第 26401 題：Taylor餘項定理

原題與已知條件

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 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 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 exact_ordered_tuple_declared_by_interface 欄序回答。

定義與適用範圍

沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。

• mko-taylor-remainder-theorem

逐步推導

source roles = [taylor-bound-boundary]

derived = [[mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary]]

Taylor ORIGINAL x minus declared center := sub(1/2, 0)

Taylor ORIGINAL x minus declared center = 1/2

(0)^0

Taylor ORIGINAL derivative0 center power0 term0 = 1

Taylor ORIGINAL derivative0 falling factor times power term0 := mul(1, 1)

Taylor ORIGINAL derivative0 falling factor times power term0 = 1

Taylor ORIGINAL derivative0 ORIGINAL coefficient term0 := mul(2, 1)

Taylor ORIGINAL derivative0 ORIGINAL coefficient term0 = 2

Taylor ORIGINAL derivative0 ascending derivative sum term0 := sum(0, 2)

Taylor ORIGINAL derivative0 ascending derivative sum term0 = 2

(0)^1

Taylor ORIGINAL derivative0 center power1 term1 = 0

Taylor ORIGINAL derivative0 falling factor times power term1 := mul(1, 0)

Taylor ORIGINAL derivative0 falling factor times power term1 = 0

Taylor ORIGINAL derivative0 ORIGINAL coefficient term1 := mul(3, 0)

Taylor ORIGINAL derivative0 ORIGINAL coefficient term1 = 0

Taylor ORIGINAL derivative0 ascending derivative sum term1 := sum(2, 0)

Taylor ORIGINAL derivative0 ascending derivative sum term1 = 2

(0)^2

Taylor ORIGINAL derivative0 center power2 term2 = 0

Taylor ORIGINAL derivative0 falling factor times power term2 := mul(1, 0)

Taylor ORIGINAL derivative0 falling factor times power term2 = 0

Taylor ORIGINAL derivative0 ORIGINAL coefficient term2 := mul(2, 0)

Taylor ORIGINAL derivative0 ORIGINAL coefficient term2 = 0

Taylor ORIGINAL derivative0 ascending derivative sum term2 := sum(2, 0)

Taylor ORIGINAL derivative0 ascending derivative sum term2 = 2

source roles = [0, 0, [2, 3, 2]]

derived = [[[0, 2, 1, 1, 2], [1, 3, 1, 0, 0], [2, 2, 1, 0, 0]], 2]

(1/2)^0

Taylor ORIGINAL displacement power0 = 1

Taylor ORIGINAL derivative times displacement0 := mul(2, 1)

Taylor ORIGINAL derivative times displacement0 = 2

Taylor ORIGINAL term0 divide factorial := div(2, 1)

Taylor ORIGINAL term0 divide factorial = 2

(0)^0

Taylor ORIGINAL derivative1 center power0 term1 = 1

Taylor ORIGINAL derivative1 falling factor times power term1 := mul(1, 1)

Taylor ORIGINAL derivative1 falling factor times power term1 = 1

Taylor ORIGINAL derivative1 ORIGINAL coefficient term1 := mul(3, 1)

Taylor ORIGINAL derivative1 ORIGINAL coefficient term1 = 3

Taylor ORIGINAL derivative1 ascending derivative sum term1 := sum(0, 3)

Taylor ORIGINAL derivative1 ascending derivative sum term1 = 3

(0)^1

Taylor ORIGINAL derivative1 center power1 term2 = 0

Taylor ORIGINAL derivative1 falling factor times power term2 := mul(2, 0)

Taylor ORIGINAL derivative1 falling factor times power term2 = 0

Taylor ORIGINAL derivative1 ORIGINAL coefficient term2 := mul(2, 0)

Taylor ORIGINAL derivative1 ORIGINAL coefficient term2 = 0

Taylor ORIGINAL derivative1 ascending derivative sum term2 := sum(3, 0)

Taylor ORIGINAL derivative1 ascending derivative sum term2 = 3

source roles = [1, 0, [2, 3, 2]]

derived = [[[1, 3, 1, 1, 3], [2, 2, 2, 0, 0]], 3]

(1/2)^1

Taylor ORIGINAL displacement power1 = 1/2

Taylor ORIGINAL derivative times displacement1 := mul(3, 1/2)

Taylor ORIGINAL derivative times displacement1 = 3/2

Taylor ORIGINAL term1 divide factorial := div(3/2, 1)

Taylor ORIGINAL term1 divide factorial = 3/2

Taylor ORIGINAL partial sum0 := sum(0, 2)

Taylor ORIGINAL partial sum0 = 2

Taylor ORIGINAL partial sum1 := sum(2, 3/2)

Taylor ORIGINAL partial sum1 = 7/2

(1/2)^0

Taylor ORIGINAL full function at x ORIGINAL argument power0 = 1

Taylor ORIGINAL full function at x coefficient term0 := mul(2, 1)

Taylor ORIGINAL full function at x coefficient term0 = 2

Taylor ORIGINAL full function at x ascending partial sum0 := sum(0, 2)

