函數列一致收斂基礎 · 脈衝 R02

本批已通過 Schema、去重、來源快照與獨立答案重算;仍未標記為人工審定。

本題庫已有對應的中英逐步詳解。原題的 candidate 與英文 ai_candidate 狀態保留;AI 覆核、有限計算與 Lean 形式證明分開標示。

batch-20261014-advance-20261104-r02-uniform-convergence-of-functions-foundations · 2026-10-14 · 100 題

欄序為來源綁定的呈現補充;原始 JSONL 與數學答案不變。

欄序與來源綁定資料

1. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[0,1],"coefficients":[1,-2],"shift":2,"epsilon":[1,10]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/10, [[0, 9], [1, 19]], 19, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/10, [[0, 9], [1, 19]], 19, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

2. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[1,2],"coefficients":[2,-3],"shift":2,"epsilon":[1,11]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/11, [[0, 21], [1, 32]], 32, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/11, [[0, 21], [1, 32]], 32, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

3. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[2,3],"coefficients":[3,-4],"shift":2,"epsilon":[1,12]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/12, [[0, 35], [1, 47]], 47, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/12, [[0, 35], [1, 47]], 47, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

4. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[3,4],"coefficients":[4,-5],"shift":2,"epsilon":[1,13]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/13, [[0, 51], [1, 64]], 64, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/13, [[0, 51], [1, 64]], 64, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

5. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[4,5],"coefficients":[5,-6],"shift":2,"epsilon":[1,14]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/14, [[0, 69], [1, 83]], 83, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/14, [[0, 69], [1, 83]], 83, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

6. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[5,6],"coefficients":[6,-7],"shift":2,"epsilon":[1,15]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/15, [[0, 89], [1, 104]], 104, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/15, [[0, 89], [1, 104]], 104, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

7. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[6,7],"coefficients":[7,-8],"shift":2,"epsilon":[1,16]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/16, [[0, 111], [1, 127]], 127, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/16, [[0, 111], [1, 127]], 127, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

8. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[7,8],"coefficients":[8,-9],"shift":2,"epsilon":[1,17]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/17, [[0, 135], [1, 152]], 152, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/17, [[0, 135], [1, 152]], 152, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

9. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[8,9],"coefficients":[9,-10],"shift":2,"epsilon":[1,18]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/18, [[0, 161], [1, 179]], 179, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/18, [[0, 161], [1, 179]], 179, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

10. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[9,10],"coefficients":[10,-11],"shift":2,"epsilon":[1,19]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/19, [[0, 189], [1, 208]], 208, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/19, [[0, 189], [1, 208]], 208, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

11. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[10,11],"coefficients":[11,-12],"shift":2,"epsilon":[1,20]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/20, [[0, 219], [1, 239]], 239, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/20, [[0, 219], [1, 239]], 239, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

12. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[11,12],"coefficients":[12,-13],"shift":2,"epsilon":[1,21]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/21, [[0, 251], [1, 272]], 272, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/21, [[0, 251], [1, 272]], 272, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

13. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[12,13],"coefficients":[13,-14],"shift":2,"epsilon":[1,22]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/22, [[0, 285], [1, 307]], 307, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/22, [[0, 285], [1, 307]], 307, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

14. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[13,14],"coefficients":[14,-15],"shift":2,"epsilon":[1,23]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/23, [[0, 321], [1, 344]], 344, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/23, [[0, 321], [1, 344]], 344, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

15. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[14,15],"coefficients":[15,-16],"shift":2,"epsilon":[1,24]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/24, [[0, 359], [1, 383]], 383, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/24, [[0, 359], [1, 383]], 383, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

16. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[15,16],"coefficients":[16,-17],"shift":2,"epsilon":[1,25]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/25, [[0, 399], [1, 424]], 424, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/25, [[0, 399], [1, 424]], 424, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

17. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[16,17],"coefficients":[17,-18],"shift":2,"epsilon":[1,26]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/26, [[0, 441], [1, 467]], 467, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/26, [[0, 441], [1, 467]], 467, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

18. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[17,18],"coefficients":[18,-19],"shift":2,"epsilon":[1,27]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/27, [[0, 485], [1, 512]], 512, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/27, [[0, 485], [1, 512]], 512, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

19. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[18,19],"coefficients":[19,-20],"shift":2,"epsilon":[1,28]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/28, [[0, 531], [1, 559]], 559, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/28, [[0, 531], [1, 559]], 559, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

20. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[19,20],"coefficients":[20,-21],"shift":2,"epsilon":[1,29]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/29, [[0, 579], [1, 608]], 608, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/29, [[0, 579], [1, 608]], 608, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

21. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[20,21],"coefficients":[21,-22],"shift":2,"epsilon":[1,30]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/30, [[0, 629], [1, 659]], 659, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/30, [[0, 629], [1, 659]], 659, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

22. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[21,22],"coefficients":[22,-23],"shift":2,"epsilon":[1,31]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/31, [[0, 681], [1, 712]], 712, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/31, [[0, 681], [1, 712]], 712, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

23. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[22,23],"coefficients":[23,-24],"shift":2,"epsilon":[1,32]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/32, [[0, 735], [1, 767]], 767, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/32, [[0, 735], [1, 767]], 767, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

24. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[23,24],"coefficients":[24,-25],"shift":2,"epsilon":[1,33]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/33, [[0, 791], [1, 824]], 824, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/33, [[0, 791], [1, 824]], 824, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

25. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-domain-witness;資料={"limit":[24,25],"coefficients":[25,-26],"shift":2,"epsilon":[1,34]}。請依 [epsilon, pointwise_witnesses, N, same_N_for_all_points, certificate] 欄序回答。

查看答案與解釋

(1/34, [[0, 849], [1, 883]], 883, 是, same_N_for_all_finite_domain_points)

Wave5第四層核對=(1/34, [[0, 849], [1, 883]], 883, 是, same_N_for_all_finite_domain_points);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-domain-witness · candidate

26. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[0,1],"coefficients":[1,-2],"shift":2,"n":1}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(1, 2/3, exact_finite_domain_supremum)

Wave5第四層核對=(1, 2/3, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

27. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[1,2],"coefficients":[2,-3],"shift":2,"n":2}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(2, 3/4, exact_finite_domain_supremum)

Wave5第四層核對=(2, 3/4, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

28. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[2,3],"coefficients":[3,-4],"shift":2,"n":3}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(3, 4/5, exact_finite_domain_supremum)

Wave5第四層核對=(3, 4/5, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

29. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[3,4],"coefficients":[4,-5],"shift":2,"n":4}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(4, 5/6, exact_finite_domain_supremum)

Wave5第四層核對=(4, 5/6, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

30. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[4,5],"coefficients":[5,-6],"shift":2,"n":5}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(5, 6/7, exact_finite_domain_supremum)

Wave5第四層核對=(5, 6/7, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

31. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[5,6],"coefficients":[6,-7],"shift":2,"n":6}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(6, 7/8, exact_finite_domain_supremum)

Wave5第四層核對=(6, 7/8, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

32. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[6,7],"coefficients":[7,-8],"shift":2,"n":7}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(7, 8/9, exact_finite_domain_supremum)

Wave5第四層核對=(7, 8/9, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

33. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[7,8],"coefficients":[8,-9],"shift":2,"n":8}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(8, 9/10, exact_finite_domain_supremum)

Wave5第四層核對=(8, 9/10, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

34. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[8,9],"coefficients":[9,-10],"shift":2,"n":9}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(9, 10/11, exact_finite_domain_supremum)

Wave5第四層核對=(9, 10/11, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

35. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[9,10],"coefficients":[10,-11],"shift":2,"n":10}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(10, 11/12, exact_finite_domain_supremum)

Wave5第四層核對=(10, 11/12, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

36. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[10,11],"coefficients":[11,-12],"shift":2,"n":11}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(11, 12/13, exact_finite_domain_supremum)

Wave5第四層核對=(11, 12/13, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

37. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[11,12],"coefficients":[12,-13],"shift":2,"n":12}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(12, 13/14, exact_finite_domain_supremum)

Wave5第四層核對=(12, 13/14, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

38. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[12,13],"coefficients":[13,-14],"shift":2,"n":13}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(13, 14/15, exact_finite_domain_supremum)

Wave5第四層核對=(13, 14/15, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

39. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[13,14],"coefficients":[14,-15],"shift":2,"n":14}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(14, 15/16, exact_finite_domain_supremum)

Wave5第四層核對=(14, 15/16, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

40. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[14,15],"coefficients":[15,-16],"shift":2,"n":15}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(15, 16/17, exact_finite_domain_supremum)

Wave5第四層核對=(15, 16/17, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

41. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[15,16],"coefficients":[16,-17],"shift":2,"n":16}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(16, 17/18, exact_finite_domain_supremum)

Wave5第四層核對=(16, 17/18, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

42. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[16,17],"coefficients":[17,-18],"shift":2,"n":17}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(17, 18/19, exact_finite_domain_supremum)

Wave5第四層核對=(17, 18/19, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

43. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[17,18],"coefficients":[18,-19],"shift":2,"n":18}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(18, 19/20, exact_finite_domain_supremum)

Wave5第四層核對=(18, 19/20, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

44. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[18,19],"coefficients":[19,-20],"shift":2,"n":19}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(19, 20/21, exact_finite_domain_supremum)

