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
A vehicle's speed is measured every 5 s (table). Estimate the distance s travelled from t = 0 to t = 30 s with Simpson's rule.
h = 5 sn = 6
s = m
2.Numerical methodsHard
Use Euler's method with h = 0.25 for y′ = t − y, y(0) = 1 to estimate y3 ≈ y(0.75).
y′ = t − yy(0) = 1h = 0.25
y3 =
3.Sequences & seriesMedium
Exactly one of the four series converges. Click it.
Recognise the series and find its sum: S = Σk=0∞ 2ᵏ/k!.
ak = 2ᵏ/k!
S =
5.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−1 2 −5; −4 1 −4; 5 3 −1]
det A =
6.Matrices & determinantsHard
Solve the system −x − 2y = −4 and −6x − 2y = −5. Find x.
−x − 2y = −4−6x − 2y = −5
x =
7.Differential equationsMedium
A forged steel part at 320 °C is left to cool in a workshop at 21 °C. After 47 min its temperature is 67 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 320 °CTa = 21 °CT(47) = 67 °C
k = min⁻¹
8.Differential equationsHard
A system obeys y″ + 2y′ + 26y = 260 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1.5); angles in radians.
y″ + 2y′ + 26y = 260y(0) = 0y′(0) = 0
y(1.5) =
3Solutions
1.Numerical methods
Answer: s = 381.7 m
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.
Weighted sum
3 + 4·8 + 2·12 + 4·15 + 2·16 + 4·16 + 14 = 229
Simpson's rule
s = (5/3)·229 = 381.7 m
Answer
s = 381.7 m
2.Numerical methods
Answer: y3 = 0.5938
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 = 1 + 0.25·(−1) = 0.75
y₂ = 0.75 + 0.25·(−0.5) = 0.625
y₃ = 0.625 + 0.25·(−0.125) = 0.59375
Graph
y3 = 0.5938
Answer
y3 = 0.5938
The exact value is y(0.75) = 0.694733.
3.Sequences & series
Answer: B: Σ 1/(2ᵏ + 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/(2ᵏ + k) → Smaller than (1/2)ᵏ, a convergent geometric series: converges by comparison.
C: Σ k!/10ᵏ → Ratio test: L = lim (k + 1)/10 = ∞, diverges.
D: Σ (1 + 1/k)ᵏ → The terms tend to e, not 0: diverges by the n-th term test.
The answer
B: Σ 1/(2ᵏ + k)
Answer
B: Σ 1/(2ᵏ + k)
4.Sequences & series
Answer: S = 7.389
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 xᵏ/k! = eˣ
Substitute
x = 2
S = e2 = 7.389
Answer
S = 7.389
5.Matrices & determinants
Answer: det A = 26
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 = 11, M₁₂ = 24, M₁₃ = −17
Cofactor expansion
det A = (−1)·11 − 2·24 + (−5)·(−17) = 26
Answer
det A = 26
6.Matrices & determinants
Answer: x = 0.2
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 = (−1)·(−2) − (−2)·(−6) = −10
Replace the x column
det Aₓ = (−4)·(−2) − (−2)·(−5) = −2
x = −2/(−10) = 0.2
Answer
x = 0.2
y = 1.9
7.Differential equations
Answer: k = 0.03983 min⁻¹
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.
Ratio of temperature differences
e−47k = 46/299
Solve for k
k = ln(299/46)/47 = 0.03983 min⁻¹
Answer
k = 0.03983 min⁻¹
8.Differential equations
Answer: y(1.5) = 8.808
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.