Taylor餘項定理基礎 · 脈衝 R03
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-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations · 2026-10-15 · 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. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[2,3],"center":0,"x":[1,2],"order":1}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([2, 3], 0, 1/2, 1, [2, 3, 0], [2, 3/2], 7/2, 7/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([2, 3], 0, 1/2, 1, [2, 3, 0], [2, 3/2], 7/2, 7/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
2. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[3,5,7],"center":0,"x":[1,2],"order":2}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([3, 5, 7], 0, 1/2, 2, [3, 5, 14, 0], [3, 5/2, 7/4], 29/4, 29/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([3, 5, 7], 0, 1/2, 2, [3, 5, 14, 0], [3, 5/2, 7/4], 29/4, 29/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
3. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[4,7,10,13],"center":0,"x":[1,2],"order":3}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([4, 7, 10, 13], 0, 1/2, 3, [4, 7, 20, 78, 0], [4, 7/2, 5/2, 13/8], 93/8, 93/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([4, 7, 10, 13], 0, 1/2, 3, [4, 7, 20, 78, 0], [4, 7/2, 5/2, 13/8], 93/8, 93/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
4. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[5,9,13,17,4],"center":0,"x":[1,2],"order":4}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([5, 9, 13, 17, 4], 0, 1/2, 4, [5, 9, 26, 102, 96, 0], [5, 9/2, 13/4, 17/8, 1/4], 121/8, 121/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([5, 9, 13, 17, 4], 0, 1/2, 4, [5, 9, 26, 102, 96, 0], [5, 9/2, 13/4, 17/8, 1/4], 121/8, 121/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
5. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[6,11,16,4,9,14],"center":0,"x":[1,2],"order":5}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([6, 11, 16, 4, 9, 14], 0, 1/2, 5, [6, 11, 32, 24, 216, 1680, 0], [6, 11/2, 4, 1/2, 9/16, 7/16], 17, 17, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([6, 11, 16, 4, 9, 14], 0, 1/2, 5, [6, 11, 32, 24, 216, 1680, 0], [6, 11/2, 4, 1/2, 9/16, 7/16], 17, 17, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
6. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[7,13],"center":0,"x":[1,2],"order":1}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([7, 13], 0, 1/2, 1, [7, 13, 0], [7, 13/2], 27/2, 27/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([7, 13], 0, 1/2, 1, [7, 13, 0], [7, 13/2], 27/2, 27/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
7. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[8,15,5],"center":0,"x":[1,2],"order":2}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([8, 15, 5], 0, 1/2, 2, [8, 15, 10, 0], [8, 15/2, 5/4], 67/4, 67/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([8, 15, 5], 0, 1/2, 2, [8, 15, 10, 0], [8, 15/2, 5/4], 67/4, 67/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
8. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[9,17,8,16],"center":0,"x":[1,2],"order":3}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([9, 17, 8, 16], 0, 1/2, 3, [9, 17, 16, 96, 0], [9, 17/2, 2, 2], 43/2, 43/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([9, 17, 8, 16], 0, 1/2, 3, [9, 17, 16, 96, 0], [9, 17/2, 2, 2], 43/2, 43/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
9. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[10,2,11,3,12],"center":0,"x":[1,2],"order":4}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([10, 2, 11, 3, 12], 0, 1/2, 4, [10, 2, 22, 18, 288, 0], [10, 1, 11/4, 3/8, 3/4], 119/8, 119/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([10, 2, 11, 3, 12], 0, 1/2, 4, [10, 2, 22, 18, 288, 0], [10, 1, 11/4, 3/8, 3/4], 119/8, 119/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
10. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[11,4,14,7,17,10],"center":0,"x":[1,2],"order":5}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([11, 4, 14, 7, 17, 10], 0, 1/2, 5, [11, 4, 28, 42, 408, 1200, 0], [11, 2, 7/2, 7/8, 17/16, 5/16], 75/4, 75/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([11, 4, 14, 7, 17, 10], 0, 1/2, 5, [11, 4, 28, 42, 408, 1200, 0], [11, 2, 7/2, 7/8, 17/16, 5/16], 75/4, 75/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
11. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[12,6],"center":0,"x":[1,2],"order":1}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([12, 6], 0, 1/2, 1, [12, 6, 0], [12, 3], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([12, 6], 0, 1/2, 1, [12, 6, 0], [12, 3], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
12. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[13,8,3],"center":0,"x":[1,2],"order":2}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([13, 8, 3], 0, 1/2, 2, [13, 8, 6, 0], [13, 4, 3/4], 71/4, 71/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([13, 8, 3], 0, 1/2, 2, [13, 8, 6, 0], [13, 4, 3/4], 71/4, 71/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
13. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[14,10,6,2],"center":0,"x":[1,2],"order":3}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([14, 10, 6, 2], 0, 1/2, 3, [14, 10, 12, 12, 0], [14, 5, 3/2, 1/4], 83/4, 83/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([14, 10, 6, 2], 0, 1/2, 3, [14, 10, 12, 12, 0], [14, 5, 3/2, 1/4], 83/4, 83/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
14. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[15,12,9,6,3],"center":0,"x":[1,2],"order":4}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([15, 12, 9, 6, 3], 0, 1/2, 4, [15, 12, 18, 36, 72, 0], [15, 6, 9/4, 3/4, 3/16], 387/16, 387/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([15, 12, 9, 6, 3], 0, 1/2, 4, [15, 12, 18, 36, 72, 0], [15, 6, 9/4, 3/4, 3/16], 387/16, 387/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
15. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[16,14,12,10,8,6],"center":0,"x":[1,2],"order":5}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([16, 14, 12, 10, 8, 6], 0, 1/2, 5, [16, 14, 24, 60, 192, 720, 0], [16, 7, 3, 5/4, 1/2, 3/16], 447/16, 447/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([16, 14, 12, 10, 8, 6], 0, 1/2, 5, [16, 14, 24, 60, 192, 720, 0], [16, 7, 3, 5/4, 1/2, 3/16], 447/16, 447/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
16. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[17,16],"center":0,"x":[1,2],"order":1}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([17, 16], 0, 1/2, 1, [17, 16, 0], [17, 8], 25, 25, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([17, 16], 0, 1/2, 1, [17, 16, 0], [17, 8], 25, 25, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
17. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[1,1,1],"center":0,"x":[1,2],"order":2}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([1, 1, 1], 0, 1/2, 2, [1, 1, 2, 0], [1, 1/2, 1/4], 7/4, 7/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([1, 1, 1], 0, 1/2, 2, [1, 1, 2, 0], [1, 1/2, 1/4], 7/4, 7/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
18. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[2,3,4,5],"center":0,"x":[1,2],"order":3}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([2, 3, 4, 5], 0, 1/2, 3, [2, 3, 8, 30, 0], [2, 3/2, 1, 5/8], 41/8, 41/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([2, 3, 4, 5], 0, 1/2, 3, [2, 3, 8, 30, 0], [2, 3/2, 1, 5/8], 41/8, 41/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
19. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[3,5,7,9,11],"center":0,"x":[1,2],"order":4}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([3, 5, 7, 9, 11], 0, 1/2, 4, [3, 5, 14, 54, 264, 0], [3, 5/2, 7/4, 9/8, 11/16], 145/16, 145/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([3, 5, 7, 9, 11], 0, 1/2, 4, [3, 5, 14, 54, 264, 0], [3, 5/2, 7/4, 9/8, 11/16], 145/16, 145/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
20. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[4,7,10,13,16,2],"center":0,"x":[1,2],"order":5}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([4, 7, 10, 13, 16, 2], 0, 1/2, 5, [4, 7, 20, 78, 384, 240, 0], [4, 7/2, 5/2, 13/8, 1, 1/16], 203/16, 203/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([4, 7, 10, 13, 16, 2], 0, 1/2, 5, [4, 7, 20, 78, 384, 240, 0], [4, 7/2, 5/2, 13/8, 1, 1/16], 203/16, 203/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
21. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[5,9],"center":0,"x":[1,2],"order":1}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([5, 9], 0, 1/2, 1, [5, 9, 0], [5, 9/2], 19/2, 19/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([5, 9], 0, 1/2, 1, [5, 9, 0], [5, 9/2], 19/2, 19/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
22. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[6,11,16],"center":0,"x":[1,2],"order":2}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([6, 11, 16], 0, 1/2, 2, [6, 11, 32, 0], [6, 11/2, 4], 31/2, 31/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([6, 11, 16], 0, 1/2, 2, [6, 11, 32, 0], [6, 11/2, 4], 31/2, 31/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
23. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[7,13,2,8],"center":0,"x":[1,2],"order":3}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([7, 13, 2, 8], 0, 1/2, 3, [7, 13, 4, 48, 0], [7, 13/2, 1/2, 1], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([7, 13, 2, 8], 0, 1/2, 3, [7, 13, 4, 48, 0], [7, 13/2, 1/2, 1], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
24. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[8,15,5,12,2],"center":0,"x":[1,2],"order":4}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([8, 15, 5, 12, 2], 0, 1/2, 4, [8, 15, 10, 72, 48, 0], [8, 15/2, 5/4, 3/2, 1/8], 147/8, 147/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([8, 15, 5, 12, 2], 0, 1/2, 4, [8, 15, 10, 72, 48, 0], [8, 15/2, 5/4, 3/2, 1/8], 147/8, 147/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
25. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-qualification；資料={"coefficients":[9,17,8,16,7,15],"center":0,"x":[1,2],"order":5}。請依 [coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary] 欄序回答。
