{"schema_version":"ocme-site-results-v1","locale":"en","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"page":"5"},"kind":"guides","total":299,"pages":13,"page":5,"entries":[{"id":"derivation-mko-discrete-exponential-growth-01","title":{"zh-Hant":"離散指數成長模型：由 x_(n+1)=r·x_n 歸納證明 x_n=x_0·r^n","en":"Discrete exponential growth: proving x_n=x_0·r^n from x_(n+1)=r·x_n by induction"},"summary":{"zh-Hant":"在固定時間步、常數因子 r≥0 的模型 x_(n+1)=r·x_n 中，以數學歸納法證明對每個自然數 n 有 x_n=x_0·r^n，並說明遞迴式與閉式描述同一數列；以 x_0=8、r=3/2 計算 x_1 到 x_4 並與閉式比對；說明 r>1、r=1、0≤r<1 與 x_0=0 的行為；判定資料 8、12、18、26 不符合任何以 x_0=8 開始的固定因子模型；並說明每步因子 3/2 不是連續成長率 1/2。","en":"In the model x_(n+1)=r·x_n with a fixed time step and constant factor r≥0, prove by induction that x_n=x_0·r^n for every natural number n, and explain that the recurrence and the closed form describe the same sequence; with x_0=8 and r=3/2 compute x_1 to x_4 and compare with the closed form; describe the behaviour for r>1, r=1, 0≤r<1 and x_0=0; decide that the data 8, 12, 18, 26 fit no fixed-factor model starting at x_0=8; and explain why the per-step factor 3/2 is not the continuous growth rate 1/2."},"mko_ids":["mko-discrete-exponential-growth"]},{"id":"derivation-mko-compound-interest-01","title":{"zh-Hant":"複利模型：本金 1000、每年 2%、三年的餘額與單利比較","en":"Compound-interest model: principal 1000 at 2% per year for three years, compared with simple interest"},"summary":{"zh-Hant":"在期別為一年、每期利率 i=1/50（2%）、期間內無存提款與費用的模型中，以歸納證明 B_n=B_0(1+i)^n，算出 B_0=1000 時三年後的餘額 132651/125（即 1061.208）與利息；與單利 B_0(1+ni)=1060 比較，並以 Bernoulli 不等式證明複利餘額對每個 n 都不小於單利；說明期別混用與期間現金流如何使單一冪次公式失效。","en":"In the model with one-year periods, rate i=1/50 (2%) per period and no deposits, withdrawals or fees, prove B_n=B_0(1+i)^n by induction, compute the balance after three years from B_0=1000, namely 132651/125 (that is 1061.208), and the interest; compare with simple interest B_0(1+ni)=1060 and prove by Bernoulli's inequality that the compound balance is never below the simple one for any n; explain how mixing period units and cash flows during the period break the single-power formula."},"mko_ids":["mko-compound-interest"]},{"id":"derivation-mko-mathematical-representation-01","title":{"zh-Hant":"數學表徵：判定文字、符號、表格與序對是否表示同一個函數","en":"Mathematical representation: deciding whether words, a formula, tables and ordered pairs represent the same function"},"summary":{"zh-Hant":"固定對象 f:D→ℤ，D={0,1,2,3}，f(x)=2x−1。判定：文字「輸入加倍後減 1」對應 2x−1；表格 T1={(0,−1),(1,1),(2,3),(3,5)} 表示 f；序對集合是 f 的圖形；由兩列 (1,1)、(3,5) 唯一恢復仿射規則；表格 T2 把 (3,5) 換成 (3,6) 後不表示 f，也不表示任何仿射規則；含重複輸入的 T3 不是函數表；定義在 ℤ 上的 g(x)=2x−1 與 f 不是同一個函數；以及表格不提供未列點的值。","en":"Fix the object f:D→ℤ with D={0,1,2,3} and f(x)=2x−1. Decide: the words 'double the input, then subtract 1' correspond to 2x−1; the table T1={(0,−1),(1,1),(2,3),(3,5)} represents f; the set of ordered pairs is the graph of f; the affine rule is recovered uniquely from the two rows (1,1) and (3,5); the table T2, with (3,5) replaced by (3,6), represents neither f nor any affine rule; T3, with a repeated input, is not a function table; g(x)=2x−1 defined on ℤ is not the same function as f; and a table supplies no values at unlisted points."