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How to prove (or to disprove) that all the roots of the polynomial of degree $n$ $$\sum_{k=0}^{k=n}(2k+1)x^k$$ belong to the disk $\{z:|z|<1\}?$ Numerical calculations confirm that, but I don't see any approach to a proof of so simply formulated statement. It would be useful in connection with an irreducibility problem.

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Let $f$ denote your polynomial. Roots of the polynomial $$g(x)=\sum_{k=0}^nx^{2k+1}=x\frac{x^{2(n+1)}-1}{x^2-1}$$ are $0$ and roots of unity. By the Gauss–Lucas theorem, the roots of $g'(x)=f(x^2)$ lie in their convex hull, and a fortiori in the disk $\{z:|z|\le1\}$. In order to get a strict inequality, it suffices to show that $g$ is square-free.

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    $\begingroup$ Could you explain a strict inequality in detail? $\endgroup$ – user64494 Jun 27 '13 at 16:03
  • $\begingroup$ It is clear that $g(x)$ is square-free, since its roots are $0$ and the $(2n+2)$-th roots of unity except $\pm 1$, each of multiplicity $1$. $\endgroup$ – GH from MO Jun 27 '13 at 16:08
  • $\begingroup$ @user64494: If $x$ is simultaneously the root of a polynomial and its derivative, then $x$ is a root with multiplicity at least $2$. $\endgroup$ – GH from MO Jun 27 '13 at 16:09
  • $\begingroup$ @GHfromMO: Yes, perhaps I should have written “observe” rather than “show”. @ user64494: The convex hull of the roots of $g$ is a polygon whose vertices have norm $1$, hence all its elements have norm $< 1$ except for the vertices. Now, the vertices are roots of $g$, which cannot be roots of $g'$ as $g$ is square-free. $\endgroup$ – Emil Jeřábek Jun 27 '13 at 16:11
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The idea is taken from this other question Polynomial with the primes as coefficients irreducible?

Show instead that $f(1/x)$ has all roots lying outside of the unit disk. For that, multiply by $(x-1)$ and equate to $0$ obtaining:

$$x^{k+1}+\sum_1^k 2x^j=2k+1$$ Take absolute values and apply the triangular inequality and one obtains:

$$|x^{k+1}|+\sum_1^k 2|x^j|\geq \left|x^{k+1}+\sum_1^k 2x^j\right|=2k+1$$ This is clearly non possible if $|x|<1$. Moreover, if $|x|=1$ there is an equality which means that all the terms are aligned. In particular $2x^2/2x$ is real, so the only possibility is $x=1,-1$. But neither is a root of $f(1/x)$ so you are done.

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The standard approach to this type of question is to use the Schur-Cohn procedure.

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