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 sin x 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 0.0005.
f(x) = sin x[a, b] = [0, 1]ε = 0.0005
n =
2.Numerical methodsHard
Use Heun's method (improved Euler) with h = 0.5 for y′ = e−ty, y(0) = 0.5 to estimate y3 ≈ y(1.5).
y′ = e−tyy(0) = 0.5h = 0.5
y3 =
3.Sequences & seriesMedium
Exactly one of the four series converges. Click it.
A: Σ (−1)ᵏ/kB: Σ 1/√kC: Σ (5/3)ᵏD: Σ 2ᵏ/k²
1A: Σ (−1)ᵏ/k
2B: Σ 1/√k
3C: Σ (5/3)ᵏ
4D: Σ 2ᵏ/k²
4.Sequences & seriesHard
Use the Maclaurin polynomial of degree 2 of f(x) = 1/(1 − 2x) to approximate f(0.25). Give P₂(0.25).
f(x) = 1/(1 − 2x)n = 2x = 0.25
P₂(0.25) =
5.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [0 2 1; 5 −4 2; 4 −2 1]
det A =
6.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [4 4; −2 −2]
λ =
7.Differential equationsMedium
A forged steel part at 360 °C is left to cool in a workshop at 27 °C. After 48 min its temperature is 76 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 218 °C?
T0 = 360 °CTa = 27 °CT(48) = 76 °CT = 218 °C
t = min
8.Differential equationsHard
A tank holds 200 L of water in which 5 kg of salt is dissolved. Brine with 0.2 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 26 kg of salt?
V = 200 Lm0 = 5 kgc = 0.2 kg/Lr = 2 L/min
t = min
3Solutions
1.Numerical methods
Answer: n = 12
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 = sin 1 = 0.8415
Solve for n
n² > 1·0.8415/(12·0.0005) = 140.2
n > 11.84 → n = 12
Answer
n = 12
2.Numerical methods
Answer: y3 = 1.251
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.
Test each series
A: Σ (−1)ᵏ/k → Alternating, with 1/k decreasing to 0: converges (conditionally).
B: Σ 1/√k → p-series with p = 1/2 ≤ 1: diverges.
C: Σ (5/3)ᵏ → Geometric with r = 5/3 > 1: the terms grow, so it diverges.
D: Σ 2ᵏ/k² → Ratio test: L = 2 > 1, diverges.
The answer
A: Σ (−1)ᵏ/k
Answer
A: Σ (−1)ᵏ/k
4.Sequences & series
Answer: P₂(0.25) = 1.75
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.
Maclaurin polynomial
P₂(x) = 1 + 2x + 4x²
Substitute
P₂(0.25) = 1 + 0.5 + 0.25 = 1.75
Answer
P₂(0.25) = 1.75
The exact value is f(0.25) = 2; the error 0.25 shrinks when you take more terms or a point closer to 0.
5.Matrices & determinants
Answer: det A = 12
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 = 0, M₁₂ = −3, M₁₃ = 6
Cofactor expansion
det A = 0·0 − 2·(−3) + 1·6 = 12
Answer
det A = 12
6.Matrices & determinants
Answer: λ = 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.
Characteristic equation
λ² − (2)λ + (0) = 0
Solve
λ = (2 ± √(4))/2 → λ = 0, 2
λ = 2
Answer
λ = 2
7.Differential equations
Answer: t = 13.92 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.
Cooling constant
k = ln(333/49)/48 = 0.03992 min⁻¹
Time to reach the target
t = ln(333/191)/0.03992 = 13.92 min
Answer
t = 13.92 min
8.Differential equations
Answer: t = 91.63 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.