{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-harmonic-oscillator-model-01",
  "mko_id": "mko-harmonic-oscillator-model",
  "title": {
    "zh-Hant": "線性簡諧振子模型：資格、k/m、多項式候選的殘差與單點能量",
    "en": "Linear harmonic oscillator: eligibility, k/m, residuals of a polynomial candidate and single-state energy"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-harmonic-oscillator-model-01",
    "source": {
      "mko_id": "mko-harmonic-oscillator-model",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "920969c8aeda483138db1c073acdd7c8acdf5a854c7cef34fb51509539f7ebb4",
      "english_sha256": "435d03c200d484abe1512ca6364253ffef8da54d800818ba816553870515ab66"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "order_basics",
      "zero_product",
      "distributive",
      "power_laws",
      "quantifier_negation",
      "contradiction",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "取 m=2、k=8。核對模型資格並算出 k/m；對多項式候選 x_c(t)=1−2t² 算出殘差 R(t)=m x_c''+k x_c，在指定點 t=0、1/2、1 求值，並證明 x_c 不是任何正長度區間上的解；證明方程 2x''+8x=0 沒有非零多項式解；最後算出單一狀態 (x,v)=(1,0) 的能量 E=4，並以能量導數恆等式說明單點能量值為何不證明守恆。",
      "en": "Take m=2 and k=8. Check that the parameters satisfy the model's requirements and compute k/m; for the polynomial candidate x_c(t)=1−2t², compute the residual R(t)=m x_c''+k x_c, evaluate it at the specified points t=0, 1/2, 1, and prove that x_c is not a solution on any interval of positive length; prove that the equation 2x''+8x=0 has no nonzero polynomial solution; finally compute the energy E=4 of the single state (x,v)=(1,0), and use the energy-derivative identity to explain why a single energy value proves no conservation."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "模型（MKO）：正質量 m 與正彈性係數 k（正有理數），位移 x 滿足 m x''+k x=0。本例 m=2、k=8。",
          "en": "Model (MKO): positive mass m and positive spring coefficient k (positive rationals), with displacement x satisfying m x''+k x=0. Here m=2 and k=8."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "多項式候選 x_c(t)=1−2t²（整係數），指定的殘差核對點為 t=0、1/2、1。",
          "en": "Polynomial candidate x_c(t)=1−2t² (integer coefficients), with the specified residual points t=0, 1/2 and 1."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "狀態能量式 E(x,v)=m v²/2+k x²/2；要求值的單一狀態為 (x,v)=(1,0)。",
          "en": "State energy expression E(x,v)=m v²/2+k x²/2; the single state to evaluate is (x,v)=(1,0)."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "資格對照用的非例：(m,k)=(0,8) 與 (m,k)=(2,−8)。",
          "en": "Non-instances for the eligibility check: (m,k)=(0,8) and (m,k)=(2,−8)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "mko_sho"
        ],
        "before_tex": "\\text{linear harmonic oscillator}",
        "after_tex": "mx''+kx=0,\\ m>0,\\ k>0",
        "relation": "definition",
        "rule": "mko_sho",
        "conditions": [
          "m>0",
          "k>0"
        ],
        "explanation": {
          "zh-Hant": "依 MKO 固定模型：正質量 m、正彈性係數 k，位移 x 滿足 m x''+k x=0。MKO 特別限定：k/m 只稱為角頻率平方候選；一般閉式解需要另外建立的三角函數與導數語義，本稿不使用。",
          "en": "Fix the model as the MKO states it: positive mass m, positive spring coefficient k, and displacement x satisfying m x''+k x=0. The MKO is explicit that k/m is only called the angular-frequency-square candidate, and that a general closed form needs separately established trigonometric and derivative semantics, which this guide does not use."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_sho"
          ],
          "explanation": {
            "zh-Hant": "此步只陳述模型，不推出新事實。",
            "en": "This step only states the model and derives no new fact."
          }
        },
        "leaf_basis": "mko_sho + def_intro"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "G1",
          "G4"
        ],
        "before_tex": "(m,k)=(2,8)",
        "after_tex": "m>0\\land k>0:\\ \\text{qualifies};\\ (0,8),\\ (2,-8)\\ \\text{do not}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "模型資格是合取 m>0 且 k>0。(2,8) 兩項都成立，合格。非例：(0,8) 的 m=0 不是正數；(2,−8) 的 k=−8<0。這兩個非例都不屬於本模型，即使它們也能寫成 m x''+k x=0 的形式（m=0 時甚至不再是二階方程）。",
          "en": "Eligibility is the conjunction m>0 and k>0. Both hold for (2,8), so it qualifies. Non-instances: (0,8) has m=0, which is not positive; (2,−8) has k=−8<0. Neither belongs to this model, even though both can be written in the form m x''+k x=0 (with m=0 it is no longer even a second-order equation)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G4",
            "and_intro_elim",
            "de_morgan_logic"
          ],
          "explanation": {
            "zh-Hant": "合格需要兩項都成立；不合格只需一項失敗。",
            "en": "Eligibility requires both conditions; failure of either rules the parameters out."
