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A polynomial problem

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#1 h Jan 24, 2025, 5:39 PM • 1 Y

Y by kiyoras_2001

Let $P(x)$ be a polynomial with real coefficients such that all its roots are real and nonzero. Prove that there are no 2 consecutive coefficients both zeros.

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#2 h Jan 25, 2025, 7:25 PM

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hef4875 wrote:

Let $P(x)$ be a polynomial with real coefficients such that all its roots are real and nonzero. Prove that there are no 2 consecutive coefficients both zeros.

bumppppp

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#3 h Feb 1, 2025, 1:17 PM • 1 Y

Y by kiyoras_2001

It is well known that if $p(x)$ is a polynomial of real coefficients that has exactly $n\ge 1$ real zeros (counting multiplicity), then $p'(x)$ has exactly $n - 1$ real zeros (counting multiplicity). To solve the original problem, we need an additional result:

Lemma 1. Let $p(x)$ be a polynomial of real coefficients. Assume that there exists $x_0\in\mathbb{R}$ such that $p(x_0)\neq 0$ and $p'(x_0) = p''(x_0) = 0$ . Then, $p(x)$ has a non-real zero.

Proof. Assume without loss of generality that $p(x)$ is monic. Let $n$ be the degree of $p(x)$ . Assume the contrary that $p(x)$ has $n$ real zeros (counting multiplicity), let them be $x_1, x_2, \dots, x_n$ , and therefore
$p(x) = \prod_{k=1}^n (x - x_k).$ Since $p(x_0)\neq 0$ , we have $x_0\neq x_k$ for all $k\in\{1, 2, \dots, n\}$ . For all $x\neq x_k, k\in\{1, 2, \dots, n\}$ , we have
$\frac{p'(x)}{p(x)} = \sum_{k=1}^n \frac{1}{x - x_k}.$ Take the derivative w. r. t $x$ , we have
$\frac{p''(x)p(x) - p'(x)^2}{p(x)^2} = -\sum_{k=1}^n \frac{1}{(x - x_k)^2}.$ But this yield contradiction, since for $x = x_0$ , the L. H. S equals $0$ while the R. H. S is strictly less than $0$ . $\square$

Return to the original problem. Let $n$ be the degree of $p(x)$ and $a_k$ be the coefficients of $x^k$ in $p(x)$ , for $k\in\{0, 1, \dots, n-1\}$ . Assume the contrary that there exists a smallest $k\in\{1, 2, \dots, n-1\}$ such that $a_k = a_{k+1} = 0$ (note that $a_0\neq 0$ since $p(0)\neq 0$ , so we can not have $k = 0$ ). By the minimality of $k$ , we must have $a_{k-1}\neq 0$ . Let $q(x) = p^{(k-1)}(x)$ , namely, the $(k-1)$ -th order derivative of $p(x)$ . Since $p(x)$ has exactly $n$ real zeros (counting multiplicity), $q(x)$ must have exactly $n - k + 1$ real zeros (counting multiplicity). And since $n - k + 1$ is the degree of $q(x)$ , we can conclude that all zeros of $q(x)$ are real. On the other hand, it is clear that $q(0) = (k-1)!a_{k-1}\neq 0$ . However, we also have $q'(0) = q''(0) = 0$ since $a_k = a_{k+1} = 0$ . According to Lemma 1, $q(x)$ must exist a non-real zero (contradiction).

This post has been edited 1 time. Last edited by bakkune, Feb 2, 2025, 5:52 AM

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#4 h Feb 1, 2025, 1:27 PM

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This is the second last problem of HUST Team Selection Test for the VMS Mathematical Olympiad for Undergraduate Students. The complete test along with detailed solutions can be seen at https://www.facebook.com/share/p/18G5rutEa4/ (in Vietnamese)

This post has been edited 1 time. Last edited by bakkune, Feb 1, 2025, 1:27 PM

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#5 h Thursday at 2:34 AM • 2 Y

Y by megarnie, ihatemath123

hmm ....

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