Lp收斂基礎 · 脈衝 R07
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-r07-lp-convergence-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. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[1,-2],"shift":2,"p":1,"n":1}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(1, 1, 1/2, exact_finite_atomic_moment)
Wave5第四層核對=(1, 1, 1/2, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
2. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[2,-3],"shift":2,"p":2,"n":2}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(2, 2, 13/32, exact_finite_atomic_moment)
Wave5第四層核對=(2, 2, 13/32, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
3. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[3,-4],"shift":2,"p":3,"n":3}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(3, 3, 91/250, exact_finite_atomic_moment)
Wave5第四層核對=(3, 3, 91/250, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
4. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[4,-5],"shift":2,"p":1,"n":4}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(1, 4, 3/4, exact_finite_atomic_moment)
Wave5第四層核對=(1, 4, 3/4, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
5. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[5,-6],"shift":2,"p":2,"n":5}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(2, 5, 61/98, exact_finite_atomic_moment)
Wave5第四層核對=(2, 5, 61/98, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
6. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[6,-7],"shift":2,"p":3,"n":6}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(3, 6, 559/1024, exact_finite_atomic_moment)
Wave5第四層核對=(3, 6, 559/1024, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
7. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[7,-8],"shift":2,"p":1,"n":7}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(1, 7, 5/6, exact_finite_atomic_moment)
Wave5第四層核對=(1, 7, 5/6, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
8. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[8,-9],"shift":2,"p":2,"n":8}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(2, 8, 29/40, exact_finite_atomic_moment)
Wave5第四層核對=(2, 8, 29/40, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
9. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[9,-10],"shift":2,"p":3,"n":9}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(3, 9, 1729/2662, exact_finite_atomic_moment)
Wave5第四層核對=(3, 9, 1729/2662, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
10. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[10,-11],"shift":2,"p":1,"n":10}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(1, 10, 7/8, exact_finite_atomic_moment)
Wave5第四層核對=(1, 10, 7/8, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
11. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[11,-12],"shift":2,"p":2,"n":11}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(2, 11, 265/338, exact_finite_atomic_moment)
Wave5第四層核對=(2, 11, 265/338, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
12. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[12,-13],"shift":2,"p":3,"n":12}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(3, 12, 3925/5488, exact_finite_atomic_moment)
Wave5第四層核對=(3, 12, 3925/5488, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
13. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[13,-14],"shift":2,"p":1,"n":13}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(1, 13, 9/10, exact_finite_atomic_moment)
Wave5第四層核對=(1, 13, 9/10, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
14. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[14,-15],"shift":2,"p":2,"n":14}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(2, 14, 421/512, exact_finite_atomic_moment)
Wave5第四層核對=(2, 14, 421/512, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
15. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[15,-16],"shift":2,"p":3,"n":15}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(3, 15, 7471/9826, exact_finite_atomic_moment)
Wave5第四層核對=(3, 15, 7471/9826, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
16. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[16,-17],"shift":2,"p":1,"n":16}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(1, 16, 11/12, exact_finite_atomic_moment)
Wave5第四層核對=(1, 16, 11/12, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
17. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[17,-18],"shift":2,"p":2,"n":17}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(2, 17, 613/722, exact_finite_atomic_moment)
Wave5第四層核對=(2, 17, 613/722, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
18. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[18,-19],"shift":2,"p":3,"n":18}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(3, 18, 12691/16000, exact_finite_atomic_moment)
Wave5第四層核對=(3, 18, 12691/16000, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
19. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[19,-20],"shift":2,"p":1,"n":19}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(1, 19, 13/14, exact_finite_atomic_moment)
Wave5第四層核對=(1, 19, 13/14, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
20. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[20,-21],"shift":2,"p":2,"n":20}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(2, 20, 841/968, exact_finite_atomic_moment)
Wave5第四層核對=(2, 20, 841/968, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
21. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[21,-22],"shift":2,"p":3,"n":21}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(3, 21, 19909/24334, exact_finite_atomic_moment)
Wave5第四層核對=(3, 21, 19909/24334, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
22. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[22,-23],"shift":2,"p":1,"n":22}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(1, 22, 15/16, exact_finite_atomic_moment)
Wave5第四層核對=(1, 22, 15/16, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
23. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[23,-24],"shift":2,"p":2,"n":23}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(2, 23, 221/250, exact_finite_atomic_moment)
Wave5第四層核對=(2, 23, 221/250, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
24. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[24,-25],"shift":2,"p":3,"n":24}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(3, 24, 29449/35152, exact_finite_atomic_moment)
Wave5第四層核對=(3, 24, 29449/35152, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
25. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-moment-at-n;資料={"masses":[[1,2],[1,2]],"coefficients":[25,-26],"shift":2,"p":1,"n":25}。請依 [p, n, p_moment, certificate] 欄序回答。
查看答案與解釋
(1, 25, 17/18, exact_finite_atomic_moment)
Wave5第四層核對=(1, 25, 17/18, exact_finite_atomic_moment);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-moment-at-n · candidate
26. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[1,-2],"shift":2,"p":1,"epsilon":[1,10]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(1, 1/10, 14, 3/32, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(1, 1/10, 14, 3/32, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
27. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[2,-3],"shift":2,"p":2,"epsilon":[1,11]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(2, 1/11, 7, 13/162, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(2, 1/11, 7, 13/162, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
28. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[3,-4],"shift":2,"p":3,"epsilon":[1,12]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(3, 1/12, 7, 91/1458, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(3, 1/12, 7, 91/1458, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
29. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[4,-5],"shift":2,"p":1,"epsilon":[1,13]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(1, 1/13, 57, 9/118, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(1, 1/13, 57, 9/118, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
30. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[5,-6],"shift":2,"p":2,"epsilon":[1,14]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(2, 1/14, 19, 61/882, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(2, 1/14, 19, 61/882, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
31. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[6,-7],"shift":2,"p":3,"epsilon":[1,15]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(3, 1/15, 15, 559/9826, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(3, 1/15, 15, 559/9826, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
32. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[7,-8],"shift":2,"p":1,"epsilon":[1,16]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(1, 1/16, 119, 15/242, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(1, 1/16, 119, 15/242, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
33. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[8,-9],"shift":2,"p":2,"epsilon":[1,17]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(2, 1/17, 34, 145/2592, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(2, 1/17, 34, 145/2592, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
34. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[9,-10],"shift":2,"p":3,"epsilon":[1,18]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(3, 1/18, 23, 1729/31250, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(3, 1/18, 23, 1729/31250, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
35. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[10,-11],"shift":2,"p":1,"epsilon":[1,19]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(1, 1/19, 198, 21/400, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(1, 1/19, 198, 21/400, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
36. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[11,-12],"shift":2,"p":2,"epsilon":[1,20]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(2, 1/20, 50, 265/5408, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(2, 1/20, 50, 265/5408, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
37. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[12,-13],"shift":2,"p":3,"epsilon":[1,21]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(3, 1/21, 33, 157/3430, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(3, 1/21, 33, 157/3430, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
38. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[13,-14],"shift":2,"p":1,"epsilon":[1,22]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(1, 1/22, 296, 27/596, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(1, 1/22, 296, 27/596, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
39. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[14,-15],"shift":2,"p":2,"epsilon":[1,23]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(2, 1/23, 68, 421/9800, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(2, 1/23, 68, 421/9800, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
40. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[15,-16],"shift":2,"p":3,"epsilon":[1,24]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(3, 1/24, 43, 7471/182250, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(3, 1/24, 43, 7471/182250, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
41. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[16,-17],"shift":2,"p":1,"epsilon":[1,25]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(1, 1/25, 411, 33/826, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(1, 1/25, 411, 33/826, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
42. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[17,-18],"shift":2,"p":2,"epsilon":[1,26]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(2, 1/26, 88, 613/16200, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(2, 1/26, 88, 613/16200, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
43. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[18,-19],"shift":2,"p":3,"epsilon":[1,27]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(3, 1/27, 54, 37/1024, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(3, 1/27, 54, 37/1024, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
44. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[19,-20],"shift":2,"p":1,"epsilon":[1,28]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(1, 1/28, 545, 39/1094, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(1, 1/28, 545, 39/1094, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
45. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[20,-21],"shift":2,"p":2,"epsilon":[1,29]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(2, 1/29, 109, 841/24642, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(2, 1/29, 109, 841/24642, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
46. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[21,-22],"shift":2,"p":3,"epsilon":[1,30]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(3, 1/30, 65, 19909/601526, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(3, 1/30, 65, 19909/601526, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
47. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[22,-23],"shift":2,"p":1,"epsilon":[1,31]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(1, 1/31, 696, 45/1396, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(1, 1/31, 696, 45/1396, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
48. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[23,-24],"shift":2,"p":2,"epsilon":[1,32]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(2, 1/32, 131, 1105/35378, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(2, 1/32, 131, 1105/35378, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
49. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[24,-25],"shift":2,"p":3,"epsilon":[1,33]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(3, 1/33, 77, 29449/986078, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(3, 1/33, 77, 29449/986078, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
50. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-epsilon-witness;資料={"masses":[[1,2],[1,2]],"coefficients":[25,-26],"shift":2,"p":1,"epsilon":[1,34]}。請依 [p, moment_tolerance, N, p_moment_at_N, strict_tail, certificate] 欄序回答。
