Is there a closed form expression for the first zero of the first Bessel function? $j_{1,1}$ 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.
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 $0&lt;{{j}_{1,1}}&lt;{{j}_{1,2}}&lt;{{j}_{1,3}}...$ then ${{\sigma }_{M}}=\sum\limits_{n=0}^{\infty }{\frac{1}{j_{1,n}^{M}}}&gt;\frac{1}{j_{1,1}^{M}}$and so $j_{1,1}^{{}}&gt;{{\left( {{\sigma }_{M}} \right)}^{-1/M}}$.  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,
$${{j}_{1,1}}\simeq 2{{\left( \frac{1}{777013} \right)}^{1/20}}{{2}^{7/10}}{{3}^{7/20}}{{5}^{3/20}}{{7}^{1/20}}{{11}^{1/20}}$$
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.