Wave5第四層核對=(19, 20/21, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

45. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[19,20],"coefficients":[20,-21],"shift":2,"n":20}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(20, 21/22, exact_finite_domain_supremum)

Wave5第四層核對=(20, 21/22, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

46. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[20,21],"coefficients":[21,-22],"shift":2,"n":21}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(21, 22/23, exact_finite_domain_supremum)

Wave5第四層核對=(21, 22/23, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

47. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[21,22],"coefficients":[22,-23],"shift":2,"n":22}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(22, 23/24, exact_finite_domain_supremum)

Wave5第四層核對=(22, 23/24, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

48. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[22,23],"coefficients":[23,-24],"shift":2,"n":23}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(23, 24/25, exact_finite_domain_supremum)

Wave5第四層核對=(23, 24/25, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

49. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[23,24],"coefficients":[24,-25],"shift":2,"n":24}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(24, 25/26, exact_finite_domain_supremum)

Wave5第四層核對=(24, 25/26, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

50. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-sup-gap;資料={"limit":[24,25],"coefficients":[25,-26],"shift":2,"n":25}。請依 [n, sup_gap, certificate] 欄序回答。

查看答案與解釋

(25, 26/27, exact_finite_domain_supremum)

Wave5第四層核對=(25, 26/27, exact_finite_domain_supremum);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-sup-gap · candidate

51. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[0,1],"coefficients":[1,-2],"shift":2,"epsilon":[1,10]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 9], [1, 19]], 19, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 9], [1, 19]], 19, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

52. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[1,2],"coefficients":[2,-3],"shift":2,"epsilon":[1,11]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 21], [1, 32]], 32, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 21], [1, 32]], 32, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

53. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[2,3],"coefficients":[3,-4],"shift":2,"epsilon":[1,12]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 35], [1, 47]], 47, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 35], [1, 47]], 47, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

54. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[3,4],"coefficients":[4,-5],"shift":2,"epsilon":[1,13]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 51], [1, 64]], 64, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 51], [1, 64]], 64, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

55. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[4,5],"coefficients":[5,-6],"shift":2,"epsilon":[1,14]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 69], [1, 83]], 83, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 69], [1, 83]], 83, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

56. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[5,6],"coefficients":[6,-7],"shift":2,"epsilon":[1,15]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 89], [1, 104]], 104, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 89], [1, 104]], 104, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

57. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[6,7],"coefficients":[7,-8],"shift":2,"epsilon":[1,16]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 111], [1, 127]], 127, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 111], [1, 127]], 127, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

58. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[7,8],"coefficients":[8,-9],"shift":2,"epsilon":[1,17]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 135], [1, 152]], 152, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 135], [1, 152]], 152, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

59. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[8,9],"coefficients":[9,-10],"shift":2,"epsilon":[1,18]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 161], [1, 179]], 179, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 161], [1, 179]], 179, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

60. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[9,10],"coefficients":[10,-11],"shift":2,"epsilon":[1,19]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 189], [1, 208]], 208, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 189], [1, 208]], 208, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

61. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[10,11],"coefficients":[11,-12],"shift":2,"epsilon":[1,20]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 219], [1, 239]], 239, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 219], [1, 239]], 239, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

62. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[11,12],"coefficients":[12,-13],"shift":2,"epsilon":[1,21]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 251], [1, 272]], 272, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 251], [1, 272]], 272, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

63. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[12,13],"coefficients":[13,-14],"shift":2,"epsilon":[1,22]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 285], [1, 307]], 307, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 285], [1, 307]], 307, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

64. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[13,14],"coefficients":[14,-15],"shift":2,"epsilon":[1,23]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 321], [1, 344]], 344, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 321], [1, 344]], 344, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

65. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[14,15],"coefficients":[15,-16],"shift":2,"epsilon":[1,24]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 359], [1, 383]], 383, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 359], [1, 383]], 383, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

66. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[15,16],"coefficients":[16,-17],"shift":2,"epsilon":[1,25]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 399], [1, 424]], 424, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 399], [1, 424]], 424, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

67. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[16,17],"coefficients":[17,-18],"shift":2,"epsilon":[1,26]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 441], [1, 467]], 467, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 441], [1, 467]], 467, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

68. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[17,18],"coefficients":[18,-19],"shift":2,"epsilon":[1,27]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 485], [1, 512]], 512, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 485], [1, 512]], 512, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

69. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[18,19],"coefficients":[19,-20],"shift":2,"epsilon":[1,28]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 531], [1, 559]], 559, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 531], [1, 559]], 559, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

70. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[19,20],"coefficients":[20,-21],"shift":2,"epsilon":[1,29]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 579], [1, 608]], 608, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 579], [1, 608]], 608, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

71. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[20,21],"coefficients":[21,-22],"shift":2,"epsilon":[1,30]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 629], [1, 659]], 659, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 629], [1, 659]], 659, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

72. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[21,22],"coefficients":[22,-23],"shift":2,"epsilon":[1,31]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 681], [1, 712]], 712, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 681], [1, 712]], 712, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

73. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[22,23],"coefficients":[23,-24],"shift":2,"epsilon":[1,32]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 735], [1, 767]], 767, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 735], [1, 767]], 767, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

74. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[23,24],"coefficients":[24,-25],"shift":2,"epsilon":[1,33]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 791], [1, 824]], 824, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 791], [1, 824]], 824, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

75. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-pointwise-max;資料={"limit":[24,25],"coefficients":[25,-26],"shift":2,"epsilon":[1,34]}。請依 [pointwise_witnesses, N, uniform_on_this_finite_domain, boundary] 欄序回答。

查看答案與解釋

([[0, 849], [1, 883]], 883, 是, finite_maximum_only_not_infinite_domain_implication)

Wave5第四層核對=([[0, 849], [1, 883]], 883, 是, finite_maximum_only_not_infinite_domain_implication);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-pointwise-max · candidate

76. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[0,1],"coefficients":[1,-2],"shift":2,"prefix_length":1}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(1, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(1, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

77. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[1,2],"coefficients":[2,-3],"shift":2,"prefix_length":2}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(2, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(2, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

78. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[2,3],"coefficients":[3,-4],"shift":2,"prefix_length":3}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(3, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(3, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

79. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[3,4],"coefficients":[4,-5],"shift":2,"prefix_length":4}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(4, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(4, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

80. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[4,5],"coefficients":[5,-6],"shift":2,"prefix_length":5}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(5, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(5, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

81. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[5,6],"coefficients":[6,-7],"shift":2,"prefix_length":6}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(6, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(6, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

82. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[6,7],"coefficients":[7,-8],"shift":2,"prefix_length":7}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(7, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(7, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

83. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[7,8],"coefficients":[8,-9],"shift":2,"prefix_length":8}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(8, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(8, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

84. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[8,9],"coefficients":[9,-10],"shift":2,"prefix_length":9}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(9, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(9, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

85. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[9,10],"coefficients":[10,-11],"shift":2,"prefix_length":10}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(10, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(10, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

86. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[10,11],"coefficients":[11,-12],"shift":2,"prefix_length":11}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(11, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(11, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

87. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[11,12],"coefficients":[12,-13],"shift":2,"prefix_length":12}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(12, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(12, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

88. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[12,13],"coefficients":[13,-14],"shift":2,"prefix_length":13}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(13, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(13, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

89. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[13,14],"coefficients":[14,-15],"shift":2,"prefix_length":14}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(14, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(14, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

90. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[14,15],"coefficients":[15,-16],"shift":2,"prefix_length":15}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(15, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(15, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

91. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[15,16],"coefficients":[16,-17],"shift":2,"prefix_length":16}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(16, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(16, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

92. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[16,17],"coefficients":[17,-18],"shift":2,"prefix_length":17}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(17, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(17, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

93. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[17,18],"coefficients":[18,-19],"shift":2,"prefix_length":18}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(18, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(18, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

94. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[18,19],"coefficients":[19,-20],"shift":2,"prefix_length":19}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(19, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(19, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

95. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[19,20],"coefficients":[20,-21],"shift":2,"prefix_length":20}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(20, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(20, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

96. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[20,21],"coefficients":[21,-22],"shift":2,"prefix_length":21}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(21, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(21, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

97. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[21,22],"coefficients":[22,-23],"shift":2,"prefix_length":22}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(22, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(22, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

98. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[22,23],"coefficients":[23,-24],"shift":2,"prefix_length":23}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(23, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(23, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

99. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[23,24],"coefficients":[24,-25],"shift":2,"prefix_length":24}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(24, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(24, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

100. 一致收斂

共同有限domain的symbolic tail與同一N(epsilon)。操作=uniform-finite-prefix-boundary;資料={"limit":[24,25],"coefficients":[25,-26],"shift":2,"prefix_length":25}。請依 [prefix_length, sufficient, boundary] 欄序回答。

查看答案與解釋

(25, 否, finite_prefix_not_uniform_convergence_certificate)

Wave5第四層核對=(25, 否, finite_prefix_not_uniform_convergence_certificate);依完整明示資料重算;同一N控制所有x且有限domain可取pointwise witnesses最大值;不可外推無限domain且finite prefix不足。

本題逐步詳解(AI 二次覆核)

函數列一致收斂 · qf-mod35-uniform-finite-prefix-boundary · candidate

原始題目 JSONL · 驗收報告