Difference between revisions of "2023 CMO Problems/Problem 2"
Anyu tsuruko (talk | contribs) (Created page with "Find the largest real number <math>C</math> such that for any positive integer <math>n</math> and any real numbers <math>x_1, x_2, \ldots, x_n</math>, the following inequality...") |
Anyu tsuruko (talk | contribs) |
||
Line 4: | Line 4: | ||
</cmath> | </cmath> | ||
== Solution 1 == | == Solution 1 == | ||
+ | Define the <math>n \times n</math> matrix <math>A_n</math> as follows: | ||
+ | <cmath> | ||
+ | A_n=\left(\begin{array}{cccccc} | ||
+ | n & n-1 & n-2 & \cdots & 2 & 1 \ | ||
+ | 1 & n & n-1 & \cdots & 3 & 2 \ | ||
+ | 2 & 1 & n & \cdots & 4 & 3 \ | ||
+ | \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \ | ||
+ | n-2 & n-3 & n-4 & \cdots & n & n-1 \ | ||
+ | n-1 & n-2 & n-3 & \cdots & 1 & n | ||
+ | \end{array}\right) | ||
+ | </cmath> | ||
+ | The problem simplifies to finding the smallest real number <math>C</math> such that for all <math>n \in \mathbb{N}^*</math> and any vector <math>x \in \mathbb{R}^n</math>, the following inequality holds: | ||
+ | <cmath> | ||
+ | x^T A_n x \geq C x^T x | ||
+ | </cmath> | ||
+ | In other words, find the real number <math>C</math> such that: | ||
+ | <cmath> | ||
+ | C \leq \inf _{n \in \mathbb{N}^*} \inf _{\substack{x \in \mathbb{R}^n \ x \neq 0}} \frac{x^T A_n x}{x^T x} | ||
+ | </cmath> | ||
+ | Given that <math>A_n</math> is not easily invertible directly, but is invertible (as it is a sparse matrix): | ||
+ | <cmath> | ||
+ | A_n^{-1}=\left(\begin{array}{cccccc} | ||
+ | \frac{n+2}{2 n+2} & \frac{1}{2} & 0 & \cdots & 0 & \frac{1}{2 n+2} \ | ||
+ | -\frac{1}{2} & 1 & \frac{1}{2} & \cdots & 0 & 0 \ | ||
+ | 0 & -\frac{1}{2} & 1 & \frac{1}{2} & \cdots & 0 \ | ||
+ | \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \ | ||
+ | 0 & 0 & \cdots & -\frac{1}{2} & 1 & \frac{1}{2} \ | ||
+ | \frac{1}{2 n+2} & 0 & \cdots & 0 & -\frac{1}{2} & \frac{n+2}{2 n+2} | ||
+ | \end{array}\right) | ||
+ | </cmath> | ||
+ | Since the inverse has non-zero entries: | ||
+ | <cmath> | ||
+ | \inf _{\substack{x \in \mathbb{R}^n \ x \neq 0}} \frac{x^T A_n x}{x^T x}=\left(\sup _{\substack{x \in \mathbb{R}^n \ x \neq 0}} \frac{x^T A_n^{-1} x}{x^T x}\right)^{-1} | ||
+ | </cmath> | ||
+ | |||
+ | For the eigenvalues <math>\mu</math> of <math>A_n</math> : | ||
+ | <cmath> | ||
+ | \frac{1}{2}+\frac{1}{2 n+2} \leq \mu-1 \leq \frac{1}{2}+\frac{1}{2 n+2} | ||
+ | </cmath> | ||
+ | |||
+ | Thus: | ||
+ | <cmath> | ||
+ | 0 \leq \mu \leq 2 | ||
+ | </cmath> | ||
+ | Given that <math>A_n^{-1}</math> is invertible, <math>\mu \neq 0</math>, therefore: | ||
+ | <cmath> | ||
+ | \begin{gathered} | ||
+ | 0<\mu \leq 2 \ | ||
+ | \sup _{\substack{x \in \mathbb{R}^n \ | ||
+ | x \neq 0}} \frac{x^T A_n^{-1} x}{x^T x} \leq 2 | ||
+ | \end{gathered} | ||
+ | </cmath> | ||
+ | |||
+ | For a specific <math>y \in \mathbb{R}^n, y=[1,-1,1,-1, \ldots]^T \in \mathbb{R}^n</math>, we have: | ||
+ | <cmath> | ||
+ | \begin{gathered} | ||
+ | y^T y=n \ | ||
+ | y^T A_n^{-1} y= \begin{cases}2(n-2)+\frac{2(n+2)}{n+1} & \text { if } n \text { is even } \ | ||
+ | 2(n-1) & \text { if } n \text { is odd }\end{cases} \ | ||
+ | \sup _n \sup _{\substack{x \in \mathbb{R}^n \ | ||
+ | x \neq 0}} \frac{x^T A_n x}{x^T x} \geq \sup _n \frac{y^T A_n y}{y^T y} \geq 2 | ||
+ | \end{gathered} | ||
+ | </cmath> | ||
+ | |||
+ | In conclusion: | ||
+ | <cmath> | ||
+ | \inf _{\substack{x \in \mathbb{R}^n \ x \neq 0}} \frac{x^T A_n x}{x^T x}=\inf _n\left(\sup _{\substack{x \in \mathbb{R}^n \ x \neq 0}} \frac{x^T A_n^{-1} x}{x^T x}\right)^{-1}=\left(\sup _{\substack{n \ n \in \mathbb{R}^n \ x \neq 0}} \frac{x^T A_n^{-1} x}{x^T x}\right)^{-1}=\frac{1}{2} | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, the maximum value of the real number <math>C</math> is <math>\frac{1}{2}</math>. | ||
+ | |||
+ | ~dalian xes|szm | ||
==See also== | ==See also== | ||
− | {{CMO box|year=2023| | + | {{CMO box|year=2023|num-b=1|num-a=3}} |
Revision as of 03:47, 25 May 2024
Find the largest real number such that for any positive integer and any real numbers , the following inequality holds:
Solution 1
Define the matrix as follows:
The problem simplifies to finding the smallest real number such that for all and any vector , the following inequality holds:
In other words, find the real number such that: Given that is not easily invertible directly, but is invertible (as it is a sparse matrix):
Since the inverse has non-zero entries:
For the eigenvalues of :
Thus: Given that is invertible, , therefore:
For a specific , we have:
In conclusion:
Therefore, the maximum value of the real number is .
~dalian xes|szm
See also
2023 CMO(CHINA) (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All CMO(CHINA) Problems and Solutions |