# Can we permute the coefficients of a polynomial so that it has NO real roots?

Let $P(x)=a_{2n}x^{2n}+a_{2n-1}x^{2n-1}+\ldots+a_{0}$ be an even degree polynomial with positive coefficients.

Is it possible to permute the coefficients of $P(x)$ so that the resulting polynomial will have NO real roots.

• In general, I really don't think that anything (non-trivial) can be said about permutations of the coefficients of a polynomial (I would be quite interested to see any interesting statement that says otherwise). – Captain Lama Mar 27 '16 at 18:33
• Lol I meant NO real roots. – Joshua Benabou Mar 27 '16 at 18:45
• It's possible for degree $2$. – joriki Mar 27 '16 at 18:53
• BTW, using \ldots instead of ... results in proper spacing (and anyway in this case \cdots would look better). – joriki Mar 27 '16 at 18:55
• Good. That means there is a permutation of the coefficients with NO real roots. :) – Joshua Benabou Mar 27 '16 at 19:14

Yes: put the $n+1$ largest coefficients on the even powers of $x$, and the $n$ smallest coefficients on the odd powers of $x$.
Clearly the polynomial will have no nonnegative roots regardless of the permutation. Changing $x$ to $-x$, it suffices to show: if $\min\{a_{2k}\} \ge \max\{a_{2k+1}\}$, then when $x>0$,$$a_{2n}x^{2n} - a_{2n-1}x^{2n-1} + \cdots + a_2x^2 -a_1x+a_0$$is always positive.
• If $x\ge1$, this follows from $$(a_{2n}x^{2n} - a_{2n-1}x^{2n-1}) + \cdots + (a_2x^2 -a_1x) +a_0 \ge 0 + \cdots + 0 + a_0 > 0.$$
• If $0<x\le1$, this follows from \begin{multline*} (a_0 - a_1x) + (a_2x^2-a_3x^3) + \cdots + (a_{2n-2}x^{2n-2}-a_{2n-1}x^{2n-1}) + a_{2n}x^{2n} \\ \ge 0 + \cdots + 0 + a_{2n}x^{2n} > 0. \end{multline*}
• For power series, the answer will depend on the radius of convergence of the power series (for example, if that radius of convergence is at most $1$, then putting the coefficients in decreasing order works). But the radius of convergence can depend upon the permutation of the coefficients...! – Greg Martin Mar 27 '16 at 21:02
• Very nice. I noticed that for a quadratic $A x^2+B x+C$ with positive $A,B,C,$ we cannot have $A^2\geq 4 B C$ and $B^2\geq 4 C A$ and $C^2\geq 4 A B ,$ else, wlog, $A=\min (A,B,C)$ and $A^2\geq 4 B C\geq 4 A^2.$ – DanielWainfleet Mar 27 '16 at 21:51
• @YoTengoUnLCD . This seems like a good Q. With a sub-question:If $f:N_0\to N_0$ is a bijection and $\sum_0^{\infty}a_n z^n$ is an entire function, what is the radius of convergence of $\sum_n a_{f(n)}z^n$ ? – DanielWainfleet Mar 27 '16 at 22:20
• @user254665. Given a numerable (infinite) series $a_n$, is not sure that any bijective function on the index that might exists with the infinite domain $N_0$ is producing (in its output $a_{f(n)}$) the whole set given by $a_n$. Think the analogy in the way is possible to enumerate the set of rational numbers $\mathbb{Q}$, but using an not carefully designed numeration despite it might be giving an infinite quantity of only elements of $\mathbb{Q}$ but even many of the infinite series in $\mathbb{Q}$ are not going to enumerate $\mathbb{Q}$ but an infinite infinitesimal of it – Marco Munari Mar 28 '16 at 20:27