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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ Good. That means there is a permutation of the coefficients with NO real roots. :) $\endgroup$ – Joshua Benabou Mar 27 '16 at 19:14
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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*}
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  • $\begingroup$ Do you have any ideas for the case of infinite polynomials, i.e power series? $\endgroup$ – YoTengoUnLCD Mar 27 '16 at 20:56
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    $\begingroup$ 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
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    $\begingroup$ @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

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