第 24066 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":[16,32,48,64,80,96],"center":3,"point":1}。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":[16,32,48,64,80,96],"center":3,"point":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-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 = [3, [16, 32, 48, 64, 80, 96]]

derived = [P(x)=sum c_k*(x-center)^k, 5]

source roles = [0]

derived = [1]

Taylor derivative0 from shifted coefficient := mul(16, 1)

Taylor derivative0 from shifted coefficient = 16

Taylor recover coefficient0 := div(16, 1)

Taylor recover coefficient0 = 16

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

Taylor k1 factorial step1 = 1

Taylor derivative1 from shifted coefficient := mul(32, 1)

Taylor derivative1 from shifted coefficient = 32

Taylor recover coefficient1 := div(32, 1)

Taylor recover coefficient1 = 32

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(48, 2)

Taylor derivative2 from shifted coefficient = 96

Taylor recover coefficient2 := div(96, 2)

Taylor recover coefficient2 = 48

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(64, 6)

Taylor derivative3 from shifted coefficient = 384

Taylor recover coefficient3 := div(384, 6)

Taylor recover coefficient3 = 64

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(80, 24)

Taylor derivative4 from shifted coefficient = 1920

Taylor recover coefficient4 := div(1920, 24)

Taylor recover coefficient4 = 80

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 derivative5 from shifted coefficient := mul(96, 120)

Taylor derivative5 from shifted coefficient = 11520

Taylor recover coefficient5 := div(11520, 120)

Taylor recover coefficient5 = 96

Taylor SAME point minus declared center := sub(1, 3)

Taylor SAME point minus declared center = -2

Taylor original Horner product5 := mul(0, -2)

Taylor original Horner product5 = 0

Taylor original Horner sum5 := sum(0, 96)

Taylor original Horner sum5 = 96

Taylor original Horner product4 := mul(96, -2)

Taylor original Horner product4 = -192

Taylor original Horner sum4 := sum(-192, 80)

Taylor original Horner sum4 = -112

Taylor original Horner product3 := mul(-112, -2)

Taylor original Horner product3 = 224

Taylor original Horner sum3 := sum(224, 64)

Taylor original Horner sum3 = 288

Taylor original Horner product2 := mul(288, -2)

Taylor original Horner product2 = -576

Taylor original Horner sum2 := sum(-576, 48)

Taylor original Horner sum2 = -528

Taylor original Horner product1 := mul(-528, -2)

Taylor original Horner product1 = 1056

Taylor original Horner sum1 := sum(1056, 32)

Taylor original Horner sum1 = 1088

Taylor original Horner product0 := mul(1088, -2)

Taylor original Horner product0 = -2176

Taylor original Horner sum0 := sum(-2176, 16)

Taylor original Horner sum0 = -2160

Taylor reconstructed Horner product5 := mul(0, -2)

Taylor reconstructed Horner product5 = 0

Taylor reconstructed Horner sum5 := sum(0, 96)

Taylor reconstructed Horner sum5 = 96

Taylor reconstructed Horner product4 := mul(96, -2)

Taylor reconstructed Horner product4 = -192

Taylor reconstructed Horner sum4 := sum(-192, 80)

Taylor reconstructed Horner sum4 = -112

Taylor reconstructed Horner product3 := mul(-112, -2)

Taylor reconstructed Horner product3 = 224

Taylor reconstructed Horner sum3 := sum(224, 64)

Taylor reconstructed Horner sum3 = 288

Taylor reconstructed Horner product2 := mul(288, -2)

Taylor reconstructed Horner product2 = -576

Taylor reconstructed Horner sum2 := sum(-576, 48)

Taylor reconstructed Horner sum2 = -528

Taylor reconstructed Horner product1 := mul(-528, -2)

Taylor reconstructed Horner product1 = 1056

Taylor reconstructed Horner sum1 := sum(1056, 32)

Taylor reconstructed Horner sum1 = 1088

Taylor reconstructed Horner product0 := mul(1088, -2)

Taylor reconstructed Horner product0 = -2176

Taylor reconstructed Horner sum0 := sum(-2176, 16)

Taylor reconstructed Horner sum0 = -2160

source roles = [[16, 32, 48, 64, 80, 96], [16, 32, 48, 64, 80, 96], 3, 1]

derived = [是, -2160, -2160, 是]

逐步反向核對

• S056 · 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。
• S055 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner sum0，不以小數近似替代。
• S054 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner product0，不以小數近似替代。
• S053 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner sum1，不以小數近似替代。
• S052 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner product1，不以小數近似替代。
• S051 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner sum2，不以小數近似替代。
• S050 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner product2，不以小數近似替代。
• S049 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner sum3，不以小數近似替代。
• S048 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner product3，不以小數近似替代。
• S047 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner sum4，不以小數近似替代。
• S046 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner product4，不以小數近似替代。
• S045 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner sum5，不以小數近似替代。
• S044 · premise_and_result_audit — 以精確有理數重算 Taylor reconstructed Horner product5，不以小數近似替代。
• S043 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner sum0，不以小數近似替代。
• S042 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner product0，不以小數近似替代。
• S041 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner sum1，不以小數近似替代。
• S040 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner product1，不以小數近似替代。
• S039 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner sum2，不以小數近似替代。
• S038 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner product2，不以小數近似替代。
• S037 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner sum3，不以小數近似替代。
• S036 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner product3，不以小數近似替代。
• S035 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner sum4，不以小數近似替代。
• S034 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner product4，不以小數近似替代。
• S033 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner sum5，不以小數近似替代。
• S032 · premise_and_result_audit — 以精確有理數重算 Taylor original Horner product5，不以小數近似替代。
• S031 · 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。
• S030 · premise_and_result_audit — 以精確有理數重算 Taylor recover coefficient5，不以小數近似替代。
• S029 · premise_and_result_audit — 以精確有理數重算 Taylor derivative5 from shifted coefficient，不以小數近似替代。
• S028 · premise_and_result_audit — 以精確有理數重算 Taylor k5 factorial step5，不以小數近似替代。
• S027 · premise_and_result_audit — 以精確有理數重算 Taylor k5 factorial step4，不以小數近似替代。
• S026 · premise_and_result_audit — 以精確有理數重算 Taylor k5 factorial step3，不以小數近似替代。
• S025 · premise_and_result_audit — 以精確有理數重算 Taylor k5 factorial step2，不以小數近似替代。
• S024 · premise_and_result_audit — 以精確有理數重算 Taylor k5 factorial step1，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

([16, 32, 48, 64, 80, 96], [16, 32, 96, 384, 1920, 11520], [16, 32, 48, 64, 80, 96], 1, -2160, -2160, 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。