text: [–][deleted]  (1 child)
  
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  [–][deleted]  (5 children)
  
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  [–] 9 points10 points  (2 children)
  
  sorry, this has been archived and can no longer be voted on
  
  I don't know why, but I do have some guesses for where I would start to look.
  
  First, looking at this animation I might guess that the sequence of polynomials f_n(z) converges to an analytic function on the open unit disk. Now, Jentzsch's theorem says that, for a power series with radius of convergence 1, every point on the circle |z| = 1 is a limit point of the zeros of the partial sums of the power series. So, these polynomials might be converging in a nice enough way so that, even though they're not partial sums of whatever power series, the proof of Jentzsch's theorem could apply to this case with small modification. Based on my experience, I would even hazard a guess that the limit function has a pole at z=1 which causes the roots of f_n(z) to "stay away".
  
  A priori, there's also the fact that the zeros of random polynomials are expected to cluster on the unit circle as the degree goes to infinity. See this paper. For this particular problem, if one could get a good enough handle on the coefficients of the polynomials in question, one might be able to directly apply Theorem 4 from that linked paper to get the result.
  
  Edit: After actually calculating some of these polynomials, it looks it's actually the reciprocal polynomials which are converging on the open unit disk. However there are two limit functions, one for n even and one for n odd. That is, if q_n(z) = zdeg(p_n) p_n(1/z), then it looks like
  
  limit as n->infinity q_(2n)(z) = 1 + 3z + 4z2 + 7z3 + 13z4 + 19z5 + 29z6 + 43z7 + ...
  
  and
  
  limit as n->infinity q_(2n+1)(z) = 2 + 2z + 6z2 + 8z3 + 14z4 + 20z5 + 34z6 + 46z7 + ...
  
  In other words, the original polynomials are converging outside the unit disk. Proving this probably isn't impossible.
  
  [–] 1 point2 points  (1 child)
  
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  This is interesting, did you see what I wrote about the coefficients in the last edit? Those numbers look very similar to what I've been seeing. I'm trying to figure out what the correcting terms are during the convergence.
  
  [–] 0 points1 point  (0 children)
  
  sorry, this has been archived and can no longer be voted on
  
  Yes, and I think looking for recurrence relations is definitely a good place to start.
  
  At this point I'll have to use LaTeX, so if the following doesn't render make sure your greasemonkey plugin is working ok.
  
  If
  
      [; \lim_{n \to \infty} q_{2n}(z) = \sum_{k=0}^{\infty} a_k z^k ;]
  
  
  then it looks like
  
      [; q_{2n}(z) = \sum_{k=0}^{n} a_k z^k + O(z^{n+1}). ;]
  
  
  Similarly, if
  
      [; \lim_{n \to \infty} q_{2n+1}(z) = \sum_{k=0}^{\infty} b_k z^k ;]
  
  
  then it looks like
  
      [; q_{2n+1}(z) = \sum_{k=0}^{n} b_k z^k + O(z^{n+1}). ;]
  
  
  If you could show that the polynomials q_n(z) look like these expressions (i.e. that they stabilize at the front) then that would at least give you that the converge on the open unit disk (since the error would go to zero as n goes to infinity).
  
  [–] -1 points0 points  (1 child)
  
  sorry, this has been archived and can no longer be voted on
  
  Rounding error.
  
  [–] 5 points6 points  (0 children)
  
  sorry, this has been archived and can no longer be voted on
  
  No no, I mean, why the solutions are all approximately of magnitude 1?
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