第 18292 題：Lp收斂

原題與已知條件

同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification；資料={"masses":[[1,2],[1,2]],"coefficients":[17,-18],"shift":2,"p":2}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 同一有限原子probability space、固定整數p大於等於1與symbolic moment tail。操作=lp-qualification；資料={"masses":[[1,2],[1,2]],"coefficients":[17,-18],"shift":2,"p":2}。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-lp-convergence

逐步推導

source operation=lp-qualification;original snapshot and mathematics retained

explicit ordered source output fields=[p, masses, coefficients, shift, qualified, certificate]

Lp complete atomic probability mass := sum(1/2, 1/2)

Lp complete atomic probability mass = 1

Lp coefficient atom0 absolute power product stage1 := mul(1, 17)

Lp coefficient atom0 absolute power product stage1 = 17

Lp coefficient atom0 absolute power product stage2 := mul(17, 17)

Lp coefficient atom0 absolute power product stage2 = 289

Lp constant weighted coefficient atom0 := mul(1/2, 289)

Lp constant weighted coefficient atom0 = 289/2

original atom=0;mass=1/2;SIGNED coefficient=17;fixed integer p=2

absolute coefficient power=289;weighted coefficient power=289/2

Lp coefficient atom1 absolute power product stage1 := mul(1, 18)

Lp coefficient atom1 absolute power product stage1 = 18

Lp coefficient atom1 absolute power product stage2 := mul(18, 18)

Lp coefficient atom1 absolute power product stage2 = 324

Lp constant weighted coefficient atom1 := mul(1/2, 324)

Lp constant weighted coefficient atom1 = 162

original atom=1;mass=1/2;SIGNED coefficient=-18;fixed integer p=2

absolute coefficient power=324;weighted coefficient power=162

Lp full weighted absolute coefficient power sum := sum(289/2, 162)

Lp full weighted absolute coefficient power sum = 613/2

SAME complete probability space;masses=[1/2, 1/2];signed coefficients=[17, -18];shift=2;fixed p=2;weighted constant=613/2

for EVERY n>=0,p-moment E|error_n|^p=(613/2)/(n+2)^2;all moments finite and tend0;this is NOT the p-th-root norm value

逐步反向核對

• 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 — 以精確有理數重算 Lp full weighted absolute coefficient power sum，不以小數近似替代。
• S010 · 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。
• S009 · premise_and_result_audit — 以精確有理數重算 Lp constant weighted coefficient atom1，不以小數近似替代。
• S008 · premise_and_result_audit — 以精確有理數重算 Lp coefficient atom1 absolute power product stage2，不以小數近似替代。
• S007 · premise_and_result_audit — 以精確有理數重算 Lp coefficient atom1 absolute power product stage1，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 Lp constant weighted coefficient atom0，不以小數近似替代。
• S004 · premise_and_result_audit — 以精確有理數重算 Lp coefficient atom0 absolute power product stage2，不以小數近似替代。
• S003 · premise_and_result_audit — 以精確有理數重算 Lp coefficient atom0 absolute power product stage1，不以小數近似替代。
• S002 · premise_and_result_audit — 以精確有理數重算 Lp complete atomic probability mass，不以小數近似替代。
• 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

(2, [1/2, 1/2], [17, -18], 2, 是, common_probability_space_and_finite_integer_p_moments)

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。