{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-initial-value-problem-01", "mko_id": "mko-initial-value-problem", "title": { "zh-Hant": "初值問題：逐一審核四個多項式候選", "en": "Initial-value problem: auditing four polynomial candidates" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-initial-value-problem-01", "source": { "mko_id": "mko-initial-value-problem", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "84ab0a91c84b74df8cfa9a3a8033b7ad83825d55e89f1d0ca847ed81f85e9925", "english_sha256": "9fff0f0029d4033fa0555d7c1431a25e5fe58386793836198e0241c073506dc0" }, "mode": "worked_example", "audience_prerequisites": [ "distributive", "eq_replace", "numeric_evaluation", "and_intro_elim", "quantifier_negation", "forall_elim" ], "goal": { "zh-Hant": "初值問題：在 ℝ 上 y''(x)=6x，且 y(0)=1、y'(0)=2。逐一審核四個整係數多項式候選 y_A=x³+2x+1、y_B=x³+1、y_C=x²+2x+1、y_D=3x⁵−15x⁴+21x³+2x+1：分開核對方程殘差與兩個初值，判定哪一個是此初值問題的解，並說明只核初值、只核一個初值、或只在有限網格上核殘差為何都不夠。", "en": "Initial-value problem: y''(x)=6x on ℝ with y(0)=1 and y'(0)=2. Audit the four integer-coefficient polynomial candidates y_A=x³+2x+1, y_B=x³+1, y_C=x²+2x+1 and y_D=3x⁵−15x⁴+21x³+2x+1 one by one: check the equation residual and the two initial values separately, decide which candidate solves this initial-value problem, and explain why checking only the initial data, only one initial value, or the residual only on a finite grid is not enough." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "初值問題：未知函數 y:ℝ→ℝ 滿足二階常微分方程 y''(x)=6x（對每個 x∈ℝ），以及初始資料 y(0)=1、y'(0)=2（x0=0、y0=1、v0=2）。", "en": "Initial-value problem: an unknown function y:ℝ→ℝ satisfies the second-order ordinary differential equation y''(x)=6x (for every x∈ℝ) together with the initial data y(0)=1 and y'(0)=2 (x0=0, y0=1, v0=2)." } }, { "id": "G2", "statement": { "zh-Hant": "候選（ℝ 上的整係數多項式）：y_A=x³+2x+1，y_B=x³+1，y_C=x²+2x+1，y_D=3x⁵−15x⁴+21x³+2x+1。", "en": "Candidates (integer-coefficient polynomials on ℝ): y_A=x³+2x+1, y_B=x³+1, y_C=x²+2x+1, y_D=3x⁵−15x⁴+21x³+2x+1." } }, { "id": "G3", "statement": { "zh-Hant": "殘差定義為 R_y(x)=y''(x)−6x；為候選 y_D 指定的有限殘差網格是 {0,1,2}。", "en": "The residual is R_y(x)=y''(x)−6x; the finite residual grid specified for candidate y_D is {0,1,2}." } } ], "steps": [ { "id": "S01", "uses": [ "G1", "mko_ivp" ], "before_tex": "\\text{initial-value problem}", "after_tex": "y''(x)=6x\\ (x\\in\\mathbb{R}),\\ y(0)=1,\\ y'(0)=2", "relation": "definition", "rule": "mko_ivp", "conditions": [], "explanation": { "zh-Hant": "依 MKO，初值問題由方程與同一未知函數在指定點的初始資料組成。這裡方程是二階的（最高導數 y''），所以初始資料給兩個：y(0)=1 與 y'(0)=2。兩者都是對同一個函數 y、在同一點 x0=0 的要求。", "en": "By the MKO, an initial-value problem consists of an equation and initial data for the same unknown function at a specified point. Here the equation is of second order (the highest derivative is y''), so two initial data are given: y(0)=1 and y'(0)=2. Both are requirements on the same function y at the same point x0=0." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "mko_ivp" ], "explanation": { "zh-Hant": "此步只寫出問題，不推出新事實；寫下初值本身並不保證有解或唯一解（MKO 第一段）。", "en": "This step only states the problem and derives no new fact; writing initial data does not by itself guarantee a solution or a unique one (MKO, first paragraph)." } }, "leaf_basis": "mko_ivp + def_intro" }, { "id": "S02", "uses": [ "S01", "G3", "mko_ode_residual" ], "before_tex": "y\\ \\text{solves the IVP}", "after_tex": "(E)\\ \\forall x\\in\\mathbb{R}:\\ R_y(x)=0\\ \\land\\ (I_1)\\ y(0)=1\\ \\land\\ (I_2)\\ y'(0)=2", "relation": "equivalent", "rule": "def_unfold", "conditions": [], "explanation": { "zh-Hant": "展開「是初值問題的解」：候選必須同時負起三個責任：方程責任 (E)，即殘差 R_y(x)=y''(x)−6x 在整個 ℝ 上為零；以及兩個初值責任 (I1)、(I2)。三者是合取，任何一項失敗，候選就不是此初值問題的解。", "en": "Unfold 'solves the initial-value problem': a candidate must meet three requirements at once: the equation requirement (E), that the residual R_y(x)=y''(x)−6x vanishes on all of ℝ, and the two initial-condition requirements (I1) and (I2). They form a conjunction, and failure of any one means the candidate does not solve this initial-value problem." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "def_unfold", "and_intro_elim" ], "explanation": { "zh-Hant": "這是定義的展開，兩向成立：三項都成立就是解，是解就三項都成立。", "en": "This unfolds the definition and holds both ways: all three requirements together make a solution, and a solution meets all three." } }, "leaf_basis": "def_unfold + and_intro_elim" }, { "id": "S03", "uses": [ "G2", "poly_deriv" ], "before_tex": "y_A=x^3+2x+1", "after_tex": "y_A'=3x^2+2,\\ y_A''=6x", "relation": "equivalent", "rule": "poly_deriv", "conditions": [], "explanation": { "zh-Hant": "用多項式導數規則逐項求 y_A 的一階與二階導數，分兩個子步。", "en": "Differentiate y_A term by term with the derivative rules for polynomials, in two sub-steps for the first and second derivatives." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "poly_deriv" ], "explanation": { "zh-Hant": "導數是唯一確定的函數，所以等式兩邊指同一個函數；N1 以差商恆等式交叉核對一階導數。", "en": "A derivative is a uniquely determined function, so both sides name the same function; N1 cross-checks the first derivative through a difference-quotient identity." } }, "leaf_basis": "poly_deriv" }, { "id": "S03.1", "uses": [ "G2", "poly_deriv" ], "before_tex": "(x^3+2x+1)'", "after_tex": "3x^2+2", "relation": "equivalent", "rule": "poly_deriv", "conditions": [], "explanation": { "zh-Hant": "和的導數是導數的和：(x³)'=3x²，(2x)'=2·(x)'=2，(1)'=0，所以 y_A'=3x²+2。交叉核對：y_A(x+h)−y_A(x)=h(3x²+2)+3xh²+h³（N1，多項式恆等式），除以 h 再令 h→0 得 3x²+2。", "en": "The derivative of a sum is the sum of the derivatives: (x³)'=3x², (2x)'=2·(x)'=2 and (1)'=0, so y_A'=3x²+2. Cross-check: y_A(x+h)−y_A(x)=h(3x²+2)+3xh²+h³ (N1, a polynomial identity); dividing by h and letting h→0 gives 3x²+2." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "poly_deriv" ], "explanation": { "zh-Hant": "每一項都直接套用規則，所得函數唯一。", "en": "Each term is a direct application of the rules, and the resulting function is unique." } }, "leaf_basis": "poly_deriv + numeric_evaluation" }, { "id": "S03.2", "uses": [ "S03.1", "poly_deriv" ], "before_tex": "(3x^2+2)'", "after_tex": "6x", "relation": "equivalent", "rule": "poly_deriv", "conditions": [], "explanation": { "zh-Hant": "再求一次導數：(3x²)'=3·2x=6x，(2)'=0，所以 y_A''=6x。", "en": "Differentiate once more: (3x²)'=3·2x=6x and (2)'=0, so y_A''=6x." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "poly_deriv" ], "explanation": { "zh-Hant": "同 S03.1，直接套用規則。", "en": "As in S03.1, a direct application of the rules." } }, "leaf_basis": "poly_deriv" }, { "id": "S04", "uses": [ "S03", "S02" ], "before_tex": "R_{y_A}(x)=y_A''(x)-6x", "after_tex": "R_{y_A}(x)=6x-6x=0\\ \\ \\forall x\\in\\mathbb{R}", "relation": "equivalent", "rule": "eq_replace", "conditions": [], "explanation": { "zh-Hant": "代入 y_A''=6x：R_{y_A}(x)=6x−6x=0。這是對每個實數 x 都成立的代數恆等式，不是在若干點上取樣的結果，所以方程責任 (E) 在整個 ℝ 上成立。", "en": "Substitute y_A''=6x: R_{y_A}(x)=6x−6x=0. This is an algebraic identity valid for every real x, not the result of sampling some points, so the equation requirement (E) holds on all of ℝ." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "eq_replace", "add_inverse" ], "explanation": { "zh-Hant": "以相等的式子代換，並用 u−u=0；對每個 x 兩邊同值。", "en": "Equal expressions are substituted and u−u=0 is used; both sides agree for every x." } }, "leaf_basis": "eq_replace + add_inverse + forall_intro" }, { "id": "S05", "uses": [ "G2", "S03" ], "before_tex": "y_A(0),\\ y_A'(0)", "after_tex": "y_A(0)=1,\\ y_A'(0)=2", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "代入 x=0：y_A(0)=0+0+1=1，y_A'(0)=3·0+2=2。兩個初值責任 (I1)、(I2) 都成立。注意 y_A'(0) 要用導數 y_A'，不是 y_A 本身。", "en": "Substitute x=0: y_A(0)=0+0+1=1 and y_A'(0)=3·0+2=2. Both initial-condition requirements (I1) and (I2) hold. Note that y_A'(0) uses the derivative y_A', not y_A itself." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "numeric_evaluation" ], "explanation": { "zh-Hant": "精確代入，兩邊同值。", "en": "Exact substitution; both sides have the same value." } }, "leaf_basis": "numeric_evaluation + forall_elim" }, { "id": "S06", "uses": [ "S02", "S04", "S05" ], "before_tex": "(E),\\ (I_1),\\ (I_2)\\ \\text{for}\\ y_A", "after_tex": "y_A\\ \\text{solves the IVP on}\\ \\mathbb{R}", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "三個責任都已證明（S04、S05），依 S02，y_A 是此初值問題在 ℝ 上的解。這同時說明此問題至少有一個解；本稿不宣稱解唯一。", "en": "All three requirements are proved (S04, S05), so by S02 y_A solves this initial-value problem on ℝ. This also shows the problem has at least one solution; uniqueness is not claimed." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "S04", "S05", "and_intro_elim" ], "explanation": { "zh-Hant": "合取需要三項全部成立；不宣稱反向的唯一性：找到一個解並不排除其他解。", "en": "The conjunction needs all three parts; no uniqueness is claimed: finding one solution does not exclude others." } }, "leaf_basis": "and_intro_elim + def_unfold" }, { "id": "S07", "uses": [ "G2", "poly_deriv", "S02" ], "before_tex": "y_B=x^3+1", "after_tex": "R_{y_B}\\equiv 0,\\ y_B(0)=1,\\ y_B'(0)=0\\neq 2", "relation": "implies", "rule": "poly_deriv", "conditions": [], "explanation": { "zh-Hant": "y_B'=3x²，y_B''=6x，所以 R_{y_B}(x)=6x−6x=0 對所有 x 成立，(E) 成立；y_B(0)=1，(I1) 成立；但 y_B'(0)=3·0=0≠2，(I2) 失敗。依 S02，y_B 滿足方程，卻不是此初值問題的解。", "en": "y_B'=3x² and y_B''=6x, so R_{y_B}(x)=6x−6x=0 for every x and (E) holds; y_B(0)=1, so (I1) holds; but y_B'(0)=3·0=0≠2, so (I2) fails. By S02, y_B satisfies the equation but does not solve this initial-value problem." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "poly_deriv", "S02", "numeric_evaluation" ], "explanation": { "zh-Hant": "否定一個合取只需一項失敗，這裡是 (I2)；不宣稱 y_B 在其他意義下不好，它確實是方程的解。", "en": "Negating a conjunction needs only one failing part, here (I2); it is not claimed that y_B is bad in any other sense, since it does solve the equation." } }, "leaf_basis": "poly_deriv + de_morgan_logic + numeric_evaluation" }, { "id": "S08", "uses": [ "S04", "S05", "S07" ], "before_tex": "y_A,\\ y_B\\ \\text{both satisfy}\\ (E),\\ (I_1)", "after_tex": "y_A(1)=4\\neq 2=y_B(1)", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "y_A 與 y_B 都滿足方程與 y(0)=1，但 y_A(1)=1+2+1=4，y_B(1)=1+1=2，兩者是不同的函數。所以只給 y(0)=1 不足以挑出一個解；第二個初值 y'(0)=2 才把 y_B 排除。這就是 MKO 說二階問題通常需要 y(x0) 與 y'(x0) 兩個資料的原因。", "en": "y_A and y_B both satisfy the equation and y(0)=1, but y_A(1)=1+2+1=4 and y_B(1)=1+1=2, so they are different functions. Hence y(0)=1 alone does not single out a solution; the second datum y'(0)=2 is what excludes y_B. This is why the MKO says a second-order problem typically needs both y(x0) and y'(x0)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S04", "S05", "S07", "numeric_evaluation" ], "explanation": { "zh-Hant": "一個具體的反例（兩個不同的函數）就足以說明一個初值不夠；不宣稱兩個初值必定足以保證唯一，那需要另外的論證。", "en": "One concrete counterexample (two different functions) suffices to show that one datum is not enough; it is not claimed that two data always guarantee uniqueness, which needs a separate argument." } }, "leaf_basis": "numeric_evaluation + exists_intro" }, { "id": "S09", "uses": [ "G2", "poly_deriv", "S02" ], "before_tex": "y_C=x^2+2x+1", "after_tex": "y_C(0)=1,\\ y_C'(0)=2,\\ R_{y_C}(0)=2\\neq 0", "relation": "implies", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "y_C'=2x+2，y_C''=2。初值：y_C(0)=1、y_C'(0)=2，兩個都成立。但 R_{y_C}(x)=2−6x，在 x=0 得 R_{y_C}(0)=2≠0。要否定「對所有 x，R(x)=0」，找到一個 x 使 R(x)≠0 就夠了；所以 (E) 失敗，y_C 不是解。只核初值的人會誤收 y_C。", "en": "y_C'=2x+2 and y_C''=2. Initial data: y_C(0)=1 and y_C'(0)=2, both hold. But R_{y_C}(x)=2−6x, and at x=0, R_{y_C}(0)=2≠0. To negate 'R(x)=0 for every x' one x with R(x)≠0 is enough, so (E) fails and y_C is not a solution. Someone who checks only the initial data would wrongly accept y_C." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "quantifier_negation", "poly_deriv", "S02" ], "explanation": { "zh-Hant": "¬∀x P(x) ⇔ ∃x ¬P(x)；見證 x=0 已明示並精確核對（N8）。", "en": "¬∀x P(x) ⇔ ∃x ¬P(x); the witness x=0 is explicit and checked exactly (N8)." } }, "leaf_basis": "quantifier_negation + poly_deriv + numeric_evaluation" }, { "id": "S10", "uses": [ "G2", "poly_deriv" ], "before_tex": "y_D=3x^5-15x^4+21x^3+2x+1", "after_tex": "y_D'=15x^4-60x^3+63x^2+2,\\ y_D''=60x^3-180x^2+126x", "relation": "equivalent", "rule": "poly_deriv", "conditions": [], "explanation": { "zh-Hant": "逐項求 y_D 的一階與二階導數，分兩個子步；係數的乘法在 N2 至 N7 精確核對。", "en": "Differentiate y_D term by term, computing the first and second derivatives in two sub-steps; the coefficient products are checked exactly in N2 to N7." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "poly_deriv" ], "explanation": { "zh-Hant": "導數唯一，兩邊指同一函數。", "en": "Derivatives are unique, so both sides name the same function." } }, "leaf_basis": "poly_deriv + numeric_evaluation" }, { "id": "S10.1", "uses": [ "G2", "poly_deriv" ], "before_tex": "(3x^5-15x^4+21x^3+2x+1)'", "after_tex": "15x^4-60x^3+63x^2+2", "relation": "equivalent", "rule": "poly_deriv", "conditions": [], "explanation": { "zh-Hant": "(3x⁵)'=5·3x⁴=15x⁴，(−15x⁴)'=−4·15x³=−60x³，(21x³)'=3·21x²=63x²，(2x)'=2，(1)'=0。", "en": "(3x⁵)'=5·3x⁴=15x⁴, (−15x⁴)'=−4·15x³=−60x³, (21x³)'=3·21x²=63x², (2x)'=2 and (1)'=0." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "poly_deriv" ], "explanation": { "zh-Hant": "逐項套用規則；係數乘法見 N2、N3、N4。", "en": "Term-by-term application of the rules; the coefficient products are N2, N3 and N4." } }, "leaf_basis": "poly_deriv + numeric_evaluation" }, { "id": "S10.2", "uses": [ "S10.1", "poly_deriv" ], "before_tex": "(15x^4-60x^3+63x^2+2)'", "after_tex": "60x^3-180x^2+126x", "relation": "equivalent", "rule": "poly_deriv", "conditions": [], "explanation": { "zh-Hant": "(15x⁴)'=4·15x³=60x³，(−60x³)'=−3·60x²=−180x²，(63x²)'=2·63x=126x，(2)'=0。", "en": "(15x⁴)'=4·15x³=60x³, (−60x³)'=−3·60x²=−180x², (63x²)'=2·63x=126x and (2)'=0." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "poly_deriv" ], "explanation": { "zh-Hant": "同 S10.1；係數乘法見 N5、N6、N7。", "en": "As in S10.1; the coefficient products are N5, N6 and N7." } }, "leaf_basis": "poly_deriv + numeric_evaluation" }, { "id": "S11", "uses": [ "S10", "G3" ], "before_tex": "R_{y_D}(x)=y_D''(x)-6x", "after_tex": "R_{y_D}(x)=60x^3-180x^2+120x=60x(x-1)(x-2)", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "R_{y_D}(x)=60x³−180x²+126x−6x=60x³−180x²+120x。提出公因式：60x(x²−3x+2)，而 x²−3x+2=(x−1)(x−2)。所以 R_{y_D}(x)=60x(x−1)(x−2)；N9 以多項式恆等式核對。", "en": "R_{y_D}(x)=60x³−180x²+126x−6x=60x³−180x²+120x. Factor out the common factor: 60x(x²−3x+2), and x²−3x+2=(x−1)(x−2). So R_{y_D}(x)=60x(x−1)(x−2); N9 checks this as a polynomial identity." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "numeric_evaluation" ], "explanation": { "zh-Hant": "反向展開 60x(x−1)(x−2) 回到 60x³−180x²+120x；兩者是同一多項式（N9 在 0 到 3 的網格上核對三次多項式恆等式）。", "en": "Backwards, expanding 60x(x−1)(x−2) returns 60x³−180x²+120x; they are the same polynomial (N9 checks the cubic identity on the grid 0 to 3)." } }, "leaf_basis": "distributive + add_assoc + numeric_evaluation" }, { "id": "S12", "uses": [ "S11", "G2", "G3", "S10" ], "before_tex": "x\\in\\{0,1,2\\}", "after_tex": "R_{y_D}(0)=R_{y_D}(1)=R_{y_D}(2)=0,\\ y_D(0)=1,\\ y_D'(0)=2", "relation": "implies", "rule": "zero_product", "conditions": [], "explanation": { "zh-Hant": "在指定網格上，60x(x−1)(x−2) 的某個因子為零，所以三點的殘差都是 0（N10）。