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2023阿里巴巴全球數(shù)學競賽預選賽題/決賽部分題個人解 (二)

2023-06-25 19:19 作者:saqatl  | 我要投稿

預選賽題 5.

(1)?這是容易的,只要考慮對角線全為?0,其它元素全為?1?的矩陣,其行列式為?(-1)%5En%20(n%20-%201)。如果?n?為奇數(shù)則任意調(diào)換矩陣的兩行即可。

(2)?先來說明怎么將?(0%2C1)?矩陣的行列式問題轉(zhuǎn)換為?(-1%2C1)?矩陣的行列式問題:對于?n%20%5Ctimes%20n?的?(0%2C1)?矩陣?X_n,我們考慮如下的步驟:

  1. 將其變?yōu)?-2X_n;

  2. 構(gòu)造矩陣?%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),其行列式與??-2X_n?相同;

  3. 將第?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?為所有元素全為?1?的?n?階方陣。那么?%7C%5Cdet%20Y_%7Bn%2B1%7D%7C%20%3D%202%5En%20%7C%5Cdet%20X_n%7C。

現(xiàn)在對?Y_%7Bn%2B1%7D?直接寫?%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,則對應的?%7C%5Cdet%20X_n%7C%20%5Cle%202%5E%7B-n%7D%20(n%2B1)%5E%7B(n%2B1)%2F2%7D。分別代入?n%20%3D%202%2C3%2C4,它們都滿足?%7C%5Cdet%20X%7C%20%5Cle%20n-1。

(3) 本問賽時我沒有任何思路,直接抄的如下論文:Clements, G. F., & Lindstr?m, B. (1965). A Sequence of (±1)-Determinants with Large Values. Proceedings of the American Mathematical Society, 16(3), 548–550.

后來發(fā)現(xiàn)可以用概率方法做。具體來說考慮隨機矩陣?Y_%7Bn%2B1%7D,其中每個元素都等可能地取?1?或?-1。寫出行列式的定義

(%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,則根據(jù)期望的線性性質(zhì)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)!

因此當?n?充分大時,存在某個?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,對應的?%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。當?n%20%3E%202023?時容易比較?%7C%5Cdet%20X_n%7C%20%3E%20n%5E%7Bn%2F4%7D。

預選賽題 6.

1. 高中題,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.?如果這樣的?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,則?%5Calpha%2C%20%5Cbeta?是方程?x%5E2%20-%206x%20%2B%207%20%3D%200?的兩根。記?%5Calpha%5En%20s%20%3D%20a_n%20%2B%20r_n,其中?a_n?為整數(shù),r_n%20%5Cin%20(-1%2F2%2C%201%2F2%5D,則?%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?充分大時?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,這樣?a_n?就能表示為?a_n%20%3D%20p%5Calpha%5En%20%2B%20q%5Cbeta%5En。

此時再次寫出?%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。但此時?a_%7Bn%2B1%7D%2Fa_n%20%3D%20%5Calpha%20%5Cnotin%20%5Cmathbb%7BQ%7D,矛盾。因此這樣的?s?不存在。

實際上直接抄?%5Cmathrm%7BPisot%7D?定理證明也可以,不過對這個問題來說多少有點小題大做。



2023阿里巴巴全球數(shù)學競賽預選賽題/決賽部分題個人解 (二)的評論 (共 條)

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