# $f$ real-rooted forbid truncated $\frac1f$ to be so?

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 $P_n(x)=\sum_{k=0}^na_kx^k$.

Question. Is it true that the polynomial $P_{2n}(x)$ has no real roots at all?

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 $r\neq\pm1$.

• Example: $1+x+x^2=0$ has no real roots; $1+x+x^2+x^3+x^4=0$ has no real roots. Did I miss something, Almaz? – T. Amdeberhan Mar 12 '17 at 16:06
• no sorry. It's all good. Please ignore my comment – almaz Mar 12 '17 at 16:06
• Okay, no problem. Thanks for checking. – T. Amdeberhan Mar 12 '17 at 16:07
• Ok. I am obviously missing something. What if f(x)=x – almaz Mar 12 '17 at 16:37
• @almaz there is no Taylor series in this case) Probably, it makes sense to write that the roots are non-zero. – Fedor Petrov Mar 12 '17 at 16:42

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.

• How does $a_{2n}\neq0$ imply non-real roots of the original $f$? BTW, I edited the question so that no $\alpha=0$. – T. Amdeberhan Mar 12 '17 at 17:09
• It is new $f(x)$, which is old $f(x)$ times $1-x/t$, where $t$ is a root in the original question. – Fedor Petrov Mar 12 '17 at 17:41
• So, Fedor, how does $a_{2n}\neq0$ imply the roots of $1/f$ are not real? – T. Amdeberhan Mar 12 '17 at 18:13
• Hm, if $P_{2n}(t)=0$, than $b_{2n}=0$, right? This is what I start with. – Fedor Petrov Mar 12 '17 at 18:17
• It would be helpful for the reader to explain that $[x^n]\frac1{f(x)(1-x/t)}$ denotes the coefficient of $x^n$ in the Taylor series of $\frac1{f(x)(1-x/t)}$. – GH from MO Mar 12 '17 at 18:57

Here is an alternative resolution motivated by Fedor's construction in his first line of argument.

Based on $f(0)=1$, it is clear that $\frac1{f(x)}=1+a_1x+a_2x^2+\cdots$ and now write $$f(x)P_{2n}(x)=f(x)(1+a_1x+a_2x^2+\cdots+a_{2n}x^{2n})= 1-a_{2n+1}x^{2n+1}+\cdots.$$ The RHS, call it $Q(x)$, is a non-constant polynomial.

More notably, observe that $Q(x)$ exhibits $2n$ (an even number of) vanishing consecutive coefficients. An application of Descartes' Rule of signs reveals $Q(x)$ has at least $2n$ non-real roots. This result is known as de Gua's Rule, which follows from Descartes' or Fourier-Budan Theorem. See the reference here, on page 28, Corollary 2.

Since $f(x)$ has only real roots, these non-real roots must come from the factor $P_{2n}(x)$ of $Q(x)$.

• @GHfromMO We do not need Descartes' Rule. If $P_{2n}$ has a real root, it has at least 2 real roots (hereafter: multiplicity included), then $fP_{2n}$ has at least $d+2$ real roots, where $d=\deg f$, $(fP_{2n})'$ has at least $d+1$ real roots, and at least $d$ non-zero real roots by Rolle. On the other hand, $(fP_{2n})'/x^{2n}$ has degree $d-1$. A contradiction. – Fedor Petrov Mar 12 '17 at 19:44
• It is particularly remarkable that Rolle's theorem is used in both proofs. – Fedor Petrov Mar 12 '17 at 19:48
• @FedorPetrov: Very nice! – GH from MO Mar 12 '17 at 19:53
• @GHfromMO How does Rolle work? It finds the root of derivative in the interval between consecutive roots. Among these intervals, all but one do not contain 0. – Fedor Petrov Mar 12 '17 at 22:20
• @GHfromMO: I've added some info in the answer to reassure your concern. – T. Amdeberhan Mar 12 '17 at 23:54