初值：y_D(0)=1（只剩常數項），y_D'(0)=2（只剩常數項 2）。y_D 通過了這份有限網格加初值的審核。", "en": "On the specified grid one factor of 60x(x−1)(x−2) is zero, so the residual is 0 at all three points (N10). Initial data: y_D(0)=1 (only the constant term remains) and y_D'(0)=2 (only the constant term 2 remains). y_D passes this finite-grid-plus-initial-data audit." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S11", "zero_product", "numeric_evaluation" ], "explanation": { "zh-Hant": "這些都是在有限個點上的事實；不宣稱反向，即不由這三點推出 (E) 在 ℝ 上成立（S13 證明它不成立）。", "en": "These are facts at finitely many points; no converse is claimed, that is, (E) on ℝ is not inferred from these three points (S13 proves it fails)." } }, "leaf_basis": "zero_product + numeric_evaluation" }, { "id": "S13", "uses": [ "S11", "S12", "S02" ], "before_tex": "R_{y_D}(3)", "after_tex": "60\\cdot 3\\cdot 2\\cdot 1=360\\neq 0\\ \\Rightarrow\\ y_D\\ \\text{not a solution}", "relation": "implies", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "在網格外取 x=3：R_{y_D}(3)=60·3·2·1=360≠0。所以「對所有 x，R(x)=0」不成立，(E) 失敗，y_D 不是此初值問題的解，雖然它通過了 S12 的有限審核。這正是 MKO 的提醒：有限點上殘差為零不是整個區間上的恆等證明。", "en": "Take x=3 outside the grid: R_{y_D}(3)=60·3·2·1=360≠0. So 'R(x)=0 for every x' fails, (E) fails and y_D does not solve this initial-value problem, even though it passed the finite audit of S12. This is the MKO's warning: a zero residual at finitely many points is not an identity on the whole interval." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S11", "quantifier_negation", "S02" ], "explanation": { "zh-Hant": "一個見證點 x=3 就足以否定全稱命題（N11）。", "en": "One witness point x=3 suffices to negate the universal statement (N11)." } }, "leaf_basis": "quantifier_negation + numeric_evaluation" }, { "id": "S14", "uses": [ "S06", "S07", "S09", "S13" ], "before_tex": "y_A,\\ y_B,\\ y_C,\\ y_D", "after_tex": "\\text{only}\\ y_A\\ \\text{solves the IVP among the four candidates}", "relation": "implies", "rule": "case_analysis", "conditions": [], "explanation": { "zh-Hant": "四個候選逐一判定：y_A 三個責任都成立（S06）；y_B 失敗於 y'(0)（S07）；y_C 失敗於方程（S09）；y_D 只通過有限網格，失敗於 x=3 的方程（S13）。結論只涵蓋這四個候選：y_A 是解，其餘不是。", "en": "Each of the four candidates is decided: y_A meets all three requirements (S06); y_B fails at y'(0) (S07); y_C fails the equation (S09); y_D passes only the finite grid and fails the equation at x=3 (S13). The conclusion covers these four candidates only: y_A is a solution and the others are not." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S06", "S07", "S09", "S13" ], "explanation": { "zh-Hant": "目標 G：四個候選的判定。足夠條件 H：S06、S07、S09、S13 四個各別判定，這四種情況涵蓋全部候選；H⇒G 由分情況論證。H 由導數規則與精確代入正向得出，沒有把 G 當作前提。", "en": "Goal G: the verdict on the four candidates. Sufficient condition H: the four separate verdicts S06, S07, S09 and S13, which cover every candidate; H⇒G by proof by cases. H is derived forwards from the derivative rules and exact substitution; G is never used as a premise." } }, "leaf_basis": "case_analysis + and_intro_elim" } ], "result": { "zh-Hant": "在四個候選中，只有 y_A=x³+2x+1 是初值問題 y''=6x、y(0)=1、y'(0)=2 在 ℝ 上的解：它的殘差恆為零，且兩個初值都成立。y_B=x³+1 滿足方程與 y(0)=1 但 y_B'(0)=0；y_C=x²+2x+1 滿足兩個初值但 R(0)=2；y_D 在網格 {0,1,2} 上殘差為零且初值相符，但 R(3)=360。