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2023阿里巴巴全球数学竞赛预选赛题/决赛部分题个人解 (二)

2023-06-25 19:19 作者:saqatl  | 我要投稿

预选赛题 5.

(1) 这是容易的,只要考虑对角线全为 0,其它元素全为 1 的矩阵,其行列式为 (-1)%5En%20(n%20-%201)。如果 n 为奇数则任意调换矩阵的两行即可。

(2) 先来说明怎么将 (0%2C1) 矩阵的行列式问题转换为 (-1%2C1) 矩阵的行列式问题:对于 n%20%5Ctimes%20n 的 (0%2C1) 矩阵 X_n,我们考虑如下的步骤:

  1. 将其变为 -2X_n

  2. 构造矩阵 %5Cleft(%20%7B%5Cbegin%7Barray%7D%7B*%7B20%7D%7Bc%7D%7D%0A%09%09%091%261%26%20%5Ccdots%20%261%5C%5C%0A%09%09%090%26%7B%7D%26%7B%7D%26%7B%7D%5C%5C%0A%09%09%09%5Cvdots%20%26%7B%7D%26%7B%20-%202%7BX_n%7D%7D%26%7B%7D%5C%5C%0A%09%09%090%26%7B%7D%26%7B%7D%26%7B%7D%0A%09%5Cend%7Barray%7D%7D%20%5Cright),其行列式与  -2X_n 相同;

  3. 将第 2%20%5Csim%20n%20%2B%201 行都加上第一行,得到 Y_%7Bn%2B1%7D%20%3D%20%5Cleft(%20%7B%5Cbegin%7Barray%7D%7B*%7B20%7D%7Bc%7D%7D%0A%09%09%091%261%26%20%5Ccdots%20%261%5C%5C%0A%09%09%091%26%7B%7D%26%7B%7D%26%7B%7D%5C%5C%0A%09%09%09%5Cvdots%20%26%7B%7D%26%7B%20-%202%7BX_n%7D%20%2B%20%7BJ_n%7D%7D%26%7B%7D%5C%5C%0A%09%09%091%26%7B%7D%26%7B%7D%26%7B%7D%0A%09%5Cend%7Barray%7D%7D%20%5Cright),其中 J_n 为所有元素全为 1 的 n 阶方阵。那么 %7C%5Cdet%20Y_%7Bn%2B1%7D%7C%20%3D%202%5En%20%7C%5Cdet%20X_n%7C

现在对 Y_%7Bn%2B1%7D 直接写 %5Cmathrm%7BHadamard%7D 不等式,得到 %7C%5Cdet%20Y_%7Bn%2B1%7D%7C%5E2%20%5Cle%20(n%20%2B%201)%5E%7Bn%20%2B%201%7D,则对应的 %7C%5Cdet%20X_n%7C%20%5Cle%202%5E%7B-n%7D%20(n%2B1)%5E%7B(n%2B1)%2F2%7D。分别代入 n%20%3D%202%2C3%2C4,它们都满足 %7C%5Cdet%20X%7C%20%5Cle%20n-1

(3) 本问赛时我没有任何思路,直接抄的如下论文:Clements, G. F., & Lindström, B. (1965). A Sequence of (±1)-Determinants with Large Values. Proceedings of the American Mathematical Society, 16(3), 548–550.

后来发现可以用概率方法做。具体来说考虑随机矩阵 Y_%7Bn%2B1%7D,其中每个元素都等可能地取 1 或 -1。写出行列式的定义

(%5Coperatorname%7Bdet%7D%20Y_%7Bn%2B1%7D)%5E2%3D%5Csum_%7B%5Csigma%2C%20%5Cpi%20%5Cin%20S_%7Bn%2B1%7D%7D(-1)%5E%7B%5Coperatorname%7Bsgn%7D%20%5Csigma%2B%5Coperatorname%7Bsgn%7D%20%5Cpi%7D%20%5Cprod_%7Bi%3D1%7D%5E%7Bn%2B1%7D%20e_%7Bi%20%5Csigma(i)%7D%20e_%7Bi%20%5Cpi(i)%7D

其中 %5Cmathbb%7BE%7D(e_i)%20%3D%200,则根据期望的线性性质有

%5Cmathbb%7BE%7D%5B(%5Cdet%20Y_%7Bn%2B1%7D)%5E2%5D%20%3D%20%5Csum%5Climits_%7B%5Csigma%20%5Cin%20S_%7Bn%2B1%7D%7D%201%20%3D%20(n%2B1)!

