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
The trapezoid rule is used for ∫02 x³ dx. The error is at most (b − a)³·M/(12n²) with M = max|f″| on [0, 2]. Find the smallest number of strips n that guarantees an error below 0.001.
f(x) = x³[a, b] = [0, 2]ε = 0.001
n =
2.Numerical methodsHard
Solve y′ = t + y, y(0) = 2 with Euler's method and step h = 0.25 up to t = 1.25. The exact solution is y = −t − 1 + (y₀ + 1)eᵗ. Find the global error |E| = |y(1.25) − y5|.
y′ = t + yy(0) = 2h = 0.25
|E| =
3.Sequences & seriesMedium
Apply the ratio test to Σk=1∞ k³·2ᵏ/5ᵏ. Find L = lim |ak+1/ak|.
ak = k³·2ᵏ/5ᵏ
L =
4.Sequences & seriesHard
Find the coefficient c₄ of (x + 1)⁴ in the Taylor series of f(x) = e−x about a = −1.
f(x) = e−xa = −1k = 4
c₄ =
5.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [2 −2; −5 −2]B = [6 −2; 2 −4]
c12 =
6.Matrices & determinantsHard
Solve the system −6x + 2y = −6 and −4x − 4y = 6. Find x.
−6x + 2y = −6−4x − 4y = 6
x =
7.Differential equationsMedium
A forged steel part at 480 °C is left to cool in a workshop at 27 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.04 min⁻¹, find its temperature T after 12 min.
T0 = 480 °CTa = 27 °Ck = 0.04 min⁻¹t = 12 min
T = °C
8.Differential equationsHard
A system obeys y″ + 2y′ + 10y = 10 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.
y″ + 2y′ + 10y = 10y(0) = 0y′(0) = 0
y(1) =
3Solutions
1.Numerical methods
Answer: n = 90
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.
Second derivative bound
M = 6·2 = 12
Solve for n
n² > 8·12/(12·0.001) = 8000
n > 89.44 → n = 90
Answer
n = 90
2.Numerical methods
Answer: |E| = 1.316
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
y5 = 6.905273
Exact value
y(1.25) = 8.221029
Global error
|E| = |8.221029 − 6.905273| = 1.316
Graph
Answer
|E| = 1.316
Halving h roughly halves the error of Euler's method.
3.Sequences & series
Answer: L = 0.4
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.
Ratio of consecutive terms
|ak+1/ak| = ((k + 1)/k)³·2/5 → 2/5
Conclusion
L = 0.4 < 1
Answer
L = 0.4
L < 1, so the series converges.
4.Sequences & series
Answer: c₄ = 0.1133
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⁽⁴⁾
f⁽⁴⁾(x) = (−1)⁴·e−x
Coefficient
c₄ = (−1)⁴·e1/4! = 0.1133
c₄ = 0.1133
Answer
c₄ = 0.1133
5.Matrices & determinants
Answer: c12 = 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
c12 = 2·(−2) + (−2)·(−4) = 4
Answer
c12 = 4
6.Matrices & determinants
Answer: x = 0.375
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)·(−4) − 2·(−4) = 32
Replace the x column
det Aₓ = (−6)·(−4) − 2·6 = 12
x = 12/32 = 0.375
Answer
x = 0.375
y = −1.875
7.Differential equations
Answer: T = 307.3 °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) = 27 + 453·e−0.04t
Substitute t
T(12) = 27 + 453·0.6188 = 307.3 °C
Answer
T = 307.3 °C
8.Differential equations
Answer: y(1) = 1.347
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.