Taylor's polynomials and loss of real roots

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?

• Would you like to share your reasons to believe that this may be true? – Iosif Pinelis May 17 '19 at 15:20
• The reason is that I've computed this for some specific polynomials that I am working with, and it all came true. – T. Amdeberhan May 17 '19 at 15:27
• Is it known to be true when p has low degree? – Pat Devlin May 17 '19 at 19:57

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 $$|x|. The first observation is well-known and almost trivial. It is enough to consider $$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.
• 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! – Pietro Majer May 20 '19 at 14:47