Iran TST 2015 Polynomial

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ATimo

228 posts

ATimo

#1 h Jun 1, 2015, 6:56 AM • 2 Y

Y by Rmasters, Adventure10

Prove that for each natural number $d$ , There is a monic and unique polynomial of degree $d$ like $P$ such that $P(1)$ ≠ $0$ and for each sequence like $a_{1}$ , $a_{2}$ , $...$ of real numbers that the recurrence relation below is true for them, there is a natural number $k$ such that $0=a_{k}=a_{k+1}= ...$ :
$P(n)a_{1} + P(n-1)a_{2} + ... + P(1)a_{n}=0$
$n>1$

This post has been edited 1 time. Last edited by ATimo, Jun 1, 2015, 6:59 AM

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Dadgarnia

164 posts

Dadgarnia

#2 h Jun 1, 2015, 7:30 AM • 2 Y

Y by Adventure10, Mango247

ATimo wrote:

Prove that for each natural number $d$ , There is a monic and unique polynomial of degree $d$ like $P$ such that $P(1)$ ≠ $0$ and for each sequence like $a_{1}$ , $a_{2}$ , $...$ of real numbers that the recurrence relation below is true for them, there is a natural number $k$ such that $0=a_{k}=a_{k+1}= ...$ :
$P(n)a_{1} + P(n-1)a_{2} + ... + P(1)a_{n}=0$
$n>1$

hint

This post has been edited 1 time. Last edited by Dadgarnia, Jun 1, 2015, 7:31 AM

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andria

824 posts

andria

#3 h Jun 1, 2015, 9:41 AM • 2 Y

Y by Adventure10, Mango247

ATimo wrote:

Prove that for each natural number $d$ , There is a monic and unique polynomial of degree $d$ like $P$ such that $P(1)$ ≠ $0$ and for each sequence like $a_{1}$ , $a_{2}$ , $...$ of real numbers that the recurrence relation below is true for them, there is a natural number $k$ such that $0=a_{k}=a_{k+1}= ...$ :
$P(n)a_{1} + P(n-1)a_{2} + ... + P(1)a_{n}=0$
$n>1$

I don't understand the last sentence is $n$ a fixed integer?

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ATimo

228 posts

ATimo

#4 h Jun 1, 2015, 1:15 PM • 1 Y

Y by Adventure10

andria wrote:

ATimo wrote:

Prove that for each natural number $d$ , There is a monic and unique polynomial of degree $d$ like $P$ such that $P(1)$ ≠ $0$ and for each sequence like $a_{1}$ , $a_{2}$ , $...$ of real numbers that the recurrence relation below is true for them, there is a natural number $k$ such that $0=a_{k}=a_{k+1}= ...$ :
$P(n)a_{1} + P(n-1)a_{2} + ... + P(1)a_{n}=0$
$n>1$

I don't understand the last sentence is $n$ a fixed integer?

No. $n$ is not a fixed integer. The relation is true for all integers $n>1$ .

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Juviette

17 posts

Juviette

#5 h Mar 8, 2018, 11:47 AM • 2 Y

Y by Adventure10, Mango247

Assume there is such $k$ : $a_k \neq 0$ and $a_i=0 \forall i>k$ . We can easily see that $G(x)=a_1P(x)+a_2P(x-1)+...+P(x-k+1)a_k$ have infinitely root $n>k, n \in \mathbb{N}$ , so $G(x)=0$ . Then $P(k-1)a_1+P(k-2)a_2+...+P(1)a_{k-1}=0=P(k-1)a_1+P(k-2)a_2+...+P(0)a_k$ so $P(0)=0$ . We can also prove that $P(k-2)=P(k-1)=...=P(0)=0$ so we have $P(n)$ is in the form $\sum f(n)\alpha _i^n$ ( $\alpha _i$ are the roots of $U(x)=a_1x^k+a_2x^{k-1}+...+a_k$ ). So all the root must equal $1$ , $deg P=k-1$ and we can assume $a_1=1$ to find other $a_i$ and there are only such P satisfied the condition

This post has been edited 3 times. Last edited by Juviette, Mar 8, 2018, 12:54 PM

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igli.2001

226 posts

igli.2001

#6 h Mar 8, 2018, 2:19 PM • 1 Y

Y by Adventure10

Juviette wrote:

Assume there is such $k$ : $a_k \neq 0$ and $a_i=0 \forall i>k$ . We can easily see that $G(x)=a_1P(x)+a_2P(x-1)+...+P(x-k+1)a_k$ have infinitely root $n>k, n \in \mathbb{N}$ , so $G(x)=0$ . Then $\color{red} P(k-1)a_1+P(k-2)a_2+...+P(1)a_{k-1}=0=P(k-1)a_1+P(k-2)a_2+...+P(0)a_k$ so $P(0)=0$ . We can also prove that $P(k-2)=P(k-1)=...=P(0)=0$ so we have $P(n)$ is in the form $\sum f(n)\alpha _i^n$ ( $\alpha _i$ are the roots of $U(x)=a_1x^k+a_2x^{k-1}+...+a_k$ ). So all the root must equal $1$ , $deg P=k-1$ and we can assume $a_1=1$ to find other $a_i$ and there are only such P satisfied the condition

How do we get this? I see that the first equality is from the condition given, but i dont see how we add the last term in the second equality

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Juviette

17 posts

Juviette

#7 h Mar 12, 2018, 11:37 AM • 2 Y

Y by Adventure10, Mango247

igli.2001 wrote:

Juviette wrote:

Assume there is such $k$ : $a_k \neq 0$ and $a_i=0 \forall i>k$ . We can easily see that $G(x)=a_1P(x)+a_2P(x-1)+...+P(x-k+1)a_k$ have infinitely root $n>k, n \in \mathbb{N}$ , so $G(x)=0$ . Then $\color{red} P(k-1)a_1+P(k-2)a_2+...+P(1)a_{k-1}=0=P(k-1)a_1+P(k-2)a_2+...+P(0)a_k$ so $P(0)=0$ . We can also prove that $P(k-2)=P(k-1)=...=P(0)=0$ so we have $P(n)$ is in the form $\sum f(n)\alpha _i^n$ ( $\alpha _i$ are the roots of $U(x)=a_1x^k+a_2x^{k-1}+...+a_k$ ). So all the root must equal $1$ , $deg P=k-1$ and we can assume $a_1=1$ to find other $a_i$ and there are only such P satisfied the condition

How do we get this? I see that the first equality is from the condition given, but i dont see how we add the last term in the second equality

$G(k-1)=0$

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Cristiano-Ezio

9 posts

Cristiano-Ezio

#8 h Jul 16, 2019, 3:17 AM • 1 Y

Y by Adventure10

Juviette wrote:

Assume there is such $k$ : $a_k \neq 0$ and $a_i=0 \forall i>k$ . We can easily see that $G(x)=a_1P(x)+a_2P(x-1)+...+P(x-k+1)a_k$ have infinitely root $n>k, n \in \mathbb{N}$ , so $G(x)=0$ . Then $P(k-1)a_1+P(k-2)a_2+...+P(1)a_{k-1}=0=P(k-1)a_1+P(k-2)a_2+...+P(0)a_k$ so $P(0)=0$ . We can also prove that $P(k-2)=P(k-1)=...=P(0)=0$ so we have $P(n)$ is in the form $\sum f(n)\alpha _i^n$ ( $\alpha _i$ are the roots of $U(x)=a_1x^k+a_2x^{k-1}+...+a_k$ ). So all the root must equal $1$ , $deg P=k-1$ and we can assume $a_1=1$ to find other $a_i$ and there are only such P satisfied the condition

why p(1)≠0?could you explain this?

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atmargi

23 posts

atmargi

#9 h Apr 16, 2020, 5:08 PM

Y by

I think this problem is wrong. Bump.

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matinyousefi

499 posts

matinyousefi

#10 h Aug 18, 2020, 6:11 PM

Y by

Here is my wording of the original problem.

Let $d$ be a positive integer. Prove that there exists a unique monic polynomial of degree $d$ such that:

(a) $P(1)\neq 0$ .
(b) For every sequence of real numbers $a_1, a_2, a_3, \dots$ which satisfies the equation $P(n)a_1+P(n-1)a_2+\dots +P(1)a_n=0$ for each positive integer $n>1$ , the sequence is eventually zero.

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