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 e−x² dx with the trapezoid rule using n = 2 strips. Give T.
f(x) = e−x²[a, b] = [0, 1]n = 2
T =
2.Numerical methodsHard
Use Euler's method with h = 0.5 for y′ = y − t², y(0) = 3 to estimate y4 ≈ y(2).
y′ = y − t²y(0) = 3h = 0.5
y4 =
3.Sequences & seriesMedium
The power series Σk=1∞ k²·2ᵏ·(x + 1)ᵏ converges on an interval around x₀ = −1. Find the left end x of that interval.
Σ k²·2ᵏ·(x + 1)ᵏx₀ = −1
x =
4.Sequences & seriesHard
f(x) = sin x is approximated by its Maclaurin polynomial of degree 3 at x = 0.2. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = 1 to find the largest possible error |R₃|.
f(x) = sin xn = 3x = 0.2M = 1
|R₃| =
5.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [3 −1 0; 0 −4 −4; 5 5 −2]
det A =
6.Matrices & determinantsHard
Solve the system 6x − 4y = −10 and −3x + 3y = −5. Find x.
6x − 4y = −10−3x + 3y = −5
x =
7.Differential equationsMedium
A can of soda at 5 °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 = 0.1 min⁻¹, find its temperature T after 5 min.
T0 = 5 °CTa = 23 °Ck = 0.1 min⁻¹t = 5 min
T = °C
8.Differential equationsHard
Solve y′ + 0.5y = 8e−3t with y(0) = 2. Find y(2).
y′ + 0.5y = 8e−3ty(0) = 2
y(2) =
3Solutions
1.Numerical methods
Answer: T = 0.7314
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.
Nodes and values
h = 0.5
Σ wᵢfᵢ = 1 + 2·0.778801 + 0.367879 = 2.925481
Trapezoid rule
T = (0.5/2)·2.925481 = 0.7314
Answer
T = 0.7314
This integral has no elementary antiderivative, which is exactly when a numerical rule is needed.
2.Numerical methods
Answer: y4 = 13.03
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 = 3 + 0.5·(3) = 4.5
y₂ = 4.5 + 0.5·(4.25) = 6.625
y₃ = 6.625 + 0.5·(5.625) = 9.4375
y₄ = 9.4375 + 0.5·(7.1875) = 13.03125
Graph
y4 = 13.03
Answer
y4 = 13.03
The exact value is y(2) = 17.389056.
3.Sequences & series
Answer: x = −1.5
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 = (1/2)·(k/(k + 1))² → 1/2
Radius
R = 0.5
Interval
−1.5 < x < −0.5
x = −1.5
Answer
x = −1.5
The ratio test says nothing at the two end points; they have to be checked separately.
4.Sequences & series
Answer: |R₃| = 6.667 × 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
|R₃| ≤ 1·0.2⁴/4! = 6.667 × 10⁻⁵
Answer
|R₃| = 6.667 × 10⁻⁵
The true error is smaller: |f(0.2) − P₃(0.2)| = 2.664 × 10⁻⁶.
5.Matrices & determinants
Answer: det A = 104
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 = 28, M₁₂ = 20, M₁₃ = 20
Cofactor expansion
det A = 3·28 − (−1)·20 + 0·20 = 104
Answer
det A = 104
6.Matrices & determinants
Answer: x = −8.333
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 − (−4)·(−3) = 6
Replace the x column
det Aₓ = (−10)·3 − (−4)·(−5) = −50
x = −50/6 = −8.333
Answer
x = −8.333
y = −10
7.Differential equations
Answer: T = 12.08 °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 + (−18)·e−0.1t
Substitute t
T(5) = 23 + (−18)·0.6065 = 12.08 °C
Answer
T = 12.08 °C
8.Differential equations
Answer: y(2) = 1.905
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.
Integrating factor
μ = e0.5t
(e0.5ty)′ = 8e−2.5t
General solution
y = 8/(−2.5)·e−3t + C·e−0.5t = −3.2·e−3t + C·e−0.5t