{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-antiderivative-01", "mko_id": "mko-antiderivative", "title": { "zh-Hant": "原函數：多項式代表原函數、常數族與候選驗收", "en": "Antiderivatives: the representative polynomial antiderivative, the constant family and checking candidates" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-antiderivative-01", "source": { "mko_id": "mko-antiderivative", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "5f2ecff6866e471d1f8da7fea1fa5e121364ae2d24b2fea405d2300fe9ff1cae", "english_sha256": "f983dd8b898bc9eac9d5237c7895e5f7dc182ac46efb5f9b58b925fc6c2ecb9d" }, "mode": "worked_example", "audience_prerequisites": [ "distributive", "power_laws", "abs_rules", "order_basics", "frac_rules", "numeric_evaluation" ], "goal": { "zh-Hant": "對 f=[3,−4,6]（f(x)=3−4x+6x²）：由定義證明四次以下多項式的求導公式；構造代表原函數 F0=[0,3,−2,2] 並驗證 F0'=f；證明 F0+C 對每個實數 C 都是原函數，且在至多五個係數的多項式中每個原函數都是 F0+C；驗收候選 G=[7,3,−2,2]（接受）、H=[0,3,−4,2]（拒絕）、K=[0,3,−3,8/3]（兩個測點相符仍拒絕）；求 F0(2)+C（C=−5），並說明原函數不是反函數。", "en": "For f=[3,−4,6] (f(x)=3−4x+6x²): prove from the definition the derivative formula for polynomials of degree at most four; build the representative antiderivative F0=[0,3,−2,2] and verify F0'=f; prove that F0+C is an antiderivative for every real C and that, among polynomials with at most five coefficients, every antiderivative is F0+C; check the candidates G=[7,3,−2,2] (accept), H=[0,3,−4,2] (reject) and K=[0,3,−3,8/3] (reject although it matches at two test points); evaluate F0(2)+C with C=−5; and explain why an antiderivative is not an inverse function." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "多項式以升冪列表表示，論域為整個 ℝ（一個區間）；末端零正規化。", "en": "Polynomials are ascending lists on all of ℝ (an interval); trailing zeros are normalized." } }, { "id": "G2", "statement": { "zh-Hant": "f=[3,−4,6]。", "en": "f=[3,−4,6]." } }, { "id": "G3", "statement": { "zh-Hant": "候選 G=[7,3,−2,2]、H=[0,3,−4,2]、K=[0,3,−3,8/3]；求值資料 x=2、C=−5。", "en": "Candidates G=[7,3,−2,2], H=[0,3,−4,2], K=[0,3,−3,8/3]; evaluation data x=2, C=−5." } } ], "steps": [ { "id": "S01", "uses": [ "G1" ], "before_tex": "F\\ \\text{antiderivative of}\\ f\\ \\text{on}\\ \\mathbb{R}", "after_tex": "F\\ \\text{differentiable on}\\ \\mathbb{R}\\ \\land\\ \\forall x\\in\\mathbb{R}:\\ F'(x)=f(x)", "relation": "definition", "rule": "mko_antideriv", "conditions": [], "explanation": { "zh-Hant": "依 MKO：原函數 F 要在區間（這裡是整個 ℝ）可微，而且導函數處處等於 f。這是關於導數的條件，與「反函數」或「導數取倒數」無關。", "en": "By the MKO: an antiderivative F must be differentiable on the interval (here all of ℝ), with derivative equal to f everywhere. This is a condition about derivatives and has nothing to do with inverse functions or reciprocals of derivatives." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "mko_antideriv" ], "explanation": { "zh-Hant": "此步只寫出定義。", "en": "This step only states the definition." } }, "leaf_basis": "mko_antideriv + def_unfold" }, { "id": "S02", "uses": [ "G2", "G1" ], "before_tex": "f=[3,-4,6]", "after_tex": "f(x)=3-4x+6x^2", "relation": "equivalent", "rule": "def_unfold", "conditions": [], "explanation": { "zh-Hant": "升冪讀法：常數 3、x 的係數 −4、x² 的係數 6。", "en": "Ascending reading: constant 3, coefficient −4 of x, coefficient 6 of x²." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "def_unfold" ], "explanation": { "zh-Hant": "讀回列表得原列表。", "en": "Reading back gives the original list." } }, "leaf_basis": "def_unfold" }, { "id": "S03", "uses": [ "mko_derivative_def" ], "before_tex": "p=[c_0,c_1,c_2,c_3,c_4]", "after_tex": "p'=[c_1,2c_2,3c_3,4c_4]\\ \\text{on}\\ \\mathbb{R}", "relation": "implies", "rule": "L_poly_deriv4", "conditions": [], "explanation": { "zh-Hant": "依引理 L_poly_deriv4，四次以下多項式處處可微，導數列表是 [c1,2c2,3c3,4c4]；常數項 c0 不出現在導數中。引理由導數定義在 S03.1 至 S03.3 證明。