MHB Problem Of The Week # 290 - Nov 24, 2017

[Ackbach]

Happy Thanksgiving, for those of you in the USA! Here is this week's POTW (not a Putnam this week!):

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If $a_0\ge a_1 \ge a_2\ge \cdots\ge a_n\ge 0,$ prove that any root $r$ of the polynomial
$$P(z)\equiv a_0 z^n+a_1 z^{n-1}+\cdots+a_n$$
satisfies $|r|\le 1$; i.e., all the roots lie inside or on the unit circle centered at the origin in the complex plane.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

Congratulations to Opalg for his correct solution to this Problem 412 of the MAA 500 Mathematical Challenges. The solution follows:

[sp]If $r$ is a root of $P(z)$ then it is also a root of $(z-1)P(z) = a_0z^{n+1} - (a_0 - a_1)z^n - (a_1 - a_2)z^{n-1} - \ldots - (a_{n-1} - a_n)z - a_n$, so that $$a_0r^{n+1} = (a_0 - a_1)r^n + (a_1 - a_2)r^{n-1} + \ldots + (a_{n-1} - a_n)r + a_n.$$ Take the absolute value of both sides and use the triangle inequality, to get $$ \begin{aligned} a_0|r|^{n+1} &= \bigl|(a_0 - a_1)r^n + (a_1 - a_2)r^{n-1} + \ldots + (a_{n-1} - a_n)r + a_n \bigr| \\ &\leqslant (a_0 - a_1)|r|^n + (a_1 - a_2)|r|^{n-1} + \ldots + (a_{n-1} - a_n)|r| + a_n.\end{aligned}$$ Now suppose that $|r| > 1$. Then $$\begin{aligned} a_0|r|^{n+1} &\leqslant (a_0 - a_1)|r|^n + (a_1 - a_2)|r|^n + \ldots + (a_{n-1} - a_n)|r|^n + a_nr^n. \\ &= \bigl( (a_0 - a_1) + (a_1 - a_2) + \ldots + (a_{n-1} - a_n) + a_n \bigr)|r|^n \\ &= a_0|r|^n <a_0|r|^{n+1}. \end{aligned}$$ That is a contradiction, and therefore each root $r$ must satisfy $|r|\leqslant 1$.[/sp]