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 10 s (table). Estimate the distance s travelled from t = 0 to t = 40 s with the trapezoid rule.
h = 10 sn = 4
s = m
2.Numerical methodsHard
Use Heun's method (improved Euler) with h = 0.2 for y′ = t² + y², y(0) = 0.5 to estimate y3 ≈ y(0.6).
y′ = t² + y²y(0) = 0.5h = 0.2
y3 =
3.Sequences & seriesMedium
Find the sum S = Σk=122 4/(k(k + 1)).
ak = 4/(k(k + 1))n = 22
S =
4.Sequences & seriesHard
Find the coefficient c₂ of (x − 8)² in the Taylor series of f(x) = ∛x about a = 8.
f(x) = ∛xa = 8k = 2
c₂ =
5.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 1 −4; 2 −1 −1; 3 −1 3]
det A =
6.Matrices & determinantsHard
Solve the system −3x − 3y = −11 and −5x − 4y = 3. Find x.
−3x − 3y = −11−5x − 4y = 3
x =
7.Differential equationsMedium
A can of soda at 5 °C is taken out of the fridge into a room at 22 °C. After 16 min its temperature is 11 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 15 °C?
T0 = 5 °CTa = 22 °CT(16) = 11 °CT = 15 °C
t = min
8.Differential equationsHard
Solve the initial value problem y′ + 2y/x = 6 for x > 0, with y(1) = 3. Find y(2).
y′ + 2y/x = 6y(1) = 3
y(2) =
3Solutions
1.Numerical methods
Answer: s = 240 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
1 + 2·4 + 2·4 + 2·9 + 13 = 48
Trapezoid rule
s = (10/2)·48 = 240 m
Answer
s = 240 m
2.Numerical methods
Answer: y3 = 0.8033
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.
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.
Partial fractions
4/(k(k + 1)) = 4·(1/k − 1/(k + 1))
Cancel
S = 4·(1 − 1/23) = 3.826
Answer
S = 3.826
4.Sequences & series
Answer: c₂ = −0.003472
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⁽²⁾
f⁽²⁾(x) = −0.2222·x1/3 − 2
Coefficient
c₂ = −0.2222·81/3 − 2/2! = −0.003472
c₂ = −0.003472
Answer
c₂ = −0.003472
5.Matrices & determinants
Answer: det A = 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.
2×2 minors of the first row
M₁₁ = −4, M₁₂ = 9, M₁₃ = 1
Cofactor expansion
det A = (−4)·(−4) − 1·9 + (−4)·1 = 3
Answer
det A = 3
6.Matrices & determinants
Answer: x = −17.67
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 = (−3)·(−4) − (−3)·(−5) = −3
Replace the x column
det Aₓ = (−11)·(−4) − (−3)·3 = 53
x = 53/(−3) = −17.67
Answer
x = −17.67
y = 21.33
7.Differential equations
Answer: t = 32.61 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/-11)/16 = 0.02721 min⁻¹
Time to reach the target
t = ln(-17/(−7))/0.02721 = 32.61 min
Answer
t = 32.61 min
8.Differential equations
Answer: y(2) = 4.25
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.