第 24063 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-exact-polynomial-reconstruction；資料={"shifted_coefficients":[13,26,39],"center":0,"point":0}。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-exact-polynomial-reconstruction；資料={"shifted_coefficients":[13,26,39],"center":0,"point":0}。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-exact-polynomial-reconstruction]

derived = [[shifted_coefficients, derivatives_at_center, reconstructed_coefficients, point, original_value, reconstructed_value, equal, certificate]]

source roles = [0, [13, 26, 39]]

derived = [P(x)=sum c_k*(x-center)^k, 2]

source roles = [0]

derived = [1]

Taylor derivative0 from shifted coefficient := mul(13, 1)

Taylor derivative0 from shifted coefficient = 13

Taylor recover coefficient0 := div(13, 1)

Taylor recover coefficient0 = 13

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

Taylor k1 factorial step1 = 1

Taylor derivative1 from shifted coefficient := mul(26, 1)

Taylor derivative1 from shifted coefficient = 26

Taylor recover coefficient1 := div(26, 1)

Taylor recover coefficient1 = 26

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 derivative2 from shifted coefficient := mul(39, 2)

Taylor derivative2 from shifted coefficient = 78

Taylor recover coefficient2 := div(78, 2)

Taylor recover coefficient2 = 39

Taylor SAME point minus declared center := sub(0, 0)

Taylor SAME point minus declared center = 0

Taylor original Horner product2 := mul(0, 0)

Taylor original Horner product2 = 0

Taylor original Horner sum2 := sum(0, 39)

Taylor original Horner sum2 = 39

Taylor original Horner product1 := mul(39, 0)

Taylor original Horner product1 = 0

Taylor original Horner sum1 := sum(0, 26)

Taylor original Horner sum1 = 26

Taylor original Horner product0 := mul(26, 0)

Taylor original Horner product0 = 0

Taylor original Horner sum0 := sum(0, 13)

Taylor original Horner sum0 = 13

Taylor reconstructed Horner product2 := mul(0, 0)

Taylor reconstructed Horner product2 = 0

Taylor reconstructed Horner sum2 := sum(0, 39)

Taylor reconstructed Horner sum2 = 39

Taylor reconstructed Horner product1 := mul(39, 0)

Taylor reconstructed Horner product1 = 0

Taylor reconstructed Horner sum1 := sum(0, 26)

Taylor reconstructed Horner sum1 = 26

Taylor reconstructed Horner product0 := mul(26, 0)

Taylor reconstructed Horner product0 = 0

Taylor reconstructed Horner sum0 := sum(0, 13)

Taylor reconstructed Horner sum0 = 13

source roles = [[13, 26, 39], [13, 26, 39], 0, 0]

derived = [是, 13, 13, 是]

逐步反向核對

• S026 · 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。
• S025 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner sum0，不以小數近似替代。
• S024 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner product0，不以小數近似替代。
• S023 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner sum1，不以小數近似替代。
• S022 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner product1，不以小數近似替代。
• S021 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner sum2，不以小數近似替代。
• S020 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner product2，不以小數近似替代。
• S019 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner sum0，不以小數近似替代。
• S018 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner product0，不以小數近似替代。
• S017 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner sum1，不以小數近似替代。
• S016 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner product1，不以小數近似替代。
• S015 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner sum2，不以小數近似替代。
• S014 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner product2，不以小數近似替代。
• S013 · premise_and_result_audit — 以精確有理數重算 Taylor SAME point minus declared center，不以小數近似替代。
• S012 · premise_and_result_audit — 以精確有理數重算 Taylor recover coefficient2，不以小數近似替代。
• S011 · premise_and_result_audit — 以精確有理數重算 Taylor derivative2 from shifted coefficient，不以小數近似替代。
• S010 · premise_and_result_audit — 以精確有理數重算 Taylor k2 factorial step2，不以小數近似替代。
• S009 · premise_and_result_audit — 以精確有理數重算 Taylor k2 factorial step1，不以小數近似替代。
• S008 · premise_and_result_audit — 以精確有理數重算 Taylor recover coefficient1，不以小數近似替代。
• S007 · premise_and_result_audit — 以精確有理數重算 Taylor derivative1 from shifted coefficient，不以小數近似替代。
• S006 · premise_and_result_audit — 以精確有理數重算 Taylor k1 factorial step1，不以小數近似替代。
• S005 · premise_and_result_audit — 以精確有理數重算 Taylor recover coefficient0，不以小數近似替代。
• S004 · premise_and_result_audit — 以精確有理數重算 Taylor derivative0 from shifted coefficient，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

([13, 26, 39], [13, 26, 78], [13, 26, 39], 0, 13, 13, 是, degree_at_most_n_polynomial_is_exactly_reconstructed)

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。