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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1$\begingroup$ Nope. Take $P(x) = -1 - x + x^2$ ; it has a root at the golden ratio, which is about $1,618\dots$ . $\endgroup$– HachinoApr 14, 2015 at 14:29
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$\begingroup$ @ Hachino, there was typo error, I have corrected the question, the coefficients are positive. $\endgroup$– SuhailApr 14, 2015 at 14:41
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3$\begingroup$ The answer is still no. Before posting such "conjectures," why not try a few simple cases first? I am emphatically voting to close. $\endgroup$– Michael RenardyApr 14, 2015 at 14:57
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2$\begingroup$ @ Michael, I did not posted any "conjecture". Kindly mention a counter example, please. $\endgroup$– SuhailApr 14, 2015 at 14:59
1 Answer
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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1$\begingroup$ This doesn't seem right. Maybe you're thinking of something else: note the exponents $n_1$, $\ldots$, $n_m$ need not be consecutive. $\endgroup$– LuciaApr 14, 2015 at 16:39
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