本稿所有求導的多項式都至多四次。", "en": "By Lemma L_poly_deriv4, a polynomial of degree at most four is differentiable everywhere with derivative list [c1,2c2,3c3,4c4]; the constant c0 does not appear in it. The lemma is proved from the definition in S03.1 to S03.3. Every polynomial differentiated here has degree at most four." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_poly_deriv4", "mko_derivative_def" ], "explanation": { "zh-Hant": "引理只涵蓋四次以下；候選至多五個係數，正好在範圍內。", "en": "The lemma covers degree at most four; candidates have at most five coefficients, exactly within range." } }, "leaf_basis": "mko_derivative_def + distributive + abs_rules + order_basics" }, { "id": "S03.1", "uses": [ "distributive", "power_laws", "frac_rules" ], "before_tex": "\\frac{p(x+h)-p(x)}{h}\\ (h\\neq 0)", "after_tex": "D+hR,\\ D=c_1+2c_2x+3c_3x^2+4c_4x^3,\\ R=c_2+c_3(3x+h)+c_4(6x^2+4xh+h^2)", "relation": "equivalent", "rule": "distributive", "conditions": [ "h\\neq 0" ], "explanation": { "zh-Hant": "逐次方展開：(x+h)²−x²=h(2x+h)，(x+h)³−x³=h(3x²+h(3x+h))，(x+h)⁴−x⁴=h(4x³+h(6x²+4xh+h²))（N1、N2、N3）。乘上係數相加、c0 相消，得 p(x+h)−p(x)=h(D+hR)；h≠0 時除以 h。", "en": "Expand power by power: (x+h)²−x²=h(2x+h), (x+h)³−x³=h(3x²+h(3x+h)), (x+h)⁴−x⁴=h(4x³+h(6x²+4xh+h²)) (N1, N2, N3). Multiplying by the coefficients and adding, with c0 cancelling, gives p(x+h)−p(x)=h(D+hR); for h≠0 divide by h." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "frac_rules" ], "explanation": { "zh-Hant": "恆等式加上 h≠0 時的除法，雙向成立。", "en": "Identities plus division by h≠0, valid both ways." } }, "leaf_basis": "distributive + power_laws + frac_rules + mul_inverse_nonzero" }, { "id": "S03.2", "uses": [ "S03.1", "abs_rules" ], "before_tex": "0<|h|\\le 1", "after_tex": "|R|\\le M=|c_2|+|c_3|(3|x|+1)+|c_4|(6x^2+4|x|+1)", "relation": "implies", "rule": "abs_rules", "conditions": [ "|h|\\le 1" ], "explanation": { "zh-Hant": "三角不等式與 |uv|=|u||v|，再用 |h|≤1、h²≤1：|3x+h|≤3|x|+1，|6x²+4xh+h²|≤6x²+4|x|+1。M 只依 x 與係數。", "en": "Triangle inequality and |uv|=|u||v|, then |h|≤1 and h²≤1: |3x+h|≤3|x|+1 and |6x²+4xh+h²|≤6x²+4|x|+1. M depends only on x and the coefficients." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "abs_rules", "order_basics" ], "explanation": { "zh-Hant": "只給上界；|h|≤1 由 S03.3 的 δ≤1 保證。", "en": "Only an upper bound; |h|≤1 is ensured by δ≤1 in S03.3." } }, "leaf_basis": "abs_rules + order_basics + square_nonneg" }, { "id": "S03.3", "uses": [ "S03.1", "S03.2", "mko_derivative_def" ], "before_tex": "\\varepsilon>0,\\ \\delta=\\min\\left(1,\\tfrac{\\varepsilon}{M+1}\\right)", "after_tex": "0<|h|<\\delta\\Rightarrow|(D+hR)-D|\\le|h|M<\\varepsilon;\\ p'(x)=D", "relation": "implies", "rule": "mko_derivative_def", "conditions": [ "M+1>0" ], "explanation": { "zh-Hant": "若 0<|h|<δ，則 |h|<1，S03.2 適用：|hR|≤|h|M≤|h|(M+1)<δ(M+1)≤ε。ε 任意，依導數定義 p'(x)=D。x 任意，所以處處可微。S03.1 至 S03.3 證得引理 L_poly_deriv4。", "en": "If 0<|h|<δ then |h|<1 and S03.2 applies: |hR|≤|h|M≤|h|(M+1)<δ(M+1)≤ε. As ε is arbitrary, the definition gives p'(x)=D. As x is arbitrary, p is differentiable everywhere. S03.1 to S03.3 prove Lemma L_poly_deriv4." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03.1", "S03.2", "mko_derivative_def", "forall_intro" ], "explanation": { "zh-Hant": "用 M+1 避免 M=0 時除以 0。", "en": "M+1 avoids division by 0 when M=0." } }, "leaf_basis": "mko_derivative_def + forall_intro + order_basics + abs_rules" }, { "id": "S04", "uses": [ "S02", "mko_antideriv" ], "before_tex": "a_k\\ \\mapsto\\ \\tfrac{a_k}{k+1}\\ \\text{as the coefficient of}\\ x^{k+1}", "after_tex": "F_0=[0,\\ \\tfrac31,\\ \\tfrac{-4}{2},\\ \\tfrac63]=[0,3,-2,2]", "relation": "definition", "rule": "mko_antideriv", "conditions": [], "explanation": { "zh-Hant": "依 MKO 的代表公式：x^k 的係數 a_k 變成 x^(k+1) 的係數 a_k/(k+1)，常數項取 0。