},"mko_ids":["mko-mathematical-representation"]},{"id":"derivation-mko-diagnostic-mathematical-assessment-01","title":{"zh-Hant":"診斷性數學評量記錄：一道分數加法題的合成記錄與判定","en":"Diagnostic assessment records: synthetic records and judgments for one fraction-addition item"},"summary":{"zh-Hant":"對題目 Q-FRAC-01「計算 3/4+1/6，答案寫成分數 a/b（a、b 為整數，b≠0）」，先精確算出期望結果 e，再寫出判定規則 R_val（作答的值等於 e 即符合）與較嚴格的 R_low（值相等且為最簡分數）。據此判定三份合成記錄：D1 作答 4/10、D2 作答 22/24、D3 作答 22/24 但規則欄空白；並證明 D1 的「不符合」不能推出任何關於真實學生能力或錯誤原因的結論。","en":"For the item Q-FRAC-01, 'compute 3/4+1/6 and write the answer as a fraction a/b with integers a, b and b≠0', first compute the expected result e exactly, then state the judgment rule R_val (a response matches when its value equals e) and the stricter rule R_low (equal value and lowest terms). Use them to judge three synthetic records: D1 with response 4/10, D2 with response 22/24, and D3 with response 22/24 but an empty rule field; and show that the label 'no match' on D1 supports no conclusion about a real student's ability or error cause."},"mko_ids":["mko-diagnostic-mathematical-assessment"]},{"id":"derivation-mko-orbit-fixed-periodic-01","title":{"zh-Hant":"軌道、固定點與週期點：完整分類 T(x)=(2x+1) mod 6 的六個狀態","en":"Orbits, fixed points and periodic points: classifying all six states of T(x)=(2x+1) mod 6"},"summary":{"zh-Hant":"在狀態空間 X={0,1,2,3,4,5} 上取自映射 T(x)=(2x+1) mod 6。求出全部固定點與最小週期 2 的週期點；證明對最小週期 p 的點，T^n(x)=x 當且僅當 p 整除 n，並用它說明 T^4(1)=1 不表示 1 的最小週期是 4；最後證明 0、2、4 不是週期點，雖然它們的軌道一步之後就進入循環。","en":"On the state space X={0,1,2,3,4,5} take the self-map T(x)=(2x+1) mod 6. Find all fixed points and all periodic points of least period 2; prove that for a point of least period p, T^n(x)=x exactly when p divides n, and use this to explain why T^4(1)=1 does not make 4 the least period of 1; finally prove that 0, 2 and 4 are not periodic, although their orbits enter a cycle after one step."},"mko_ids":["mko-orbit-fixed-periodic"]},{"id":"derivation-mko-initial-value-problem-01","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"},"summary":{"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."},"mko_ids":["mko-initial-value-problem"]},{"id":"derivation-mko-bisection-method-01","title":{"zh-Hant":"二分法：四步 bracket 更新、不變量與根存在的獨立責任","en":"Bisection: four bracket updates, the invariant and the separate proof obligation for root existence"},"summary":{"zh-Hant":"對 f(x)=x²−2，從 [1,2] 與所提供的端點符號 f(1)<0、f(2)>0 出發，依所提供的中點符號執行四步二分，追蹤 bracket；證明「端點異號且 a<b」是不變量、n 步後寬度為 1/2^n、在每次所需符號可取得且更新有效的延伸執行中，寬度停止準則必在有限步內終止；再證明這些 bracket 資料本身不證明有根，並另外引用中間值定理，對連續的 f 得出最終 bracket 內有根。","en":"For f(x)=x²−2, start from [1,2] with the supplied endpoint signs f(1)<0 and f(2)>0, run four bisection steps with the supplied midpoint signs and trace the brackets; prove that 'opposite endpoint signs and a<b' is an invariant, that the width after n steps is 1/2^n, and that, in an extended run with every needed sign available and effective updates, a width-based stopping rule terminates after finitely many steps; then show that the bracket data alone do not prove a root exists, and separately cite the intermediate value theorem to conclude, for the continuous f, that the final bracket contains a root."