How many power series of the form $1+\sum_{k=1}^{\infty} a_{k}x^{k}$ with $a_{k}\in \{-1,0,1 \}$, that have a double zero $f(x)=f'(x)=0$ in $(0,1)$, are there. Ok, there are many ways to understand the question: set theoretical, topological, measure theoretical. I would be especially interested in the Bernoulli measures of the coefficient space $C\subseteq \{-1,0,1\}^{\mathbb{N}}$ of such series.

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    $\begingroup$ Is it known whether there is a polynomial of this kind with a double zero in $(0,1)$? $\endgroup$ – Noam D. Elkies Aug 30 '12 at 1:05
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    $\begingroup$ @Noam Nice question! I gave it a quick stab, looking for a double root at $(\sqrt{5}-1)/2$. The LLL found no solutions with degree $\leq 30$. $\endgroup$ – David E Speyer Aug 30 '12 at 14:45
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    $\begingroup$ @David Speyer Thanks. LLL is a good idea. $(\sqrt5-1)/2$ is too small but the next one works quickly: $1-x-x^2+x^4-x^5+x^7+x^8 = (1-x+x^2) (1-x^2-x^3)^2$. $\endgroup$ – Noam D. Elkies Aug 30 '12 at 15:19
  • $\begingroup$ The only other example with degree $\le 10$ is $1-z-{z}^{3}-{z}^{5}+{z}^{6}+{z}^{8}+{z}^{9}+{z}^{10}= \left( {z}^{2}+1 \right) \left( {z}^{2}-z+1 \right) \left( {z}^{3}+{z}^{2}-1 \right) ^{2} $ $\endgroup$ – Robert Israel Aug 30 '12 at 17:54
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    $\begingroup$ LLL??????? Take a monic $p\in\mathbb Z[x]$. Consider $P(x)=\sum_{k=1}^n a_kx^k$ with $a_k=0,\pm 1$. Look at its remainder modulo $p^2$ in $\mathbb Z[x]$. Its coefficients are of the kind $\sum_z \sum_k a_k(c_z z^k+d_zkz^k)$ where $z$ runs over the roots of $p$ (interpolation). We need just to have $\sum_k a_k z^k$ and $\sum_k a_k kz^k$ less than fixed $\delta>0$ for all $z$ to get zero remainder. $\{-1,0,1\}$ is the difference set of $\{0,1\}$. We have $2^n$ options and $Cn\max(|z|,1)^{2n}$ boxes for each $z$ plus conjugation symmetry, so for $p(z)=z^3+z^2-1$ we win just by Dirichlet. $\endgroup$ – fedja Aug 31 '12 at 1:11

At least the set-theoretical question can be answered: the are the cardinality of the continuum many such series, as can be deduced from the results in this paper (not all of them attributed by the authors to themselves):

MR2293600 (2007k:30003) Reviewed Shmerkin, Pablo(FIN-JVS-MS); Solomyak, Boris(1-WA) Zeros of {−1,0,1} power series and connectedness loci for self-affine sets. (English summary) Experiment. Math. 15 (2006), no. 4, 499–511.

  • $\begingroup$ I would guess that you probably mean, more speciically, continuum many? $\endgroup$ – Joel David Hamkins Aug 30 '12 at 2:06
  • $\begingroup$ @Joel: Yes, that is a more precise statement, will fix. $\endgroup$ – Igor Rivin Aug 30 '12 at 2:19
  • $\begingroup$ Thank You Igor Rivin. I have thought so. But the measuretheortical question remains open, so I do not click the question as answered. I guess the measure of the set is small (0?) but I have no idea to prove this. $\endgroup$ – Jörg Neunhäuserer Aug 30 '12 at 13:09

Some more examples with polynomials:

