For a root: check the sign change for bisection, or compute f and f′ for Newton, and keep six or more digits in a table of iterates.
For a table value: interpolate linearly between the two neighbouring points.
For an integral: find h = (b − a)/n, list f at every node, apply the weights (1, 2, …, 2, 1 for trapezoid; 1, 4, 2, …, 4, 1 for Simpson with even n) and multiply by h/2 or h/3.
For an ODE: step from t₀ with yₙ₊₁ = yₙ + h·f(tₙ, yₙ) (Euler) or the averaged slope (Heun), one row per step, until you reach the requested t.
Sequences & series
Arithmetic sequence
aₙ = a₁ + (n − 1)d, Sₙ = n(a₁ + aₙ)/2
Geometric sequence
aₙ = a₁rⁿ⁻¹, Sₙ = a₁(1 − rⁿ)/(1 − r)
Infinite geometric series
S = a₁/(1 − r), |r| < 1
p-series
Σ1/kᵖ converges only for p > 1
Ratio test
L = lim |ak+1/ak|: L < 1 converges, L > 1 diverges
Radius of convergence
R = lim |ck/ck+1|
Taylor polynomial
Pₙ(x) = Σ f⁽ᵏ⁾(a)(x − a)ᵏ/k!
Maclaurin series
eˣ = Σxᵏ/k!, sin x = x − x³/3! + …, cos x = 1 − x²/2! + …, ln(1 + x) = x − x²/2 + …
Lagrange error bound
|Rₙ| ≤ M·|x − a|ⁿ⁺¹/(n + 1)!
Method
Identify the type: arithmetic (constant difference), geometric (constant ratio), a standard series, a power series or a Taylor polynomial.
For sums use Sₙ = n(a₁ + aₙ)/2, Sₙ = a₁(1 − rⁿ)/(1 − r), or S = a₁/(1 − r) when |r| < 1, with a₁ the first term of the sum.
For convergence: n-th term test first, then compare with Σ1/kᵖ or a geometric series, and use the ratio or root test for factorials and k-th powers.
For Taylor polynomials list f(a), f′(a), f″(a), …, divide the k-th derivative by k!, and bound the error with M·|x − a|ⁿ⁺¹/(n + 1)!.
Matrices & determinants
2×2 determinant
det = ad − bc
2×2 inverse
A⁻¹ = (1/det)·[d −b; −c a]
Product
(AB)ij = Σk AikBkj
Eigenvalues
det(A − λI) = 0
Cramer’s rule
xi = det(Ai)/det(A)
Differential equations
Exponential growth/decay
dy/dt = ky → y = y0ekt
Half-life
k = ln 2 / t½
Newton's law of cooling
T(t) = Ta + (T0 − Ta)e−kt
Integrating factor
y′ + p(x)y = q(x): μ = e∫p dx, (μy)′ = μq
Mixing tank
dm/dt = r·c − (r/V)·m
2Practice problems
1.Numerical methodsMedium
Approximate ∫01 eˣ dx with Simpson's rule, n = 2, and compare it with the exact value. Find the absolute error |E| = |S − I|.
f(x) = eˣ[a, b] = [0, 1]n = 2
|E| =
2.Numerical methodsHard
Use Euler's method with h = 0.2 for y′ = 1 + sin(ty), y(0) = 0.5 to estimate y5 ≈ y(1).
y′ = 1 + sin(ty)y(0) = 0.5h = 0.2
y5 =
3.Sequences & seriesMedium
Exactly one of the four series converges. Click it.
Recognise the series and find its sum: S = Σk=0∞ (−1)ᵏ·22k+1/(2k + 1)!.
ak = (−1)ᵏ·22k+1/(2k + 1)!
S =
5.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−1 0 −5; 1 0 −1; 0 −5 −5]
det A =
6.Matrices & determinantsHard
Solve the system −2x + 5y = −6 and 6x + y = 6. Find x.
−2x + 5y = −66x + y = 6
x =
7.Differential equationsMedium
A cup of coffee at 80 °C is left in a room at 19 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.12 min⁻¹, find its temperature T after 13 min.
T0 = 80 °CTa = 19 °Ck = 0.12 min⁻¹t = 13 min
T = °C
8.Differential equationsHard
Solve y′ + y = 7e−2t with y(0) = 5. Find y(2).
y′ + y = 7e−2ty(0) = 5
y(2) =
3Solutions
1.Numerical methods
Answer: |E| = 0.0005793
Numerical integration: split [a, b] into n strips of width h = (b − a)/n with nodes x₀ … xₙ. Trapezoid rule: T = (h/2)[f₀ + 2f₁ + 2f₂ + … + 2fn−1 + fₙ]. Simpson's rule (n even): S = (h/3)[f₀ + 4f₁ + 2f₂ + 4f₃ + … + 4fn−1 + fₙ]. The trapezoid error shrinks like h² (bound (b − a)h²·M/12 with M = max|f″|), Simpson's like h⁴, so Simpson is usually far more accurate for the same nodes.
Exact value
I = 1.718282
Simpson's rule
S = (0.5/3)·(1 + 4·1.648721 + 2.718282) = 1.718861
Error
|E| = |1.718861 − 1.718282| = 0.0005793
Answer
|E| = 0.0005793
Relative error: 0.03372%.
2.Numerical methods
Answer: y5 = 1.921
Numerical solution of y′ = f(t, y): Euler's method steps along the slope at the start of each step: yn+1 = yₙ + h·f(tₙ, yₙ). Heun's method (improved Euler) averages the slope at the start and at an Euler prediction: k₁ = f(tₙ, yₙ), k₂ = f(tₙ + h, yₙ + h·k₁), yn+1 = yₙ + (h/2)(k₁ + k₂). The classical Runge–Kutta method (RK4) combines four slopes: yn+1 = yₙ + (h/6)(k₁ + 2k₂ + 2k₃ + k₄). The global error of Euler is proportional to h, Heun's to h², RK4's to h⁴.
Euler steps
y₁ = 0.5 + 0.2·(1) = 0.7
y₂ = 0.7 + 0.2·(1.139543) = 0.927909
y₃ = 0.927909 + 0.2·(1.3627) = 1.200449
y₄ = 1.200449 + 0.2·(1.659587) = 1.532366
y₅ = 1.532366 + 0.2·(1.941108) = 1.920588
Graph
y5 = 1.921
Answer
y5 = 1.921
3.Sequences & series
Answer: A: Σ (1/3)ᵏ
Convergence: if the terms do not tend to 0 the series diverges (n-th term test). Σ1/kᵖ converges only for p > 1, and Σrᵏ only for |r| < 1. Compare a series with one of these when its terms behave like them for large k. Ratio test: L = lim |ak+1/ak|; root test: L = lim |ak|1/k. L < 1 converges, L > 1 diverges, L = 1 decides nothing. A power series Σck(x − x₀)ᵏ converges for |x − x₀| < R with R = lim |ck/ck+1|; an alternating series with terms decreasing to 0 converges.
Test each series
A: Σ (1/3)ᵏ → Geometric with r = 1/3, |r| < 1: converges.
B: Σ 3/k → A multiple of the harmonic series (p = 1): diverges.
C: Σ (k + 5)/k² → Behaves like 1/k for large k: diverges by comparison with the harmonic series.
D: Σ (4k + 1)/(k + 2) → The terms tend to 4, not 0: diverges by the n-th term test.
The answer
A: Σ (1/3)ᵏ
Answer
A: Σ (1/3)ᵏ
4.Sequences & series
Answer: S = 0.9093
Taylor polynomials: near x = a a smooth function is approximated by Pₙ(x) = f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + … + f⁽ⁿ⁾(a)(x − a)ⁿ/n!; for a = 0 this is the Maclaurin polynomial. Standard series: eˣ = Σxᵏ/k!, sin x = x − x³/3! + x⁵/5! − …, cos x = 1 − x²/2! + x⁴/4! − …, ln(1 + x) = x − x²/2 + x³/3 − …, 1/(1 − x) = Σxᵏ. Lagrange error bound: |f(x) − Pₙ(x)| ≤ M·|x − a|ⁿ⁺¹/(n + 1)!, where M is the largest value of |f⁽ⁿ⁺¹⁾| between a and x.
Standard series
Σ (−1)ᵏx2k+1/(2k + 1)! = sin x
Substitute
x = 2
S = sin 2 = 0.9093
Answer
S = 0.9093
5.Matrices & determinants
Answer: det A = 30
Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.
2×2 minors of the first row
M₁₁ = −5, M₁₂ = −5, M₁₃ = −5
Cofactor expansion
det A = (−1)·(−5) − 0·(−5) + (−5)·(−5) = 30
Answer
det A = 30
6.Matrices & determinants
Answer: x = 1.125
Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.
Determinant of the coefficient matrix
det A = (−2)·1 − 5·6 = −32
Replace the x column
det Aₓ = (−6)·1 − 5·6 = −36
x = −36/(−32) = 1.125
Answer
x = 1.125
y = −0.75
7.Differential equations
Answer: T = 31.82 °C
First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.
Solution of the cooling law
T(t) = 19 + 61·e−0.12t
Substitute t
T(13) = 19 + 61·0.2101 = 31.82 °C
Answer
T = 31.82 °C
8.Differential equations
Answer: y(2) = 1.496
First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.