Taylor ORIGINAL full function at x ascending partial sum0 = 2

(1/2)^1

Taylor ORIGINAL full function at x ORIGINAL argument power1 = 1/2

Taylor ORIGINAL full function at x coefficient term1 := mul(3, 1/2)

Taylor ORIGINAL full function at x coefficient term1 = 3/2

Taylor ORIGINAL full function at x ascending partial sum1 := sum(2, 3/2)

Taylor ORIGINAL full function at x ascending partial sum1 = 7/2

(1/2)^2

Taylor ORIGINAL full function at x ORIGINAL argument power2 = 1/4

Taylor ORIGINAL full function at x coefficient term2 := mul(2, 1/4)

Taylor ORIGINAL full function at x coefficient term2 = 1/2

Taylor ORIGINAL full function at x ascending partial sum2 := sum(7/2, 1/2)

Taylor ORIGINAL full function at x ascending partial sum2 = 4

Taylor ORIGINAL f(x) minus Taylor value := sub(4, 7/2)

Taylor ORIGINAL f(x) minus Taylor value = 1/2

source roles = [[2, 3, 2], 0, 1/2, 1, [2, 3], [2, 3/2]]

derived = [7/2, 4, 1/2, 2]

(0)^0

Taylor ORIGINAL next derivative center power0 term2 = 1

Taylor ORIGINAL next derivative falling factor times power term2 := mul(2, 1)

Taylor ORIGINAL next derivative falling factor times power term2 = 2

Taylor ORIGINAL next derivative ORIGINAL coefficient term2 := mul(2, 2)

Taylor ORIGINAL next derivative ORIGINAL coefficient term2 = 4

Taylor ORIGINAL next derivative ascending derivative sum term2 := sum(0, 4)

Taylor ORIGINAL next derivative ascending derivative sum term2 = 4

source roles = [2, 0, [2, 3, 2]]

derived = [[[2, 2, 2, 1, 4]], 4]

(1/2)^2

Taylor bound ORIGINAL absolute displacement power = 1/4

Taylor bound supplied M times power := mul(4, 1/4)

Taylor bound supplied M times power = 1

Taylor bound divide next factorial := div(1, 2)

Taylor bound divide next factorial = 1/2

source roles = [valid_bound, 1/2, 4, 4, 1/2]

derived = [1/2, 是, 是, 否]

逐步反向核對

• S052 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S051 · premise_and_result_audit — 以精確有理數重算 Taylor bound divide next factorial，不以小數近似替代。
• S050 · premise_and_result_audit — 以精確有理數重算 Taylor bound supplied M times power，不以小數近似替代。
• S049 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S048 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S047 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL next derivative ascending derivative sum term2，不以小數近似替代。
• S046 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL next derivative ORIGINAL coefficient term2，不以小數近似替代。
• S045 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL next derivative falling factor times power term2，不以小數近似替代。
• S044 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S043 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S042 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL f(x) minus Taylor value，不以小數近似替代。
• S041 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL full function at x ascending partial sum2，不以小數近似替代。
• S040 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL full function at x coefficient term2，不以小數近似替代。
• S039 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S038 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL full function at x ascending partial sum1，不以小數近似替代。
• S037 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL full function at x coefficient term1，不以小數近似替代。
• S036 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S035 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL full function at x ascending partial sum0，不以小數近似替代。
• S034 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL full function at x coefficient term0，不以小數近似替代。
• S033 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S032 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL partial sum1，不以小數近似替代。
• S031 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL partial sum0，不以小數近似替代。
• S030 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL term1 divide factorial，不以小數近似替代。
• S029 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative times displacement1，不以小數近似替代。
• S028 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S027 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S026 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative1 ascending derivative sum term2，不以小數近似替代。
• S025 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative1 ORIGINAL coefficient term2，不以小數近似替代。
• S024 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative1 falling factor times power term2，不以小數近似替代。
• S023 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S022 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative1 ascending derivative sum term1，不以小數近似替代。
• S021 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative1 ORIGINAL coefficient term1，不以小數近似替代。
• S020 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative1 falling factor times power term1，不以小數近似替代。
• S019 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S018 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL term0 divide factorial，不以小數近似替代。
• S017 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative times displacement0，不以小數近似替代。
• S016 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S015 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S014 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative0 ascending derivative sum term2，不以小數近似替代。
• S013 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative0 ORIGINAL coefficient term2，不以小數近似替代。
• S012 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative0 falling factor times power term2，不以小數近似替代。
• S011 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S010 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative0 ascending derivative sum term1，不以小數近似替代。
• S009 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative0 ORIGINAL coefficient term1，不以小數近似替代。
• S008 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative0 falling factor times power term1，不以小數近似替代。
• S007 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S006 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative0 ascending derivative sum term0，不以小數近似替代。
• S005 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative0 ORIGINAL coefficient term0，不以小數近似替代。
• S004 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL derivative0 falling factor times power term0，不以小數近似替代。
• S003 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S002 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL x minus declared center，不以小數近似替代。
• S001 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。

結論

(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)

依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。

本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。