第 24024 題：Taylor多項式

原題與已知條件

固定center、nonnegative order與完整zero-through-n derivative jet。操作=taylor-coefficient-construction；資料={"derivatives":[24,48,72,96,120,144],"center":11,"order":5}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 固定center、nonnegative order與完整zero-through-n derivative jet。操作=taylor-coefficient-construction；資料={"derivatives":[24,48,72,96,120,144],"center":11,"order":5}。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-polynomial

逐步推導

source roles = [taylor-coefficient-construction]

derived = [[center, order, shifted_coefficients, indexed_coefficients, certificate]]

source roles = [11, 5, [24, 48, 72, 96, 120, 144]]

derived = [[0, 1, 2, 3, 4, 5], 6, 是]

source roles = [0]

derived = [1]

Taylor coefficient0 derivative over factorial := div(24, 1)

Taylor coefficient0 derivative over factorial = 24

Taylor k1 factorial step1 := mul(1, 1)

Taylor k1 factorial step1 = 1

Taylor coefficient1 derivative over factorial := div(48, 1)

Taylor coefficient1 derivative over factorial = 48

Taylor k2 factorial step1 := mul(1, 1)

Taylor k2 factorial step1 = 1

Taylor k2 factorial step2 := mul(1, 2)

Taylor k2 factorial step2 = 2

Taylor coefficient2 derivative over factorial := div(72, 2)

Taylor coefficient2 derivative over factorial = 36

Taylor k3 factorial step1 := mul(1, 1)

Taylor k3 factorial step1 = 1

Taylor k3 factorial step2 := mul(1, 2)

Taylor k3 factorial step2 = 2

Taylor k3 factorial step3 := mul(2, 3)

Taylor k3 factorial step3 = 6

Taylor coefficient3 derivative over factorial := div(96, 6)

Taylor coefficient3 derivative over factorial = 16

Taylor k4 factorial step1 := mul(1, 1)

Taylor k4 factorial step1 = 1

Taylor k4 factorial step2 := mul(1, 2)

Taylor k4 factorial step2 = 2

Taylor k4 factorial step3 := mul(2, 3)

Taylor k4 factorial step3 = 6

Taylor k4 factorial step4 := mul(6, 4)

Taylor k4 factorial step4 = 24

Taylor coefficient4 derivative over factorial := div(120, 24)

Taylor coefficient4 derivative over factorial = 5

Taylor k5 factorial step1 := mul(1, 1)

Taylor k5 factorial step1 = 1

Taylor k5 factorial step2 := mul(1, 2)

Taylor k5 factorial step2 = 2

Taylor k5 factorial step3 := mul(2, 3)

Taylor k5 factorial step3 = 6

Taylor k5 factorial step4 := mul(6, 4)

Taylor k5 factorial step4 = 24

Taylor k5 factorial step5 := mul(24, 5)

Taylor k5 factorial step5 = 120

Taylor coefficient5 derivative over factorial := div(144, 120)

Taylor coefficient5 derivative over factorial = 6/5

source roles = [11, 5, [24, 48, 36, 16, 5, 6/5]]

derived = [T_n(x)=sum c_k*(x-center)^k, T_n^(j)(center)=declared derivative[j] for EVERY0<=j<=n, no remainder or infinite-series certificate]

逐步反向核對

• 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 coefficient5 derivative over factorial，不以小數近似替代。
• S023 · premise_and_result_audit — 以精確有理數重算 Taylor k5 factorial step5，不以小數近似替代。
• S022 · premise_and_result_audit — 以精確有理數重算 Taylor k5 factorial step4，不以小數近似替代。
• S021 · premise_and_result_audit — 以精確有理數重算 Taylor k5 factorial step3，不以小數近似替代。
• S020 · premise_and_result_audit — 以精確有理數重算 Taylor k5 factorial step2，不以小數近似替代。
• S019 · premise_and_result_audit — 以精確有理數重算 Taylor k5 factorial step1，不以小數近似替代。
• S018 · premise_and_result_audit — 以精確有理數重算 Taylor coefficient4 derivative over factorial，不以小數近似替代。
• S017 · premise_and_result_audit — 以精確有理數重算 Taylor k4 factorial step4，不以小數近似替代。
• S016 · premise_and_result_audit — 以精確有理數重算 Taylor k4 factorial step3，不以小數近似替代。
• S015 · premise_and_result_audit — 以精確有理數重算 Taylor k4 factorial step2，不以小數近似替代。
• S014 · premise_and_result_audit — 以精確有理數重算 Taylor k4 factorial step1，不以小數近似替代。
• S013 · premise_and_result_audit — 以精確有理數重算 Taylor coefficient3 derivative over factorial，不以小數近似替代。
• S012 · premise_and_result_audit — 以精確有理數重算 Taylor k3 factorial step3，不以小數近似替代。
• S011 · premise_and_result_audit — 以精確有理數重算 Taylor k3 factorial step2，不以小數近似替代。
• S010 · premise_and_result_audit — 以精確有理數重算 Taylor k3 factorial step1，不以小數近似替代。
• S009 · premise_and_result_audit — 以精確有理數重算 Taylor coefficient2 derivative over factorial，不以小數近似替代。
• S008 · premise_and_result_audit — 以精確有理數重算 Taylor k2 factorial step2，不以小數近似替代。
• S007 · premise_and_result_audit — 以精確有理數重算 Taylor k2 factorial step1，不以小數近似替代。
• S006 · premise_and_result_audit — 以精確有理數重算 Taylor coefficient1 derivative over factorial，不以小數近似替代。
• S005 · premise_and_result_audit — 以精確有理數重算 Taylor k1 factorial step1，不以小數近似替代。
• S004 · premise_and_result_audit — 以精確有理數重算 Taylor coefficient0 derivative over factorial，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 — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• 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

(11, 5, [24, 48, 36, 16, 5, 6/5], [[0, 24], [1, 48], [2, 36], [3, 16], [4, 5], [5, 6/5]], coefficient_k_equals_derivative_k_over_k_factorial)

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。