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Let $\sum_{k=0}^{n}a_{k}z^{k}$ be a polynomial of degree $n$ with real coefficients satisfying $$a_{0}>a_{1}>....>a_{n-1}>a_{n}>0$$ Prove that $p(z)=0$ implies $\left|z\right|>1$.

I have seen similar questions here but none of them proves that solutions can't exist on the unit circle.

Links for similar questions: Showing that the roots of a polynomial with descending positive coefficients lie in the unit disc.

Let $a_n$ be a decreasing sequence. Prove that the power series $\sum a_n x^n$ has no roots in $A=\{z\in C:|z|<1\}$

Km356
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    In the summation by parts here, we have the strict inequality $$\lvert (1-z)p(z)\rvert > a_0 - \sum_{k = 1}^n (a_{k-1} - a_k)\lvert z\rvert^k - a_n \lvert z\rvert^{n+1}$$ for all $z$ with $\lvert z\rvert = 1$ except for $z = 1$. And that $z = 1$ isn't a zero follows by positivity of the coefficients. – Daniel Fischer Feb 06 '20 at 14:31
  • How is that inequality derived in the first place? The derivation here seems much cleaner. – Fimpellizzeri Feb 06 '20 at 14:38
  • Hint: If $a,b\in\Bbb C $ and $|a|>|b|$ then $a+b\ne0$. – David C. Ullrich Feb 06 '20 at 14:41

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The proof from Let $a_n$ be a decreasing sequence. Prove that the power series $\sum a_n x^n$ has no roots in $A=\{z\in C:|z|<1\}$ can be modified slightly to show that $|z| \le 1$ is not possible for a zero of $p$:

If $p(z) = 0$ and $|z| \le 1$ then $$ \begin{align} a_0 &= \lvert a_0 - (1-z)p(z) \rvert = \left\lvert \sum_{k = 1}^{n} (a_{k-1} - a_k)z^k + a_n z^{n+1} \right\rvert \\ &\underset{(*)}{\le} \sum_{k = 1}^{n} (a_{k-1} - a_k) |z^k| + a_n |z| ^{n+1} \\ &\underset{(**)}{\le} \sum_{k = 1}^{n} (a_{k-1} - a_k) + a_n = a_0 \, . \end{align} $$ So equality holds

The positive factors do not affect the argument, therefore $z, z^2, \ldots, z^n$ must all have the same argument. In particular, $z = z^2/z$ is real and positive, with absolute value $1$, so that $z=1$. But $p(1) \ne 0$, obviously.

Martin R
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  • I don't understand this part "which implies that the ratios of all terms $(a_{k-1} - a_k)z^k$ are real and non-negative, and that is only possible for $z=0$ or $z=1$." – Km356 Feb 06 '20 at 14:53
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    In order for us to have equality in the first triangle inequality, all complex number 'vectors' $(a_{k-1} -a_k)z^k$ and $a_nz^{n+1}$ must point in the same direction. The coefficient of each $z^i$ is real and positive, and hence each of the $z^i$ itself must point in the same direction. This is only possible if $z$ is real and nonnegative. – Fimpellizzeri Feb 06 '20 at 14:55
  • Thanks for the explanation – Km356 Feb 06 '20 at 15:01
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    @Km356: I have tried to explain it better. – Martin R Feb 06 '20 at 15:02