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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Taylor's polynomials and loss of real roots
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.


