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.

  • $\begingroup$ 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). $\endgroup$ – Captain Lama Mar 27 '16 at 18:33
  • $\begingroup$ Lol I meant NO real roots. $\endgroup$ – Joshua Benabou Mar 27 '16 at 18:45
  • 2
    $\begingroup$ It's possible for degree $2$. $\endgroup$ – joriki Mar 27 '16 at 18:53
  • $\begingroup$ BTW, using \ldots instead of ... results in proper spacing (and anyway in this case \cdots would look better). $\endgroup$ – joriki Mar 27 '16 at 18:55
  • 1
    $\begingroup$ Good. That means there is a permutation of the coefficients with NO real roots. :) $\endgroup$ – 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*}
| cite | improve this answer | |
  • $\begingroup$ Do you have any ideas for the case of infinite polynomials, i.e power series? $\endgroup$ – YoTengoUnLCD Mar 27 '16 at 20:56
  • 1
    $\begingroup$ 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...! $\endgroup$ – Greg Martin Mar 27 '16 at 21:02
  • $\begingroup$ 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 $A^2\geq 4 B C\geq 4 A^2.$ $\endgroup$ – DanielWainfleet Mar 27 '16 at 21:51
  • 3
    $\begingroup$ @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 $\sum_n a_{f(n)}z^n$ ? $\endgroup$ – DanielWainfleet Mar 27 '16 at 22:20
  • $\begingroup$ @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 $\endgroup$ – Marco Munari Mar 28 '16 at 20:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.