S001 · implies
操作前
domain=R; coefficient lists are in ascending powers
操作後
constant term first; rational pairs=[numerator,denominator]
條件:無
先固定係數次序,不能把升冪清單當降冪,也不能把分子分母交換。
步驟識別與依賴
規則:polynomial_representation · 依賴:G001
R0001-01317 · q-20261007-advance-20261008-r06-antiderivative-0017
全實數論域,以下函數/點均以明示實數模型解讀。f(x)=17x^4,n為0至5。求常數項固定零的原函數升冪有理係數,整數也保留 /1。
G001 — 全實數論域,以下函數/點均以明示實數模型解讀。f(x)=17x^4,n為0至5。求常數項固定零的原函數升冪有理係數,整數也保留 /1。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。
操作前
domain=R; coefficient lists are in ascending powers
操作後
constant term first; rational pairs=[numerator,denominator]
條件:無
先固定係數次序,不能把升冪清單當降冪,也不能把分子分母交換。
規則:polynomial_representation · 依賴:G001
操作前
f coefficients ascending=[0, 0, 0, 0, 17]
操作後
seek F₀ with constant coefficient0 and F₀′=f
條件:無
原函數並不一般唯一;本題額外固定常數項為0。
規則:canonical_primitive_domain · 依賴:G001, S001
操作前
primitive coefficient at degree1 := div(0, 1)
操作後
primitive coefficient at degree1 = 0
條件:無
由(x^(k+1))′=(k+1)x^k,係數須除以非零正整數k+1。
規則:exact_rational_arithmetic · 依賴:G001, S002
操作前
differentiate back coefficient0 := mul(0, 1)
操作後
differentiate back coefficient0 = 0
條件:無
微分乘回k+1,應恢復原多項式對應係數。
規則:exact_rational_arithmetic · 依賴:G001, S003
操作前
primitive coefficient at degree2 := div(0, 2)
操作後
primitive coefficient at degree2 = 0
條件:無
由(x^(k+1))′=(k+1)x^k,係數須除以非零正整數k+1。
規則:exact_rational_arithmetic · 依賴:G001, S004
操作前
differentiate back coefficient1 := mul(0, 2)
操作後
differentiate back coefficient1 = 0
條件:無
微分乘回k+1,應恢復原多項式對應係數。
規則:exact_rational_arithmetic · 依賴:G001, S005
操作前
primitive coefficient at degree3 := div(0, 3)
操作後
primitive coefficient at degree3 = 0
條件:無
由(x^(k+1))′=(k+1)x^k,係數須除以非零正整數k+1。
規則:exact_rational_arithmetic · 依賴:G001, S006
操作前
differentiate back coefficient2 := mul(0, 3)
操作後
differentiate back coefficient2 = 0
條件:無
微分乘回k+1,應恢復原多項式對應係數。
規則:exact_rational_arithmetic · 依賴:G001, S007
操作前
primitive coefficient at degree4 := div(0, 4)
操作後
primitive coefficient at degree4 = 0
條件:無
由(x^(k+1))′=(k+1)x^k,係數須除以非零正整數k+1。
規則:exact_rational_arithmetic · 依賴:G001, S008
操作前
differentiate back coefficient3 := mul(0, 4)
操作後
differentiate back coefficient3 = 0
條件:無
微分乘回k+1,應恢復原多項式對應係數。
規則:exact_rational_arithmetic · 依賴:G001, S009
操作前
primitive coefficient at degree5 := div(17, 5)
操作後
primitive coefficient at degree5 = 17/5
條件:無
由(x^(k+1))′=(k+1)x^k,係數須除以非零正整數k+1。
規則:exact_rational_arithmetic · 依賴:G001, S010
操作前
differentiate back coefficient4 := mul(17/5, 5)
操作後
differentiate back coefficient4 = 17
條件:無
微分乘回k+1,應恢復原多項式對應係數。
規則:exact_rational_arithmetic · 依賴:G001, S011
操作前
[0, 0, 0, 0, 0, 17/5]
操作後
all derivative coefficients agree with f
條件:無
係數逐項相同證明全實數上的多項式恆等,不是只在幾個點相等。
規則:global_polynomial_identity · 依賴:G001, S012
操作前
[0, 0, 0, 0, 0, 17/5]
操作後
reduced coefficients;integer coefficients retain /1
條件:無
輸出升冪係數清單;多餘最高次零係數去掉,但零多項式保留[0/1]。
規則:coefficient_output_protocol · 依賴:G001, S013
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 — 以精確有理數重算 differentiate back coefficient4,不以小數近似替代。S011 · premise_and_result_audit — 以精確有理數重算 primitive coefficient at degree5,不以小數近似替代。S010 · premise_and_result_audit — 以精確有理數重算 differentiate back coefficient3,不以小數近似替代。S009 · premise_and_result_audit — 以精確有理數重算 primitive coefficient at degree4,不以小數近似替代。S008 · premise_and_result_audit — 以精確有理數重算 differentiate back coefficient2,不以小數近似替代。S007 · premise_and_result_audit — 以精確有理數重算 primitive coefficient at degree3,不以小數近似替代。S006 · premise_and_result_audit — 以精確有理數重算 differentiate back coefficient1,不以小數近似替代。S005 · premise_and_result_audit — 以精確有理數重算 primitive coefficient at degree2,不以小數近似替代。S004 · premise_and_result_audit — 以精確有理數重算 differentiate back coefficient0,不以小數近似替代。S003 · premise_and_result_audit — 以精確有理數重算 primitive coefficient at degree1,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。[0/1, 0/1, 0/1, 0/1, 0/1, 17/5]
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。
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。
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