

With this approximation of the constant for $x=3000$ since there only so many calculated zeros of Bessel functions.

One must have a discrete limit because $x\in\Bbb N^+$. There was no possible closed form for this constant using closed form finders either since the accuracy was too low or since the number has no closed form itself. Sorry if there was not much else that I could find out, but I will keep working on this constant.

What is an alternate form of this limit as a sum, integral or other operator besides a limit? Please correct me and give me feedback!

Answer 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 $\tan w=w$.

Answer 2

The following was found using the Spherical Bessel function of the first kind $\text j_n(x)$ and Bessel function of the first kind $\text J_n(x)$ after seeing the OEIS Link in @Gary’s great answer which said:

Also the first root of the spherical Bessel function of the 1st kind, j_1(x). - Stanislav Sykora, Nov 14 2013

Since $$\text j_n(x)=\sqrt{\frac\pi{2x}}\,\text J_{n+\frac12}(x)\implies \text j_1(x)= \sqrt{\frac\pi{2x}}\,\text J_{\frac32}(x)=0 \implies x=j_{\frac 32,n\in \Bbb N}$$

therefore:

$$\lim_{n\to\infty }(\text j_{n,n}-\text y_{n,n})=-\frac\pi2\csc\left(\text j_{\frac 32,1}\right)$$

and the fixed point of $\tan(x)$ in closed form using Mathematics functions is:

$$\tan(x)=x\implies x=\text j_{\frac 32,n},n\in\Bbb N$$

as seen in this computation. Please correct me and give me feedback!