          }
        },
        "leaf_basis": "and_intro_elim + de_morgan_logic + order_basics"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "mko_sho"
        ],
        "before_tex": "\\tfrac{k}{m}",
        "after_tex": "\\tfrac{8}{2}=4",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [
          "m\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "k/m=8/2=4，m≠0 所以可除。依 MKO，4 只稱為角頻率平方候選；把它解讀成「角頻率 2」或「頻率」需要三角函數語義，本稿不做。",
          "en": "k/m=8/2=4, and m≠0 allows the division. By the MKO, 4 is called only the angular-frequency-square candidate; reading it as 'angular frequency 2' or as a 'frequency' needs trigonometric semantics, which this guide does not supply."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "反向：4·2=8；精確有理數計算。",
            "en": "Backward: 4·2=8; an exact rational computation."
          }
        },
        "leaf_basis": "numeric_evaluation + frac_rules"
      },
      {
        "id": "S04",
        "uses": [
          "G2",
          "poly_deriv"
        ],
        "before_tex": "x_c(t)=1-2t^2",
        "after_tex": "x_c'(t)=-4t,\\ x_c''(t)=-4",
        "relation": "equivalent",
        "rule": "poly_deriv",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(1)'=0，(−2t²)'=−2·2t=−4t；再微分：(−4t)'=−4。交叉核對：x_c(t+h)−x_c(t)=−4t·h−2h²（N10，多項式恆等式），除以 h 再令 h→0 得 −4t。",
          "en": "(1)'=0 and (−2t²)'=−2·2t=−4t; differentiating again, (−4t)'=−4. Cross-check: x_c(t+h)−x_c(t)=−4t·h−2h² (N10, a polynomial identity); dividing by h and letting h→0 gives −4t."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "poly_deriv"
          ],
          "explanation": {
            "zh-Hant": "導數唯一確定，兩邊指同一函數。",
            "en": "Derivatives are uniquely determined, so both sides name the same function."
          }
        },
        "leaf_basis": "poly_deriv + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "S04",
          "G1"
        ],
        "before_tex": "R(t)=m x_c''(t)+k x_c(t)",
        "after_tex": "R(t)=-16t^2",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代入 m=2、k=8 與 S04 的導數，展開化簡（子步 S05.1）。殘差是 t 的多項式 −16t²，對每個 t 都有確定的值。",
          "en": "Substitute m=2, k=8 and the derivatives of S04, then expand and simplify (sub-step S05.1). The residual is the polynomial −16t², with a definite value for every t."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "代換與展開都是等值改寫；N2 以多項式恆等式核對。",
            "en": "Substitution and expansion are rewritings of equals; N2 checks the result as a polynomial identity."
          }
        },
        "leaf_basis": "eq_replace + distributive"
      },
      {
        "id": "S05.1",
        "uses": [
          "S04",
          "G1",
          "distributive"
        ],
        "before_tex": "2\\cdot(-4)+8(1-2t^2)",
        "after_tex": "-8+8-16t^2=-16t^2",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "2·(−4)=−8；8(1−2t²)=8−16t²（分配律）；−8+8=0，剩下 −16t²。",
          "en": "2·(−4)=−8; 8(1−2t²)=8−16t² by distributivity; −8+8=0, leaving −16t²."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "每一步都是對所有 t 成立的恆等式。",
            "en": "Each move is an identity valid for every t."
          }
        },
        "leaf_basis": "distributive + add_inverse + add_zero"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "G2"
        ],
        "before_tex": "t\\in\\{0,\\tfrac{1}{2},1\\}",
        "after_tex": "R(0)=0,\\ R(\\tfrac{1}{2})=-4,\\ R(1)=-16",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在指定點求殘差：R(0)=0，R(1/2)=−16·1/4=−4，R(1)=−16（N3、N4）。只有 t=0 的殘差為零。",
          "en": "Evaluate the residual at the specified points: R(0)=0, R(1/2)=−16·1/4=−4 and R(1)=−16 (N3, N4). Only at t=0 is the residual zero."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確代入，可逐一重算。",
            "en": "Exact substitution, recomputable one by one."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S05",
          "S06",
          "quantifier_negation"
        ],
        "before_tex": "R(t)=-16t^2",
        "after_tex": "x_c\\ \\text{solves the equation on no interval of positive length}",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "t≠0 時 t²>0，所以 R(t)=−16t²<0≠0。任何正長度區間都含有 t≠0 的點，在那裡殘差不為零，所以「區間上每個 t 都有 R(t)=0」不成立。R(0)=0 只是單點事實，不能使 x_c 成為解。",
          "en": "For t≠0, t²>0, so R(t)=−16t²<0≠0. Every interval of positive length contains a point t≠0, where the residual is nonzero, so 'R(t)=0 for every t in the interval' fails. R(0)=0 is a single-point fact and cannot make x_c a solution."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "quantifier_negation",
            "square_nonneg",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "否定全稱只需一個見證點；這裡每個 t≠0 都是見證。",
            "en": "Negating a universal statement needs one witness; here every t≠0 is a witness."
