In my paper with É. Fouvry and Ph. Michel where we find upper bounds for the number of certain sheaves on the affine line over a finite field with bounded ramification, the combinatorial part of the argument involves spherical codes and the method of Kabatjanski and Levenshtein, and turns out to depend on the rather recondite question of knowing a lower bound on the size of the largest zero of the -th Hermite polynomial , which is defined for integers by
This is a classical orthogonal polynomial (which implies in particular that all zeros of are real and simple). The standard reference for such questions seems to still be Szegö’s book, in which one can read the following rather remarkable asymptotic formula:
where is the first (real) zero of the function
which is a close cousin of the Airy function (see formula (6.32.8) in Szegö’s book, noting that he observes the Peano paragraphing rule, according to which section 6.32 comes before 6.4).
(Incidentally, if — like me — you tend to trust any random PDF you download to check a formula like that, you might end up with a version containing a typo: the cube root of is, in some printings, replaced by a square root…)
Szegö references work of a number of people (Zernike, Hahn. Korous, Bottema, Van Veen and Spencer), and sketches a proof based on ideas of Sturm on comparison of solutions of two differential equations.
As it happens, it is better for our purposes to have explicit inequalities, and there is an elementary proof of the estimate
which is only asymptotically weaker by a factor from the previous formula. This is also explained by Szegö, and since the argument is rather cute and short, I will give a sketch of it.
Besides the fact that the zeros of are real and simple, we will use the easy facts that , and that is an even function for even, and an odd function for odd, and most importantly (since all other properties are rather generic!) that they satisfy the differential equation
The crucial lemma is the following result of Laguerre:
Let be a polynomial of degree . Let be a simple zero of , and let
Then if is any line or circle passing through and , either all zeros of are in , or both components of contain at least one zero of .
Before explaining the proof of this, let’s see how it gives the desired lower bound on the largest zero of . We apply Laguerre’s result with and . Using the differential equation, we obtain
Now consider the circle such that the segment is a diameter of .
Now note that is the smallest zero of (as we observed above, is either odd or even). We can not have : if that were the case, the unbounded component of the complement of the circle would not contain any zero, and neither would contain all zeros (since ), contradicting the conclusion of Laguerre's Lemma. Hence we get
and this implies
as claimed. (Note that if , one deduces easily that the inequality is strict, but there is equality for .)
Now for the proof of the Lemma. One defines a polynomial by
so that has degree and has zero set formed of the zeros of different from (since the latter is assumed to be simple). Using the definition, we have
We now compute the value at of the logarithmic derivative of , which is well-defined: we have
which becomes, by the above formulas and the definition of , the identity
where is a Möbius transformation.
Recalling that , this means that is the average of the . It is then elementary that for line , either is contained in , or intersects both components of the complement of . Now apply to this assertion: one gets that either is contained in , or intersects both components of the complement of . We are now done, after observing that the lines passing through are precisely the images under of the circles and lines passing through and through (because , and each line passes through in the projective line.)