因此当 n 充分大时,存在某个 Y_%7Bn%2B1%7D 使得 %7C%5Cdet%20Y_%7Bn%2B1%7D%7C%20%5Cge%20%5Csqrt%7B(n%2B1)!%7D%20%5Cge%20e%5E%7B-(n%2B1)%2F2%20%2B%20o(n%2B1)%7D%20(n%2B1)%5E%7B(n%2B1)%2F2%7D,对应的 %7C%5Cdet%20X_n%7C%20%5Cge%202%5E%7B-n%7D%20e%5E%7B-(n%2B1)%2F2%20%2B%20o(n%20%2B%201)%7D%20(n%2B1)%5E%7B(n%2B1)%2F2%7D。当 n%20%3E%202023 时容易比较 %7C%5Cdet%20X_n%7C%20%3E%20n%5E%7Bn%2F4%7D

预选赛题 6.

1. 高中题,直接取 s%20%3D%201 并注意到 (1%20%2B%20%5Csqrt%7B2%7D)%5En%20%2B%20(1%20-%20%5Csqrt%7B2%7D)%5En%20%5Cin%20%5Cmathbb%7BN%7D 即可得到 %5C%7C(1%20%2B%20%5Csqrt%7B2%7D)%5En%5C%7C%20%3D%20(%5Csqrt%7B2%7D%20-%201)%5En%20%5Cto%200

2. 如果这样的 s 存在,令 %5Calpha%20%3D%203%20%2B%20%5Csqrt%7B2%7D%2C%20%5Cbeta%20%3D%203%20-%20%5Csqrt%7B2%7D,则 %5Calpha%2C%20%5Cbeta 是方程 x%5E2%20-%206x%20%2B%207%20%3D%200 的两根。记 %5Calpha%5En%20s%20%3D%20a_n%20%2B%20r_n,其中 a_n 为整数,r_n%20%5Cin%20(-1%2F2%2C%201%2F2%5D,则 %5Clim%5Climits_%7Bn%20%5Cto%20%5Cinfty%7D%20r_n%20%3D%200。而

0%20%3D%20(%5Calpha%5E%7Bn%2B2%7D%20-%206%5Calpha%5E%7Bn%2B1%7D%20%2B%207%5Calpha%5En)s%20%3D%20(a_%7Bn%2B2%7D%20-%206a_%7Bn%2B1%7D%20%2B%207a_n)%20%2B%20(r_%7Bn%2B2%7D%20-%206r_%7Bn%2B1%7D%20%2B%207r_n)

且 n 充分大时 r_%7Bn%2B2%7D%20-%206r_%7Bn%2B1%7D%20%2B%207r_n%20%5Cto%200,因此 a_%7Bn%2B2%7D%20-%206a_%7Bn%2B1%7D%20%2B%207a_n%20%5Cto%200,这样 a_n 就能表示为 a_n%20%3D%20p%5Calpha%5En%20%2B%20q%5Cbeta%5En

此时再次写出 %5Clim%5Climits_%7Bn%20%5Cto%20%5Cinfty%7D%20r_n%20%3D%20%5Clim%5Climits_%7Bn%20%5Cto%20%5Cinfty%7D%20(%5Calpha%5En%20s%20-%20a_n)%20%3D%20%5Clim%5Climits_%7Bn%20%5Cto%20%5Cinfty%7D%20((s%20-%20p)%5Calpha%5En%20%2B%20q%5Cbeta%5En)%20%3D%200。由于 %5Calpha%2C%20%5Cbeta%20%3E%201,因此 s%20%3D%20p%2C%20q%20%3D%200。但此时 a_%7Bn%2B1%7D%2Fa_n%20%3D%20%5Calpha%20%5Cnotin%20%5Cmathbb%7BQ%7D,矛盾。因此这样的 s 不存在。

实际上直接抄 %5Cmathrm%7BPisot%7D 定理证明也可以,不过对这个问题来说多少有点小题大做。



2023阿里巴巴全球数学竞赛预选赛题/决赛部分题个人解 (二)的评论 (共 条)

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