Let $f(x)$ be a polynomial in the ring $\mathbb{R}[x]$, the roots are all real and $f(0)=1$. Write the Taylor series of $1/f(x)$ around the origin as
$$\frac1{f(x)}=\sum\_{k=0}^{\infty}a\_kx^k,$$
and denote $P\_n(x)=\sum\_{k=0}^na\_kx^k$.



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> **Question.** Is it true that the polynomial $P\_{2n}(x)$ has no real roots at all?
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*A prototypical example:* If we choose $f(x)=1-x$ then $P\_{2n}(x)=\sum\_{k=0}^{2n}x^k=\frac{1-x^{2n+1}}{1-x}$ has all roots $r$ on the unit circle and $r\neq\pm1$. 



$f$ real-rooted forbid truncated $\frac1f$ to be so?
We use a standard notation $[x^n]h(x)$ for a coefficient of $x^n$ in the series of $h(x)$. 


Assume that $t$ is a real root of $P\_n$. First of all, note that $$0=a\_0+a\_1t+\dots+a\_nt^n=t^n[x^n]\frac1{f(x)(1-x/t)},$$
thus for $g(x)=f(x)(1-x/t)$ we just need to prove that all coefficients $b\_{2n}=[x^{2n}]1/g(x)$ are non-zero. We may write $g(x)=\prod\_{i=1}^m(1-\alpha\_ix)$, suppose for a moment that $\alpha$'s are different. Then $1/g=\sum c\_i/(1-\alpha\_ix)$, where $c\_i=\alpha\_i^{m-1}/\prod\_{j\ne i}(\alpha\_i-\alpha\_j)$. Therefore $$b\_{2n}=\sum\_{i=1}^m\frac{\alpha\_i^{m-1+2n}}{\prod\_{j\ne i}(\alpha\_i-\alpha\_j)}.$$
It is a Schur's function corresponding to a partition $(2n,0,\dots,0)$, and this fact holds true even when some $\alpha$'s coincide (by continuity). I believe it should be well known that the value of this Schur function is always positive (when not all $\alpha$'s are equal to 0). Below is some proof.


We induct on $m$, base $m=1$ is clear. Again assume that $\alpha$'s are different. Note that $b\_{2n}$ is a coefficient of $t^{m-1}$ in the polynomial $h(t)$ of degree at most $m-1$ satisfying $h(\alpha\_i)=H(\alpha\_i)$, where $H(\alpha)=\alpha^{m-1+2n}$. By Rolle's theorem for the function $h-H$ there exist $\beta\_1,\dots,\beta\_{m-1}$ between consecutive $\alpha$'s for which $h'(\beta\_i)=H'(\beta\_i)$. Therefore for $m$ distinct $\alpha$'s we may find $m-1$ distinct $\beta$'s, and this procedure allows a limit case in which some $\alpha's$ coincide (in this case some $\beta$'s also may coincide, but not all of them are equal to 0.) Thus we reduce $m$ to $m-1$ and may induct.