},"mko_ids":["mko-bisection-method"]},{"id":"derivation-mko-linear-programming-01","title":{"zh-Hant":"線性規劃：在明示候選中求最佳，並說明它不是連續最優","en":"Linear programming: the best supplied candidate, and why it is not the continuous optimum"},"summary":{"zh-Hant":"限制 3x+2y≤12、x+2y≤6、−x≤0、−y≤0，目標為最大化 L(x,y)=x+y。對明示的七個候選 P1=(0,0)、P2=(4,0)、P3=(3,1)、P4=(2,2)、P5=(0,3)、P6=(4,1)、P7=(3,2)，逐限制核對可行性，只在可行候選間比較目標值，求出候選中的最大值與全部最佳點；再以可行點 Q=(3,3/2) 與一個不等式證書證明連續可行域上的最大值是 9/2，而不是候選中的 4。","en":"Constraints 3x+2y≤12, x+2y≤6, −x≤0, −y≤0; objective: maximize L(x,y)=x+y. For the seven supplied candidates P1=(0,0), P2=(4,0), P3=(3,1), P4=(2,2), P5=(0,3), P6=(4,1), P7=(3,2), check feasibility constraint by constraint, compare objective values only among feasible candidates, and find the best value among the candidates and all points attaining it; then use the feasible point Q=(3,3/2) and an inequality certificate to prove that the maximum over the continuous feasible region is 9/2, not the candidate value 4."},"mko_ids":["mko-linear-programming"]},{"id":"derivation-mko-discrete-logistic-growth-01","title":{"zh-Hant":"離散 logistic 增量模型：[0,K] 的不變性、固定點與一條短軌道","en":"Discrete logistic increment model: invariance of [0,K], fixed points and a short orbit"},"summary":{"zh-Hant":"對模型 G(x)=x+r x(1−x/K)（K>0、0≤r≤1），證明：若 0≤x≤K，則 0≤G(x)≤K，因此從 [0,K] 出發的整條軌道都留在 [0,K]；0 與 K 是固定點，且 r>0 時它們是僅有的固定點。再以 K=8、r=1、x0=2 精確算出 x1、x2、x3，並以 r=3 的例子說明條件 r≤1 不可省。","en":"For the model G(x)=x+r x(1−x/K) with K>0 and 0≤r≤1, prove: if 0≤x≤K then 0≤G(x)≤K, so the whole orbit starting in [0,K] stays in [0,K]; 0 and K are fixed points, and for r>0 they are the only ones. Then compute x1, x2, x3 exactly for K=8, r=1, x0=2, and use an example with r=3 to show that the condition r≤1 cannot be dropped."},"mko_ids":["mko-discrete-logistic-growth"]},{"id":"derivation-mko-harmonic-oscillator-model-01","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"},"summary":{"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."},"mko_ids":["mko-harmonic-oscillator-model"]},{"id":"derivation-mko-present-value-discounting-01","title":{"zh-Hant":"現值與折現：證明折現與複利互逆，並算出 1331 的三年現值","en":"Present value and discounting: proving that discounting inverts compounding, and the three-year present value of 1331"},"summary":{"zh-Hant":"在固定每期利率 i（1+i>0）與期數 n 下，證明 (1+i)^n>0，因此現值 PV_n(F)=F/(1+i)^n 有定義；證明 PV_n 與複利 A_n(P)=P(1+i)^n 互為反函數。再以 i=1/10（每年）、n=3、F=1331 算出現值 1000，用逐年折現與 roundtrip 交叉核對，並說明把折現寫成乘法或混用年率與月率為何錯誤。","en":"For a fixed rate i per period (1+i>0) and a period count n, prove that (1+i)^n>0, so the present value PV_n(F)=F/(1+i)^n is defined; prove that PV_n and compounding A_n(P)=P(1+i)^n are inverse functions of each other. Then, with i=1/10 per year, n=3 and F=1331, compute the present value 1000, cross-check by year-by-year discounting and a round trip, and explain why writing discounting as a multiplication, or mixing annual and monthly rates, is wrong."