原參數 epsilon 在此界定 p-moment:E|誤差|^p < epsilon,並非直接對其 p 次方根範數設同一界。
查看答案與解釋
(1, 1/34, 866, 51/1736, 是, forall_n_ge_N_symbolic_monotone_tail)
Wave5第四層核對=(1, 1/34, 866, 51/1736, 是, forall_n_ge_N_symbolic_monotone_tail);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-epsilon-witness · candidate
51. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[1,-2],"shift":2,"p":1}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(1, [1/2, 1/2], [1, -2], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(1, [1/2, 1/2], [1, -2], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
52. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[2,-3],"shift":2,"p":2}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(2, [1/2, 1/2], [2, -3], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(2, [1/2, 1/2], [2, -3], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
53. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[3,-4],"shift":2,"p":3}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(3, [1/2, 1/2], [3, -4], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(3, [1/2, 1/2], [3, -4], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
54. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[4,-5],"shift":2,"p":1}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(1, [1/2, 1/2], [4, -5], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(1, [1/2, 1/2], [4, -5], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
55. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[5,-6],"shift":2,"p":2}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(2, [1/2, 1/2], [5, -6], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(2, [1/2, 1/2], [5, -6], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
56. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[6,-7],"shift":2,"p":3}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(3, [1/2, 1/2], [6, -7], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(3, [1/2, 1/2], [6, -7], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
57. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[7,-8],"shift":2,"p":1}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(1, [1/2, 1/2], [7, -8], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(1, [1/2, 1/2], [7, -8], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
58. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[8,-9],"shift":2,"p":2}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(2, [1/2, 1/2], [8, -9], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(2, [1/2, 1/2], [8, -9], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
59. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[9,-10],"shift":2,"p":3}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(3, [1/2, 1/2], [9, -10], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(3, [1/2, 1/2], [9, -10], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
60. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[10,-11],"shift":2,"p":1}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(1, [1/2, 1/2], [10, -11], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(1, [1/2, 1/2], [10, -11], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
61. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[11,-12],"shift":2,"p":2}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(2, [1/2, 1/2], [11, -12], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(2, [1/2, 1/2], [11, -12], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
62. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[12,-13],"shift":2,"p":3}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(3, [1/2, 1/2], [12, -13], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(3, [1/2, 1/2], [12, -13], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
63. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[13,-14],"shift":2,"p":1}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(1, [1/2, 1/2], [13, -14], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(1, [1/2, 1/2], [13, -14], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
64. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[14,-15],"shift":2,"p":2}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(2, [1/2, 1/2], [14, -15], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(2, [1/2, 1/2], [14, -15], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
65. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[15,-16],"shift":2,"p":3}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(3, [1/2, 1/2], [15, -16], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(3, [1/2, 1/2], [15, -16], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
66. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[16,-17],"shift":2,"p":1}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(1, [1/2, 1/2], [16, -17], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(1, [1/2, 1/2], [16, -17], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
67. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[17,-18],"shift":2,"p":2}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(2, [1/2, 1/2], [17, -18], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(2, [1/2, 1/2], [17, -18], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
68. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[18,-19],"shift":2,"p":3}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(3, [1/2, 1/2], [18, -19], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(3, [1/2, 1/2], [18, -19], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
69. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[19,-20],"shift":2,"p":1}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(1, [1/2, 1/2], [19, -20], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(1, [1/2, 1/2], [19, -20], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
70. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[20,-21],"shift":2,"p":2}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(2, [1/2, 1/2], [20, -21], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(2, [1/2, 1/2], [20, -21], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
71. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[21,-22],"shift":2,"p":3}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(3, [1/2, 1/2], [21, -22], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(3, [1/2, 1/2], [21, -22], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
72. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[22,-23],"shift":2,"p":1}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(1, [1/2, 1/2], [22, -23], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(1, [1/2, 1/2], [22, -23], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
73. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[23,-24],"shift":2,"p":2}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(2, [1/2, 1/2], [23, -24], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(2, [1/2, 1/2], [23, -24], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
74. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[24,-25],"shift":2,"p":3}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(3, [1/2, 1/2], [24, -25], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(3, [1/2, 1/2], [24, -25], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
75. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification;資料={"masses":[[1,2],[1,2]],"coefficients":[25,-26],"shift":2,"p":1}。請依 [p, masses, coefficients, shift, qualified, certificate] 欄序回答。
查看答案與解釋
(1, [1/2, 1/2], [25, -26], 2, 是, common_probability_space_and_finite_integer_p_moments)
Wave5第四層核對=(1, [1/2, 1/2], [25, -26], 2, 是, common_probability_space_and_finite_integer_p_moments);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-qualification · candidate
76. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[1,-2],"shift":2,"p":1,"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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(1, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
77. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[2,-3],"shift":2,"p":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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(2, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
78. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[3,-4],"shift":2,"p":3,"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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(3, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
79. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[4,-5],"shift":2,"p":1,"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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(4, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
80. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[5,-6],"shift":2,"p":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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(5, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
81. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[6,-7],"shift":2,"p":3,"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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(6, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
82. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[7,-8],"shift":2,"p":1,"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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(7, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
83. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[8,-9],"shift":2,"p":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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(8, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
84. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[9,-10],"shift":2,"p":3,"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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(9, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
85. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[10,-11],"shift":2,"p":1,"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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(10, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
86. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[11,-12],"shift":2,"p":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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(11, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
87. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[12,-13],"shift":2,"p":3,"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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(12, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
88. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[13,-14],"shift":2,"p":1,"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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(13, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
89. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[14,-15],"shift":2,"p":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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(14, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
90. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[15,-16],"shift":2,"p":3,"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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(15, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
91. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[16,-17],"shift":2,"p":1,"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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(16, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
92. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[17,-18],"shift":2,"p":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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(17, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
93. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[18,-19],"shift":2,"p":3,"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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(18, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
94. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[19,-20],"shift":2,"p":1,"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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(19, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
95. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[20,-21],"shift":2,"p":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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(20, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
96. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[21,-22],"shift":2,"p":3,"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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(21, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
97. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[22,-23],"shift":2,"p":1,"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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(22, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
98. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[23,-24],"shift":2,"p":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_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(23, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
99. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[24,-25],"shift":2,"p":3,"prefix_length":24}。請依 [prefix_length, sufficient, boundary] 欄序回答。
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(24, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(24, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
100. Lp收斂
同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-finite-prefix-boundary;資料={"masses":[[1,2],[1,2]],"coefficients":[25,-26],"shift":2,"p":1,"prefix_length":25}。請依 [prefix_length, sufficient, boundary] 欄序回答。
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(25, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication)
Wave5第四層核對=(25, 否, finite_prefix_not_Lp_convergence_certificate_or_mode_implication);依完整明示資料重算;核E absolute error power p趨零且moments有限;finite prefix不足,a.s. probability或其他Lq implications另需假設。
本題逐步詳解(AI 二次覆核)
Lp收斂 · qf-mod35-lp-finite-prefix-boundary · candidate
原始題目 JSONL · 驗收報告