$$\matrix{\left( {z}^{6}+{z}^{5}-{z}^{3}+z+1 \right) \left( z+{z}^{4}-1 \right) ^{2}\cr \left( {z}^{8}+{z}^{7}-{z}^{5}-{z}^{4}-{z}^{3}+z+1 \right) \left( z+ {z}^{6}-1 \right) ^{2}\cr \left( {z}^{9}+{z}^{8}-{z}^{6}-{z}^{5}-{z}^{4}-{z}^{3}+z+1 \right) \left( z+{z}^{7}-1 \right) ^{2}\cr \left( {z}^{4}-{z}^{3}+{z}^{2}-z+1 \right) \left( {z}^{2}+{z}^{5}-1 \right) ^{2}\cr \left( {z}^{6}-{z}^{5}+{z}^{4}-{z}^{3}+{z}^{2}-z+1 \right) \left( {z }^{2}+{z}^{7}-1 \right) ^{2}\cr \left( {z}^{6}-{z}^{5}+{z}^{3}-z+1 \right) \left( {z}^{3}+{z}^{4}-1 \right) ^{2}\cr \left( {z}^{7}-{z}^{5}+{z}^{4}+{z}^{3}-{z}^{2}+1 \right) \left( {z}^ {3}+{z}^{5}-1 \right) ^{2}\cr \left( -{z}^{10}+{z}^{8}-{z}^{7}+{z}^{6}+{z}^{5}-2\;{z}^{4}+{z}^{3}-z +1 \right) \left( {z}^{3}+{z}^{7}-1 \right) ^{2}\cr \left( {z}^{4}-{z}^{3}+{z}^{2}-z+1 \right) \left( {z}^{4}+{z}^{5}-1 \right) ^{2}\cr \left( {z}^{4}-{z}^{2}+1 \right) \left( {z}^{4}+{z}^{6}-1 \right) ^{ 2}\cr \left( {z}^{6}-{z}^{5}+{z}^{4}-{z}^{3}+{z}^{2}-z+1 \right) \left( {z }^{4}+{z}^{7}-1 \right) ^{2}\cr \left( {z}^{8}-{z}^{7}+{z}^{5}-{z}^{4}+{z}^{3}-z+1 \right) \left( {z }^{5}+{z}^{6}-1 \right) ^{2}\cr \left( {z}^{6}-{z}^{5}+{z}^{4}-{z}^{3}+{z}^{2}-z+1 \right) \left( {z }^{6}+{z}^{7}-1 \right) ^{2}\cr \left( -{z}^{14}-{z}^{13}-2\;{z}^{12}-{z}^{11}+{z}^{9}+2\;{z}^{8}-2\;{z}^{5}+2\;{z}^{2}+z+1 \right) \left( {z}^{5}-{z}^{3}+{z}^{2}+z-1 \right) ^{2}\cr \left( {z}^{5}+{z}^{4}-{z}^{3}-{z}^{2}+z+1 \right) \left( {z}^{5}+{z }^{3}-{z}^{2}+z-1 \right) ^{2}\cr \left( {z}^{15}+{z}^{14}-{z}^{11}-{z}^{10}+{z}^{9}+{z}^{8}+{z}^{7}+{z }^{6}-{z}^{5}-{z}^{4}+z+1 \right) \left( {z}^{5}-{z}^{4}+{z}^{3}+z-1 \right) ^{2}\cr }$$

  • $\begingroup$ With LLL or some other method? Any luck finding a triple root? $\endgroup$ – Noam D. Elkies Aug 31 '12 at 0:08
  • $\begingroup$ See my remark above plus en.wikipedia.org/wiki/Lehmer%27s_conjecture $\endgroup$ – fedja Aug 31 '12 at 1:21
  • $\begingroup$ @Noam Modify your earlier result to get as many roots as you want: $p(y) = (1-x+x^2) (1-x^2-x^3)^2$ where $x=y^n$. $\endgroup$ – Marc Chamberland Aug 31 '12 at 2:10
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    $\begingroup$ LLL finally turned up a few triple roots, namely multiples of $(1-x^2-x^3)^3$ such as $$ 1 - x - x^2 - x^3 + x^4 + x^5 + x^6 + x^8 - x^9 + x^{10} - x^{11} - x^{14} - x^{16} + x^{17} - x^{18} - x^{19} - x^{21} - x^{22} \phantom{etc} $$ $$ \phantom{etc} - x^{23} - x^{25} + x^{26} + x^{27} - x^{28} - x^{30} - x^{31} - x^{32} - x^{33} - x^{34} - x^{35} - x^{36} - x^{37} + x^{38} - x^{40} - x^{41}. $$ $\endgroup$ – Noam D. Elkies Aug 31 '12 at 12:31
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    $\begingroup$ What I was using was integer linear programming. Given polynomial $p(z)$ of degree $m$, find integers $a_1,\ldots,a_n$ that satisfy the constraints that the coefficients of each power of $z$ in $p(z)(1 + a_1 z + \ldots + a_n z^n$ are all $\ge -1$ and $\le 1$. $\endgroup$ – Robert Israel Aug 31 '12 at 17:10

I am going to address the question for $\mathrm{Bernoulli}(1/2)$ measures, using probabilistic language. This is not a complete answer, but I am trying to relate your question to the properties of the distribution of $f(x)$. Clearly, for $x<1/2$ we never even reach zero, but my guess is that for $x>1/2$ this distribution is absolutely continuous, though I am unable to prove this at the moment.

So formally, at least,

$$\displaystyle \mathsf{E} \, \sum_{f(x)=0} \mathsf{1}\{|f^\prime(x)| < \epsilon\} = \intop_0^1 \mathsf{E} \, \delta(f(x)) \mathsf{1}\{|f(x)|<\epsilon\} |f^\prime(x)| dx \le \epsilon \intop_0^1 \mathsf{E} \, \delta(f(x)) dx.$$

$\mathsf{E} \, \delta$ is the density at zero, and it can be made perfect sense of, provided that the law of $f(x)$ has continuous density at zero. I don't know whether it has continuous density, but if we manage to prove that $f(x)$ has at least bounded density for $x>1/2$, then we can write inequalities with approximations of $\delta$ to get the same results...

  • $\begingroup$ Dear Alexander Shamov, thansk for Your idea, but i do not see clear in the moment if it works this way... Best Jörg Neunhäuserer $\endgroup$ – Jörg Neunhäuserer Sep 2 '12 at 20:12

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