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If $n_1<n_2<n_3\cdots<n_m$ are positive integers. Does the polynomial $a_0+a_1z^{n_1}+a_2z^{n_2}+\cdots+a_mz^{n_m}$ satisfying $$ 0<a_0\leq a_1\leq \cdots\leq a_m $$ has all its zeros inside $|z|\leq 1$?

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    $\begingroup$ Nope. Take $P(x) = -1 - x + x^2$ ; it has a root at the golden ratio, which is about $1,618\dots$ . $\endgroup$
    – Hachino
    Apr 14, 2015 at 14:29
  • $\begingroup$ @ Hachino, there was typo error, I have corrected the question, the coefficients are positive. $\endgroup$
    – Suhail
    Apr 14, 2015 at 14:41
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    $\begingroup$ The answer is still no. Before posting such "conjectures," why not try a few simple cases first? I am emphatically voting to close. $\endgroup$ Apr 14, 2015 at 14:57
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    $\begingroup$ @ Michael, I did not posted any "conjecture". Kindly mention a counter example, please. $\endgroup$
    – Suhail
    Apr 14, 2015 at 14:59

1 Answer 1

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Edit: The following answer only works if the exponents $n_i$ are consecutive. Otherwise there is still a counterexample $x^3+x+1=0$ (see the comment below. It has a unique real root between -1 and 0, so the other two conjugate complex roots must have modulus greater than 1 (by Vieta's theorem).

The answer to the modified question is yes. Multiply the equation by $x-1$. Then you will get a polynomial whose leading coefficient dominates the sum of the absolute value of the other coefficients. It's easy to show all roots of such a polynomial have absolute value at most 1.

I vaguely remember this is one of the problems in the Chinese mathematical Olympiad, which suggests it's probably not research level mathematics.

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    $\begingroup$ This doesn't seem right. Maybe you're thinking of something else: note the exponents $n_1$, $\ldots$, $n_m$ need not be consecutive. $\endgroup$
    – Lucia
    Apr 14, 2015 at 16:39
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    $\begingroup$ $x^3+x+1$ is a counterexample. $\endgroup$ Apr 14, 2015 at 16:58
  • $\begingroup$ Ah, I missed that. Sorry. $\endgroup$
    – Fan Zheng
    Apr 14, 2015 at 18:54

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