What do roots of random polynomials look like?

Answer:  they look awesome!

I found this picture, made by Sam Derbyshire, on John Baez’s This Week in Mathematical Physics, progenitor of all math blogs and still teaching me things 16 years into its run.  It depicts the roots of polynomials of degree at most 24 with all coefficients \pm 1.  The image above is a closeup of a neighborhood of (1/2)e^{i/5}.  (Or so Baez says; judging by the picture I wonder if he meant (4/5)e^{i/5}.)  The full picture looks like this:

You can see why there’s a big hole around zero:  if |z| is small, \pm 1 \pm z \pm z^2 \pm \ldots will have a hard time being very far from 1.

I wonder what this picture looks like as the maximum degree of the polynomial gets higher and higher?  I suppose this would have something to do with the amount of time spent near zero by the random walk which starts at 0, then moves by either z^k or -z^k at time k.  When |z| < 1/2, the limiting measure of this random walk is supported on a kind of Cantor set:  in any event, it never returns to 0.  When |z| > 1, the random walk is going to get farther and farther away; again, we should not expect to return to 0.

When 1/2 < |z| < 1, I would have thought this random walk was better understood.  But not so!  Boris Solonyak proved in 1995 (JSTOR link)  that the random walk converges to an absolutely continuous measure m_z for almost all real z in (1/2,1).  It’s conjectured that this is the case for all but countably many z, but this remains out of reach.  There are certainly exceptions:  Erdos showed, for instance, that when z = (1/2)(1+\sqrt{5}), the limiting measure is purely singular.  (Let’s hope Dan Brown doesn’t find out about this.)

All this reminds me of a beautiful lecture Balint Virag gave here at UW a few months ago, about the roots of random complex power series

\sum_{i=0}^\infty a_i z^i

where the coefficients a_i are independent complex Gaussians with mean 0 and variance 1.  In this much smoother case, the density of zeros in the open unit disc is very simple:  it’s (1/\pi) (1-|z|^2){-2}.  In a 2005 paper, Virag and Yuval Peres do much more:  they show that the joint intensity function for an n-tuple z_1, …. z_n in the unit disc is

\pi^{-n} det [(1-z_i \bar{z}_j))^{-2}]_{i,j}

In fact, the distribution of zeroes turns out to be invariant under the whole group of Mobius transformations on the disc!

Actually, I lied — what Virag talked about in Madison was more recent work about random power series of the form

\sum_{i=0}^\infty a_i (1/\sqrt{i!}) z^i.

These guys converge on the whole complex plane, not just on the open unit disc:  and it turns out that the point process measuring the locations of their zeroes is now invariant under complex translations!  Virag and his collaborators have a very precise understanding of this process:  the repulsion between zeroes, the distribution of the deviation from the expected number of zeroes in larger and larger dilates of a nice region, etc.

Of course I had to ask myself:  what is the p-adic version of all this?  And here I got a bit stuck.  One might of course study power series of the form

P = \sum_{i=0}^\infty a_i z^i

where now a_i is distributed on Haar measure on Z_p.  But of course this is more like a power series with real Gaussian coefficient, which, Balint tells me, is not nearly as nice as the complex case.  Unfortunately, Q_p doesn’t have an algebraic closure that’s locally compact!  So I’m not sure what, if anything, the p-adic analogue of the complex Gaussian could possibly be.

Still, it seems fun to understand the zeroes of a polynomial P as a above, with random coefficients in Z_p.  The distances of the zeroes from 0, for instance, is governed by the Newton polygon of P.  Is there a nice description of the resulting distribution on the set of Newton polygons?  It’s easy to see that the total number of zeroes in the unit disc — that is, the total length of the Newton polygon — is geometrically distributed with parameter 1/p.

Is there a p-adic analogue of 1/\sqrt{i!}?  That is, can you build a decay into the coefficients of P to get a random p-adic power series defined on all of C_p, and which, let’s say, is invariant under translation by Q_p?

3 thoughts on “What do roots of random polynomials look like?

  1. David says:

    What is that little black hole near the upper right of the picture?

  2. Sam says:

    I am amazed by the appearance of the Heighway Dragon around the edges! However it is may not be so suprising since the set of all fractions about 0 using digits (0,1) in the complex base(1+i) form the Heighway Dragon, with one centered at each of the Gaussian integers. The Dragon will resolve as a 2 dimensional space filling curve when using 2 digits and using any complex base of magnitude sqrt(2). The resulting sets using another base z will have dimension d = log(2)/log(|z|).


  3. Bjorn Poonen apparently has a paper explaining some of the fractal-like appearance (but I was unable to find it on his web page). At least, that was what I gathered when this subject came up at the JMM MathOverflow meetup last month.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


Get every new post delivered to your Inbox.

Join 426 other followers

%d bloggers like this: