Power series with double zeros 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. 
 A: 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. 
A: Some more examples with polynomials:
$$\matrix{\left( {z}^{6}+{z}^{5}-{z}^{3}+z+1 \right)  \left( z+{z}^{4}-1 \right) ^{2}\cr
 \left( {z}^{8}+{z}^{7}-{z}^{5}-{z}^{4}-{z}^{3}+z+1 \right)  \left( z+
{z}^{6}-1 \right) ^{2}\cr
 \left( {z}^{9}+{z}^{8}-{z}^{6}-{z}^{5}-{z}^{4}-{z}^{3}+z+1 \right) 
 \left( z+{z}^{7}-1 \right) ^{2}\cr
 \left( {z}^{4}-{z}^{3}+{z}^{2}-z+1 \right)  \left( {z}^{2}+{z}^{5}-1
 \right) ^{2}\cr
 \left( {z}^{6}-{z}^{5}+{z}^{4}-{z}^{3}+{z}^{2}-z+1 \right)  \left( {z
}^{2}+{z}^{7}-1 \right) ^{2}\cr
 \left( {z}^{6}-{z}^{5}+{z}^{3}-z+1 \right)  \left( {z}^{3}+{z}^{4}-1
 \right) ^{2}\cr
 \left( {z}^{7}-{z}^{5}+{z}^{4}+{z}^{3}-{z}^{2}+1 \right)  \left( {z}^
{3}+{z}^{5}-1 \right) ^{2}\cr
 \left( -{z}^{10}+{z}^{8}-{z}^{7}+{z}^{6}+{z}^{5}-2\;{z}^{4}+{z}^{3}-z
+1 \right)  \left( {z}^{3}+{z}^{7}-1 \right) ^{2}\cr
 \left( {z}^{4}-{z}^{3}+{z}^{2}-z+1 \right)  \left( {z}^{4}+{z}^{5}-1
 \right) ^{2}\cr
 \left( {z}^{4}-{z}^{2}+1 \right)  \left( {z}^{4}+{z}^{6}-1 \right) ^{
2}\cr
 \left( {z}^{6}-{z}^{5}+{z}^{4}-{z}^{3}+{z}^{2}-z+1 \right)  \left( {z
}^{4}+{z}^{7}-1 \right) ^{2}\cr
 \left( {z}^{8}-{z}^{7}+{z}^{5}-{z}^{4}+{z}^{3}-z+1 \right)  \left( {z
}^{5}+{z}^{6}-1 \right) ^{2}\cr
 \left( {z}^{6}-{z}^{5}+{z}^{4}-{z}^{3}+{z}^{2}-z+1 \right)  \left( {z
}^{6}+{z}^{7}-1 \right) ^{2}\cr
\left( -{z}^{14}-{z}^{13}-2\;{z}^{12}-{z}^{11}+{z}^{9}+2\;{z}^{8}-2\;{z}^{5}+2\;{z}^{2}+z+1 \right)  \left( {z}^{5}-{z}^{3}+{z}^{2}+z-1
 \right) ^{2}\cr
 \left( {z}^{5}+{z}^{4}-{z}^{3}-{z}^{2}+z+1 \right)  \left( {z}^{5}+{z
}^{3}-{z}^{2}+z-1 \right) ^{2}\cr
 \left( {z}^{15}+{z}^{14}-{z}^{11}-{z}^{10}+{z}^{9}+{z}^{8}+{z}^{7}+{z
}^{6}-{z}^{5}-{z}^{4}+z+1 \right)  \left( {z}^{5}-{z}^{4}+{z}^{3}+z-1
 \right) ^{2}\cr
}$$
A: 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...