},"mko_ids":["mko-present-value-discounting"]},{"id":"derivation-mko-complex-number-01","title":{"zh-Hant":"複數：以有序對計算 (2+3i)(4−5i) 並判定相等","en":"Complex numbers: computing (2+3i)(4−5i) as an ordered pair and deciding equality"},"summary":{"zh-Hant":"設 z=2+3i、w=4−5i。以有序對（實部, 虛部係數）求 z+w、z−w 與 z·w。乘法先由分配律與 i²=−1 推出一般公式 (a+bi)(c+di)=(ac−bd)+(ad+bc)i，再代入數值，並以直接展開交叉核對；最後判定 z·w 是否等於實數 23。","en":"Let z=2+3i and w=4−5i. Find z+w, z−w and z·w as ordered pairs (real part, imaginary coefficient). For the product, first derive the general formula (a+bi)(c+di)=(ac−bd)+(ad+bc)i from distributivity and i²=−1, then substitute the numbers and cross-check by direct expansion; finally decide whether z·w equals the real number 23."},"mko_ids":["mko-complex-number"]},{"id":"derivation-mko-real-vector-space-01","title":{"zh-Hant":"實向量空間：在 ℝ³ 核對公理實例並證明一條分配律","en":"Real vector space: auditing axiom instances in ℝ³ and proving one distributive law"},"summary":{"zh-Hant":"在 ℝ³ 中取 u=(1,−2,3)、v=(4,0,−1)、a=2、b=−3。（一）確認維度相同；（二）精確核對 a(u+v)=au+av、(a+b)u=au+bu、(ab)u=a(bu)、1u=u、u+(−u)=0 五個公理實例；（三）對所有實數 a 與所有 u、v∈ℝ³ 證明 a(u+v)=au+av；（四）以 ℝ² 上改造的純量乘法說明實例通過不等於公理成立。","en":"In ℝ³ take u=(1,−2,3), v=(4,0,−1), a=2, b=−3. (1) Confirm the dimensions agree; (2) check exactly the five axiom instances a(u+v)=au+av, (a+b)u=au+bu, (ab)u=a(bu), 1u=u and u+(−u)=0; (3) prove a(u+v)=au+av for every real a and all u, v∈ℝ³; (4) use a modified scalar multiplication on ℝ² to show that successful instances do not establish an axiom."},"mko_ids":["mko-real-vector-space"]},{"id":"derivation-mko-finite-group-01","title":{"zh-Hant":"有限群：以完整 Cayley 表核對群公理","en":"Finite group: auditing the group axioms from a complete Cayley table"},"summary":{"zh-Hant":"給兩個完整運算表。（甲）G={0,1,2,3} 配上表 T，證明 (G,*) 是有限群：封閉、全部 4³=64 個三元組的結合律、雙側單位元、雙側逆元。（乙）K={0,1,2} 配上表 ∘：它對稱、有雙側單位元、每個元素都有雙側逆元，卻在三元組 (1,1,2) 違反結合律，因此不是群；同時說明含 0 的 19 個三元組全部通過，抽樣會誤判。","en":"Two complete operation tables are given. (A) For G={0,1,2,3} with table T, prove that (G,*) is a finite group: closure, associativity for all 4³=64 triples, a two-sided identity and two-sided inverses. (B) K={0,1,2} with table ∘ is symmetric, has a two-sided identity and two-sided inverses for every element, yet violates associativity at the triple (1,1,2), so it is not a group; also show that all 19 triples containing 0 pass, so sampling would misjudge it."},"mko_ids":["mko-finite-group"]},{"id":"derivation-mko-metric-space-01","title":{"zh-Hant":"度量空間：逐格與逐三元組核對有限距離表","en":"Metric space: auditing a finite distance table cell by cell and triple by triple"},"summary":{"zh-Hant":"給三張完整整數距離表。（甲）X={0,1,2,3} 上的表 d（四點環上的步數距離）：逐格核非負、對角線為零、非對角線不為零、對稱，並以涵蓋全部 4³=64 個三元組的分情形論證核三角不等式，得出 (X,d) 是度量空間。（乙）Y={0,1,2} 上的表 e 通過前四項卻在 (0,1,2) 違反三角不等式。