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Values of p(1)

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Source: Baltic way 2009

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uNc

#1 h Nov 11, 2009, 1:47 PM • 3 Y

Y by Adventure10, Mango247, Spiritpalm

A polynomial $p(x)$ of degree $n\ge 2$ has exactly $n$ real roots, counted with multiplicity. We know that the coefficient of $x^n$ is $1$ , all the roots are less than or equal to $1$ , and $p(2)=3^n$ . What values can $p(1)$ take?

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9 posts

#2 h Nov 12, 2009, 9:59 AM • 2 Y

Y by Adventure10, Mango247

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jgnr

#3 h Dec 4, 2009, 6:03 AM • 1 Y

Y by Adventure10

How do you find $max(f)$ by AM-GM?

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11 posts

L-b

#4 h Dec 13, 2009, 7:00 PM • 2 Y

Y by Adventure10, Mango247

Well, I think the point of AM-GM is, that $\left(\frac {a}{\lambda} - 1\right)(a\lambda - 1) \leq (a - 1)(a - 1)$ for every $\lambda > 1$ , which can be easily proved by the obvious inequality $\frac {1}{\lambda} + \lambda\geq 2$ . Therefore maximum of $f$ is, when all $x_{i}$ are equal.

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211 posts

#6 h Feb 11, 2024, 3:34 AM

Y by

Sorry what's the theory ? :sleeping:

:sleeping:

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#7 h Mar 19, 2024, 3:34 AM

Y by

britishprobe17 wrote:

Sorry what's the theory ? :sleeping:

:sleeping:

wdym

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#8 h Mar 19, 2024, 3:37 AM

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I think L-b is wrong, as the variable ( the line ) is not bigger then 1, but is 1.

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118 posts

#9 h Sep 12, 2024, 3:11 AM

Y by

Umm. I did not say that

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#10 h Sep 18, 2024, 6:39 AM

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peelybonehead wrote:

britishprobe17 wrote:

Sorry what's the theory ? :sleeping:

:sleeping:

wdym

i got that, my bad

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1007 posts

#11 h Sep 18, 2024, 11:51 AM

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Rephrase the problem as follows:
Given $b_1, b_2, \cdots, b_n \in \mathbb{R}^+$ with $\prod_i (b_i + 1) = 3^n$ , find the possible values of $\prod_i b_i$ .
Now note that by taking $b_2$ till $b_n$ tending to 0, and $b_1$ tending to $3^n - 1$ , $b_1 \cdots b_n > 0$ , where we can get arbitrarily close to 0.
Further, by AM-GM, we can establish that $b_1 \cdots b_n \le 2^n$ , with equality holding at $b_1 = \cdots = b_n = 2$ .By continuity, the answer is $(0, 2^n]$ .

This post has been edited 1 time. Last edited by AshAuktober, Sep 18, 2024, 11:51 AM

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#12 h Sep 30, 2024, 1:47 PM • 2 Y

Y by kisah_sangjuara, Nari_Tom

Solved with KISAH_SANGJUARA.

All the possible values that $p(1)$ can take is in the interval $[0,2^n]$ . First we prove that $p(1) \in [0,2^n]$ .
Since all the roots of $p(x)$ is $\le 1$ , we can write
$p(x)=(x+x_1-1)(x+x_2-1)\cdots(x+x_n-1)$ for some nonnegative values $x_1,x_2,\cdots x_n$ . We have
$p(2)=(1+x_1)(1+x_2)\cdots(1+x_n) \Rightarrow 3^n=(1+x_1)(1+x_2)\cdots(1+x_n)$ By AM-GM, we have
$1+x_i = 1+\frac{x_i}{2}+\frac{x_i}{2}\ge 3\sqrt[3]{\frac{x^2_i}{4}}$ . Multiplying this for $i=1,2,\cdots,n$ yields
$3^n=(1+x_1)(1+x_2)\cdots(1+x_n) \ge 3^n\sqrt[3]{\frac{(x_1x_2\cdots x_n)^2}{4^n}}$ $\Rightarrow 2^n \ge x_1x_2\cdots x_n$ . Clearly $p(1)=x_1x_2\cdots x_n \ge 0$ , hence $p(1) \in [0,2^n]$ . Now we prove that every value in $[0,2^n]$ is achievable.

Lemma. There exist nonnegative reals $x_1,x_2$ satisfying
$(1+x_1)(1+x_2)=9$ and $x_1x_2=p$ for all $p \in [0,4]$ .
Proof. Notice that the equation
$p+x_1+\frac{p}{x_1}=8 \Rightarrow x^2_1+(p-8)x_1+p=0$ has a positive root. Indeed, just choose
$x_1=\frac{8-p+\sqrt{(p-16)(p-4)}}{2}$ . It follows that $x_2=\frac{p}{x_1}$ is nonnegative, also $x_1,x_2$ satisfies the required condition.

For $n=2$ , it follows from the lemma that every value in $[0,2^n]$ is achievable
For $n>2$ , choose $x_i=2$ for every $i=3,\cdots,n$ . This reduces the case to $n=2$ , then it also follows from the lemma that every value in $[0,2^n]$ is achievable. As desired.

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#13 h Apr 9, 2025, 8:12 AM

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We can also use Holder's inequality to prove the upper found.

$3^n=(1+x_1)(1+x_2)\cdots(1+x_n) \ge (1+\sqrt[n]{x_1x_2 \cdots x_n})^n$ $\implies$ $2^n \ge x_1x_2 \cdots x_n$ .

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