Power series with double zeros - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T18:37:09Z http://mathoverflow.net/feeds/question/105899 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105899/power-series-with-double-zeros Power series with double zeros Jörg Neunhäuserer 2012-08-30T00:51:22Z 2012-08-31T02:25:36Z <p>How many power series of the form $1+\sum_{k=1}^{\infty} a_{k}x^{k}$ with <code>$a_{k}\in \{-1,0,1 \}$</code>, 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 <code>$C\subseteq \{-1,0,1\}^{\mathbb{N}}$</code> of such series. </p> http://mathoverflow.net/questions/105899/power-series-with-double-zeros/105902#105902 Answer by Igor Rivin for Power series with double zeros Igor Rivin 2012-08-30T01:39:21Z 2012-08-30T02:20:13Z <p>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):</p> <p>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. </p> http://mathoverflow.net/questions/105899/power-series-with-double-zeros/105953#105953 Answer by Alexander Shamov for Power series with double zeros Alexander Shamov 2012-08-30T15:08:59Z 2012-08-30T17:02:22Z <p>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 $x&lt;1/2$ 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.</p> <p>So formally, at least,</p> <p>$\displaystyle \mathsf{E} \, \sum_{f(x)=0} \mathsf{1}\{|f^\prime(x)| &lt; \epsilon\} = \intop_0^1 \mathsf{E} \, \delta(f(x)) \mathsf{1}\{|f(x)|&lt;\epsilon\} |f^\prime(x)| dx \le \epsilon \intop_0^1 \mathsf{E} \, \delta(f(x)) dx$.</p> <p>$\mathsf{E} \, \delta$ 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...</p> http://mathoverflow.net/questions/105899/power-series-with-double-zeros/106007#106007 Answer by Robert Israel for Power series with double zeros Robert Israel 2012-08-30T23:56:20Z 2012-08-30T23:56:20Z <p>Some more examples with polynomials:</p> <p>$$\matrix{\left( {z}^{6}+{z}^{5}-{z}^{3}+z+1 \right) \left( z+{z}^{4}-1 \right) ^{2}\cr \left( {z}^{8}+{z}^{7}-{z}^{5}-{z}^{4}-{z}^{3}+z+1 \right) \left( z+ {z}^{6}-1 \right) ^{2}\cr \left( {z}^{9}+{z}^{8}-{z}^{6}-{z}^{5}-{z}^{4}-{z}^{3}+z+1 \right) \left( z+{z}^{7}-1 \right) ^{2}\cr \left( {z}^{4}-{z}^{3}+{z}^{2}-z+1 \right) \left( {z}^{2}+{z}^{5}-1 \right) ^{2}\cr \left( {z}^{6}-{z}^{5}+{z}^{4}-{z}^{3}+{z}^{2}-z+1 \right) \left( {z }^{2}+{z}^{7}-1 \right) ^{2}\cr \left( {z}^{6}-{z}^{5}+{z}^{3}-z+1 \right) \left( {z}^{3}+{z}^{4}-1 \right) ^{2}\cr \left( {z}^{7}-{z}^{5}+{z}^{4}+{z}^{3}-{z}^{2}+1 \right) \left( {z}^ {3}+{z}^{5}-1 \right) ^{2}\cr \left( -{z}^{10}+{z}^{8}-{z}^{7}+{z}^{6}+{z}^{5}-2\;{z}^{4}+{z}^{3}-z +1 \right) \left( {z}^{3}+{z}^{7}-1 \right) ^{2}\cr \left( {z}^{4}-{z}^{3}+{z}^{2}-z+1 \right) \left( {z}^{4}+{z}^{5}-1 \right) ^{2}\cr \left( {z}^{4}-{z}^{2}+1 \right) \left( {z}^{4}+{z}^{6}-1 \right) ^{ 2}\cr \left( {z}^{6}-{z}^{5}+{z}^{4}-{z}^{3}+{z}^{2}-z+1 \right) \left( {z }^{4}+{z}^{7}-1 \right) ^{2}\cr \left( {z}^{8}-{z}^{7}+{z}^{5}-{z}^{4}+{z}^{3}-z+1 \right) \left( {z }^{5}+{z}^{6}-1 \right) ^{2}\cr \left( {z}^{6}-{z}^{5}+{z}^{4}-{z}^{3}+{z}^{2}-z+1 \right) \left( {z }^{6}+{z}^{7}-1 \right) ^{2}\cr \left( -{z}^{14}-{z}^{13}-2\;{z}^{12}-{z}^{11}+{z}^{9}+2\;{z}^{8}-2\;{z}^{5}+2\;{z}^{2}+z+1 \right) \left( {z}^{5}-{z}^{3}+{z}^{2}+z-1 \right) ^{2}\cr \left( {z}^{5}+{z}^{4}-{z}^{3}-{z}^{2}+z+1 \right) \left( {z}^{5}+{z }^{3}-{z}^{2}+z-1 \right) ^{2}\cr \left( {z}^{15}+{z}^{14}-{z}^{11}-{z}^{10}+{z}^{9}+{z}^{8}+{z}^{7}+{z }^{6}-{z}^{5}-{z}^{4}+z+1 \right) \left( {z}^{5}-{z}^{4}+{z}^{3}+z-1 \right) ^{2}\cr }$$</p>