Is there a closed form expression for the first zero of the first Bessel function?

Question

$j_{1,1}$ denotes the first zero of the first Bessel function of the first kind. (That's a lot of firsts!) It's approximately equal to $3.83$. My question is, is there any closed form expression for its value? Even a infinite series or infinite product that yields it would be good.

I ask because this value is used in physics, in the context of diffraction of light through a circular aperture, and students often make the mistake of thinking that the number just pops out of nowhere.

Answer 1

The series expansion is:

$$J_{1}\left(x\right)=\sum_{n=0}^\infty\frac{x^{2n+1}}{2^{2n+1} n! (n+1)!}$$ $$=\frac{x}{2} - \frac{x^{3}}{16} + \frac{x^{5}}{384} - \frac{x^{7}}{18432} + …$$

The denominators are Sloan Sequence A002474.

Sloan Sequence A115369 is the first zero.

Answer 2

If you want an approximation of the first root without invoking at any time the Bessel functions, what you could consider is an $[2m+1,2n]$ Padé approximation of $J_1(x)$ built at $x=0$. It will be looking like $$J_1(x)\approx\frac{x \left(\frac12+\sum_{j=1}^m a_j \,x^{2j}\right)}{1+\sum_{k=1}^n b_k\, x^{2k} }$$ The coefficients are "easy" to compute since, using the infinite series given in Geremia's answer, we know exactly all required derivatives of $J_1(x)$ at $x=0$. These approximations are equivalent to Taylor expansions to $O\left(x^{2(m+n+1})\right)$.

In order to keep it simple (solution of a quadratic equation in $x^2$), select $m=2$. The following table gives the equation to be solved, its first root and its decimal representation. $$\left( \begin{array}{cccc} n & \text{equation} & \text{root} & \text{value} \\ 1 & x^4-40 x^2+384=0 & 4 & 4.00000 \\ 2 & 11 x^4-552 x^2+5760=0 & \sqrt{\frac{12}{11} \left(23-\sqrt{89}\right)} & 3.84698 \\ 3 & 23 x^4-1296 x^2+14080=0 & \sqrt{\frac{8}{23} \left(81-\sqrt{1501}\right)} & 3.83382 \\ 4 & 97 x^4-5810 x^2+64400=0 & \sqrt{\frac{5}{97} \left(581-\sqrt{87689}\right)} & 3.83202 \end{array} \right)$$ while the "exact" value is $3.83171$.

Just for the fun of it, plot on the same graph, for $0 \leq x \leq 4$, $J_1(x)$ and $\frac{x(11 x^4-552 x^2+5760) }{ 4 x^4+336 x^2+11520}$

Answer 3

I like Claude's approach and approximation. However for variety (and possibly interest) here is another approach which i pinched from Watson who pinched from Rayleigh who pinched from Euler. And while Claude's approximation converges from above, here we'll approach from below. First we can write the Bessel function as an infinite product with the form $${{J}_{1}}\left( z \right)=\frac{1}{2}z\prod\limits_{n=0}^{\infty }{\left( 1-\frac{{{z}^{2}}}{j_{1,n}^{2}} \right)}$$ Taking the logarithmic derivative, we have then $$\frac{1}{{{J}_{1}}\left( z \right)}{{J}_{1}}'\left( z \right)=\frac{1}{z}-2z\sum\limits_{n=0}^{\infty }{\frac{1}{j_{1,n}^{2}}\frac{1}{1-{{z}^{2}}/j_{1,n}^{2}}}$$ Converting this into a series we find $$\frac{1}{{{J}_{1}}\left( z \right)}{{J}_{1}}'\left( z \right)=\frac{1}{z}-2z\sum\limits_{n=0}^{\infty }{\frac{1}{j_{1,n}^{2}}\sum\limits_{m=0}^{\infty }{\frac{{{z}^{2m}}}{j_{1,n}^{2m}}}}$$ And so we recover the definition of Rayleigh’s function $$\frac{1}{{{J}_{1}}\left( z \right)}{{J}_{1}}'\left( z \right)=\frac{1}{z}-2z\sum\limits_{m=0}^{\infty }{{{z}^{2m}}{{\sigma }_{2m+2}}}$$ where

$${{\sigma }_{2m+2}}=\sum\limits_{n=0}^{\infty }{\frac{1}{j_{1,n}^{2m+2}}}$$ Using well known recurrence relationships we can simplify this to $${{J}_{2}}\left( z \right)=2{{J}_{1}}\left( z \right)\sum\limits_{m=0}^{\infty }{{{z}^{2m+1}}{{\sigma }_{2m+2}}}$$ Substituting the series representation for the Bessel functions we have $$\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{z}^{2k}}}{{{2}^{2k+2}}k!\left( k+2 \right)!}}=\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{2}^{2n}}n!\left( n+1 \right)!}}\sum\limits_{m=0}^{\infty }{{{\sigma }_{2m+2}}{{z}^{2m+2n}}}$$ And reversing the summation, i.e. $\sum\limits_{k=0}^{\infty }{\sum\limits_{j=0}^{\infty }{{{a}_{k,j}}}}=\sum\limits_{j=0}^{\infty }{\sum\limits_{k=0}^{j}{{{a}_{k,j-k}}}}$ one finds $$\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{z}^{2k}}}{{{2}^{2k+2}}k!\left( k+2 \right)!}}=\sum\limits_{m=0}^{\infty }{{{z}^{2m}}}\sum\limits_{n=0}^{m}{\frac{{{\left( -1 \right)}^{n}}}{{{2}^{2n}}n!\left( n+1 \right)!}{{\sigma }_{2m-2n+2}}}$$ Equating coefficients $$\sum\limits_{n=0}^{k}{\frac{{{\left( -1 \right)}^{n}}{{\sigma }_{2k-2n+2}}}{{{2}^{2n}}n!\left( n+1 \right)!}}=\frac{{{\left( -1 \right)}^{k}}}{{{2}^{2k+2}}k!\left( k+2 \right)!}$$ It’s very easy to create the sigma numbers in mathematica (and even by hand I suppose), so explicitly we have $${{\sigma }_{2}}=\frac{1}{8}, {{\sigma }_{4}}=\frac{1}{192}, {{\sigma }_{6}}=\frac{1}{3072}, {{\sigma }_{8}}=\frac{1}{46080},…,{{\sigma }_{20}}=\frac{777013}{361634098839552000}$$ Now since $0<{{j}_{1,1}}<{{j}_{1,2}}<{{j}_{1,3}}...$ then ${{\sigma }_{M}}=\sum\limits_{n=0}^{\infty }{\frac{1}{j_{1,n}^{M}}}>\frac{1}{j_{1,1}^{M}}$and so $j_{1,1}^{{}}>{{\left( {{\sigma }_{M}} \right)}^{-1/M}}$. You can bound this the other way as well (see Watson) but what we have is enough to get good approximations. So for a somewhat reasonable example (about 2 decimal places - when rounded)

$${{j}_{1,1}}\simeq {{2}^{5/4}}{{3}^{1/4}}{{5}^{1/8}}$$

a more unreasonable example, that yields about 6 places, $${{j}_{1,1}}\simeq 2{{\left( \frac{1}{777013} \right)}^{1/20}}{{2}^{7/10}}{{3}^{7/20}}{{5}^{3/20}}{{7}^{1/20}}{{11}^{1/20}}$$