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 2 s (table). Estimate the distance s travelled from t = 0 to t = 12 s with Simpson's rule.
h = 2 sn = 6
s = m
2.Numerical methodsHard
Use Euler's method with h = 0.1 for y′ = t² + y², y(0) = 1 to estimate y4 ≈ y(0.4).
y′ = t² + y²y(0) = 1h = 0.1
y4 =
3.Sequences & seriesMedium
Exactly one of the four series converges. Click it.
Use the Maclaurin polynomial of degree 5 of f(x) = sin(−x) to approximate f(2). Give P₅(2).
f(x) = sin(−x)n = 5x = 2
P₅(2) =
5.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [−2 −2; −2 5]B = [1 0; 1 −5]
c11 =
6.Matrices & determinantsHard
Solve the system −5x + 4y = 1 and 6x − 5y = −12. Find x.
−5x + 4y = 16x − 5y = −12
x =
7.Differential equationsMedium
A can of soda at 6 °C is taken out of the fridge into a room at 23 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.04 min⁻¹, find its temperature T after 20 min.
T0 = 6 °CTa = 23 °Ck = 0.04 min⁻¹t = 20 min
T = °C
8.Differential equationsHard
Solve y″ + y′ + 9.25y = 0 with y(0) = 2 and y′(0) = 0 (released from rest). Find y(0.25); angles in radians.
y″ + y′ + 9.25y = 0y(0) = 2y′(0) = 0
y(0.25) =
3Solutions
1.Numerical methods
Answer: s = 158 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
6 + 4·10 + 2·10 + 4·11 + 2·15 + 4·19 + 21 = 237
Simpson's rule
s = (2/3)·237 = 158 m
Answer
s = 158 m
2.Numerical methods
Answer: y4 = 1.573
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·(1) = 1.1
y₂ = 1.1 + 0.1·(1.22) = 1.222
y₃ = 1.222 + 0.1·(1.533284) = 1.375328
y₄ = 1.375328 + 0.1·(1.981528) = 1.573481
Graph
y4 = 1.573
Answer
y4 = 1.573
3.Sequences & series
Answer: A: Σ (−1)ᵏ/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 (−1)ᵏ/k → Alternating, with 1/k decreasing to 0: converges (conditionally).
B: Σ (5/3)ᵏ → Geometric with r = 5/3 > 1: the terms grow, so it diverges.
C: Σ (1 + 1/k)ᵏ → The terms tend to e, not 0: diverges by the n-th term test.
D: Σ (k + 2)/k² → Behaves like 1/k for large k: diverges by comparison with the harmonic series.
The answer
A: Σ (−1)ᵏ/k
Answer
A: Σ (−1)ᵏ/k
4.Sequences & series
Answer: P₅(2) = −0.9333
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
P₅(x) = −x + 0.1667x³ − 0.008333x⁵
Substitute
P₅(2) = −2 + 1.333333 − 0.266667 = −0.9333
Answer
P₅(2) = −0.9333
The exact value is f(2) = −0.909297; the error 0.0240359 shrinks when you take more terms or a point closer to 0.
5.Matrices & determinants
Answer: c11 = −4
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
c11 = (−2)·1 + (−2)·1 = −4
Answer
c11 = −4
6.Matrices & determinants
Answer: x = 43
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 = (−5)·(−5) − 4·6 = 1
Replace the x column
det Aₓ = 1·(−5) − 4·(−12) = 43
x = 43/1 = 43
Answer
x = 43
y = 54
7.Differential equations
Answer: T = 15.36 °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) = 23 + (−17)·e−0.04t
Substitute t
T(20) = 23 + (−17)·0.4493 = 15.36 °C
Answer
T = 15.36 °C
8.Differential equations
Answer: y(0.25) = 1.492
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.