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 ∫02 √x·e−x dx with the trapezoid rule using n = 5 strips. Give T.
f(x) = √x·e−x[a, b] = [0, 2]n = 5
T =
2.Numerical methodsHard
Use Euler's method with h = 0.5 for y′ = e−ty, y(0) = 0 to estimate y3 ≈ y(1.5).
y′ = e−tyy(0) = 0h = 0.5
y3 =
3.Sequences & seriesMedium
Find the radius of convergence R of the power series Σk=1∞ (x + 2)ᵏ/(k·2ᵏ).
Σ (x + 2)ᵏ/(k·2ᵏ)x₀ = −2
R =
4.Sequences & seriesHard
f(x) = cos x is approximated at x = 1.5 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 1, find the smallest degree n that guarantees an error below 0.0005.
f(x) = cos xx = 1.5M = 1ε = 0.0005
n =
5.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [−6 −1; 2 1]B = [5 0; 5 −2]
c11 =
6.Matrices & determinantsHard
Solve the system 6x − 6y = 0 and −5x + 3y = 6. Find x.
6x − 6y = 0−5x + 3y = 6
x =
7.Differential equationsMedium
Solve y″ + 5y′ + 6y = 0 with y(0) = 1 and y′(0) = -2. Find y(2).
y″ + 5y′ + 6y = 0y(0) = 1y′(0) = -2
y(2) =
8.Differential equationsHard
A system obeys y″ + 3y′ + 18.25y = 36.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.25); angles in radians.
y″ + 3y′ + 18.25y = 36.5y(0) = 0y′(0) = 0
y(0.25) =
3Solutions
1.Numerical methods
Answer: T = 0.6027
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.
This integral has no elementary antiderivative, which is exactly when a numerical rule is needed.
2.Numerical methods
Answer: y3 = 1.095
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 = 0 + 0.5·(1) = 0.5
y₂ = 0.5 + 0.5·(0.778801) = 0.8894
y₃ = 0.8894 + 0.5·(0.410902) = 1.094851
Graph
y3 = 1.095
Answer
y3 = 1.095
3.Sequences & series
Answer: R = 2
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.
Coefficients
ck/ck+1 = 2·((k + 1)/k) → 2
Radius
R = 2
Answer
R = 2
The ratio test says nothing at the two end points; they have to be checked separately.
4.Sequences & series
Answer: n = 8
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.
Bounds for increasing n
n = 7: 0.0006356 ≥ 0.0005
n = 8: 0.0001059 < 0.0005
Smallest degree
n = 8
Answer
n = 8
5.Matrices & determinants
Answer: c11 = −35
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 = (−6)·5 + (−1)·5 = −35
Answer
c11 = −35
6.Matrices & determinants
Answer: x = −3
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 = 6·3 − (−6)·(−5) = −12
Replace the x column
det Aₓ = 0·3 − (−6)·6 = 36
x = 36/(−12) = −3
Answer
x = −3
y = −3
7.Differential equations
Answer: y(2) = 0.01832
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.
Characteristic roots
(r + 2)(r + 3) = 0 → r = −2, −3
Constants from the initial values
A + B = 1, −2A − 3B = -2
A = 1, B = 0
Evaluate
y(2) = 1·e−4 + 0·e−6 = 0.01832
Answer
y(2) = 0.01832
8.Differential equations
Answer: y(0.25) = 0.8236
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.