Numerical Methods: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

1Formulas and methods

Numerical methods

Bisection: c = (a + b)/2, width after n steps (b − a)/2ⁿ
Newton–Raphson: x_n+1 = xₙ − f(xₙ)/f′(xₙ)
Linear interpolation: y = y₁ + (y₂ − y₁)(x − x₁)/(x₂ − x₁)
Trapezoid rule: T = (h/2)[f₀ + 2f₁ + … + 2f_n−1 + fₙ], h = (b − a)/n
Simpson’s rule (n even): S = (h/3)[f₀ + 4f₁ + 2f₂ + … + 4f_n−1 + fₙ]
Trapezoid error bound: |E| ≤ (b − a)h²M/12, M = max|f″|
Euler: y_n+1 = yₙ + h·f(tₙ, yₙ)
Heun: y_n+1 = yₙ + (h/2)(k₁ + k₂), k₂ = f(tₙ + h, yₙ + hk₁)
RK4: y_n+1 = yₙ + (h/6)(k₁ + 2k₂ + 2k₃ + k₄)

Method

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 |a_k+1/a_k|: L < 1 converges, L > 1 diverges
Radius of convergence: R = lim |c_k/c_k+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 = Σ_k A_ikB_kj
Eigenvalues: det(A − λI) = 0
Cramer’s rule: x_i = det(A_i)/det(A)

Differential equations

Exponential growth/decay: dy/dt = ky → y = y₀e^kt
Half-life: k = ln 2 / t_½
Newton's law of cooling: T(t) = T_a + (T₀ − T_a)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 ∫₁⁴ 1/x dx with the trapezoid rule, n = 4, and compare it with the exact value. Find the absolute error |E| = |T − I|.

f(x) = 1/x[a, b] = [1, 4]n = 4

|E| =

2.Numerical methodsHard

Use Euler's method with h = 0.2 for y′ = t + y, y(0) = 1 to estimate y₄ ≈ y(0.8).

y′ = t + yy(0) = 1h = 0.2

y₄ =

3.Sequences & seriesMedium

Find the sum S = Σ_k=1^∞ 1/(k(k + 3)).

a_k = 1/(k(k + 3))

S =

4.Sequences & seriesHard

Recognise the series and find its sum: S = Σ_k=0^∞ 1.5ᵏ/k!.

a_k = 1.5ᵏ/k!

S =

5.Matrices & determinantsMedium

Find the determinant det A of the 3×3 matrix A shown.

A = [−1 4 4; 2 −5 0; 0 3 0]

det A =

6.Matrices & determinantsHard

Solve the system −4x + y = 8 and 2x − 2y = −9. Find x.

−4x + y = 82x − 2y = −9

x =

7.Differential equationsMedium

A cup of coffee at 84 °C is left in a room at 24 °C. After 8 min its temperature is 71 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T₀ = 84 °CT_a = 24 °CT(8) = 71 °C

k = min⁻¹

8.Differential equationsHard

Solve y′ + y = 8e^−2t with y(0) = 5. Find y(1).

y′ + y = 8e^−2ty(0) = 5

y(1) =

3Solutions

1.Numerical methods

Answer: |E| = 0.0418

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 + … + 2f_n−1 + fₙ]. Simpson's rule (n even): S = (h/3)[f₀ + 4f₁ + 2f₂ + 4f₃ + … + 4f_n−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.

Exact value

I = 1.386294

Trapezoid rule

T = (0.75/2)·(1 + 2·0.571429 + 2·0.4 + 2·0.307692 + 0.25) = 1.428091

Error

|E| = |1.428091 − 1.386294| = 0.0418

Answer

|E| = 0.0418

Relative error: 3.015%.

2.Numerical methods

Answer: y₄ = 2.347

Numerical solution of y′ = f(t, y): Euler's method steps along the slope at the start of each step: y_n+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), y_n+1 = yₙ + (h/2)(k₁ + k₂). The classical Runge–Kutta method (RK4) combines four slopes: y_n+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 = 1 + 0.2·(1) = 1.2

y₂ = 1.2 + 0.2·(1.4) = 1.48

y₃ = 1.48 + 0.2·(1.88) = 1.856

y₄ = 1.856 + 0.2·(2.456) = 2.3472

Graph

y₄ = 2.347

Answer

y₄ = 2.347

The exact value is y(0.8) = 2.651082.

3.Sequences & series

Answer: S = 0.6111

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 |a_k+1/a_k|; root test: L = lim |a_k|^1/k. L < 1 converges, L > 1 diverges, L = 1 decides nothing. A power series Σc_k(x − x₀)ᵏ converges for |x − x₀| < R with R = lim |c_k/c_k+1|; an alternating series with terms decreasing to 0 converges.

Partial fractions

1/(k(k + 3)) = (1/3)·(1/k − 1/(k + 3))

Cancel

S = (1/3)·(1 + 1/2 + 1/3) = 0.6111

Answer

S = 0.6111

4.Sequences & series

Answer: S = 4.482

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.

Standard series

Σ xᵏ/k! = eˣ

Substitute

x = 1.5

S = e^1.5 = 4.482

Answer

S = 4.482

5.Matrices & determinants

Answer: det A = 24

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 = 0, M₁₂ = 0, M₁₃ = 6

Cofactor expansion

det A = (−1)·0 − 4·0 + 4·6 = 24

Answer

det A = 24

6.Matrices & determinants

Answer: x = −1.167

Determinant of the coefficient matrix

det A = (−4)·(−2) − 1·2 = 6

Replace the x column

det Aₓ = 8·(−2) − 1·(−9) = −7

x = −7/6 = −1.167

Answer

x = −1.167

y = 3.333

7.Differential equations

Answer: k = 0.03052 min⁻¹

First-order differential equations: dy/dt = k·y has the solution y = y₀·e^kt (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.

Ratio of temperature differences

e^−8k = 47/60

Solve for k

k = ln(60/47)/8 = 0.03052 min⁻¹

Answer

k = 0.03052 min⁻¹

8.Differential equations

Answer: y(1) = 3.7

Integrating factor

μ = e^t

(e^ty)′ = 8e^−t

General solution

y = 8/(−1)·e^−2t + C·e^−t = −8·e^−2t + C·e^−t

Initial condition

5 = −8 + C → C = 13

Evaluate

y(1) = −8·0.1353 + 13·0.3679 = 3.7

Answer

y(1) = 3.7

ExamProblems · examproblems.com/courses/numerical-methods