第 26471 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-qualification；資料={"coefficients":[5,9],"center":0,"x":[1,2],"order":1}。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-qualification；資料={"coefficients":[5,9],"center":0,"x":[1,2],"order":1}。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-qualification]

derived = [[coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, 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(5, 1)

Taylor ORIGINAL derivative0 ORIGINAL coefficient term0 = 5

Taylor ORIGINAL derivative0 ascending derivative sum term0 := sum(0, 5)

Taylor ORIGINAL derivative0 ascending derivative sum term0 = 5

(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(9, 0)

Taylor ORIGINAL derivative0 ORIGINAL coefficient term1 = 0

Taylor ORIGINAL derivative0 ascending derivative sum term1 := sum(5, 0)

Taylor ORIGINAL derivative0 ascending derivative sum term1 = 5

source roles = [0, 0, [5, 9]]

derived = [[[0, 5, 1, 1, 5], [1, 9, 1, 0, 0]], 5]

(1/2)^0

Taylor ORIGINAL displacement power0 = 1

Taylor ORIGINAL derivative times displacement0 := mul(5, 1)

Taylor ORIGINAL derivative times displacement0 = 5

Taylor ORIGINAL term0 divide factorial := div(5, 1)

Taylor ORIGINAL term0 divide factorial = 5

(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(9, 1)

Taylor ORIGINAL derivative1 ORIGINAL coefficient term1 = 9

Taylor ORIGINAL derivative1 ascending derivative sum term1 := sum(0, 9)

Taylor ORIGINAL derivative1 ascending derivative sum term1 = 9

source roles = [1, 0, [5, 9]]

derived = [[[1, 9, 1, 1, 9]], 9]

(1/2)^1

Taylor ORIGINAL displacement power1 = 1/2

Taylor ORIGINAL derivative times displacement1 := mul(9, 1/2)

Taylor ORIGINAL derivative times displacement1 = 9/2

Taylor ORIGINAL term1 divide factorial := div(9/2, 1)

Taylor ORIGINAL term1 divide factorial = 9/2

Taylor ORIGINAL partial sum0 := sum(0, 5)

Taylor ORIGINAL partial sum0 = 5

Taylor ORIGINAL partial sum1 := sum(5, 9/2)

Taylor ORIGINAL partial sum1 = 19/2

(1/2)^0

Taylor ORIGINAL full function at x ORIGINAL argument power0 = 1

Taylor ORIGINAL full function at x coefficient term0 := mul(5, 1)

Taylor ORIGINAL full function at x coefficient term0 = 5

Taylor ORIGINAL full function at x ascending partial sum0 := sum(0, 5)

Taylor ORIGINAL full function at x ascending partial sum0 = 5

(1/2)^1

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

Taylor ORIGINAL full function at x coefficient term1 := mul(9, 1/2)

Taylor ORIGINAL full function at x coefficient term1 = 9/2

Taylor ORIGINAL full function at x ascending partial sum1 := sum(5, 9/2)

Taylor ORIGINAL full function at x ascending partial sum1 = 19/2

Taylor ORIGINAL f(x) minus Taylor value := sub(19/2, 19/2)

Taylor ORIGINAL f(x) minus Taylor value = 0

source roles = [[5, 9], 0, 1/2, 1, [5, 9], [5, 9/2]]

derived = [19/2, 19/2, 0, 1]

Taylor ORIGINAL next derivative EMPTY derivative sum := sum()

Taylor ORIGINAL next derivative EMPTY derivative sum = 0

source roles = [2, 0, [5, 9]]

derived = [[], 0]

逐步反向核對

• S034 · 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。
• S033 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL next derivative EMPTY derivative sum，不以小數近似替代。
• S032 · 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。
• S031 · 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。
• S030 · 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。
• S029 · 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。
• 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 — 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。
• S026 · 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。
• S025 · 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。
• S024 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL partial sum1，不以小數近似替代。
• S023 · premise_and_result_audit — 以精確有理數重算 Taylor ORIGINAL partial sum0，不以小數近似替代。
• S022 · 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。
• S021 · 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。
• S020 · 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。
• 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 derivative1 ascending derivative sum term1，不以小數近似替代。
• S017 · 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。
• S016 · 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。
• 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 term0 divide factorial，不以小數近似替代。
• S013 · 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。
• S012 · 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。
• 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

([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。