查看答案與解釋
([9, 17, 8, 16, 7, 15], 0, 1/2, 5, [9, 17, 16, 96, 168, 1800, 0], [9, 17/2, 2, 2, 7/16, 15/32], 717/32, 717/32, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)
Wave6第十層核對=([9, 17, 8, 16, 7, 15], 0, 1/2, 5, [9, 17, 16, 96, 168, 1800, 0], [9, 17/2, 2, 2, 7/16, 15/32], 717/32, 717/32, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-qualification · candidate
26. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[2,3],"center":0,"x":[1,2],"order":1}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(1, 1, [2, 3/2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(1, 1, [2, 3/2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
27. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[3,5,7],"center":0,"x":[1,2],"order":2}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(2, 2, [3, 5/2, 7/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(2, 2, [3, 5/2, 7/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
28. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[4,7,10,13],"center":0,"x":[1,2],"order":3}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(3, 3, [4, 7/2, 5/2, 13/8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(3, 3, [4, 7/2, 5/2, 13/8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
29. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[5,9,13,17,4],"center":0,"x":[1,2],"order":4}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(4, 4, [5, 9/2, 13/4, 17/8, 1/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(4, 4, [5, 9/2, 13/4, 17/8, 1/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
30. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[6,11,16,4,9,14],"center":0,"x":[1,2],"order":5}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(5, 5, [6, 11/2, 4, 1/2, 9/16, 7/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(5, 5, [6, 11/2, 4, 1/2, 9/16, 7/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
31. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[7,13],"center":0,"x":[1,2],"order":1}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(1, 1, [7, 13/2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(1, 1, [7, 13/2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
32. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[8,15,5],"center":0,"x":[1,2],"order":2}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(2, 2, [8, 15/2, 5/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(2, 2, [8, 15/2, 5/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
33. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[9,17,8,16],"center":0,"x":[1,2],"order":3}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(3, 3, [9, 17/2, 2, 2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(3, 3, [9, 17/2, 2, 2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
34. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[10,2,11,3,12],"center":0,"x":[1,2],"order":4}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(4, 4, [10, 1, 11/4, 3/8, 3/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(4, 4, [10, 1, 11/4, 3/8, 3/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
35. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[11,4,14,7,17,10],"center":0,"x":[1,2],"order":5}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(5, 5, [11, 2, 7/2, 7/8, 17/16, 5/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(5, 5, [11, 2, 7/2, 7/8, 17/16, 5/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
36. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[12,6],"center":0,"x":[1,2],"order":1}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(1, 1, [12, 3], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(1, 1, [12, 3], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
37. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[13,8,3],"center":0,"x":[1,2],"order":2}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(2, 2, [13, 4, 3/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(2, 2, [13, 4, 3/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
38. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[14,10,6,2],"center":0,"x":[1,2],"order":3}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(3, 3, [14, 5, 3/2, 1/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(3, 3, [14, 5, 3/2, 1/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
39. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[15,12,9,6,3],"center":0,"x":[1,2],"order":4}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(4, 4, [15, 6, 9/4, 3/4, 3/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(4, 4, [15, 6, 9/4, 3/4, 3/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
40. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[16,14,12,10,8,6],"center":0,"x":[1,2],"order":5}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(5, 5, [16, 7, 3, 5/4, 1/2, 3/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(5, 5, [16, 7, 3, 5/4, 1/2, 3/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
41. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[17,16],"center":0,"x":[1,2],"order":1}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(1, 1, [17, 8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(1, 1, [17, 8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
42. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[1,1,1],"center":0,"x":[1,2],"order":2}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(2, 2, [1, 1/2, 1/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(2, 2, [1, 1/2, 1/4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
43. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[2,3,4,5],"center":0,"x":[1,2],"order":3}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(3, 3, [2, 3/2, 1, 5/8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(3, 3, [2, 3/2, 1, 5/8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
44. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[3,5,7,9,11],"center":0,"x":[1,2],"order":4}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(4, 4, [3, 5/2, 7/4, 9/8, 11/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(4, 4, [3, 5/2, 7/4, 9/8, 11/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
45. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[4,7,10,13,16,2],"center":0,"x":[1,2],"order":5}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(5, 5, [4, 7/2, 5/2, 13/8, 1, 1/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(5, 5, [4, 7/2, 5/2, 13/8, 1, 1/16], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
46. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[5,9],"center":0,"x":[1,2],"order":1}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(1, 1, [5, 9/2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(1, 1, [5, 9/2], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
47. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[6,11,16],"center":0,"x":[1,2],"order":2}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(2, 2, [6, 11/2, 4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(2, 2, [6, 11/2, 4], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
48. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[7,13,2,8],"center":0,"x":[1,2],"order":3}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(3, 3, [7, 13/2, 1/2, 1], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(3, 3, [7, 13/2, 1/2, 1], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
49. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[8,15,5,12,2],"center":0,"x":[1,2],"order":4}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(4, 4, [8, 15/2, 5/4, 3/2, 1/8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(4, 4, [8, 15/2, 5/4, 3/2, 1/8], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
50. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-zero-remainder；資料={"coefficients":[9,17,8,16,7,15],"center":0,"x":[1,2],"order":5}。請依 [degree, order, Taylor_terms, remainder, zero_remainder, certificate] 欄序回答。
查看答案與解釋
(5, 5, [9, 17/2, 2, 2, 7/16, 15/32], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)
Wave6第十層核對=(5, 5, [9, 17/2, 2, 2, 7/16, 15/32], 0, 是, degree_at_most_n_polynomial_has_zero_remainder)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-zero-remainder · candidate
51. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[2,3,2],"center":0,"x":[1,2],"order":1}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(2, 1, 4, 1/4, 是, 1/2, 1/2, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(2, 1, 4, 1/4, 是, 1/2, 1/2, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
52. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[3,5,7,3],"center":0,"x":[1,2],"order":2}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(3, 2, 18, 1/4, 是, 3/8, 3/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(3, 2, 18, 1/4, 是, 3/8, 3/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
53. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[4,7,10,13,4],"center":0,"x":[1,2],"order":3}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(4, 3, 96, 1/4, 是, 1/4, 1/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(4, 3, 96, 1/4, 是, 1/4, 1/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
54. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[5,9,13,17,4,5],"center":0,"x":[1,2],"order":4}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(5, 4, 600, 1/4, 是, 5/32, 5/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(5, 4, 600, 1/4, 是, 5/32, 5/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
55. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[6,11,16,4,9,14,6],"center":0,"x":[1,2],"order":5}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(6, 5, 4320, 1/4, 是, 3/32, 3/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(6, 5, 4320, 1/4, 是, 3/32, 3/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
56. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[7,13,7],"center":0,"x":[1,2],"order":1}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(2, 1, 14, 1/4, 是, 7/4, 7/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(2, 1, 14, 1/4, 是, 7/4, 7/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
57. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[8,15,5,8],"center":0,"x":[1,2],"order":2}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(3, 2, 48, 1/4, 是, 1, 1, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(3, 2, 48, 1/4, 是, 1, 1, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
58. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[9,17,8,16,9],"center":0,"x":[1,2],"order":3}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(4, 3, 216, 1/4, 是, 9/16, 9/16, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(4, 3, 216, 1/4, 是, 9/16, 9/16, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
59. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[10,2,11,3,12,10],"center":0,"x":[1,2],"order":4}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(5, 4, 1200, 1/4, 是, 5/16, 5/16, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(5, 4, 1200, 1/4, 是, 5/16, 5/16, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
60. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[11,4,14,7,17,10,11],"center":0,"x":[1,2],"order":5}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(6, 5, 7920, 1/4, 是, 11/64, 11/64, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(6, 5, 7920, 1/4, 是, 11/64, 11/64, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
61. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[12,6,12],"center":0,"x":[1,2],"order":1}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(2, 1, 24, 1/4, 是, 3, 3, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(2, 1, 24, 1/4, 是, 3, 3, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
62. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[13,8,3,13],"center":0,"x":[1,2],"order":2}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(3, 2, 78, 1/4, 是, 13/8, 13/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(3, 2, 78, 1/4, 是, 13/8, 13/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
63. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[14,10,6,2,14],"center":0,"x":[1,2],"order":3}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(4, 3, 336, 1/4, 是, 7/8, 7/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(4, 3, 336, 1/4, 是, 7/8, 7/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
64. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[15,12,9,6,3,15],"center":0,"x":[1,2],"order":4}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(5, 4, 1800, 1/4, 是, 15/32, 15/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(5, 4, 1800, 1/4, 是, 15/32, 15/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
65. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[16,14,12,10,8,6,16],"center":0,"x":[1,2],"order":5}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(6, 5, 11520, 1/4, 是, 1/4, 1/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(6, 5, 11520, 1/4, 是, 1/4, 1/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
66. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[17,16,17],"center":0,"x":[1,2],"order":1}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(2, 1, 34, 1/4, 是, 17/4, 17/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(2, 1, 34, 1/4, 是, 17/4, 17/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
67. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[1,1,1,18],"center":0,"x":[1,2],"order":2}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(3, 2, 108, 1/4, 是, 9/4, 9/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(3, 2, 108, 1/4, 是, 9/4, 9/4, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
68. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[2,3,4,5,19],"center":0,"x":[1,2],"order":3}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(4, 3, 456, 1/4, 是, 19/16, 19/16, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(4, 3, 456, 1/4, 是, 19/16, 19/16, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
69. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[3,5,7,9,11,20],"center":0,"x":[1,2],"order":4}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(5, 4, 2400, 1/4, 是, 5/8, 5/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(5, 4, 2400, 1/4, 是, 5/8, 5/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
70. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[4,7,10,13,16,2,21],"center":0,"x":[1,2],"order":5}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(6, 5, 15120, 1/4, 是, 21/64, 21/64, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(6, 5, 15120, 1/4, 是, 21/64, 21/64, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
71. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[5,9,22],"center":0,"x":[1,2],"order":1}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(2, 1, 44, 1/4, 是, 11/2, 11/2, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(2, 1, 44, 1/4, 是, 11/2, 11/2, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
72. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[6,11,16,23],"center":0,"x":[1,2],"order":2}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(3, 2, 138, 1/4, 是, 23/8, 23/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(3, 2, 138, 1/4, 是, 23/8, 23/8, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
73. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[7,13,2,8,24],"center":0,"x":[1,2],"order":3}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(4, 3, 576, 1/4, 是, 3/2, 3/2, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(4, 3, 576, 1/4, 是, 3/2, 3/2, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
74. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[8,15,5,12,2,25],"center":0,"x":[1,2],"order":4}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(5, 4, 3000, 1/4, 是, 25/32, 25/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(5, 4, 3000, 1/4, 是, 25/32, 25/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
75. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-lagrange-witness；資料={"coefficients":[9,17,8,16,7,15,26],"center":0,"x":[1,2],"order":5}。請依 [degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate] 欄序回答。
查看答案與解釋
(6, 5, 18720, 1/4, 是, 13/32, 13/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)
Wave6第十層核對=(6, 5, 18720, 1/4, 是, 13/32, 13/32, 是, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-lagrange-witness · candidate
76. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[2,3,2],"center":0,"x":[1,2],"order":1,"M":4}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(valid_bound, 1/2, 4, 4, 1/2, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(valid_bound, 1/2, 4, 4, 1/2, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
77. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[3,5,7,3],"center":0,"x":[1,2],"order":2,"M":17}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(undersized_M, 3/8, 18, 17, 17/48, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(undersized_M, 3/8, 18, 17, 17/48, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
78. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[4,7,10,13,4],"center":0,"x":[1,2],"order":3,"M":96}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(valid_bound, 1/4, 96, 96, 1/4, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(valid_bound, 1/4, 96, 96, 1/4, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
79. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[5,9,13,17,4,5],"center":0,"x":[1,2],"order":4,"M":599}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(undersized_M, 5/32, 600, 599, 599/3840, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(undersized_M, 5/32, 600, 599, 599/3840, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
80. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[6,11,16,4,9,14,6],"center":0,"x":[1,2],"order":5,"M":4320}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(valid_bound, 3/32, 4320, 4320, 3/32, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(valid_bound, 3/32, 4320, 4320, 3/32, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
81. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[7,13,7],"center":0,"x":[1,2],"order":1,"M":13}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(undersized_M, 7/4, 14, 13, 13/8, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(undersized_M, 7/4, 14, 13, 13/8, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
82. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[8,15,5,8],"center":0,"x":[1,2],"order":2,"M":48}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(valid_bound, 1, 48, 48, 1, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(valid_bound, 1, 48, 48, 1, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
83. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[9,17,8,16,9],"center":0,"x":[1,2],"order":3,"M":215}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(undersized_M, 9/16, 216, 215, 215/384, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(undersized_M, 9/16, 216, 215, 215/384, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
84. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[10,2,11,3,12,10],"center":0,"x":[1,2],"order":4,"M":1200}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(valid_bound, 5/16, 1200, 1200, 5/16, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(valid_bound, 5/16, 1200, 1200, 5/16, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
85. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[11,4,14,7,17,10,11],"center":0,"x":[1,2],"order":5,"M":7919}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(undersized_M, 11/64, 7920, 7919, 7919/46080, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(undersized_M, 11/64, 7920, 7919, 7919/46080, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
86. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[12,6,12],"center":0,"x":[1,2],"order":1,"M":24}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(valid_bound, 3, 24, 24, 3, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(valid_bound, 3, 24, 24, 3, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
87. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[13,8,3,13],"center":0,"x":[1,2],"order":2,"M":77}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(undersized_M, 13/8, 78, 77, 77/48, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(undersized_M, 13/8, 78, 77, 77/48, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
88. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[14,10,6,2,14],"center":0,"x":[1,2],"order":3,"M":336}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(valid_bound, 7/8, 336, 336, 7/8, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(valid_bound, 7/8, 336, 336, 7/8, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
89. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[15,12,9,6,3,15],"center":0,"x":[1,2],"order":4,"M":1799}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(undersized_M, 15/32, 1800, 1799, 1799/3840, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(undersized_M, 15/32, 1800, 1799, 1799/3840, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
90. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[16,14,12,10,8,6,16],"center":0,"x":[1,2],"order":5,"M":11520}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(valid_bound, 1/4, 11520, 11520, 1/4, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(valid_bound, 1/4, 11520, 11520, 1/4, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
91. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[17,16,17],"center":0,"x":[1,2],"order":1,"M":33}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(undersized_M, 17/4, 34, 33, 33/8, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(undersized_M, 17/4, 34, 33, 33/8, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
92. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[1,1,1,18],"center":0,"x":[1,2],"order":2,"M":108}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(valid_bound, 9/4, 108, 108, 9/4, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(valid_bound, 9/4, 108, 108, 9/4, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
93. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[2,3,4,5,19],"center":0,"x":[1,2],"order":3,"M":455}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(undersized_M, 19/16, 456, 455, 455/384, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(undersized_M, 19/16, 456, 455, 455/384, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
94. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[3,5,7,9,11,20],"center":0,"x":[1,2],"order":4,"M":2400}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(valid_bound, 5/8, 2400, 2400, 5/8, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(valid_bound, 5/8, 2400, 2400, 5/8, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
95. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[4,7,10,13,16,2,21],"center":0,"x":[1,2],"order":5,"M":15119}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(undersized_M, 21/64, 15120, 15119, 15119/46080, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(undersized_M, 21/64, 15120, 15119, 15119/46080, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
96. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[5,9,22],"center":0,"x":[1,2],"order":1,"M":44}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(valid_bound, 11/2, 44, 44, 11/2, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(valid_bound, 11/2, 44, 44, 11/2, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
97. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[6,11,16,23],"center":0,"x":[1,2],"order":2,"M":137}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(undersized_M, 23/8, 138, 137, 137/48, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(undersized_M, 23/8, 138, 137, 137/48, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
98. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[7,13,2,8,24],"center":0,"x":[1,2],"order":3,"M":576}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(valid_bound, 3/2, 576, 576, 3/2, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(valid_bound, 3/2, 576, 576, 3/2, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
99. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"undersized_M","coefficients":[8,15,5,12,2,25],"center":0,"x":[1,2],"order":4,"M":2999}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(undersized_M, 25/32, 3000, 2999, 2999/3840, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(undersized_M, 25/32, 3000, 2999, 2999/3840, 否, 否, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
100. Taylor餘項定理
center、order、interval-wide n+1 differentiability與Taylor remainder責任。操作=taylor-bound-boundary；資料={"mode":"valid_bound","coefficients":[9,17,8,16,7,15,26],"center":0,"x":[1,2],"order":5,"M":18720}。請依 [mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary] 欄序回答。
actual_error_inequality與qualified_bound_certificate分開；若x=center，即使M不足，0≤0仍可能成立，不能據此認證derivative bound。
查看答案與解釋
(valid_bound, 13/32, 18720, 18720, 13/32, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)
Wave6第十層核對=(valid_bound, 13/32, 18720, 18720, 13/32, 是, 是, 否, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)；依完整明示資料獨立重算；polynomial zero/exact Lagrange cases與sup-derivative bound分開；finite jet不證一般remainder、Taylor-series equality或analyticity。
本題逐步詳解（AI 二次覆核）
Taylor餘項定理 · qf-mod47-taylor-bound-boundary · candidate
原始題目 JSONL · 驗收報告