# Concept derivation guides 299 concept teaching guides, separate from the 36,000 individual question derivations. [Search all questions & derivations →](/en/explore/) [← Previous](/en/derivations/?page=4#results)[1](/en/derivations/?page=1#results) … [4](/en/derivations/?page=4#results) 5 [6](/en/derivations/?page=6#results) … [13](/en/derivations/?page=13#results)[Next →](/en/derivations/?page=6#results) 299 results · Page 5 of 13 Concept guide ## [Discrete exponential growth: proving x_n=x_0·r^n from x_(n+1)=r·x_n by induction](/en/derivations/derivation-mko-discrete-exponential-growth-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Discrete exponential growth model derivation-mko-discrete-exponential-growth-01 Concept guide ## [Compound-interest model: principal 1000 at 2% per year for three years, compared with simple interest](/en/derivations/derivation-mko-compound-interest-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Compound-interest model derivation-mko-compound-interest-01 Concept guide ## [Mathematical representation: deciding whether words, a formula, tables and ordered pairs represent the same function](/en/derivations/derivation-mko-mathematical-representation-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Mathematical representation derivation-mko-mathematical-representation-01 Concept guide ## [Diagnostic assessment records: synthetic records and judgments for one fraction-addition item](/en/derivations/derivation-mko-diagnostic-mathematical-assessment-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Diagnostic mathematical assessment records derivation-mko-diagnostic-mathematical-assessment-01 Concept guide ## [Orbits, fixed points and periodic points: classifying all six states of T(x)=(2x+1) mod 6](/en/derivations/derivation-mko-orbit-fixed-periodic-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Orbits, fixed points and periodic points derivation-mko-orbit-fixed-periodic-01 Concept guide ## [Initial-value problem: auditing four polynomial candidates](/en/derivations/derivation-mko-initial-value-problem-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Initial-value problem derivation-mko-initial-value-problem-01 Concept guide ## [Bisection: four bracket updates, the invariant and the separate proof obligation for root existence](/en/derivations/derivation-mko-bisection-method-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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 a0 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. Discrete logistic increment model derivation-mko-discrete-logistic-growth-01 Concept guide ## [Linear harmonic oscillator: eligibility, k/m, residuals of a polynomial candidate and single-state energy](/en/derivations/derivation-mko-harmonic-oscillator-model-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Linear harmonic-oscillator model derivation-mko-harmonic-oscillator-model-01 Concept guide ## [Present value and discounting: proving that discounting inverts compounding, and the three-year present value of 1331](/en/derivations/derivation-mko-present-value-discounting-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Present value and discounting derivation-mko-present-value-discounting-01 Concept guide ## [Complex numbers: computing (2+3i)(4−5i) as an ordered pair and deciding equality](/en/derivations/derivation-mko-complex-number-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Complex number derivation-mko-complex-number-01 Concept guide ## [Real vector space: auditing axiom instances in ℝ³ and proving one distributive law](/en/derivations/derivation-mko-real-vector-space-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Real vector space derivation-mko-real-vector-space-01 Concept guide ## [Finite group: auditing the group axioms from a complete Cayley table](/en/derivations/derivation-mko-finite-group-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Finite group derivation-mko-finite-group-01 Concept guide ## [Metric space: auditing a finite distance table cell by cell and triple by triple](/en/derivations/derivation-mko-metric-space-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Metric space derivation-mko-metric-space-01 Concept guide ## [Finite discrete probability distribution: validating a PMF, event probability, expectation and pushforward](/en/derivations/derivation-mko-finite-discrete-probability-distribution-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Finite discrete probability distribution derivation-mko-finite-discrete-probability-distribution-01 Concept guide ## [Complex modulus and conjugate: proving z·z̄=|z|² and that conjugation preserves the operations](/en/derivations/derivation-mko-complex-modulus-conjugate-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Complex modulus and conjugation derivation-mko-complex-modulus-conjugate-01 Concept guide ## [Linear combinations and span: deciding membership using finite coefficient lists as witnesses](/en/derivations/derivation-mko-linear-combination-span-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) (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. Linear combinations and span derivation-mko-linear-combination-span-01 Concept guide ## [Subgroups and homomorphisms: auditing on the complete tables of ℤ/4 and ℤ/2](/en/derivations/derivation-mko-subgroup-homomorphism-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Subgroups and group homomorphisms derivation-mko-subgroup-homomorphism-01 Concept guide ## [Convergence in a metric space: limit and uniqueness for an ultimately constant sequence on a finite metric table](/en/derivations/derivation-mko-metric-convergence-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Convergence in a metric space derivation-mko-metric-convergence-01 Concept guide ## [Sigma-algebras and measure spaces: an atom-generated sigma-algebra and a probability measure on a four-point carrier](/en/derivations/derivation-mko-sigma-algebra-measure-space-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Sigma-algebras and measure spaces derivation-mko-sigma-algebra-measure-space-01 Concept guide ## [Discrete-time stochastic process: marginals and two-time joint laws from a complete path distribution, with Markov and stationarity checks](/en/derivations/derivation-mko-discrete-stochastic-process-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Discrete-time stochastic process derivation-mko-discrete-stochastic-process-01 Concept guide ## [Multivariable real-valued function: coefficient order, declared domain, slices and point values of a bivariate quadratic](/en/derivations/derivation-mko-multivariable-real-function-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Multivariable real-valued function derivation-mko-multivariable-real-function-01 Concept guide ## [Linear independence and bases: proving an ordered basis of ℝ² by exact elimination, with two counterexamples that separate the two obligations](/en/derivations/derivation-mko-linear-independence-basis-01/?from=%2Fen%2Fderivations%2F%3Fpage%3D5) 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. Linear independence and bases derivation-mko-linear-independence-basis-01 [← Previous](/en/derivations/?page=4#results)[1](/en/derivations/?page=1#results) … [4](/en/derivations/?page=4#results) 5 [6](/en/derivations/?page=6#results) … [13](/en/derivations/?page=13#results)[Next →](/en/derivations/?page=6#results) 299 results · Page 5 of 13