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 10 s (table). Estimate the distance s travelled from t = 0 to t = 60 s with the trapezoid rule.
h = 10 sn = 6
s = m
2.Numerical methodsHard
Take one step of the classical Runge–Kutta method (RK4) with h = 0.1 for y′ = 1 + sin(ty), y(0) = 1. Find y₁ ≈ y(0.1).
y′ = 1 + sin(ty)y(0) = 1h = 0.1
y1 =
3.Sequences & seriesMedium
Exactly one of the four series converges. Click it.
f(x) = cos x is approximated by its Maclaurin polynomial of degree 2 at x = −0.5. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = 1 to find the largest possible error |R₂|.
f(x) = cos xn = 2x = −0.5M = 1
|R₂| =
5.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [6 2; 5 −2]B = [−4 −4; −6 −5]
c12 =
6.Matrices & determinantsHard
Solve the system 4x + 5y = −3 and 5x + y = −4. Find x.
4x + 5y = −35x + y = −4
x =
7.Differential equationsMedium
A can of soda at 3 °C is taken out of the fridge into a room at 29 °C. After 16 min its temperature is 22 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 3 °CTa = 29 °CT(16) = 22 °C
k = min⁻¹
8.Differential equationsHard
Solve the initial value problem y′ + 3y/x = 2 for x > 0, with y(1) = 5. Find y(2).
y′ + 3y/x = 2y(1) = 5
y(2) =
3Solutions
1.Numerical methods
Answer: s = 805 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
7 + 2·10 + 2·10 + 2·12 + 2·18 + 2·18 + 18 = 161
Trapezoid rule
s = (10/2)·161 = 805 m
Answer
s = 805 m
2.Numerical methods
Answer: y1 = 1.105
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 → p-series with p = 1/2 ≤ 1: diverges.
B: Σ (k + 1)/k² → Behaves like 1/k for large k: diverges by comparison with the harmonic series.
C: Σ 1/(k² + 9) → Smaller than 1/k², a convergent p-series: converges by comparison.
D: Σ (5/4)ᵏ → Geometric with r = 5/4 > 1: the terms grow, so it diverges.
The answer
C: Σ 1/(k² + 9)
Answer
C: Σ 1/(k² + 9)
4.Sequences & series
Answer: |R₂| = 0.02083
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.
Bound on the derivative
M = 1
Lagrange bound
|R₂| ≤ 1·0.5³/3! = 0.02083
Answer
|R₂| = 0.02083
The true error is smaller: |f(−0.5) − P₂(−0.5)| = 0.002583.
5.Matrices & determinants
Answer: c12 = −34
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.
Row times column
c12 = 6·(−4) + 2·(−5) = −34
Answer
c12 = −34
6.Matrices & determinants
Answer: x = −0.8095
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 = 4·1 − 5·5 = −21
Replace the x column
det Aₓ = (−3)·1 − 5·(−4) = 17
x = 17/(−21) = −0.8095
Answer
x = −0.8095
y = 0.04762
7.Differential equations
Answer: k = 0.08201 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−16k = -7/-26
Solve for k
k = ln(-26/-7)/16 = 0.08201 min⁻¹
Answer
k = 0.08201 min⁻¹
8.Differential equations
Answer: y(2) = 1.563
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.