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S Aug 12, 2016 at 20:23 history suggested Hans CC BY-SA 3.0
Better LaTeX script.
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S Aug 12, 2016 at 20:23
May 14, 2015 at 0:05 comment added David S. Newman Can the first factor of the product be (1-x-x^2-x^3-...) that is can all the coefficients be minus one and still satisfy the conditions for the series, that its coefficients be from {-1,0,+1}?
Nov 10, 2013 at 1:54 comment added David S. Newman @fedja You are correct that a product of the sort that you describe would suffice to answer my question. I searched for such products. If I remember correctly there were none with coefficients of the required sort that went beyond x^19 or thereabouts. This was a relatively easy computation. But the search space for what I am looking for is much larger.
Oct 31, 2013 at 6:42 comment added Aaron Meyerowitz @Fedja I don't see why that suffices, although I'm not saying that it is hard. However your product would include $\epsilon_1\epsilon_7+\epsilon_3\epsilon_5)x^8$ the coefficient being even must be zero so...$\epsilon_1\epsilon_3\epsilon_5\epsilon_7=-1.$ I could be wrong, but I think that similar considerations easily force $\epsilon_3=\epsilon_5=\cdots=\epsilon_{25}=-\epsilon_1$ but then $\epsilon_{27}=\epsilon_1$ and then one ends up with $2x^{28}.$
Oct 30, 2013 at 12:17 comment added Brendan McKay @Konstantinos: I don't think it makes a difference. Just negate the ones that don't.
Oct 30, 2013 at 12:13 comment added fedja It would suffice to find a product $\prod_{k\rm{\ odd}}(1+\varepsilon_k x^k)$ ($\varepsilon_k=\pm 1$) whose Taylor expansion has coefficients in $\{-1,0,1\}$. Is there any obvious reason why that is impossible?
Oct 30, 2013 at 10:22 comment added Konstantinos Gaitanas Must all the parentheses begin with +1?
Oct 30, 2013 at 5:18 review Suggested edits
Oct 30, 2013 at 7:48
Oct 30, 2013 at 3:55 history edited darij grinberg
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Oct 30, 2013 at 11:50
Oct 30, 2013 at 3:43 review First posts
Oct 30, 2013 at 8:34
Oct 30, 2013 at 3:28 history asked David S. Newman CC BY-SA 3.0