When does $\sum_{k=0}^M a_k x^{2k}$, $a_k \in \mathbb{R}$, $M \in \mathbb{N}$ have no roots in $[0,1]$? There is nothing special about the $1$, the question can be generalized to $[0,c]$ but that would add potential unnecessary complications.

Obviously since if $a_0 = 0$ we always have a root at $x=0$ that is a necessary condition but it is also obviously not sufficient.

A really "overpowered" sufficient condition can easily be found, such as $|a_0| - |\sum_{k\neq0} a_k| > 0$ but this is obviously not necessary.

I'm looking ideally for necessary and sufficient conditions but also realize this may be a difficult question, so any better ideas for necessary (but not sufficient) or sufficient (but not necessary) conditions on the $a_k$ would also be of interest to me.

  • $\begingroup$ Depending on your purpose, you may find Sturm's theorem useful. $\endgroup$ – jgon May 20 at 19:27
  • $\begingroup$ It is my understanding of Sturm's theorem that it would be very useful for the case with explicit order $M$ and explicit coefficients $a_k$ but that it won't provide me with general conditions on said coefficients if I don't wish to specify what they are? $\endgroup$ – MM8 May 20 at 19:33
  • $\begingroup$ yes, if the $a_k$ are variables, it would be very difficult to apply Sturm's theorem. I wasn't entirely sure of your purpose, and thus my comment. $\endgroup$ – jgon May 20 at 19:38
  • $\begingroup$ I did check Sturm's theorem before so your intuition was not far off at all. :) Well, they are not variables per-se, they are constants of course since this is supposed to be a polynomial in each case. But I am interested in conditions for the general case, so I can't use explicit values for the coefficients, making Euclidean division and thus Sturm's theorem not really doable (at least not that I can see). $\endgroup$ – MM8 May 20 at 19:41

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