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
The trapezoid rule is used for ∫01 x³ dx. The error is at most (b − a)³·M/(12n²) with M = max|f″| on [0, 1]. Find the smallest number of strips n that guarantees an error below 0.0001.
f(x) = x³[a, b] = [0, 1]ε = 0.0001
n =
2.Numerical methodsHard
Take one step of the classical Runge–Kutta method (RK4) with h = 0.5 for y′ = 1 + sin(ty), y(0) = 1. Find y₁ ≈ y(0.5).
y′ = 1 + sin(ty)y(0) = 1h = 0.5
y1 =
3.Sequences & seriesMedium
Find the sum S = Σk=1∞ 4/(k(k + 1)).
ak = 4/(k(k + 1))
S =
4.Sequences & seriesHard
Recognise the series and find its sum: S = Σk=0∞ (−1)ᵏ/(2k)!.
ak = (−1)ᵏ/(2k)!
S =
5.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [4 2 −5; 5 −1 −4; 3 −2 −2]
det A =
6.Matrices & determinantsHard
Solve the system −3x + 3y = 3 and −x + 3y = −7. Find x.
−3x + 3y = 3−x + 3y = −7
x =
7.Differential equationsMedium
A cup of coffee at 86 °C is left in a room at 21 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.07 min⁻¹, find its temperature T after 14 min.
T0 = 86 °CTa = 21 °Ck = 0.07 min⁻¹t = 14 min
T = °C
8.Differential equationsHard
A tank holds 400 L of water in which 20 kg of salt is dissolved. Brine with 0.25 kg of salt per litre flows in at 4 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 15 min?
V = 400 Lm0 = 20 kgc = 0.25 kg/Lr = 4 L/min
m = kg
3Solutions
1.Numerical methods
Answer: n = 71
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.
Second derivative bound
M = 6·1 = 6
Solve for n
n² > 1·6/(12·0.0001) = 5000
n > 70.71 → n = 71
Answer
n = 71
2.Numerical methods
Answer: y1 = 1.667
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.
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.
Partial fractions
4/(k(k + 1)) = 4·(1/k − 1/(k + 1))
Cancel
S = 4·1 = 4
Answer
S = 4
4.Sequences & series
Answer: S = 0.5403
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/(2k)! = cos x
Substitute
x = 1
S = cos 1 = 0.5403
Answer
S = 0.5403
5.Matrices & determinants
Answer: det A = 7
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 = −6, M₁₂ = 2, M₁₃ = −7
Cofactor expansion
det A = 4·(−6) − 2·2 + (−5)·(−7) = 7
Answer
det A = 7
6.Matrices & determinants
Answer: x = −5
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 = (−3)·3 − 3·(−1) = −6
Replace the x column
det Aₓ = 3·3 − 3·(−7) = 30
x = 30/(−6) = −5
Answer
x = −5
y = −4
7.Differential equations
Answer: T = 45.4 °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) = 21 + 65·e−0.07t
Substitute t
T(14) = 21 + 65·0.3753 = 45.4 °C
Answer
T = 45.4 °C
8.Differential equations
Answer: m = 31.14 kg
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.
Set up the equation
dm/dt = 4·0.25 − (4/400)·m = 1 − 0.01·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.