          }
        },
        "leaf_basis": "quantifier_negation + square_nonneg + order_basics"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "poly_deriv",
          "poly_roots"
        ],
        "before_tex": "x\\ \\text{polynomial},\\ 2x''+8x=0\\ \\text{on an interval}",
        "after_tex": "x=0",
        "relation": "implies",
        "rule": "L_no_poly",
        "conditions": [
          "m\\neq 0",
          "k\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "證明引理 L_no_poly：本方程沒有非零多項式解。這說明 S07 的失敗不是選錯係數：任何多項式候選（零除外）都不可能是解；非零解只能來自多項式以外的函數，例如另需建立語義的三角函數。分三個子步，以首項係數比較導出矛盾。",
          "en": "Prove Lemma L_no_poly: this equation has no nonzero polynomial solution. This shows that the failure in S07 is not a matter of choosing the wrong coefficients: no polynomial candidate other than zero can be a solution, and nonzero solutions must come from non-polynomial functions, such as trigonometric functions whose semantics must be established separately. Three sub-steps reach a contradiction by comparing leading coefficients."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_no_poly",
            "poly_deriv",
            "poly_roots"
          ],
          "explanation": {
            "zh-Hant": "引理只用 k≠0、多項式導數的次數規則與根數上界；x=0 確實是解，所以結論是「只有零多項式」。",
            "en": "The lemma uses only k≠0, the degree rule for derivatives of polynomials and the bound on the number of roots; x=0 is indeed a solution, so the conclusion is 'only the zero polynomial'."
          }
        },
        "leaf_basis": "contradiction + poly_deriv + poly_roots + zero_product"
      },
      {
        "id": "S08.1",
        "uses": [
          "poly_deriv",
          "zero_product"
        ],
        "before_tex": "x(t)=\\sum_{j=0}^{d}a_jt^j,\\ a_d\\neq 0",
        "after_tex": "[t^d]\\,(kx)=k\\,a_d\\neq 0",
        "relation": "implies",
        "rule": "zero_product",
        "conditions": [
          "k\\neq 0",
          "a_d\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "反設 x 不是零多項式，次數為 d、首項係數 a_d≠0。k x 的 t^d 係數是 k a_d；因 k≠0、a_d≠0，由零乘積性質 k a_d≠0。",
          "en": "Suppose, for contradiction, that x is not the zero polynomial, with degree d and leading coefficient a_d≠0. The t^d coefficient of k x is k a_d; since k≠0 and a_d≠0, the zero-product property gives k a_d≠0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "zero_product"
          ],
          "explanation": {
            "zh-Hant": "需要 k≠0；本模型 k>0 所以成立。",
            "en": "It needs k≠0, which holds since k>0 in this model."
          }
        },
        "leaf_basis": "zero_product + distributive"
      },
      {
        "id": "S08.2",
        "uses": [
          "S08.1",
          "poly_deriv"
        ],
        "before_tex": "x\\ \\text{of degree}\\ d",
        "after_tex": "[t^d]\\,(mx'')=0",
        "relation": "implies",
        "rule": "poly_deriv",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若 d≤1，x''是零多項式，m x''=0；若 d≥2，兩次微分使次數降為 d−2<d，所以 m x'' 沒有 t^d 項。兩種情況下 m x'' 的 t^d 係數都是 0。",
          "en": "If d≤1, x'' is the zero polynomial and m x''=0; if d≥2, differentiating twice lowers the degree to d−2<d, so m x'' has no t^d term. In both cases the t^d coefficient of m x'' is 0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "poly_deriv",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "兩種情況涵蓋所有 d≥0。",
            "en": "The two cases cover every d≥0."
          }
        },
        "leaf_basis": "poly_deriv + case_analysis"
      },
      {
        "id": "S08.3",
        "uses": [
          "S08.1",
          "S08.2",
          "poly_roots"
        ],
        "before_tex": "[t^d]\\,(mx''+kx)=0+k\\,a_d\\neq 0",
        "after_tex": "mx''+kx\\neq 0\\ \\text{on any interval}\\ \\Rightarrow\\ \\bot",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "多項式相加時同次係數相加，所以 m x''+k x 的 t^d 係數是 0+k a_d≠0，它是非零多項式，至多有 d 個根，不可能在正長度區間上處處為零。這與「x 在該區間上是解」矛盾，所以 x 是零多項式。S08.1 至 S08.3 證得引理 L_no_poly。",
          "en": "Adding polynomials adds coefficients of equal degree, so the t^d coefficient of m x''+k x is 0+k a_d≠0; it is a nonzero polynomial with at most d roots and cannot vanish everywhere on an interval of positive length. This contradicts 'x solves the equation on that interval', so x is the zero polynomial. S08.1 to S08.3 prove Lemma L_no_poly."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "poly_roots",
            "contradiction",
            "S08.1",
            "S08.2"
          ],
          "explanation": {
            "zh-Hant": "矛盾需要「非零多項式根數有限」這一前置事實，以及區間長度為正（含無窮多點）。",
            "en": "The contradiction needs the prerequisite that a nonzero polynomial has finitely many roots, and an interval of positive length (with infinitely many points)."
          }
        },
        "leaf_basis": "contradiction + poly_roots + add_assoc"
      },
      {
        "id": "S09",
        "uses": [
          "S03",
          "S08",
          "mko_sho"
        ],
        "before_tex": "\\text{nonzero solution}\\ x",
        "after_tex": "x\\ \\text{is not a polynomial}",
        "relation": "implies",
        "rule": "contrapositive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "L_no_poly 的逆否：非零解不是多項式。MKO 指出一般閉式解需要另外建立的三角函數及導數語義；本稿沒有建立它們，所以不宣稱任何具體的非零解公式，也不宣稱非零解存在。",
          "en": "The contrapositive of L_no_poly: a nonzero solution is not a polynomial. The MKO says a general closed form needs separately established trigonometric and derivative semantics; this guide establishes neither, so it claims no specific formula for a nonzero solution and does not even claim that one exists."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "contrapositive",
            "mko_sho"
          ],
          "explanation": {
            "zh-Hant": "逆否與原命題等價；這一步刻意不往存在性推進。",
            "en": "A contrapositive is equivalent to the original; this step deliberately does not move on to existence."
          }
        },
        "leaf_basis": "contrapositive + mko_sho"
      },
      {
        "id": "S10",
        "uses": [
          "G3",
          "G1"
        ],
        "before_tex": "E(1,0)",
        "after_tex": "\\tfrac{2\\cdot 0^2}{2}+\\tfrac{8\\cdot 1^2}{2}=4",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，能量式可作為單一狀態的量值：E(1,0)=2·0/2+8·1/2=4（N5）。這是一個狀態的一個數，不涉及任何時間演化。",
          "en": "By the MKO, the energy expression can be used as a single-state quantity: E(1,0)=2·0/2+8·1/2=4 (N5). It is one number for one state and involves no evolution in time."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確代入。",
            "en": "Exact substitution."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S04",
          "G3",
          "G1"
        ],
        "before_tex": "E(t)=E(x_c(t),x_c'(t))",
        "after_tex": "E(t)=16t^2+4(1-2t^2)^2=4+16t^4",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "沿候選 x_c 計算能量：m x_c'²/2=(2/2)(16t²)=16t²，k x_c²/2=4(1−4t²+4t⁴)=4−16t²+16t⁴；相加得 E(t)=4+16t⁴（N6）。所以 E(0)=4（與 S10 相同的狀態 (1,0)），但 E(1)=20（N7）：沿這條非解的曲線，能量不是常數。",
          "en": "Compute the energy along the candidate x_c: m x_c'²/2=(2/2)(16t²)=16t² and k x_c²/2=4(1−4t²+4t⁴)=4−16t²+16t⁴; the sum is E(t)=4+16t⁴ (N6). So E(0)=4 (the same state (1,0) as in S10), but E(1)=20 (N7): along this non-solution curve the energy is not constant."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "power_laws"
          ],
          "explanation": {
            "zh-Hant": "展開是對所有 t 成立的恆等式（N6 以次數 ≤4 的網格核對）。",
            "en": "The expansion is an identity for all t (N6 checks it on a grid for degree ≤4)."
          }
        },
        "leaf_basis": "distributive + power_laws + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "mko_deriv_rules"
        ],
        "before_tex": "E(t)=\\tfrac{m}{2}x'(t)^2+\\tfrac{k}{2}x(t)^2",
        "after_tex": "E'(t)=x'(t)\\big(mx''(t)+kx(t)\\big)",
        "relation": "implies",
        "rule": "L_energy",
        "conditions": [
          "x\\ \\text{twice differentiable}"
        ],
        "explanation": {
          "zh-Hant": "證明引理 L_energy：能量沿任一二次可微曲線的變化率，等於速度乘以方程的殘差。殘差沿整條曲線恆為零，足以推出 E' 恆為零；反向一般不成立，因為乘積也可能因速度為零而為零。單點能量值本身不提供這些全區間資訊。",
          "en": "Prove Lemma L_energy: along any twice-differentiable curve, the rate of change of the energy equals the velocity times the residual. An identically zero residual is sufficient for E' to vanish identically; the converse generally fails because the product can also vanish when the velocity is zero. One energy value provides no such interval-wide information."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_energy",
            "mko_deriv_rules"
          ],
          "explanation": {
            "zh-Hant": "需要 x 二次可微，才能對 x' 再微分。",
            "en": "It needs x twice differentiable so that x' can be differentiated again."
          }
        },
        "leaf_basis": "mko_deriv_rules"
      },
      {
        "id": "S12.1",
        "uses": [
          "mko_deriv_rules"
        ],
        "before_tex": "\\big(\\tfrac{m}{2}x'^2+\\tfrac{k}{2}x^2\\big)'",
        "after_tex": "\\tfrac{m}{2}\\cdot 2x'x''+\\tfrac{k}{2}\\cdot 2xx'=x'(mx''+kx)",
        "relation": "equivalent",
        "rule": "mko_deriv_rules",
        "conditions": [
          "x\\ \\text{twice differentiable}"
        ],
        "explanation": {
          "zh-Hant": "依和與常數倍規則逐項微分；由乘積規則，(x'·x')'=x''x'+x'x''=2x'x''，(x·x)'=2xx'。所以 E'=(m/2)·2x'x''+(k/2)·2xx'=m x'x''+k x x'=x'(m x''+k x)（分配律）。這證得引理 L_energy。",
          "en": "Differentiate term by term with the sum and constant-multiple rules; by the product rule, (x'·x')'=x''x'+x'x''=2x'x'' and (x·x)'=2xx'. So E'=(m/2)·2x'x''+(k/2)·2xx'=m x'x''+k x x'=x'(m x''+k x) by distributivity. This proves Lemma L_energy."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_deriv_rules",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "導數唯一，兩邊是同一函數的兩種寫法。",
            "en": "Derivatives are unique, so both sides are two ways of writing the same function."
          }
        },
        "leaf_basis": "mko_deriv_rules + distributive + mul_comm"
      },
      {
        "id": "S13",
        "uses": [
          "S05",
          "S11",
          "S12",
          "S04"
        ],
        "before_tex": "x_c'(t)R(t)=(-4t)(-16t^2)",
        "after_tex": "64t^3=\\tfrac{d}{dt}(4+16t^4)",
        "relation": "implies",
        "rule": "L_energy",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對候選驗證引理：x_c'·R=(−4t)(−16t²)=64t³（N8），而 S11 的 E(t)=4+16t⁴ 的導數是 64t³（N9 以差商恆等式核對），兩者一致。能量變化正來自非零殘差。對真正的解，殘差恆為零，所以 E'(t)=0 在每一點成立；要由此推出 E 為常數，還需要「導數恆為零的函數在區間上為常數」的定理，本稿沒有使用。單一狀態的 E=4 不證明任何守恆。",
          "en": "Check the lemma on the candidate: x_c'·R=(−4t)(−16t²)=64t³ (N8), and the derivative of E(t)=4+16t⁴ from S11 is 64t³ (N9 checks this through a difference-quotient identity); they agree. The change in energy comes exactly from the nonzero residual. For a genuine solution the residual is identically zero, so E'(t)=0 at every point; concluding that E is constant would further need the theorem that a function with zero derivative on an interval is constant, which this guide does not use. The single-state value E=4 proves no conservation."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_energy",
            "S05",
            "S11"
          ],
          "explanation": {
            "zh-Hant": "例子是對引理的一致性核對，不是引理的證明（證明在 S12.1）；守恆需要的定理明確未用。",
            "en": "The example is a consistency check of the lemma, not its proof (which is S12.1); the theorem conservation would need is explicitly not used."
          }
        },
        "leaf_basis": "L_energy + poly_deriv + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S02",
          "S03",
          "S07",
          "S08",
          "S10",
          "S13"
        ],
        "before_tex": "m=2,\\ k=8,\\ x_c=1-2t^2,\\ (x,v)=(1,0)",
        "after_tex": "\\text{qualifies};\\ \\tfrac{k}{m}=4;\\ x_c\\ \\text{not a solution};\\ \\text{no nonzero polynomial solution};\\ E(1,0)=4",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整：(2,8) 合格（S02），k/m=4 是角頻率平方候選（S03）；x_c 只在 t=0 殘差為零，不是任何正長度區間上的解（S06、S07）；沒有非零多項式解（S08）；單一狀態能量 E(1,0)=4（S10），對真正的解，殘差恆零使 E' 恆零（S12、S13），這是充分方向；單點能量值不證守恆，也不由 E' 恆零反推殘差恆零。",
          "en": "Summary: (2,8) qualifies (S02) and k/m=4 is the angular-frequency-square candidate (S03); x_c has zero residual only at t=0 and solves the equation on no interval of positive length (S06, S07); there is no nonzero polynomial solution (S08); the single-state energy is E(1,0)=4 (S10), and for a genuine solution the identically zero residual makes E' identically zero (S12, S13), a sufficient direction. A single energy value proves no conservation, and E'≡0 alone does not imply a zero residual."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S02",
            "S03",
            "S07",
            "S08",
            "S10",
            "S13"
          ],
          "explanation": {
            "zh-Hant": "目標 G：五項結論。足夠條件 H：S02、S03、S07、S08、S10、S13 各自成立；H⇒G 由合取。H 都由定義、導數規則與多項式事實正向得出，沒有把 G 當作前提。",
            "en": "Goal G: the five conclusions. Sufficient condition H: S02, S03, S07, S08, S10 and S13 each hold; H⇒G by conjunction. H is derived forwards from the definitions, the derivative rules and facts about polynomials; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "m=2、k=8 的模型合格，k/m=4 是角頻率平方候選。多項式候選 x_c=1−2t² 的殘差為 R(t)=−16t²：R(0)=0、R(1/2)=−4、R(1)=−16，所以它不是任何正長度區間上的解；事實上 2x''+8x=0 沒有非零多項式解。單一狀態 (1,0) 的能量是 4；由 E'=x'(m x''+k x)，真正解的殘差恆零足以使 E' 恆零；由此推出區間上能量恆定仍需零導數常數定理，本稿未使用。反向不成立，例如曲線 x(t)=1 有 E(t)=4 恆定但殘差為8。單點能量值不證明守恆。",
      "en": "The model with m=2 and k=8 qualifies, and k/m=4 is the angular-frequency-square candidate. The polynomial candidate x_c=1−2t² has residual R(t)=−16t²: R(0)=0, R(1/2)=−4 and R(1)=−16, so it solves the equation on no interval of positive length; in fact 2x''+8x=0 has no nonzero polynomial solution. The single state (1,0) has energy 4; since E'=x'(m x''+k x), an identically zero residual of a genuine solution is sufficient for E'≡0. Concluding constant energy on an interval still requires the zero-derivative theorem, which is not used here. The converse fails: x(t)=1 has constant E(t)=4 but residual 8. A single energy value proves no conservation."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "R(0)=0，所以 x_c=1−2t² 是 2x''+8x=0 的解",
          "en": "R(0)=0, so x_c=1−2t² solves 2x''+8x=0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "R(1)=−16≠0；R(t)=−16t² 只在 t=0 為零。",
          "en": "R(1)=−16≠0; R(t)=−16t² vanishes only at t=0."
        }
      },
      {
        "claim": {
          "zh-Hant": "在 (x,v)=(1,0) 算得 E=4，所以能量沿軌道守恆",
          "en": "E=4 at (x,v)=(1,0), so energy is conserved along the trajectory"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "單點值不涉及時間演化；沿 x_c 有 E(0)=4、E(1)=20。",
          "en": "A single value involves no evolution in time; along x_c, E(0)=4 and E(1)=20."
        }
      },
      {
        "claim": {
          "zh-Hant": "(m,k)=(0,8) 也是線性簡諧振子模型",
          "en": "(m,k)=(0,8) is also a linear harmonic oscillator model"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "模型要求 m>0；m=0 時方程變成 8x=0，不再是二階方程。",
          "en": "The model requires m>0; with m=0 the equation becomes 8x=0 and is no longer of second order."
        }
      },
      {
        "claim": {
          "zh-Hant": "取次數夠高的多項式，一定能找到 2x''+8x=0 的非零多項式解",
          "en": "With a polynomial of high enough degree, a nonzero polynomial solution of 2x''+8x=0 can always be found"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "L_no_poly：t^d 係數為 8a_d≠0，任何非零多項式都不是解。",
          "en": "L_no_poly: the t^d coefficient is 8a_d≠0, so no nonzero polynomial is a solution."
        }
      },
      {
        "claim": {
          "zh-Hant": "k/m=4，所以振子的頻率是 4",
          "en": "k/m=4, so the oscillator's frequency is 4"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "k/m 只是角頻率平方候選；把它轉成角頻率或頻率需要本稿未建立的三角函數語義。",
          "en": "k/m is only the angular-frequency-square candidate; turning it into an angular frequency or a frequency needs trigonometric semantics not established here."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "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": "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": "power_laws",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。",
          "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1."
        }
      },
      {
        "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": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "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": "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": "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": "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": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "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": "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": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "id": "square_nonneg",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數平方非負：對所有實數 u，u²≥0，且 u²=0 ⇔ u=0。",
          "en": "Squares are nonnegative: for every real u, u²≥0, and u²=0 ⇔ u=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": "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": "contrapositive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逆否命題：P⇒Q 與 ¬Q⇒¬P 等價。逆命題 Q⇒P 一般不與 P⇒Q 等價。",
          "en": "Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q."
        }
      },
      {
        "id": "mul_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。",
          "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)."
        }
      },
      {
        "id": "mko_sho",
        "kind": "cited",
        "source": {
          "mko_id": "mko-harmonic-oscillator-model",
          "sha256": "920969c8aeda483138db1c073acdd7c8acdf5a854c7cef34fb51509539f7ebb4",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "線性簡諧振子模型由正質量 m 與正彈性係數 k 組成，位移 x 滿足 m x''+k x=0。比值 k/m 是角頻率平方候選；只有另有三角函數及導數語義時才能建立一般閉式解。",
          "en": "A linear harmonic oscillator has positive mass m and positive spring coefficient k, with displacement x satisfying m*x″+k*x=0. The ratio k/m is the angular-frequency-square candidate. A general closed-form solution requires separately established trigonometric-function and derivative semantics."
        }
      },
      {
        "id": "mko_deriv_rules",
        "kind": "cited",
        "source": {
          "mko_id": "mko-derivative-rules",
          "sha256": "4756c1f6226c8cecc330975377ab5d42ed0e461b0fe039c05959fc60ab7c628f",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在相應點可微的實函數有 (f+g)'=f'+g'、(cf)'=cf'、(fg)'=f'g+fg'；若 g 在 x 可微且 f 在 g(x) 可微，(f∘g)'(x)=f'(g(x))g'(x)。本輪 f、g 是全 R 多項式，以常數項起算的升冪係數列表表示。",
          "en": "For real functions differentiable at the relevant points, (f+g)′=f′+g′, (cf)′=cf′, and (fg)′=f′g+fg′. If g is differentiable at x and f at g(x), then (f∘g)′(x)=f′(g(x))g′(x). Here f,g are polynomials defined on all of ℝ represented by ascending coefficients starting with the constant term."
        }
      },
      {
        "id": "poly_deriv",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "多項式導數：(t^j)'=j t^(j−1)（j≥1），常數的導數為 0，導數對和與常數倍線性。因此次數 d≥1 的多項式的導數次數為 d−1，常數的導數為零多項式。",
          "en": "Derivatives of polynomials: (t^j)'=j t^(j−1) for j≥1, constants have derivative 0, and differentiation is linear over sums and constant multiples. Hence the derivative of a polynomial of degree d≥1 has degree d−1, and the derivative of a constant is the zero polynomial."
        }
      },
      {
        "id": "poly_roots",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "次數 d 的非零多項式至多有 d 個實根；因此在一個正長度區間上處處為零的多項式是零多項式（所有係數為 0）。",
          "en": "A nonzero polynomial of degree d has at most d real roots; hence a polynomial that vanishes everywhere on an interval of positive length is the zero polynomial (all coefficients 0)."
        }
      },
      {
        "id": "L_no_poly",
        "kind": "lemma",
        "proved_in": [
          "S08.1",
          "S08.2",
          "S08.3"
        ],
        "statement": {
          "zh-Hant": "設 m、k 為非零常數。若多項式 x 在某個正長度區間上滿足 m x''+k x=0，則 x 是零多項式。",
          "en": "Let m and k be nonzero constants. If a polynomial x satisfies m x''+k x=0 on some interval of positive length, then x is the zero polynomial."
        }
      },
      {
        "id": "L_energy",
        "kind": "lemma",
        "proved_in": [
          "S12.1"
        ],
        "statement": {
          "zh-Hant": "若 x 在某區間上二次可微，令 E(t)=m x'(t)²/2+k x(t)²/2，則 E'(t)=x'(t)·(m x''(t)+k x(t))。",
          "en": "If x is twice differentiable on an interval and E(t)=m x'(t)²/2+k x(t)²/2, then E'(t)=x'(t)·(m x''(t)+k x(t))."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "8/2",
        "rhs": "4"
      },
      {
        "id": "N2",
        "step": "S05.1",
        "vars": [
          "t"
        ],
        "degree": 2,
        "lhs": "2*(-4)+8*(1-2*t**2)",
        "rhs": "-16*t**2"
      },
      {
        "id": "N3",
        "step": "S06",
        "lhs": "-16*(1/2)**2",
        "rhs": "-4"
      },
      {
        "id": "N4",
        "step": "S06",
        "lhs": "-16*1**2",
        "rhs": "-16"
      },
      {
        "id": "N5",
        "step": "S10",
        "lhs": "2*0**2/2+8*1**2/2",
        "rhs": "4"
      },
      {
        "id": "N6",
        "step": "S11",
        "vars": [
          "t"
        ],
        "degree": 4,
        "lhs": "2*(-4*t)**2/2+8*(1-2*t**2)**2/2",
        "rhs": "4+16*t**4"
      },
      {
        "id": "N7",
        "step": "S11",
        "lhs": "4+16*1**4",
        "rhs": "20"
      },
      {
        "id": "N8",
        "step": "S13",
        "vars": [
          "t"
        ],
        "degree": 3,
        "lhs": "(-4*t)*(-16*t**2)",
        "rhs": "64*t**3"
      },
      {
        "id": "N9",
        "step": "S13",
        "vars": [
          "t",
          "h"
        ],
        "degree": 4,
        "lhs": "(4+16*(t+h)**4)-(4+16*t**4)",
        "rhs": "h*(64*t**3)+h**2*(96*t**2)+h**3*(64*t)+16*h**4"
      },
      {
        "id": "N10",
        "step": "S04",
        "vars": [
          "t",
          "h"
        ],
        "degree": 2,
        "lhs": "(1-2*(t+h)**2)-(1-2*t**2)",
        "rhs": "h*(-4*t)-2*h**2"
      },
      {
        "id": "N11",
        "step": "S02",
        "lhs": "min(2,8)",
        "relation": ">",