3/1=3、−4/2=−2、6/3=2（N4、N5），所以 F0=3x−2x²+2x³。係數以最簡分數寫出，分母為正。", "en": "By the MKO's representative formula, the coefficient a_k of x^k becomes the coefficient a_k/(k+1) of x^(k+1), with constant term 0. 3/1=3, −4/2=−2, 6/3=2 (N4, N5), so F0=3x−2x²+2x³. Coefficients are written as reduced fractions with positive denominators." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "mko_antideriv" ], "explanation": { "zh-Hant": "此步只依公式構造候選；是否為原函數由 S05 驗證。", "en": "This step only builds the candidate by the formula; whether it is an antiderivative is verified in S05." } }, "leaf_basis": "mko_antideriv + frac_rules + numeric_evaluation" }, { "id": "S05", "uses": [ "S03", "S04", "S02", "S01" ], "before_tex": "F_0=[0,3,-2,2]", "after_tex": "F_0'=[3,\\ 2\\cdot(-2),\\ 3\\cdot 2]=[3,-4,6]=f;\\ F_0\\ \\text{is an antiderivative}", "relation": "implies", "rule": "L_poly_deriv4", "conditions": [], "explanation": { "zh-Hant": "依引理，F0 處處可微，導數列表 [3,−4,6]，即 3−4x+6x²=f(x)，對每個 x 成立（N6 為多項式恆等式）。依 S01，F0 是 f 在 ℝ 上的原函數。", "en": "By the lemma, F0 is differentiable everywhere with derivative list [3,−4,6], that is 3−4x+6x²=f(x) at every x (N6 is a polynomial identity). By S01, F0 is an antiderivative of f on ℝ." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S01", "S03", "S04" ], "explanation": { "zh-Hant": "目標 G：F0 是原函數。足夠條件 H：F0 可微且導數列表等於 f 的列表；H⇒G 由定義。H 由引理逐係數正向證成，沒有用有限測點。", "en": "Goal G: F0 is an antiderivative. Sufficient condition H: F0 is differentiable with derivative list equal to the list of f; H⇒G by the definition. H is proved forwards coefficient by coefficient from the lemma, without finite test points." } }, "leaf_basis": "L_poly_deriv4 + mko_antideriv + numeric_evaluation" }, { "id": "S06", "uses": [ "S03", "S05" ], "before_tex": "F_0+C=[C,3,-2,2],\\ C\\in\\mathbb{R}", "after_tex": "(F_0+C)'=[3,-4,6]=f\\ \\text{for every}\\ C", "relation": "implies", "rule": "L_poly_deriv4", "conditions": [], "explanation": { "zh-Hant": "F0+C 的列表只在常數項不同；引理的導數列表不含 c0，所以導數仍是 [3,−4,6]=f。C 任意，依全稱推廣，整個常數族 F0+C 都是原函數。給出一個原函數 F0 與給出全部原函數不同。", "en": "The list of F0+C differs only in the constant term; the lemma's derivative list does not contain c0, so the derivative is still [3,−4,6]=f. As C is arbitrary, generalization gives that the whole family F0+C consists of antiderivatives. Giving one antiderivative F0 differs from giving all of them." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "S05", "forall_intro" ], "explanation": { "zh-Hant": "此步只證「F0+C 都是原函數」；「沒有其他」在 S09 證明。", "en": "This step proves only that every F0+C is an antiderivative; that there are no others is proved in S09." } }, "leaf_basis": "L_poly_deriv4 + forall_intro" }, { "id": "S07", "uses": [ "G3", "S06" ], "before_tex": "G=[7,3,-2,2]", "after_tex": "G=F_0+7,\\ G'=f:\\ \\text{accept}", "relation": "implies", "rule": "L_poly_deriv4", "conditions": [], "explanation": { "zh-Hant": "G 與 F0 只在常數項不同（7 對 0），所以 G=F0+7；由 S06 取 C=7，G'=f，接受。候選的常數項不影響驗收。", "en": "G differs from F0 only in the constant term (7 versus 0), so G=F0+7; by S06 with C=7, G'=f, and G is accepted. A candidate's constant term does not affect acceptance." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S06" ], "explanation": { "zh-Hant": "逐係數比較 G 與 F0 的非常數項。", "en": "Compare the nonconstant coefficients of G and F0 one by one." } }, "leaf_basis": "L_poly_deriv4 + forall_elim" }, { "id": "S08", "uses": [ "G3", "S03", "S02" ], "before_tex": "H=[0,3,-4,2]", "after_tex": "H'=[3,-8,6]\\neq[3,-4,6]=f:\\ \\text{reject}", "relation": "implies", "rule": "L_poly_deriv4", "conditions": [], "explanation": { "zh-Hant": "依引理，H'=[3,2·(−4),3·2]=[3,−8,6]。索引 1 的係數（由 0 起算） −8≠−4，所以 H'−f=−4x，不是零多項式；例如 x=1 時 H'(1)=1、f(1)=5（N7、N8）。