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Real-rootedness, log-concavity, and unimodality are intertwined properties. It's in this light that I was prompted to ask the question below.

Suppose the roots of a polynomial $p(x)$ are all real and $p(0)>0$. Fix an integer $k\geq0$ and consider the function $f=\frac1{p^{2k+1}}$. I like to consider the partial sums (polynomial) $$\sum_{j=0}^{2k}\frac{f^{(j)}(0)}{j!}x^j; \tag1$$ where $f^{(j)}$ means the $j$-th derivative.

QUESTION. Is it true that the polynomial in (1) has no real roots?

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    $\begingroup$ Would you like to share your reasons to believe that this may be true? $\endgroup$ – Iosif Pinelis May 17 '19 at 15:20
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    $\begingroup$ The reason is that I've computed this for some specific polynomials that I am working with, and it all came true. $\endgroup$ – T. Amdeberhan May 17 '19 at 15:27
  • $\begingroup$ Is it known to be true when p has low degree? $\endgroup$ – Pat Devlin May 17 '19 at 19:57
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It immediately follows from the observations that every even order Taylor polynomial of $e^{ax}$ is strictly positive for any $a\in\mathbb R$ and that $\frac 1{b-x}=\int_0^\infty e^{ax}e^{-ab}\,da$ and $\frac 1{b+x}=\int_0^\infty e^{-ax}e^{-ab}\,da$ for $b>0$ and $|x|<b$. The first observation is well-known and almost trivial. It is enough to consider $a=1$. Let $P_n$ be the Taylor polynomial of $e^x$. Since all coefficients are positive, the real zeroes (if they exist) must be negative. But if $z$ is a negative real zero closest to the origin, we have $P'(z)=P(z)-\frac {z^n}{n!}=-\frac{z^n}{n!}<0$, so there must be a closer to the origin zero - a contradiction. The power $2k+1$ on $p$ is just a red herring. Any polynomial with all zeroes real will work.

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  • $\begingroup$ btw, for negative $x$, we also have $P_n(x)>e^x>0$ because of the Lagrange remainder formula. But I like a lot the ode argument, that seems to be applicable to solutions to many linear ode. Thank you! $\endgroup$ – Pietro Majer May 20 '19 at 14:47
  • $\begingroup$ @PietroMajer Indeed, good point! $\endgroup$ – fedja May 20 '19 at 22:48

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