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.

4$\begingroup$ 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

1$\begingroup$ @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

4$\begingroup$ @David Speyer Thanks. LLL is a good idea. $(\sqrt51)/2$ is too small but the next one works quickly: $1xx^2+x^4x^5+x^7+x^8 = (1x+x^2) (1x^2x^3)^2$. $\endgroup$ – Noam D. Elkies Aug 30 '12 at 15:19

$\begingroup$ The only other example with degree $\le 10$ is $1z{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

2$\begingroup$ 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 $p(z)=z^3+z^21$ we win just by Dirichlet. $\endgroup$ – fedja Aug 31 '12 at 1:11
At least the settheoretical 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(FINJVSMS); Solomyak, Boris(1WA) Zeros of {−1,0,1} power series and connectedness loci for selfaffine 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:
$$\matrix{\left( {z}^{6}+{z}^{5}{z}^{3}+z+1 \right) \left( z+{z}^{4}1 \right) ^{2}\cr \left( {z}^{8}+{z}^{7}{z}^{5}{z}^{4}{z}^{3}+z+1 \right) \left( z+ {z}^{6}1 \right) ^{2}\cr \left( {z}^{9}+{z}^{8}{z}^{6}{z}^{5}{z}^{4}{z}^{3}+z+1 \right) \left( z+{z}^{7}1 \right) ^{2}\cr \left( {z}^{4}{z}^{3}+{z}^{2}z+1 \right) \left( {z}^{2}+{z}^{5}1 \right) ^{2}\cr \left( {z}^{6}{z}^{5}+{z}^{4}{z}^{3}+{z}^{2}z+1 \right) \left( {z }^{2}+{z}^{7}1 \right) ^{2}\cr \left( {z}^{6}{z}^{5}+{z}^{3}z+1 \right) \left( {z}^{3}+{z}^{4}1 \right) ^{2}\cr \left( {z}^{7}{z}^{5}+{z}^{4}+{z}^{3}{z}^{2}+1 \right) \left( {z}^ {3}+{z}^{5}1 \right) ^{2}\cr \left( {z}^{10}+{z}^{8}{z}^{7}+{z}^{6}+{z}^{5}2\;{z}^{4}+{z}^{3}z +1 \right) \left( {z}^{3}+{z}^{7}1 \right) ^{2}\cr \left( {z}^{4}{z}^{3}+{z}^{2}z+1 \right) \left( {z}^{4}+{z}^{5}1 \right) ^{2}\cr \left( {z}^{4}{z}^{2}+1 \right) \left( {z}^{4}+{z}^{6}1 \right) ^{ 2}\cr \left( {z}^{6}{z}^{5}+{z}^{4}{z}^{3}+{z}^{2}z+1 \right) \left( {z }^{4}+{z}^{7}1 \right) ^{2}\cr \left( {z}^{8}{z}^{7}+{z}^{5}{z}^{4}+{z}^{3}z+1 \right) \left( {z }^{5}+{z}^{6}1 \right) ^{2}\cr \left( {z}^{6}{z}^{5}+{z}^{4}{z}^{3}+{z}^{2}z+1 \right) \left( {z }^{6}+{z}^{7}1 \right) ^{2}\cr \left( {z}^{14}{z}^{13}2\;{z}^{12}{z}^{11}+{z}^{9}+2\;{z}^{8}2\;{z}^{5}+2\;{z}^{2}+z+1 \right) \left( {z}^{5}{z}^{3}+{z}^{2}+z1 \right) ^{2}\cr \left( {z}^{5}+{z}^{4}{z}^{3}{z}^{2}+z+1 \right) \left( {z}^{5}+{z }^{3}{z}^{2}+z1 \right) ^{2}\cr \left( {z}^{15}+{z}^{14}{z}^{11}{z}^{10}+{z}^{9}+{z}^{8}+{z}^{7}+{z }^{6}{z}^{5}{z}^{4}+z+1 \right) \left( {z}^{5}{z}^{4}+{z}^{3}+z1 \right) ^{2}\cr }$$

$\begingroup$ 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: $p(y) = (1x+x^2) (1x^2x^3)^2$ where $x=y^n$. $\endgroup$ – Marc Chamberland Aug 31 '12 at 2:10

1$\begingroup$ LLL finally turned up a few triple roots, namely multiples of $(1x^2x^3)^3$ 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

1$\begingroup$ 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