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Jan 23, 2019 at 23:05 comment added Zach Teitler Right, I forgot about that $1/C$. Anyway the obvious source of extra decompositions is by scaling each $x_i$ on the right hand side by some factor $t_i$, so you end up with $(t_1 x_1)^{k_1} \dotsm (t_n x_n)^{k_n}$; as long as the $t$'s multiply out to $1$, you got a different decomposition. I haven't thought about how to get $t$'s that minimize the $c_i$'s (or $\sum |c_i|$), hmm. And unfortunately there are other decompositions, these scalings don't hit all possible decompositions. I'm not sure how to describe the other ones, let alone minimize them. Interesting question.
Jan 23, 2019 at 22:59 comment added RBega2 @ZachTeitler One question about how you go the bounds on $c_i$ from the article you linked to. If I am understanding equation (8) correctly, then there is a coefficient $1/C$ that should be pretty large, possibly making the bounds one gets on on $c_i$ smaller than what you wrote.
Jan 23, 2019 at 22:51 comment added RBega2 @ZachTeitler Yes I mean monomial--this is very far from my area of research so I am a bit out of my element even with terminology. Thank you for the reference! It certainly looks like the sort of result I am interested in.
Jan 23, 2019 at 22:37 comment added Zach Teitler When you say $P = x_1^{k_1} \dotsm x_n^{k_n}$ is a homogeneous polynomial, do you actually intend that $P$ is a monomial? If so, then you might wish to look at arxiv.org/abs/1201.2922 (sorry for self-promoting :( ). The paper gives a simple explicit decomposition of $P$ in which each $\mathbf{v}_i$ has roots of unity as entries, so I guess length $\sqrt{n}$, and each $c_i$ is also a root of unity; if you scale to use unit vectors then I guess $|c_i|=n^{k/2}$ if I'm understanding right. And $N = \left(\prod(k_i+1)\right)/(\min\{k_i+1\})$. The decompositions are not unique though.
Jan 23, 2019 at 21:41 history asked RBega2 CC BY-SA 4.0