H 被拒絕。", "en": "By the lemma, H'=[3,2·(−4),3·2]=[3,−8,6]. The coefficient at index 1 (counting from 0) is −8≠−4, so H'−f=−4x, not the zero polynomial; for example at x=1, H'(1)=1 and f(1)=5 (N7, N8). H is rejected." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "quantifier_negation" ], "explanation": { "zh-Hant": "一點不等即可否定 H'=f。", "en": "One point of disagreement refutes H'=f." } }, "leaf_basis": "L_poly_deriv4 + quantifier_negation + numeric_evaluation" }, { "id": "S09", "uses": [ "S03", "S04" ], "before_tex": "P=[p_0,p_1,p_2,p_3,p_4],\\ P'=f\\ \\text{on}\\ \\mathbb{R}", "after_tex": "P=F_0+p_0", "relation": "implies", "rule": "L_same_deriv", "conditions": [], "explanation": { "zh-Hant": "依引理 L_same_deriv，在至多五個係數的多項式中，任何原函數都與 F0 只差常數項。這正是 MKO 所說「導數係數相同時，非零次係數相同」。引理在 S09.1、S09.2 證明。", "en": "By Lemma L_same_deriv, among polynomials with at most five coefficients, every antiderivative differs from F0 only in the constant term. This is the MKO's 'identical derivative coefficients force identical positive-degree coefficients'. The lemma is proved in S09.1 and S09.2." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_same_deriv" ], "explanation": { "zh-Hant": "結論限於多項式模型；不宣稱一般可微函數的區間版本。", "en": "The conclusion is limited to the polynomial model; the interval version for general differentiable functions is not claimed." } }, "leaf_basis": "poly_coeff_identity + frac_rules + mul_inverse_nonzero" }, { "id": "S09.1", "uses": [ "S03", "S02", "poly_coeff_identity" ], "before_tex": "P'(x)=f(x)\\ \\forall x", "after_tex": "[p_1,2p_2,3p_3,4p_4]=[3,-4,6,0]", "relation": "implies", "rule": "poly_coeff_identity", "conditions": [], "explanation": { "zh-Hant": "由 S03，P' 的列表為 [p1,2p2,3p3,4p4]；f 補一個末端零為 [3,−4,6,0]（同一多項式）。兩者在每個 x 取值相同，依 poly_coeff_identity，係數逐項相同。", "en": "By S03, P' has list [p1,2p2,3p3,4p4]; f padded with a trailing zero is [3,−4,6,0] (the same polynomial). They agree at every x, so by poly_coeff_identity the coefficients agree term by term." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "poly_coeff_identity" ], "explanation": { "zh-Hant": "從函數恆等到係數相等使用 poly_coeff_identity；僅像 S11 那樣核對兩個點仍不足以證明本題的恆等式。", "en": "Passing from identical polynomial functions to equal coefficients uses poly_coeff_identity; checking only two points as in S11 is not enough to establish the identity here." } }, "leaf_basis": "poly_coeff_identity + def_unfold" }, { "id": "S09.2", "uses": [ "S09.1", "frac_rules" ], "before_tex": "kp_k=a_{k-1}\\ (k=1,2,3,4)", "after_tex": "p_1=3,\\ p_2=-2,\\ p_3=2,\\ p_4=0;\\ P=[p_0,3,-2,2]=F_0+p_0", "relation": "implies", "rule": "frac_rules", "conditions": [ "k\\neq 0" ], "explanation": { "zh-Hant": "k=1,2,3,4 都不為 0，可以除：p1=3/1=3，p2=−4/2=−2，p3=6/3=2，p4=0/4=0。所以 P 的非常數係數與 F0 完全相同，P=F0+p0，常數 p0 自由。S09.1、S09.2 證得引理 L_same_deriv。", "en": "k=1, 2, 3, 4 are all nonzero, so we may divide: p1=3/1=3, p2=−4/2=−2, p3=6/3=2, p4=0/4=0. So the nonconstant coefficients of P are exactly those of F0, P=F0+p0, and the constant p0 is free. S09.1 and S09.2 prove Lemma L_same_deriv." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S09.1", "frac_rules", "mul_inverse_nonzero" ], "explanation": { "zh-Hant": "k=0 的位置（常數項）沒有方程，所以 p0 不受限制。", "en": "There is no equation at position k=0 (the constant term), so p0 is unconstrained." } }, "leaf_basis": "frac_rules + mul_inverse_nonzero + numeric_evaluation" }, { "id": "S10", "uses": [ "S04", "G3" ], "before_tex": "F_0(2)+C,\\ C=-5", "after_tex": "(3\\cdot 2-2\\cdot 2^2+2\\cdot 2^3)+(-5)=14-5=9", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "F0(2)=6−8+16=14（N9），加 C=−5 得 9（N10）。求導會消去 C，但求值不會：C 與 x 都是真正的資料，不能因為導數不含 C 就把它刪掉。", "en": "F0(2)=6−8+16=14 (N9), and adding C=−5 gives 9 (N10). Differentiation removes C, but evaluation does not: both C and x are genuine inputs and C may not be dropped just because the derivative does not contain it." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "numeric_evaluation" ], "explanation": { "zh-Hant": "精確求值。", "en": "Exact evaluation." } }, "leaf_basis": "numeric_evaluation" }, { "id": "S11", "uses": [ "G3", "S03", "S02" ], "before_tex": "K=[0,3,-3,\\tfrac83]", "after_tex": "K'=[3,-6,8];\\ K'(0)=f(0),\\ K'(1)=f(1),\\ K'(2)=23\\neq 19=f(2):\\ \\text{reject}", "relation": "implies", "rule": "L_poly_deriv4", "conditions": [], "explanation": { "zh-Hant": "依引理，K'=[3,2·(−3),3·(8/3)]=[3,−6,8]。K'−f=2x²−2x=2x(x−1)（N14），在 x=0、x=1 恰為 0，所以只在這兩點抽樣會誤判（N11、N12）；但它不是零多項式，x=2 時 K'(2)=23、f(2)=19（N13）。逐係數比較（−6≠−4）直接拒絕 K。", "en": "By the lemma, K'=[3,2·(−3),3·(8/3)]=[3,−6,8]. K'−f=2x²−2x=2x(x−1) (N14), which vanishes at x=0 and x=1, so sampling only those points would mislead (N11, N12); but it is not the zero polynomial, and at x=2, K'(2)=23 while f(2)=19 (N13). Comparing coefficients (−6≠−4) rejects K directly." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "quantifier_negation" ], "explanation": { "zh-Hant": "兩個相符的測點不能證明恆等；一個不符的係數或點就足以拒絕。", "en": "Two matching test points cannot prove an identity; one disagreeing coefficient or point suffices to reject." } }, "leaf_basis": "L_poly_deriv4 + quantifier_negation + numeric_evaluation" }, { "id": "S12", "uses": [ "S03", "mko_antideriv" ], "before_tex": "5x^0,\\ 0", "after_tex": "5x^0=5\\ (\\text{also at}\\ x=0)\\ \\to\\ [0,5];\\ 0\\ \\to\\ [0]", "relation": "implies", "rule": "mko_antideriv", "conditions": [], "explanation": { "zh-Hant": "單項式邊界：在多項式中 x^0 恆為 1，包括 x=0，所以 5x^0 是常數 5，代表原函數為 5x，即 [0,5]（導數 [5]，由引理）。係數 c=0 時函數為零，代表原函數正規化為 [0]，不保留無意義的高次零項；其原函數族是所有常數。", "en": "Monomial boundary: in polynomials x^0 is always 1, including at x=0, so 5x^0 is the constant 5 and its representative antiderivative is 5x, that is [0,5] (derivative [5] by the lemma). With coefficient c=0 the function is zero, and its representative antiderivative is normalized to [0], without meaningless higher zero terms; its family of antiderivatives is all constants." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "mko_antideriv" ], "explanation": { "zh-Hant": "依 MKO 假設 a2 的慣例；以引理核對導數。", "en": "Follows the convention of MKO assumption a2; the derivative is checked by the lemma." } }, "leaf_basis": "mko_antideriv + L_poly_deriv4 + power_laws" }, { "id": "S13", "uses": [ "S03", "S01" ], "before_tex": "p(x)=2x", "after_tex": "\\text{antiderivative}\\ x^2;\\ \\text{inverse}\\ y\\mapsto\\tfrac y2;\\ 1/p'=\\tfrac12:\\ \\text{three different objects}", "relation": "implies", "rule": "mko_antideriv", "conditions": [], "explanation": { "zh-Hant": "對 p(x)=2x：原函數 x²（[0,0,1]，由引理導數為 [0,2]=p）；反函數是 y↦y/2；導數的倒數是 1/2。在輸入 4，x² 給 16，反函數給 2（N15）。三者完全不同，原函數不是反函數，也不是導數取倒數。", "en": "For p(x)=2x: an antiderivative is x² ([0,0,1], whose derivative is [0,2]=p by the lemma); the inverse function is y↦y/2; the reciprocal of the derivative is 1/2. At input 4, x² gives 16 while the inverse gives 2 (N15). The three are entirely different: an antiderivative is neither an inverse function nor a reciprocal of the derivative." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "S01" ], "explanation": { "zh-Hant": "一個輸入的值不同就說明兩函數不同。", "en": "Different values at one input show the two functions differ." } }, "leaf_basis": "mko_antideriv + L_poly_deriv4 + quantifier_negation" }, { "id": "S14", "uses": [ "S05", "S06", "S09" ], "before_tex": "\\text{S05, S06, S09}", "after_tex": "\\{P:\\ P'=f,\\ P\\ \\text{at most five coefficients}\\}=\\{F_0+C:\\ C\\in\\mathbb{R}\\}", "relation": "implies", "rule": "set_extensionality", "conditions": [], "explanation": { "zh-Hant": "S06 給出 ⊇（每個 F0+C 都是原函數），S09 給出 ⊆（每個這樣的 P 都是 F0+p0）。