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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 (四)

2023-06-25 19:20 作者:saqatl  | 我要投稿

分析题 3. 是否存在 %5Cmathbb%7BC%7D 上的全纯函数 f 使得对任意 z%20%5Cin%20%5Cmathbb%7BC%7D 都有 f%5E%7B(2023)%7D(z)%20%3D%20e%5E%7B2023z%7D?其中 f%5E%7B(n)%7D(x) 表示 f(x) 的 n 次迭代。

假设这样的 f 存在,首先证明对任意 k%20%5Cin%20%5C%7B1%2C2%2C%5Cldots%2C2023%5C%7Df%5E%7B(k)%7D(z) 可以取到 %5Cmathbb%7BC%7D%20%5Cbackslash%20%5C%7B0%5C%7D 的所有数。一方面,f 一定不能取到全体复数:若 f(%5Cmathbb%7BC%7D)%20%3D%20%5Cmathbb%7BC%7D,则 f%5E%7B(2023)%7D(%5Cmathbb%7BC%7D)%20%3D%20%5Cmathbb%7BC%7D,但 f%5E%7B(2023)%7D(z)%20%5Cne%200,矛盾。另一方面,如果 f 取不到 %5Cmathbb%7BC%7D 中的两个点,则根据 %5Cmathrm%7BPicard%7D 小定理,f 则为常数,从而 f%5E%7B(2023)%7D 也是常数,矛盾。因此 f 仅恰好取不到 0。同理,对各 f%5E%7B(k)%7D 使用 %5Cmathrm%7BPicard%7D 小定理即得到对任意 k%20%5Cin%20%5C%7B1%2C2%2C%5Cldots%2C2023%5C%7Df%5E%7B(k)%7D 均仅取不到 0

因此,存在全纯函数 h 使得 f%20%3D%20e%5Eh。则 f%5E%7B(2023)%7D(z)%20%3D%20e%5E%7Bh(f%5E%7B(2022)%7D(z))%7D%20%3D%20e%5E%7B2023z%7D,故存在 C%20%3D%202%5Cpi%20ik 使得 h(f%5E%7B(2022)%7D(z))%20%3D%202023z%20%2B%20C,因此 h(%5Cmathbb%7BC%7D)%20%3D%20%5Cmathbb%7BC%7D。故存在 m%2Cn%20%5Cin%20%5Cmathbb%7BN%7D 使得 h(0)%20%5Cne%202m%5Cpi%20i%2C%202n%5Cpi%20i 以及存在 w_1%20%5Cne%20w_2 使得 h(w_1)%20%3D%202m%5Cpi%20i%2C%20h(w_2)%20%3D%202n%5Cpi%20i。并且由于 w_1%2C%20w_2%20%5Cne%200,存在 z_1%20%5Cne%20z_2 使得 f%5E%7B(2021)%7D(z_1)%20%3D%20w_1%2C%20f%5E%7B(2021)%7D(z_2)%20%3D%20w_2

此时

%5Cbegin%7Baligned%7D%0A%092023z_1%20%2B%20C%20%3D%20h(f%5E%7B(2022)%7D(z_1))%20%3D%20h(e%5E%7Bh(f%5E%7B(2021)%7D(z_1))%7D)%20%3D%20h(e%5E%7Bh(w_1)%7D)%20%3D%20h(e%5E%7B2m%5Cpi%20i%7D)%20%3D%20h(1)%5C%5C%0A%092023z_2%20%2B%20C%20%3D%20h(f%5E%7B(2022)%7D(z_2))%20%3D%20h(e%5E%7Bh(f%5E%7B(2021)%7D(z_2))%7D)%20%3D%20h(e%5E%7Bh(w_2)%7D)%20%3D%20h(e%5E%7B2n%5Cpi%20i%7D)%20%3D%20h(1)%0A%5Cend%7Baligned%7D

