第 23084 題：高階導數

原題與已知條件

f^(0)=f且f^(n+1)=(f^n)'的逐階存在與n in natural including zero。操作=higher-zero-beyond-degree；資料={"coefficients":[9,18,27,36],"order":6,"point":1}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — f^(0)=f且f^(n+1)=(f^n)'的逐階存在與n in natural including zero。操作=higher-zero-beyond-degree；資料={"coefficients":[9,18,27,36],"order":6,"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-higher-order-derivative

逐步推導

source operation=higher-zero-beyond-degree;original snapshot and mathematics retained

explicit ordered source output fields=[declared_degree_bound, order, derivative_coefficients, point, value, zero_derivative_certificate, certificate]

ORIGINAL low-to-high coefficients=[9, 18, 27, 36];declared order=6

declared degree BOUND=3=ORIGINAL length-1;initialize derivative order0 with the UNTRIMMED original list

Higher derivative level1 ORIGINAL power1 coefficient product := mul(18, 1)

Higher derivative level1 ORIGINAL power1 coefficient product = 18

Higher derivative level1 ORIGINAL power2 coefficient product := mul(27, 2)

Higher derivative level1 ORIGINAL power2 coefficient product = 54

Higher derivative level1 ORIGINAL power3 coefficient product := mul(36, 3)

Higher derivative level1 ORIGINAL power3 coefficient product = 108

previous derivative order=0;actual low-to-high list=[9, 18, 27, 36];drop ONLY its constant term

ALL [old_power,old_coefficient,multiplier,new_coefficient] rows=[[1, 18, 1, 18], [2, 27, 2, 54], [3, 36, 3, 108]];next raw list=[18, 54, 108];all intermediate products SAFE

Higher derivative level2 ORIGINAL power1 coefficient product := mul(54, 1)

Higher derivative level2 ORIGINAL power1 coefficient product = 54

Higher derivative level2 ORIGINAL power2 coefficient product := mul(108, 2)

Higher derivative level2 ORIGINAL power2 coefficient product = 216

previous derivative order=1;actual low-to-high list=[18, 54, 108];drop ONLY its constant term

ALL [old_power,old_coefficient,multiplier,new_coefficient] rows=[[1, 54, 1, 54], [2, 108, 2, 216]];next raw list=[54, 216];all intermediate products SAFE

Higher derivative level3 ORIGINAL power1 coefficient product := mul(216, 1)

Higher derivative level3 ORIGINAL power1 coefficient product = 216

previous derivative order=2;actual low-to-high list=[54, 216];drop ONLY its constant term

ALL [old_power,old_coefficient,multiplier,new_coefficient] rows=[[1, 216, 1, 216]];next raw list=[216];all intermediate products SAFE

previous derivative order=3;actual low-to-high list=[216];drop ONLY its constant term

ALL [old_power,old_coefficient,multiplier,new_coefficient] rows=[];next raw list=[];all intermediate products SAFE

previous derivative order=4;actual low-to-high list=[];drop ONLY its constant term

ALL [old_power,old_coefficient,multiplier,new_coefficient] rows=[];next raw list=[];all intermediate products SAFE

previous derivative order=5;actual low-to-high list=[];drop ONLY its constant term

ALL [old_power,old_coefficient,multiplier,new_coefficient] rows=[];next raw list=[];all intermediate products SAFE

ORIGINAL coefficients=[9, 18, 27, 36];order=6;declared degree bound=3

final derivative coefficients=[0];for each original power k>=order multiplier=k*(k-1)*...*(k-order+1);lower powers vanish

Higher derivative Horner power0 previous value times ORIGINAL point := mul(0, 1)

Higher derivative Horner power0 previous value times ORIGINAL point = 0

Higher derivative Horner power0 plus ORIGINAL derivative coefficient := sum(0, 0)

Higher derivative Horner power0 plus ORIGINAL derivative coefficient = 0

逐步反向核對

• S017 · premise_and_result_audit — 以精確有理數重算 Higher derivative Horner power0 plus ORIGINAL derivative coefficient，不以小數近似替代。
• S016 · premise_and_result_audit — 以精確有理數重算 Higher derivative Horner power0 previous value times ORIGINAL point，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 — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S013 · 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。
• 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 Higher derivative level3 ORIGINAL power1 coefficient product，不以小數近似替代。
• S009 · 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。
• S008 · premise_and_result_audit — 以精確有理數重算 Higher derivative level2 ORIGINAL power2 coefficient product，不以小數近似替代。
• S007 · premise_and_result_audit — 以精確有理數重算 Higher derivative level2 ORIGINAL power1 coefficient product，不以小數近似替代。
• S006 · 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。
• S005 · premise_and_result_audit — 以精確有理數重算 Higher derivative level1 ORIGINAL power3 coefficient product，不以小數近似替代。
• S004 · premise_and_result_audit — 以精確有理數重算 Higher derivative level1 ORIGINAL power2 coefficient product，不以小數近似替代。
• S003 · premise_and_result_audit — 以精確有理數重算 Higher derivative level1 ORIGINAL power1 coefficient product，不以小數近似替代。
• 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

(3, 6, [0], 1, 0, 是, orders_above_polynomial_degree_are_zero_without_analyticity_claim)

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。