（丙）Z={0,1,2} 上的表 f 非負、對稱、對角線為零，卻有非對角零值，違反分離性。","en":"Three complete integer distance tables are given. (A) On X={0,1,2,3}, table d (step distance on a 4-cycle): check nonnegativity, zero diagonal, nonzero off-diagonal entries and symmetry cell by cell, and the triangle inequality by a case argument covering all 4³=64 triples, concluding that (X,d) is a metric space. (B) Table e on Y={0,1,2} passes the first four checks but violates the triangle inequality at (0,1,2). (C) Table f on Z={0,1,2} is nonnegative, symmetric and zero on the diagonal, yet has an off-diagonal zero and violates separation."},"mko_ids":["mko-metric-space"]},{"id":"derivation-mko-finite-discrete-probability-distribution-01","title":{"zh-Hant":"有限離散機率分布：核 PMF 資格、事件機率、期望與推前分布","en":"Finite discrete probability distribution: validating a PMF, event probability, expectation and pushforward"},"summary":{"zh-Hant":"隨機變數 X 取值 −1、0、1、2，質量 p(−1)=1/6、p(0)=1/3、p(1)=1/4、p(2)=1/4。（一）先核 p 是否為合格 PMF；（二）求 P(X≥1)；（三）求 E[X]；（四）求 Y=X² 的推前分布、證明它仍是 PMF，求 E[Y] 並以 Σx²p(x) 交叉核對；（五）判定兩個候選 q、r 不是 PMF：q 每項介於 0 與 1 但總和不為 1，r 總和為 1 但有負項。","en":"A random variable X takes the values −1, 0, 1, 2 with masses p(−1)=1/6, p(0)=1/3, p(1)=1/4, p(2)=1/4. (1) First check that p is a valid PMF; (2) find P(X≥1); (3) find E[X]; (4) find the pushforward distribution of Y=X², prove it is again a PMF, find E[Y] and cross-check it against Σx²p(x); (5) decide that two candidates q and r are not PMFs: every mass of q lies between 0 and 1 but the total is not 1, while r sums to 1 but has a negative mass."},"mko_ids":["mko-finite-discrete-probability-distribution"]},{"id":"derivation-mko-complex-modulus-conjugate-01","title":{"zh-Hant":"複數模與共軛：證明 z·z̄=|z|² 與共軛保持運算","en":"Complex modulus and conjugate: proving z·z̄=|z|² and that conjugation preserves the operations"},"summary":{"zh-Hant":"對 z=a+bi、w=c+di（a、b、c、d 為實數）證明：（一）z·z̄=(a²+b²)+0i，即模平方的實數嵌入；（二）共軛保持加法：(z+w)‾=z̄+w̄；（三）共軛保持乘法：(zw)‾=z̄·w̄；（四）由此得 |zw|²=|z|²|w|²。再以 z=3−4i、w=1+2i 精確核對，並區分模平方與模。","en":"For z=a+bi and w=c+di (a, b, c, d real) prove: (1) z·z̄=(a²+b²)+0i, the real embedding of the modulus squared; (2) conjugation preserves addition: (z+w)‾=z̄+w̄; (3) conjugation preserves multiplication: (zw)‾=z̄·w̄; (4) hence |zw|²=|z|²|w|². Then check everything exactly for z=3−4i and w=1+2i, distinguishing the modulus squared from the modulus."},"mko_ids":["mko-complex-modulus-conjugate"]},{"id":"derivation-mko-linear-combination-span-01","title":{"zh-Hant":"線性組合與生成：以有限係數 witness 判定 span 成員","en":"Linear combinations and span: deciding membership using finite coefficient lists as witnesses"},"summary":{"zh-Hant":"（一）在 ℝ² 中，v₁=(1,2)、v₂=(3,−1)，判定 b=(7,0) 是否屬於 span{v₁,v₂}：先證明兩個向量的生成集恰為 {c₁v₁+c₂v₂}，再解係數、證明唯一，並由 witness 重建 b。（二）在 ℝ³ 中，u₁=(1,0,1)、u₂=(0,1,1)，證明 t=(2,3,4) 不屬於 span{u₁,u₂}，而 t'=(2,3,5) 屬於：同為三維並不決定成員資格。","en":"(1) In ℝ², with v₁=(1,2) and v₂=(3,−1), decide whether b=(7,0) lies in span{v₁,v₂}: first prove that the span of two vectors is exactly {c₁v₁+c₂v₂}, then solve for the coefficients, prove they are unique, and rebuild b from the witness. (2) In ℝ³, with u₁=(1,0,1) and u₂=(0,1,1), prove that t=(2,3,4) is not in span{u₁,u₂} while t'=(2,3,5) is: having three coordinates does not decide membership."