第 24061 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":[11,22,33,44,55],"center":-2,"point":-4}。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":[11,22,33,44,55],"center":-2,"point":-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-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 = [-2, [11, 22, 33, 44, 55]]

derived = [P(x)=sum c_k*(x-center)^k, 4]

source roles = [0]

derived = [1]

Taylor derivative0 from shifted coefficient := mul(11, 1)

Taylor derivative0 from shifted coefficient = 11

Taylor recover coefficient0 := div(11, 1)

Taylor recover coefficient0 = 11

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

Taylor k1 factorial step1 = 1

Taylor derivative1 from shifted coefficient := mul(22, 1)

Taylor derivative1 from shifted coefficient = 22

Taylor recover coefficient1 := div(22, 1)

Taylor recover coefficient1 = 22

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(33, 2)

Taylor derivative2 from shifted coefficient = 66

Taylor recover coefficient2 := div(66, 2)

Taylor recover coefficient2 = 33

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 derivative3 from shifted coefficient := mul(44, 6)

Taylor derivative3 from shifted coefficient = 264

Taylor recover coefficient3 := div(264, 6)

Taylor recover coefficient3 = 44

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 derivative4 from shifted coefficient := mul(55, 24)

Taylor derivative4 from shifted coefficient = 1320

Taylor recover coefficient4 := div(1320, 24)

Taylor recover coefficient4 = 55

Taylor SAME point minus declared center := sub(-4, -2)

Taylor SAME point minus declared center = -2

Taylor original Horner product4 := mul(0, -2)

Taylor original Horner product4 = 0

Taylor original Horner sum4 := sum(0, 55)

Taylor original Horner sum4 = 55

Taylor original Horner product3 := mul(55, -2)

Taylor original Horner product3 = -110

Taylor original Horner sum3 := sum(-110, 44)

Taylor original Horner sum3 = -66

Taylor original Horner product2 := mul(-66, -2)

Taylor original Horner product2 = 132

Taylor original Horner sum2 := sum(132, 33)

Taylor original Horner sum2 = 165

Taylor original Horner product1 := mul(165, -2)

Taylor original Horner product1 = -330

Taylor original Horner sum1 := sum(-330, 22)

Taylor original Horner sum1 = -308

Taylor original Horner product0 := mul(-308, -2)

Taylor original Horner product0 = 616

Taylor original Horner sum0 := sum(616, 11)

Taylor original Horner sum0 = 627

Taylor reconstructed Horner product4 := mul(0, -2)

Taylor reconstructed Horner product4 = 0

Taylor reconstructed Horner sum4 := sum(0, 55)

Taylor reconstructed Horner sum4 = 55

Taylor reconstructed Horner product3 := mul(55, -2)

Taylor reconstructed Horner product3 = -110

Taylor reconstructed Horner sum3 := sum(-110, 44)

Taylor reconstructed Horner sum3 = -66

Taylor reconstructed Horner product2 := mul(-66, -2)

Taylor reconstructed Horner product2 = 132

Taylor reconstructed Horner sum2 := sum(132, 33)

Taylor reconstructed Horner sum2 = 165

Taylor reconstructed Horner product1 := mul(165, -2)

Taylor reconstructed Horner product1 = -330

Taylor reconstructed Horner sum1 := sum(-330, 22)

Taylor reconstructed Horner sum1 = -308

Taylor reconstructed Horner product0 := mul(-308, -2)

Taylor reconstructed Horner product0 = 616

Taylor reconstructed Horner sum0 := sum(616, 11)

Taylor reconstructed Horner sum0 = 627

source roles = [[11, 22, 33, 44, 55], [11, 22, 33, 44, 55], -2, -4]

derived = [是, 627, 627, 是]

逐步反向核對

• S045 · 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。
• S044 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner sum0，不以小數近似替代。
• S043 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner product0，不以小數近似替代。
• S042 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner sum1，不以小數近似替代。
• S041 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner product1，不以小數近似替代。
• S040 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner sum2，不以小數近似替代。
• S039 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner product2，不以小數近似替代。
• S038 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner sum3，不以小數近似替代。
• S037 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner product3，不以小數近似替代。
• S036 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner sum4，不以小數近似替代。
• S035 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner product4，不以小數近似替代。
• S034 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner sum0，不以小數近似替代。
• S033 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner product0，不以小數近似替代。
• S032 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner sum1，不以小數近似替代。
• S031 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner product1，不以小數近似替代。
• S030 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner sum2，不以小數近似替代。
• S029 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner product2，不以小數近似替代。
• S028 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner sum3，不以小數近似替代。
• S027 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner product3，不以小數近似替代。
• S026 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner sum4，不以小數近似替代。
• S025 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner product4，不以小數近似替代。
• S024 · 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。
• S023 · premise_and_result_audit — 以精確有理數重算 Taylor recover coefficient4，不以小數近似替代。
• S022 · premise_and_result_audit — 以精確有理數重算 Taylor derivative4 from shifted coefficient，不以小數近似替代。
• S021 · premise_and_result_audit — 以精確有理數重算 Taylor k4 factorial step4，不以小數近似替代。
• S020 · premise_and_result_audit — 以精確有理數重算 Taylor k4 factorial step3，不以小數近似替代。
• S019 · premise_and_result_audit — 以精確有理數重算 Taylor k4 factorial step2，不以小數近似替代。
• S018 · premise_and_result_audit — 以精確有理數重算 Taylor k4 factorial step1，不以小數近似替代。
• S017 · premise_and_result_audit — 以精確有理數重算 Taylor recover coefficient3，不以小數近似替代。
• S016 · premise_and_result_audit — 以精確有理數重算 Taylor derivative3 from shifted coefficient，不以小數近似替代。
• S015 · premise_and_result_audit — 以精確有理數重算 Taylor k3 factorial step3，不以小數近似替代。
• S014 · premise_and_result_audit — 以精確有理數重算 Taylor k3 factorial step2，不以小數近似替代。
• S013 · premise_and_result_audit — 以精確有理數重算 Taylor k3 factorial step1，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

([11, 22, 33, 44, 55], [11, 22, 66, 264, 1320], [11, 22, 33, 44, 55], -4, 627, 627, 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。