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.

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 $\frac{x(11 x^4-552 x^2+5760) }{ 4 x^4+336 x^2+11520}$