},"mko_ids":["mko-linear-combination-span"]},{"id":"derivation-mko-subgroup-homomorphism-01","title":{"zh-Hant":"子群與群同態：在 ℤ/4 與 ℤ/2 的完整表上核對","en":"Subgroups and homomorphisms: auditing on the complete tables of ℤ/4 and ℤ/2"},"summary":{"zh-Hant":"G={0,1,2,3} 以 (a+b) mod 4 為運算，K={0,1} 以 (a+b) mod 2 為運算，兩者都給出完整 Cayley 表。（一）證明 H={0,2} 是 G 的子群，H'={0,1} 不是；（二）證明映射表 f(0)=0、f(1)=1、f(2)=0、f(3)=1 是同態 G→K（涵蓋全部 16 個輸入對）；（三）求 kernel 與 image，說明 f 滿射但不單射，因此不是同構；（四）說明常數映射 h≡1 不是同態。","en":"G={0,1,2,3} with operation (a+b) mod 4 and K={0,1} with operation (a+b) mod 2 are given by complete Cayley tables. (1) Prove that H={0,2} is a subgroup of G and H'={0,1} is not; (2) prove that the mapping table f(0)=0, f(1)=1, f(2)=0, f(3)=1 is a homomorphism G→K, covering all 16 input pairs; (3) find the kernel and the image, and show that f is surjective but not injective, hence not an isomorphism; (4) show that the constant map h≡1 is not a homomorphism."},"mko_ids":["mko-subgroup-homomorphism"]},{"id":"derivation-mko-metric-convergence-01","title":{"zh-Hant":"度量空間中的收斂：有限度量表上最終常值序列的極限與唯一性","en":"Convergence in a metric space: limit and uniqueness for an ultimately constant sequence on a finite metric table"},"summary":{"zh-Hant":"在三點集合 X={a,b,c} 上給定完整度量表 d(a,b)=1、d(a,c)=2、d(b,c)=3/2。序列 x_0=a、x_1=c、x_2=a，並明示對所有 n≥3 有 x_n=b。(1) 核對 d 是度量；(2) 以 ε-N 定義證明 x_n→b，並找出各 ε 下最小的有效 N；(3) 證明度量空間中極限唯一，並直接給出 x_n 不收斂到 a、也不收斂到 c 的見證；(4) 用一個在有限尾段上與 x_n 完全一致、極限卻不同的序列，說明只核有限尾段不足以判定極限是否為指定的 b。","en":"On the three-point set X={a,b,c} a complete metric table is given: d(a,b)=1, d(a,c)=2, d(b,c)=3/2. The sequence has x_0=a, x_1=c, x_2=a, and it is declared that x_n=b for every n≥3. (1) Check that d is a metric; (2) prove x_n→b from the ε-N definition and find the least valid N for each ε; (3) prove that limits in a metric space are unique, and give direct witnesses that x_n converges neither to a nor to c; (4) using a sequence that agrees with x_n on a finite tail but has a different limit, show that checking a finite tail cannot decide whether the limit is the specified b."},"mko_ids":["mko-metric-convergence"]},{"id":"derivation-mko-sigma-algebra-measure-space-01","title":{"zh-Hant":"σ-代數與測度空間：四點載體上由原子生成的 σ-代數與一個機率測度","en":"Sigma-algebras and measure spaces: an atom-generated sigma-algebra and a probability measure on a four-point carrier"},"summary":{"zh-Hant":"在 X={0,1,2,3} 上，以原子 A_1={0}、A_2={1,2}、A_3={3} 生成的 8 個集合構成 Σ，並以 m_1=1/2、m_2=1/3、m_3=1/6 定義 μ(∪_{i∈I}A_i)=Σ_{i∈I}m_i。證明 Σ 是 σ-代數（含可數聯集的有限歸約），證明 μ 是測度（含可數可加性的有限歸約）且為機率測度；再說明 𝒢={∅,{0},{1},X} 不是 σ-代數，以及只改一個值的 ν 不是測度。","en":"On X={0,1,2,3}, the 8 sets generated by the atoms A_1={0}, A_2={1,2}, A_3={3} form Σ, and μ(∪_{i∈I}A_i)=Σ_{i∈I}m_i with m_1=1/2, m_2=1/3, m_3=1/6. Prove that Σ is a sigma-algebra (with the finite reduction of countable unions), that μ is a measure (with the finite reduction of countable additivity) and a probability measure; then show that 𝒢={∅,{0},{1},X} is not a sigma-algebra and that ν, which changes a single value, is not a measure."},"mko_ids":["mko-sigma-algebra-measure-space"]},{"id":"derivation-mko-discrete-stochastic-process-01","title":{"zh-Hant":"離散隨機過程：由完整路徑分布導出邊際與兩時刻聯合分布，並檢驗 Markov 性與平穩性","en":"Discrete-time stochastic process: marginals and two-time joint laws from a complete path distribution, with Markov and stationarity checks"},"summary":{"zh-Hant":"時域 {0,1,2}、狀態 {0,1}。完整路徑支撐與精確 PMF 為 (0,0,0):1/4、(0,1,1):1/4、(1,0,1):1/8、(1,1,0):1/8、(1,1,1):1/4。(1) 核對支撐與 PMF，建立機率空間並說明 X_0、X_1、X_2 是同一空間上的隨機變數；(2) 求三個時刻的邊際分布與 (X_1,X_2) 的聯合分布；(3) 證明這個過程不平穩、不具 Markov 性；(4) 構造一個邊際相同但聯合分布不同的路徑律，說明邊際表不等於完整過程。","en":"Horizon {0,1,2}, states {0,1}. The complete path support with exact PMF is (0,0,0):1/4, (0,1,1):1/4, (1,0,1):1/8, (1,1,0):1/8, (1,1,1):1/4. (1) Check the support and the PMF, build the probability space and explain why X_0, X_1, X_2 are random variables on the same space; (2) find the marginal law at each time and the joint law of (X_1,X_2); (3) prove that this process is neither stationary nor Markov; (4) construct a path law with the same marginals but a different joint law, showing that the marginal tables are not the whole process."},"mko_ids":["mko-discrete-stochastic-process"]},{"id":"derivation-mko-multivariable-real-function-01","title":{"zh-Hant":"多變數實函數：二變數二次多項式的係數次序、宣告定義域、切片與點值","en":"Multivariable real-valued function: coefficient order, declared domain, slices and point values of a bivariate quadratic"},"summary":{"zh-Hant":"係數向量依 [1,x,y,x²,xy,y²] 次序為 c=(1,−2,3,1,2,−1)，宣告定義域為閉矩形 R=[−1,2]×[0,3]。(1) 寫出 f 並確認 f:R→ℝ 是函數；(2) 分別核對 x、y 邊界，判定 (2,3)、(3,1)、(0,−1) 是否在 R 中；(3) 計算 f(2,1) 與 f(1,2)，並證明交換 x、y 係數得到的 g 是不同的函數；(4) 求 y=1 的 x-切片，並說明給定的 y=1 切片與指定四個點值各自都不足以決定 f。","en":"The coefficient vector in the order [1,x,y,x²,xy,y²] is c=(1,−2,3,1,2,−1), and the declared domain is the closed rectangle R=[−1,2]×[0,3]. (1) Write f and confirm that f:R→ℝ is a function; (2) check the x and y bounds separately to decide whether (2,3), (3,1), (0,−1) lie in R; (3) compute f(2,1) and f(1,2), and prove that g, obtained by swapping the x and y coefficients, is a different function; (4) find the x-slice at y=1 and show that neither the given y=1 slice nor the four specified point values determine f."},"mko_ids":["mko-multivariable-real-function"]},{"id":"derivation-mko-linear-independence-basis-01","title":{"zh-Hant":"線性獨立與基底：以精確消去證明 ℝ² 的一組有序基底，並以兩個反例分開兩條責任","en":"Linear independence and bases: proving an ordered basis of ℝ² by exact elimination, with two counterexamples that separate the two obligations"},"summary":{"zh-Hant":"在 ℝ² 中令 v_1=(1,2)、v_2=(3,5)、v_3=(1,1)、w=(4,7)。(1) 以精確消去證明 (v_1,v_2) 線性獨立；(2) 對任意 (p,q) 求出係數，證明 span(v_1,v_2)=ℝ²，因而 (v_1,v_2) 是有序基底；(3) 求 w 的唯一座標；(4) 證明 (v_1) 獨立但不生成 ℝ²，而 (v_1,v_2,v_3) 生成 ℝ² 但不獨立，且 w 在其中的表示不唯一。","en":"In ℝ² let v_1=(1,2), v_2=(3,5), v_3=(1,1), w=(4,7). (1) Prove by exact elimination that (v_1,v_2) is linearly independent; (2) find the coefficients for an arbitrary (p,q), proving span(v_1,v_2)=ℝ², so that (v_1,v_2) is an ordered basis; (3) find the unique coordinates of w; (4) prove that (v_1) is independent but does not span ℝ², while (v_1,v_2,v_3) spans ℝ² but is not independent, and w has more than one representation in it."},"mko_ids":["mko-linear-independence-basis"]}],"title":"Concept derivation guides"}