依集合外延性，兩集合相等：在多項式模型中，f 的原函數恰是常數族 F0+C。", "en": "S06 gives ⊇ (every F0+C is an antiderivative) and S09 gives ⊆ (every such P is F0+p0). By set extensionality the sets are equal: in the polynomial model, the antiderivatives of f are exactly the constant family F0+C." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S06", "S09", "set_extensionality" ], "explanation": { "zh-Hant": "目標 G：兩集合相等。足夠條件 H：雙向包含；H⇒G 由外延性。兩個包含分別由 S06、S09 正向證成。", "en": "Goal G: the two sets are equal. Sufficient condition H: inclusion both ways; H⇒G by extensionality. The two inclusions are proved forwards in S06 and S09." } }, "leaf_basis": "set_extensionality + subset_def" } ], "result": { "zh-Hant": "f=[3,−4,6] 的代表原函數是 F0=[0,3,−2,2]=3x−2x²+2x³，F0'=f。對每個實數 C，F0+C 都是原函數；在至多五個係數的多項式中，原函數恰是 F0+C。G=[7,3,−2,2]=F0+7 接受；H 因 H'=[3,−8,6] 拒絕；K′ 雖在 x=0、1 與 f 的值相符，但 K'=[3,−6,8]≠f，拒絕。F0(2)+C 在 C=−5 時為 9。原函數不是反函數，也不是導數的倒數。", "en": "The representative antiderivative of f=[3,−4,6] is F0=[0,3,−2,2]=3x−2x²+2x³, with F0'=f. For every real C, F0+C is an antiderivative; among polynomials with at most five coefficients, the antiderivatives are exactly F0+C. G=[7,3,−2,2]=F0+7 is accepted; H is rejected because H'=[3,−8,6]; K′ matches f at x=0 and 1, but K'=[3,−6,8]≠f, so it is rejected. F0(2)+C equals 9 for C=−5. An antiderivative is neither an inverse function nor a reciprocal of the derivative." }, "negative_controls": [ { "claim": { "zh-Hant": "F0 是 f 唯一的原函數", "en": "F0 is the only antiderivative of f" }, "expected": "reject", "reason": { "zh-Hant": "F0+7=G 也是（S07）；全部原函數是 F0+C。", "en": "F0+7=G is one too (S07); all antiderivatives are F0+C." } }, { "claim": { "zh-Hant": "K'(0)=f(0) 且 K'(1)=f(1)，所以 K 是原函數", "en": "K'(0)=f(0) and K'(1)=f(1), so K is an antiderivative" }, "expected": "reject", "reason": { "zh-Hant": "K'(2)=23≠19=f(2)；K'=[3,−6,8]≠[3,−4,6]。", "en": "K'(2)=23≠19=f(2); K'=[3,−6,8]≠[3,−4,6]." } }, { "claim": { "zh-Hant": "求導會消去 C，所以 C=−5 時 F0(2)+C=14", "en": "Differentiation removes C, so F0(2)+C=14 when C=−5" }, "expected": "reject", "reason": { "zh-Hant": "求值保留 C：14+(−5)=9。", "en": "Evaluation keeps C: 14+(−5)=9." } }, { "claim": { "zh-Hant": "2x 的原函數是它的反函數 x/2", "en": "The antiderivative of 2x is its inverse function x/2" }, "expected": "reject", "reason": { "zh-Hant": "(x/2)'=1/2≠2x；原函數是 x²+C。", "en": "(x/2)'=1/2≠2x; the antiderivatives are x²+C." } }, { "claim": { "zh-Hant": "5x^0 在 x=0 是 0^0，所以它的原函數在 0 沒有定義", "en": "5x^0 is 0^0 at x=0, so its antiderivative is undefined at 0" }, "expected": "reject", "reason": { "zh-Hant": "多項式慣例 x^0=1（包括 x=0），5x^0 是常數 5，原函數 5x 處處有定義。", "en": "The polynomial convention is x^0=1, including at x=0; 5x^0 is the constant 5, with antiderivative 5x defined everywhere." } } ], "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": "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": "abs_rules", "kind": "prerequisite", "statement": { "zh-Hant": "絕對值（實數）：|u|=u（u≥0）、|u|=−u（u<0）；|u|≥0；|uv|=|u||v|；三角不等式 |u+v|≤|u|+|v|；|u|v 之一；u0 推出 uwvw。", "en": "Order of the reals: exactly one of uv holds; u0 give uwvw." } }, { "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": "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_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": "mul_inverse_nonzero", "kind": "prerequisite", "statement": { "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。", "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined." } }, { "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": "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": "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": "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": "set_extensionality", "kind": "prerequisite", "statement": { "zh-Hant": "集合相等：A=B 