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 √x·e−x dx with Simpson's rule using n = 6 strips. Give S.
f(x) = √x·e−x[a, b] = [0, 1]n = 6
S =
2.Numerical methodsHard
Solve y′ = t − y, y(0) = 3 with Euler's method and step h = 0.5 up to t = 2. The exact solution is y = t − 1 + (y₀ + 1)e−t. Find the global error |E| = |y(2) − y4|.
y′ = t − yy(0) = 3h = 0.5
|E| =
3.Sequences & seriesMedium
Exactly one of the four series diverges. Click it.
A: Σ 3ᵏ/k!B: Σ 1/(4k + 2)C: Σ (−1)ᵏ/√kD: Σ (4/5)ᵏ
1A: Σ 3ᵏ/k!
2B: Σ 1/(4k + 2)
3C: Σ (−1)ᵏ/√k
4D: Σ (4/5)ᵏ
4.Sequences & seriesHard
f(x) = cos x is approximated by its Maclaurin polynomial of degree 4 at x = 0.3. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = 1 to find the largest possible error |R₄|.
f(x) = cos xn = 4x = 0.3M = 1
|R₄| =
5.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 −4 −1; −1 5 −1; −3 2 −3]
det A =
6.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−3 3; −8 7]
λ =
7.Differential equationsMedium
Solve y″ + 7y′ + 12y = 0 with y(0) = 4 and y′(0) = 0. Find y(1.5).
y″ + 7y′ + 12y = 0y(0) = 4y′(0) = 0
y(1.5) =
8.Differential equationsHard
A system obeys y″ + 2y′ + 10y = 20 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 2y′ + 10y = 20y(0) = 0y′(0) = 0
ymax =
3Solutions
1.Numerical methods
Answer: S = 0.3732
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: |E| = 0.2913
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
y4 = 1.25
Exact value
y(2) = 1.541341
Global error
|E| = |1.541341 − 1.25| = 0.2913
Graph
Answer
|E| = 0.2913
Halving h roughly halves the error of Euler's method.
3.Sequences & series
Answer: B: Σ 1/(4k + 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.
Test each series
A: Σ 3ᵏ/k! → Ratio test: L = lim 3/(k + 1) = 0 < 1, converges.
B: Σ 1/(4k + 2) → Behaves like 1/(4k) for large k; compared with the harmonic series it diverges.
C: Σ (−1)ᵏ/√k → Alternating, with 1/√k decreasing to 0: converges (conditionally).
D: Σ (4/5)ᵏ → Geometric with r = 4/5, |r| < 1: converges.
The answer
B: Σ 1/(4k + 2)
Answer
B: Σ 1/(4k + 2)
4.Sequences & series
Answer: |R₄| = 2.025 × 10⁻⁵
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
|R₄| ≤ 1·0.3⁵/5! = 2.025 × 10⁻⁵
Answer
|R₄| = 2.025 × 10⁻⁵
The true error is smaller: |f(0.3) − P₄(0.3)| = 1.011 × 10⁻⁶.
5.Matrices & determinants
Answer: det A = 39
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 = −13, M₁₂ = 0, M₁₃ = 13
Cofactor expansion
det A = (−4)·(−13) − (−4)·0 + (−1)·13 = 39
Answer
det A = 39
6.Matrices & determinants
Answer: λ = 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.
Characteristic equation
λ² − (4)λ + (3) = 0
Solve
λ = (4 ± √(4))/2 → λ = 1, 3
λ = 3
Answer
λ = 3
7.Differential equations
Answer: y(1.5) = 0.148
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 + 3)(r + 4) = 0 → r = −3, −4
Constants from the initial values
A + B = 4, −3A − 4B = 0
A = 16, B = −12
Evaluate
y(1.5) = 16·e−4.5 + (−12)·e−6 = 0.148
Answer
y(1.5) = 0.148
8.Differential equations
Answer: ymax = 2.702
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.