由此得z_1%20%3D%20z_2,矛盾。故这样的 f 不存在。

应用与计算数学题 1. %5Cmathrm%7BGauss-Seidel%7D 迭代

考虑线性方程组 Ax%20%3D%20b,其中 A%20%3D%20(A_%7Bij%7D)%20%5Cin%20%5Cmathbb%7BR%7D%5E%7Bn%20%5Ctimes%20n%7D 对称半正定且对角元都是正数,b%20%3D%20(b_i)%20%5Cin%20%5Cmathbb%7BR%7D%5En。假设该方程组有解,%5Cmathrm%7BGauss-Seidel%7D 迭代的求解格式为

x_i%5E%7B(k%2B1)%7D%20%3A%3D%20%5Cfrac%7B1%7D%7BA_%7Bii%7D%7D%20%5Cleft(b_i%20-%20%5Csum%5Climits_%7Bj%3D1%7D%5E%7Bi-1%7D%20A_%7Bij%7D%20x_j%5E%7B(k%2B1)%7D%20-%20%5Csum%5Climits_%7Bj%3Di%2B1%7D%5E%7Bn%7D%20A_%7Bij%7Dx_j%5E%7B(k)%7D%5Cright)%2C%20%5C%20i%20%3D%201%2C%20%5Cldots%2C%20n%2C%20%5C%20k%20%5Cin%20%5Cmathbb%7BN%7D

(a) 证明:%5Cmathrm%7BGauss-Seidel%7D 迭代从任意初始点出发都收敛。

(b) 若 A 的秩为 1,证明从任意初始点出发,%5Cmathrm%7BGauss-Seidel%7D 迭代一步即可收敛。

(a) 直接写 Golob 书上的证明。

即验证 G_%7BGS%7D%20%3D%20-(D_A%20%2B%20L_A)%5E%7B-1%7D%20L_A%5ET 的特征值均在单位圆内。该矩阵与

G%20%3D%20D_A%5E%7B1%2F2%7D%20G_%7BGS%7D%20D_A%5E%7B-1%2F2%7D%20%3D%20-(I%20%2B%20L)%5E%7B-1%7D%20L%5ET%2C%20%5Cquad%20L%20%3D%20D_A%5E%7B-1%2F2%7D%20L_A%20D_A%5E%7B-1%2F2%7D

的特征值相同。考虑

-(I%20%2B%20L)%5E%7B-1%7D%20L%5ET%20v%20%3D%20%5Clambda%20v%2C%20%5Cquad%20v%5E%7BH%7Dv%20%3D%201

则 -v%5EH%20L%5EH%20v%20%3D%20%5Clambda(1%20%2B%20v%5EH%20%20Lv)。若 v%5EH%20%20Lv%20%3D%20a%20%2B%20bi,则可以算得

%7C%5Clambda%7C%5E2%20%3D%20%5Cleft%7C%5Cfrac%7B-a%20%2B%20bi%7D%7B1%20%2B%20a%20%2B%20bi%7D%5Cright%7C%20%3D%20%5Cfrac%7Ba%5E2%20%2B%20b%5E2%7D%7B1%20%2B%202a%20%2B%20a%5E2%20%2B%20b%5E2%7D

然而由于 D_A%5E%7B-1%2F2%7D%20A%20D_A%5E%7B-1%2F2%7D%20%3D%201%20%2B%20L%20%2B%20L%5ET 正定,故 0%20%3C%201%20%2B%20v%5EH%20Lv%20%2B%20v%5EH%20L%5ET%20v%20%3D%201%20%2B%202a。因此 %7C%5Clambda%7C%20%3C%201

(b) 由题意可设 A%20%3D%20cww%5EH,通过计算不难得到当 i%20%5Cge%202%2C%20n%20%5Cge%201 时

x_i%5E%7B(n)%7D%20%3D%20x_i%5E%7B(n-1)%7D%20%2B%20%5Cfrac%7B1%7D%7Bcw_i%5E2%7D%20%5Cleft(nb_i%20-%20%5Cfrac%7Bw_i%7D%7Bw_%7Bi-1%7D%7Db_%7Bi-1%7D%5Cright)

自然如此迭代一步即可收敛。


2023阿里巴巴全球数学竞赛预选赛题/决赛部分题个人解 (四)的评论 (共 条)

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