        "rhs": "0"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：m、k 為正有理數；只核模型資格、k/m、多項式候選的指定點殘差與單一狀態能量式（假設 a1）。另證兩個與多項式及導數規則有關的引理：沒有非零多項式解，以及能量導數恆等式。不執行 sin/cos 閉式解、不證守恆定理或穩定性（假設 a2）；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: m and k are positive rationals; only model eligibility, k/m, residuals of a polynomial candidate at specified points and a single-state energy expression are checked (assumption a1). Two lemmas about polynomials and the derivative rules are also proved: there is no nonzero polynomial solution, and the energy-derivative identity. No sin/cos closed-form solution, conservation theorem or stability proof is carried out (assumption a2); the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "m,\\ k",
        "質量與彈性係數，皆為正有理數",
        "mass and spring coefficient, positive rationals"
      ],
      [
        "x(t)",
        "位移",
        "displacement"
      ],
      [
        "x_c(t)=1-2t^2",
        "多項式候選",
        "the polynomial candidate"
      ],
      [
        "R(t)=mx_c''(t)+kx_c(t)",
        "候選的殘差",
        "the residual of the candidate"
      ],
      [
        "E(x,v)=\\tfrac{m v^2}{2}+\\tfrac{k x^2}{2}",
        "狀態能量式",
        "the state energy expression"
      ],
      [
        "\\tfrac{k}{m}",
        "角頻率平方候選",
        "the angular-frequency-square candidate"
      ]
    ],
    "md_error_controls": [
      {
        "title": "沒有三角函數語義就不宣稱 sin/cos 解 / No sin/cos claim without trigonometric semantics",
        "zh": "MKO 指出的誤解之一是未有三角函數導數支援就直接宣稱一般 sin/cos 解。本稿只用多項式與導數規則：x_c=1−2t² 看似某個餘弦的截斷，但殘差是 −16t²（S05）；而且 L_no_poly 證明沒有非零多項式解（S08）。非零解必須來自另外建立語義的函數，本稿不建立。",
        "en": "One mistake named by the MKO is to declare the general sin/cos solution without support for trigonometric derivatives. This guide uses only polynomials and the derivative rules: x_c=1−2t² looks like a truncated cosine, but its residual is −16t² (S05), and L_no_poly proves there is no nonzero polynomial solution (S08). Nonzero solutions must come from functions whose semantics are established separately, which this guide does not do."
      },
      {
        "title": "單點能量不是守恆定理 / A single energy value is not a conservation theorem",
        "zh": "MKO 指出的另一誤解。E(1,0)=4 是一個狀態的數。沿候選 x_c，能量從 E(0)=4 變到 E(1)=20（S11）。由 L_energy，E'=x'·R，殘差沿整條曲線為零，足以使 E' 恆為零；再用「零導數函數為常數」定理可得能量恆定（S13）。殘差恆零不是任意候選曲線能量恆定的必要條件：x(t)=1 的速度為零，E(t)=4 恆定但殘差是 8。",
        "en": "The other mistake named by the MKO. E(1,0)=4 is a number for one state. Along the candidate x_c the energy changes from E(0)=4 to E(1)=20 (S11). By L_energy, E'=x'·R, so an identically zero residual is sufficient for E'≡0; the theorem that a zero-derivative function is constant then gives constant energy (S13). Zero residual is not necessary for constant energy along an arbitrary candidate curve: x(t)=1 has zero velocity and constant E(t)=4, but residual 8."
      },
      {
        "title": "單點殘差為零不是解 / Zero residual at one point is not a solution",
        "zh": "R(0)=0，但 R(1/2)=−4、R(1)=−16。是否為解要看整個區間上的殘差；有限個點為零也不夠，何況這裡只有一點。",
        "en": "R(0)=0, but R(1/2)=−4 and R(1)=−16. Whether a candidate is a solution depends on the residual on the whole interval; zero at finitely many points is not enough, let alone at one point as here."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "一般閉式解（例如以 cos 2t、sin 2t 表示）、能量守恆定理與穩定性屬於 MKO 的 formal next，需要三角函數導數與「零導數函數為常數」等定理，本稿未做。",
        "en": "The general closed-form solution (for example in terms of cos 2t and sin 2t), the energy conservation theorem and stability belong to the MKO's formal next step; they need trigonometric derivatives and theorems such as 'a function with zero derivative is constant', and are not done here."
      },
      {
        "zh-Hant": "poly_roots（非零多項式根數有限）以前置知識使用，未在本稿證明；N2、N6、N8、N9、N10 是多項式恆等式的網格核對。",
        "en": "poly_roots (a nonzero polynomial has finitely many roots) is used as a prerequisite and not proved here; N2, N6, N8, N9 and N10 are grid checks of polynomial identities."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "b4e705c85a4a975387cba8996aea06dba3740d4a6e351d15e2887c3eea0e3093",
    "original_source_sha256": "554e271c524ba8ce10183cce3c7dd0413596c2f0895d68b7bdb617f4bf4ec344",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M050"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}