本稿不宣稱解唯一。", "en": "Of the four candidates, only y_A=x³+2x+1 solves the initial-value problem y''=6x, y(0)=1, y'(0)=2 on ℝ: its residual is identically zero and both initial values hold. y_B=x³+1 satisfies the equation and y(0)=1 but y_B'(0)=0; y_C=x²+2x+1 satisfies both initial values but R(0)=2; y_D has zero residual on the grid {0,1,2} and correct initial values, but R(3)=360. Uniqueness is not claimed." }, "negative_controls": [ { "claim": { "zh-Hant": "y_C(0)=1 且 y_C'(0)=2，所以 y_C 是此初值問題的解", "en": "y_C(0)=1 and y_C'(0)=2, so y_C solves this initial-value problem" }, "expected": "reject", "reason": { "zh-Hant": "R_{y_C}(x)=2−6x，R_{y_C}(0)=2≠0，方程不成立。", "en": "R_{y_C}(x)=2−6x and R_{y_C}(0)=2≠0, so the equation fails." } }, { "claim": { "zh-Hant": "y_D 在 0、1、2 的殘差都是 0 且初值相符，所以 y_D 是解", "en": "y_D has zero residual at 0, 1 and 2 and correct initial values, so y_D is a solution" }, "expected": "reject", "reason": { "zh-Hant": "R_{y_D}(3)=60·3·2·1=360≠0。", "en": "R_{y_D}(3)=60·3·2·1=360≠0." } }, { "claim": { "zh-Hant": "y_B 滿足 y''=6x 與 y(0)=1，所以是此初值問題的解", "en": "y_B satisfies y''=6x and y(0)=1, so it solves this initial-value problem" }, "expected": "reject", "reason": { "zh-Hant": "y_B'(0)=0≠2，第二個初值不成立。", "en": "y_B'(0)=0≠2, so the second initial value fails." } }, { "claim": { "zh-Hant": "找到一個滿足方程與初值的候選，就證明了解唯一", "en": "Finding one candidate that meets the equation and the initial data proves that the solution is unique" }, "expected": "reject", "reason": { "zh-Hant": "存在一個解不推出沒有其他解；只給 y(0)=1 時 y_A 與 y_B 就是兩個不同的解。唯一性需要另外的論證，本稿沒有給。", "en": "One solution existing does not exclude others; with only y(0)=1, y_A and y_B are two different solutions. Uniqueness needs a separate argument, which this guide does not give." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "distributive", "kind": "prerequisite", "statement": { "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。", "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu." } }, { "id": "eq_replace", "kind": "prerequisite", "statement": { "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。", "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true." } }, { "id": "numeric_evaluation", "kind": "prerequisite", "statement": { "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。", "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable." } }, { "id": "and_intro_elim", "kind": "prerequisite", "statement": { "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。", "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q." } }, { "id": "quantifier_negation", "kind": "prerequisite", "statement": { "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。", "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)." } }, { "id": "forall_elim", "kind": "prerequisite", "statement": { "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。", "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds." } }, { "id": "def_intro", "kind": "prerequisite", "statement": { "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。", "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable." } }, { "id": "def_unfold", "kind": "prerequisite", "statement": { "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。", "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true." } }, { "id": "add_inverse", "kind": "prerequisite", "statement": { "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。", "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)." } }, { "id": "forall_intro", "kind": "prerequisite", "statement": { "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。", "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain." } }, { "id": "de_morgan_logic", "kind": "prerequisite", "statement": { "zh-Hant": "邏輯的德摩根律：¬(P∧Q) ⇔ ¬P∨¬Q；¬(P∨Q) ⇔ ¬P∧¬Q。", "en": "De Morgan's laws for logic: ¬(P∧Q) ⇔ ¬P∨¬Q and ¬(P∨Q) ⇔ ¬P∧¬Q." } }, { "id": "exists_intro", "kind": "prerequisite", "statement": { "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。", "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)." } }, { "id": "add_assoc", "kind": "prerequisite", "statement": { "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。", "en": "Associativity of addition: (u+v)+w=u+(v+w)." } }, { "id": "zero_product", "kind": "prerequisite", "statement": { "zh-Hant": "零乘積性質：在 ℤ、ℚ、ℝ、ℂ（或任何整域）中，uv=0 ⇔ u=0 或 v=0。", "en": "Zero-product property: in ℤ, ℚ, ℝ, ℂ (or any integral domain), uv=0 ⇔ u=0 or v=0." } }, { "id": "case_analysis", "kind": "prerequisite", "statement": { "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。", "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility." } }, { "id": "mko_ivp", "kind": "cited", "source": { "mko_id": "mko-initial-value-problem", "sha256": "84ab0a91c84b74df8cfa9a3a8033b7ad83825d55e89f1d0ca847ed81f85e9925", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "初值問題由常微分方程及同一未知函數在指定點的初始資料組成。一階問題常給 y(x0)=y0；二階問題通常需 y(x0)=y0 與 y'(x0)=v0。候選解必須同時滿足方程與初值。", "en": "An initial-value problem consists of an ordinary differential equation and initial data for the same unknown function at a specified point. A first-order problem commonly gives y(x0)=y0; a second-order problem typically needs y(x0)=y0 and y'(x0)=v0. A candidate solution must satisfy both the equation and the initial data." } }, { "id": "mko_ode_residual", "kind": "cited", "source": { "mko_id": "mko-ordinary-differential-equation", "sha256": "78a220ac6f5f781865fa9d3cdc17fd9d5f931ff83cfdd5890bfaae2b91641a13", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "常微分方程是在指定區間上，關聯未知單變數函數 y 與其有限階導數的方程。方程中出現的最高導數階數稱為階數；把候選 y 代入並得零殘差，只核對該候選在指定範圍是否滿足方程。", "en": "On a specified interval, an ordinary differential equation relates an unknown single-variable function y and its finite-order derivatives. Its order is the highest derivative order occurring in the equation. A zero substitution residual checks only whether a given candidate satisfies the relation over the specified domain." } }, { "id": "poly_deriv", "kind": "prerequisite", "statement": { "zh-Hant": "多項式導數：多項式在 ℝ 上處處可微；(x^k)'=k x^(k−1)（k≥1），常數的導數為 0，且 (f+g)'=f'+g'、(cf)'=cf'。", "en": "Derivatives of polynomials: polynomials are differentiable everywhere on ℝ; (x^k)'=k x^(k−1) for k≥1, the derivative of a constant is 0, and (f+g)'=f'+g', (cf)'=cf'." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S03.1", "vars": [ "x", "h" ], "degree": 3, "lhs": "((x+h)**3+2*(x+h)+1)-(x**3+2*x+1)", "rhs": "h*(3*x**2+2)+3*x*h**2+h**3" }, { "id": "N2", "step": "S10.1", "lhs": "5*3", "rhs": "15" }, { "id": "N3", "step": "S10.1", "lhs": "4*15", "rhs": "60" }, { "id": "N4", "step": "S10.1", "lhs": "3*21", "rhs": "63" }, { "id": "N5", "step": "S10.2", "lhs": "4*15", "rhs": "60" }, { "id": "N6", "step": "S10.2", "lhs": "3*60", "rhs": "180" }, { "id": "N7", "step": "S10.2", "lhs": "2*63", "rhs": "126" }, { "id": "N8", "step": "S09", "lhs": "2-6*0", "rhs": "2" }, { "id": "N9", "step": "S11", "vars": [ "x" ], "degree": 3, "lhs": "(60*x**3-180*x**2+126*x)-6*x", "rhs": "60*x*(x-1)*(x-2)" }, { "id": "N10", "step": "S12", "vars": [ "x" ], "range": [ 0, 2 ], "lhs": "60*x**3-180*x**2+120*x", "rhs": "0" }, { "id": "N11", "step": "S13", "lhs": "60*3**3-180*3**2+120*3", "rhs": "360" }, { "id": "N12", "step": "S08", "lhs": "1**3+2*1+1", "rhs": "4" }, { "id": "N13", "step": "S08", "lhs": "1**3+1", "rhs": "2" }, { "id": "N14", "step": "S12", "lhs": "3*0**5-15*0**4+21*0**3+2*0+1", "rhs": "1" } ] }, "presentation": { "md_scope": { "zh-Hant": "在 MKO 的範圍內工作：整係數多項式候選、明示初值與指定的有限殘差網格（假設 a1）。候選 y_A 的殘差以多項式導數規則符號化算出，因此在整個 ℝ 上為零；y_D 示範有限網格核對的限度。本稿不宣稱解唯一、不求一般解、不討論最大解區間（假設 a2）；不改動 canonical MKO 或英文 edition。", "en": "Works inside the MKO's scope: integer-coefficient polynomial candidates, explicit initial values and a specified finite residual grid (assumption a1). The residual of candidate y_A is computed symbolically with the derivative rules for polynomials and is therefore zero on all of ℝ; y_D shows the limits of a finite-grid check. The guide does not claim uniqueness, find a general solution or discuss maximal intervals of existence (assumption a2), and does not change the canonical MKO or its English edition." }, "md_symbols": [ [ "y_A,\\ y_B,\\ y_C,\\ y_D", "四個多項式候選", "the four polynomial candidates" ], [ "R_y(x)=y''(x)-6x", "候選 y 的方程殘差", "the equation residual of a candidate y" ], [ "(E)", "方程責任：對所有 x∈ℝ，R_y(x)=0", "the equation requirement: R_y(x)=0 for every x∈ℝ" ], [ "(I_1),\\ (I_2)", "初值責任：y(0)=1、y'(0)=2", "the initial-condition requirements: y(0)=1 and y'(0)=2" ], [ "\\{0,1,2\\}", "為 y_D 指定的有限殘差網格", "the finite residual grid specified for y_D" ] ], "md_error_controls": [ { "title": "只核初值不核方程 / Checking the initial data but not the equation", "zh": "MKO 指出的第一個誤解。y_C 滿足 y(0)=1 與 y'(0)=2，但 y_C''=2，殘差 2−6x 在 x=0 就是 2（S09）。方程責任與初值責任要分開核對，任何一項失敗就不是候選解。", "en": "The first mistake named by the MKO. y_C satisfies y(0)=1 and y'(0)=2, but y_C''=2 and the residual 2−6x is already 2 at x=0 (S09). The equation requirement and the initial-condition requirements must be checked separately, and failure of either rules the candidate out." }, { "title": "有限網格的零殘差不證明全區間解 / Zero residual on a finite grid does not establish a solution on the whole interval", "zh": "MKO 指出的第二個誤解。y_D 的殘差 60x(x−1)(x−2) 在網格 {0,1,2} 上全為零，初值也相符，但 R(3)=360（S12、S13）。有限採樣只能否定（找到非零點），不能單獨證明恆等於零；y_A 的殘差是以代數恆等式 6x−6x=0 證明對所有 x 為零。", "en": "The second mistake named by the MKO. The residual 60x(x−1)(x−2) of y_D vanishes on the whole grid {0,1,2} and the initial values match, but R(3)=360 (S12, S13). Finite sampling can refute (by finding a nonzero point) but cannot by itself prove a residual is identically zero; for y_A, the residual is shown to vanish for all x by the algebraic identity 6x−6x=0." }, { "title": "二階問題需要兩個初值 / A second-order problem needs two initial values", "zh": "y_A 與 y_B 都滿足 y''=6x 與 y(0)=1，卻是不同的函數（S08）。只核 y(0) 會把 y_B 誤當成解；y'(0)=2 是必要的第二個資料。核 y'(0) 時要代入導數 y'，不是 y 本身。", "en": "y_A and y_B both satisfy y''=6x and y(0)=1, yet they are different functions (S08). Checking only y(0) would wrongly accept y_B; y'(0)=2 is the necessary second datum. When checking y'(0), substitute into the derivative y', not into y itself." } ], "md_open_items": [ { "zh-Hant": "本稿不證明此初值問題的解唯一。一條可能的路線是：兩解之差 w 滿足 w''=0、w(0)=0、w'(0)=0，再用「導數恆為零的函數在區間上為常數」（均值定理的推論）；這需要引用該定理，本稿未做。", "en": "This guide does not prove that the solution of this initial-value problem is unique. One possible route: the difference w of two solutions satisfies w''=0, w(0)=0 and w'(0)=0, and then 'a function with zero derivative on an interval is constant' (a corollary of the mean value theorem) applies; that needs a citation of the theorem, which is not done here." }, { "zh-Hant": "N1 與 N9 是多項式恆等式的網格核對（對每個變數的次數上限已給定），N10 只是三個點的有限核對；本稿的結論不依賴它們，它們是交叉核對。", "en": "N1 and N9 are grid checks of polynomial identities (with a stated degree bound per variable), and N10 is a finite check at three points only; the conclusions do not depend on them, they are cross-checks." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "9ab9d67dde3d67d0467dfdb56a74490ed9fc771c9c18aba72aa959ee7d220859", "original_source_sha256": "9ab9d67dde3d67d0467dfdb56a74490ed9fc771c9c18aba72aa959ee7d220859", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }