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 ∫02 eˣ dx with the trapezoid rule, n = 4, and compare it with the exact value. Find the absolute error |E| = |T − I|.
f(x) = eˣ[a, b] = [0, 2]n = 4
|E| =
2.Numerical methodsHard
Use Heun's method (improved Euler) with h = 0.1 for y′ = −2ty, y(0) = 2 to estimate y2 ≈ y(0.2).
y′ = −2tyy(0) = 2h = 0.1
y2 =
3.Sequences & seriesMedium
Exactly one of the four series diverges. Click it.
A: Σ ln k / kB: Σ (−1)ᵏ/kC: Σ 5/k³D: Σ (2/5)ᵏ
1A: Σ ln k / k
2B: Σ (−1)ᵏ/k
3C: Σ 5/k³
4D: Σ (2/5)ᵏ
4.Sequences & seriesHard
f(x) = cos x is approximated at x = 1 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 1, find the smallest degree n that guarantees an error below 1 × 10⁻⁵.
f(x) = cos xx = 1M = 1ε = 1 × 10⁻⁵
n =
5.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−1 4 4; −3 3 −5; −4 −2 3]
det A =
6.Matrices & determinantsHard
Solve the system −6x − 6y = −2 and 4x − 4y = 7. Find x.
−6x − 6y = −24x − 4y = 7
x =
7.Differential equationsMedium
A can of soda at 6 °C is taken out of the fridge into a room at 23 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.09 min⁻¹, find its temperature T after 17 min.
T0 = 6 °CTa = 23 °Ck = 0.09 min⁻¹t = 17 min
T = °C
8.Differential equationsHard
A tank holds 500 L of water in which 10 kg of salt is dissolved. Brine with 0.05 kg of salt per litre flows in at 20 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 21 kg of salt?
V = 500 Lm0 = 10 kgc = 0.05 kg/Lr = 20 L/min
t = min
3Solutions
1.Numerical methods
Answer: |E| = 0.1326
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.
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.
Heun steps
y₁ = 2 + (0.1/2)(0 − 0.4) = 1.98
y₂ = 1.98 + (0.1/2)(−0.396 − 0.77616) = 1.921392
Graph
y2 = 1.921
Answer
y2 = 1.921
The exact value is y(0.2) = 1.921579.
3.Sequences & series
Answer: A: Σ ln k / 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 ln k / k → Larger than 1/k for k ≥ 3: diverges by comparison with the harmonic series.
B: Σ (−1)ᵏ/k → Alternating, with 1/k decreasing to 0: converges (conditionally).
C: Σ 5/k³ → p-series with p = 3 > 1: converges.
D: Σ (2/5)ᵏ → Geometric with r = 2/5, |r| < 1: converges.
The answer
A: Σ ln k / k
Answer
A: Σ ln k / k
4.Sequences & series
Answer: n = 8
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.
Bounds for increasing n
n = 7: 2.48 × 10⁻⁵ ≥ 1 × 10⁻⁵
n = 8: 2.756 × 10⁻⁶ < 1 × 10⁻⁵
Smallest degree
n = 8
Answer
n = 8
5.Matrices & determinants
Answer: det A = 189
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 = −1, M₁₂ = −29, M₁₃ = 18
Cofactor expansion
det A = (−1)·(−1) − 4·(−29) + 4·18 = 189
Answer
det A = 189
6.Matrices & determinants
Answer: x = 1.042
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 = (−6)·(−4) − (−6)·4 = 48
Replace the x column
det Aₓ = (−2)·(−4) − (−6)·7 = 50
x = 50/48 = 1.042
Answer
x = 1.042
y = −0.7083
7.Differential equations
Answer: T = 19.32 °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) = 23 + (−17)·e−0.09t
Substitute t
T(17) = 23 + (−17)·0.2165 = 19.32 °C
Answer
T = 19.32 °C
8.Differential equations
Answer: t = 33.04 min
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.