當且僅當 A⊆B 且 B⊆A，即兩者有相同元素。", "en": "Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements." } }, { "id": "subset_def", "kind": "prerequisite", "statement": { "zh-Hant": "子集：A⊆B 表示 A 的每個元素都是 B 的元素，即 ∀x (x∈A⇒x∈B)。", "en": "Subset: A⊆B means every element of A is an element of B, i.e. ∀x (x∈A⇒x∈B)." } }, { "id": "mko_antideriv", "kind": "cited", "source": { "mko_id": "mko-antiderivative", "sha256": "5f2ecff6866e471d1f8da7fea1fa5e121364ae2d24b2fea405d2300fe9ff1cae", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "F 是 f 在區間上的原函數，指 F 在該區間可微且 F'=f。全 ℝ 多項式 f(x)=Σa_k x^k 有代表 F0(x)=Σa_k x^(k+1)/(k+1)，常數項固定 0；F0+C 仍為原函數。多項式原函數的導數係數相同時，非零次係數相同，只差常數項。", "en": "F is an antiderivative of f on an interval when F is differentiable there and F'=f. For the polynomial defined on all of ℝ, f(x)=Σa_k x^k, choose F0(x)=Σa_k x^(k+1)/(k+1) with constant term zero; F0+C is also an antiderivative. Within the polynomial model, identical derivative coefficients force identical positive-degree coefficients, leaving only the constant term free." } }, { "id": "mko_derivative_def", "kind": "cited", "source": { "mko_id": "mko-derivative", "sha256": "b523a35b9e5c20d66a7346f84ac5bece2c4167b17d46208fdb9b40115deeaf64", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "對 f:ℝ→ℝ，f 在 x 的導數是 lim_(h→0,h≠0)(f(x+h)−f(x))/h，若此兩側有限極限存在。", "en": "For f:ℝ→ℝ, the derivative at x is lim_(h→0,h≠0)(f(x+h)−f(x))/h when this finite two-sided limit exists." } }, { "id": "poly_coeff_identity", "kind": "prerequisite", "statement": { "zh-Hant": "兩個實係數多項式若在每個實數 x 取值相同，則正規化後的係數列表相同（非零多項式只有有限個根）。", "en": "Two real polynomials that take the same value at every real x have the same normalized coefficient lists (a nonzero polynomial has only finitely many roots)." } }, { "id": "L_poly_deriv4", "kind": "lemma", "proved_in": [ "S03.1", "S03.2", "S03.3" ], "statement": { "zh-Hant": "對 p(x)=c0+c1x+c2x²+c3x³+c4x⁴（實係數）與每個實數 x：p 在 x 可微且 p'(x)=c1+2c2x+3c3x²+4c4x³；即 [c0,c1,c2,c3,c4] 的導數列表為 [c1,2c2,3c3,4c4]。", "en": "For p(x)=c0+c1x+c2x²+c3x³+c4x⁴ with real coefficients and every real x: p is differentiable at x and p'(x)=c1+2c2x+3c3x²+4c4x³; that is, [c0,c1,c2,c3,c4] has derivative list [c1,2c2,3c3,4c4]." } }, { "id": "L_same_deriv", "kind": "lemma", "proved_in": [ "S09.1", "S09.2" ], "statement": { "zh-Hant": "若 P=[p0,p1,p2,p3,p4] 在 ℝ 上滿足 P'=f，則 p1=3、p2=−2、p3=2、p4=0，即 P=F0+p0。", "en": "If P=[p0,p1,p2,p3,p4] satisfies P'=f on ℝ, then p1=3, p2=−2, p3=2, p4=0, that is P=F0+p0." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S03.1", "vars": [ "x", "h" ], "degree": 2, "lhs": "(x+h)**2-x**2", "rhs": "h*(2*x+h)" }, { "id": "N2", "step": "S03.1", "vars": [ "x", "h" ], "degree": 3, "lhs": "(x+h)**3-x**3", "rhs": "h*(3*x**2+h*(3*x+h))" }, { "id": "N3", "step": "S03.1", "vars": [ "x", "h" ], "degree": 4, "lhs": "(x+h)**4-x**4", "rhs": "h*(4*x**3+h*(6*x**2+4*x*h+h**2))" }, { "id": "N4", "step": "S04", "lhs": "-4/2", "rhs": "-2" }, { "id": "N5", "step": "S04", "lhs": "6/3", "rhs": "2" }, { "id": "N6", "step": "S05", "vars": [ "x" ], "degree": 2, "lhs": "3+2*(-2)*x+3*2*x**2", "rhs": "3-4*x+6*x**2" }, { "id": "N7", "step": "S08", "lhs": "3+2*(-4)*1+3*2*1**2", "rhs": "1" }, { "id": "N8", "step": "S08", "lhs": "3-4*1+6*1**2", "rhs": "5" }, { "id": "N9", "step": "S10", "lhs": "3*2-2*2**2+2*2**3", "rhs": "14" }, { "id": "N10", "step": "S10", "lhs": "14+(-5)", "rhs": "9" }, { "id": "N11", "step": "S11", "lhs": "3+2*(-3)*0+3*(8/3)*0**2", "rhs": "3-4*0+6*0**2" }, { "id": "N12", "step": "S11", "lhs": "3+2*(-3)*1+3*(8/3)*1**2", "rhs": "3-4*1+6*1**2" }, { "id": "N13", "step": "S11", "lhs": "3+2*(-3)*2+3*(8/3)*2**2", "relation": "!=", "rhs": "3-4*2+6*2**2" }, { "id": "N14", "step": "S11", "vars": [ "x" ], "degree": 2, "lhs": "(3-6*x+8*x**2)-(3-4*x+6*x**2)", "rhs": "2*x*(x-1)" }, { "id": "N15", "step": "S13", "lhs": "4**2", "relation": "!=", "rhs": "4/2" }, { "id": "N16", "lhs": "8/3", "rhs": "2+2/3" } ] }, "presentation": { "md_scope": { "zh-Hant": "只在 MKO 的全實數多項式模型內工作：f 的具體輸入至多四個整數係數，G、H、K 等驗收候選至多五個有理係數，均用升冪列表；一般原函數族 F0+C 與係數證明則容許實數 C 及實係數多項式。「只差常數」只在多項式模型內以係數證明；一般可微函數在區間上的版本需要平均值定理，不在本稿證明。不改動 canonical MKO 或英文 edition。", "en": "Works only inside the MKO's polynomial model on all of ℝ: the concrete input f has at most four integer coefficients and exercise candidates such as G,H,K have at most five rational coefficients in ascending lists; the general family F0+C and its coefficient proof allow real C and real polynomial coefficients. 'Differ only by a constant' is proved by coefficients within the polynomial model only; the version for general differentiable functions on an interval needs the mean value theorem and is not proved here. It does not change the canonical MKO or its English edition." }, "md_symbols": [ [ "[a_0,a_1,a_2,a_3]", "升冪係數列表", "an ascending coefficient list" ], [ "F_0", "常數項為 0 的代表原函數", "the representative antiderivative with constant term 0" ], [ "F_0+C", "原函數常數族", "the constant family of antiderivatives" ], [ "D,\\ R,\\ M", "差商的主項、餘項與餘項的界", "the main term, remainder and remainder bound of the quotient" ] ], "md_error_controls": [ { "title": "不要漏掉常數族 / Do not drop the constant family", "zh": "MKO 的誤解之一是漏掉常數族。F0 只是選定的代表；全部原函數是 F0+C（S06、S14）。C 在求導時消失，但在求值時有作用：F0(2)+C 在 C=−5 時是 9，不是 14（S10）。", "en": "One misconception the MKO lists is dropping the constant family. F0 is only the chosen representative; all antiderivatives are F0+C (S06, S14). C disappears under differentiation but matters in evaluation: F0(2)+C is 9 for C=−5, not 14 (S10)." }, { "title": "有限測點不能代替多項式恆等 / Finite test points cannot replace a polynomial identity", "zh": "另一個誤解是以幾個相等測點代替 F'=f。K' 與 f 在 x=0、1 相等，因為差 2x(x−1) 在這兩點為零；但 x=2 時 23≠19（S11）。驗收要逐係數比較導數列表（S09.1 說明為何係數比較才對應函數恆等）。", "en": "Another misconception is replacing F'=f by a few equal test points. K' and f agree at x=0 and 1 because their difference 2x(x−1) vanishes there; but at x=2, 23≠19 (S11). Acceptance must compare derivative lists coefficient by coefficient (S09.1 explains why coefficient comparison matches the functional identity)." }, { "title": "原函數不是反函數 / An antiderivative is not an inverse function", "zh": "第三個誤解是把原函數當反函數。p(x)=2x 的原函數是 x²，反函數是 y/2，導數的倒數是 1/2，三者不同（S13）。原函數的定義只涉及 F'=f（S01）。", "en": "A third misconception is treating an antiderivative as an inverse function. For p(x)=2x the antiderivative is x², the inverse is y/2, and the reciprocal of the derivative is 1/2, three different objects (S13). The definition of an antiderivative involves only F'=f (S01)." } ], "md_open_items": [ { "zh-Hant": "一般可微函數在區間上「兩個原函數只差常數」需要平均值定理；本稿只在多項式模型內以係數證明，不外推到一般函數或不連通論域。", "en": "For general differentiable functions on an interval, 'two antiderivatives differ by a constant' needs the mean value theorem; this guide proves it only by coefficients inside the polynomial model and does not extend it to general functions or disconnected domains." }, { "zh-Hant": "poly_coeff_identity（處處相等的多項式係數相同）是本稿的前提，未在本稿中證明。", "en": "poly_coeff_identity (polynomials equal everywhere have equal coefficients) is a premise of this guide and is not proved here." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "0a7ad7b92f3eb8cf9c92f59c031efe93847f4df94d0f2541c2d6a9ae98227411", "original_source_sha256": "2ed700aa9d9c5a319e9f4d0ebcba193c2e7bdf4f88f31d9da27c9e22df7547ef", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [ "BR-M027", "BR-M028" ], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }