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 ∫01.5 1/(1 + x³) dx with Simpson's rule using n = 2 strips. Give S.
f(x) = 1/(1 + x³)[a, b] = [0, 1.5]n = 2
S =
2.Numerical methodsHard
Use Euler's method with h = 0.1 for y′ = cos t − y, y(0) = 1 to estimate y3 ≈ y(0.3).
y′ = cos t − yy(0) = 1h = 0.1
y3 =
3.Sequences & seriesMedium
Exactly one of the four series converges. Click it.
Find the coefficient c₃ of (x − 1)³ in the Taylor series of f(x) = eˣ about a = 1.
f(x) = eˣa = 1k = 3
c₃ =
5.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [4 5; −4 −5]B = [−3 4; −6 2]
c12 =
6.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [3 1; 4 3]
λ =
7.Differential equationsMedium
A can of soda at 6 °C is taken out of the fridge into a room at 23 °C. After 19 min its temperature is 13 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 17 °C?
T0 = 6 °CTa = 23 °CT(19) = 13 °CT = 17 °C
t = min
8.Differential equationsHard
Solve the initial value problem y′ + 3y/x = 4x2 for x > 0, with y(1) = 1. Find y(3).
y′ + 3y/x = 4x2y(1) = 1
y(3) =
3Solutions
1.Numerical methods
Answer: S = 1.01
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.
Nodes and values
h = 0.75
Σ wᵢfᵢ = 1 + 4·0.703297 + 0.228571 = 4.041758
Simpson's rule
S = (0.75/3)·4.041758 = 1.01
Answer
S = 1.01
This integral has no elementary antiderivative, which is exactly when a numerical rule is needed.
2.Numerical methods
Answer: y3 = 0.9976
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 + 0.1·(0) = 1
y₂ = 1 + 0.1·(−0.00499583) = 0.9995
y₃ = 0.9995 + 0.1·(−0.0194338) = 0.997557
Graph
y3 = 0.9976
Answer
y3 = 0.9976
3.Sequences & series
Answer: D: Σ k²/5ᵏ
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 (5/4)ᵏ → Geometric with r = 5/4 > 1: the terms grow, so it diverges.
B: Σ (5k + 1)/(k + 2) → The terms tend to 5, not 0: diverges by the n-th term test.
C: Σ (k + 5)/k² → Behaves like 1/k for large k: diverges by comparison with the harmonic series.
D: Σ k²/5ᵏ → Ratio test: L = 1/5 < 1, converges.
The answer
D: Σ k²/5ᵏ
Answer
D: Σ k²/5ᵏ
4.Sequences & series
Answer: c₃ = 0.453
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.
Derivative f⁽³⁾
f⁽³⁾(x) = eˣ
Coefficient
c₃ = e1/3! = 0.453
c₃ = 0.453
Answer
c₃ = 0.453
5.Matrices & determinants
Answer: c12 = 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.
Row times column
c12 = 4·4 + 5·2 = 26
Answer
c12 = 26
6.Matrices & determinants
Answer: λ = 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.
Characteristic equation
λ² − (6)λ + (5) = 0
Solve
λ = (6 ± √(16))/2 → λ = 1, 5
λ = 5
Answer
λ = 5
7.Differential equations
Answer: t = 37.29 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(-17/-10)/19 = 0.02793 min⁻¹
Time to reach the target
t = ln(-17/(−6))/0.02793 = 37.29 min
Answer
t = 37.29 min
8.Differential equations
Answer: y(3) = 18.01
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.