We call a polynomial $P(x)=a_dx^d+...+a_0$ of degree $d$ nice if $$\frac{2021(|a_d|+...+|a_0|)}{2022}<max_{0 \le i \le d}|a_i|$$Initially Shayan has a sequence of $d$ distinct real numbers; $r_1,...,r_d \neq \pm 1$. At each step he choose a positive integer $N>1$ and raises the $d$ numbers he has to the exponent of $N$, then delete the previous $d$ numbers and constructs a monic polynomial of degree $d$ with these number as roots, then examine whether it is nice or not. Prove that after some steps, all the polynomials that shayan produces would be nice polynomials Proposed by Navid Safaei
Problem
Source: Iranian RMM TST 2021 Day2 P3
Tags: algebra, polynomial
Mr.C
16.04.2021 09:29
BOBTHEGR8 wrote:
Claim - Given distinct numbers $a_0,a_1,\cdots,a_d$ there exists a number $N$ such that for $n>N$ we have that any polynomial formed using $a_0^n,a_1^n,\cdots,a_d^n$ is nice.
Assume $|a_0|$ is max , then each of $\left |\dfrac{a_i^n}{a_0^n}\right |$ tends to $0$
And hence we can find a $N$ such that if $n>N$ then $\sum_{i=1}^d\left |\dfrac{a_i^n}{a_0^n} \right | < \frac{1}{2021} $
$$\implies\sum_{i=1}^d\left |a_i^n\right | < \frac{|a_0^n|}{2021} \implies \sum_{i=0}^d\left |a_i^n\right | < \frac{2022|a_0^n|}{2021}\implies \frac{2021(\sum_{i=0}^d\left |a_i^n\right |)}{2022}<|a_0^n| $$Hence the polynomial is nice .
Now simply take $a_0 =1$ and $a_k = r_k$ to solve the problem .Note that each step the exponents must at least double and so we can find $m$ steps such that $2^m>N$ when exponent becomes $>N$
sorry if i was unclear. the new $d$ numbers are the roots of the polynomial
Rg230403
19.04.2021 11:52
Solved with p_square This seems very silly for a TST 3 problem(all you say is some term is very large) but here goes- In case there is some $r_i$ such that $|r_i|>1$- WLOG, let $|r_1|\ge \cdots \ge |r_k|>1$ and $1>r_{k+1}\ge \cdots |r_d|\ge 0$. Now, for all large enough $n$, we have $|r_k|^n>10000^d$ and $|r_{k+1}|^n<10000^d$. Now, $|r_1\cdots r_k|^n>10000^d(|P|)$ where $P$ is the product of any arbitrary subset of $\{r_1,\cdots r_d\}$ and not equal to $r_1\cdots r_k$. Thus, we want to show that for these $n$, the polynomial is nice. We want $|a_{d-k}|>2021(|a_0|+\cdots |a_d|-a_{d-k})$. Observe that $|a_{d-k}|>|r_1\cdots r_k|^n-\frac{2^d}{10000^d}(|r_1\cdots r_k|^n)>\frac{|r_1\cdots r_k|^n}{2}$. Now, $2021(|a_0|+|a_1|+\cdots |a_d|-a_{d-k})<2021(\frac{2^d}{10000^d}(|r_1\cdots r_k|^n))<2021\frac{|r_1\cdots r_k|^n}{5000}<\frac{|r_1\cdots r_k|^n}{2}$ and we are done. Now, if no $|r_i|>1$ then for large enough $n$, all $a_i$ are abitrarily small and $a_d=1$ and we are done.
We cannot remove the $a_d$ term from the summation for exactly this reason as otherwise we can have very simple cases like $(0.5,0.25,-0.5)$ cause us trouble.
bumjoooon
27.05.2021 18:56
Mr.C wrote: We call a polynomial $P(x)=a_dx^d+...+a_0$ of degree $d$ nice if $$\frac{2021(|a_d|+...+|a_0|)}{2022}<max_{0 \le i \le d}|a_i|$$Initially Shayan has a sequence of $d$ distinct real numbers; $r_1,...,r_d \neq \pm 1$. At each step he choose a positive integer $N>1$ and raises the $d$ numbers he has to the exponent of $N$, then delete the previous $d$ numbers and constructs a monic polynomial of degree $d$ with these number as roots, then examine whether it is nice or not. Prove that after some steps, all the polynomials that shayan produces would be nice polynomials Proposed by Navid Safaei I'm confused understanding the problem. So when $r_1 , ... ,r_d$ provided at the first moment, then after steps these $d$ numbers become just $r_1^M , ... ,r_d^M$ for some $M$? I hope the new $d$ numbers are new coefficient(except for 1) which would make the problem more exciting.