Numerical Methods: cram pack

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

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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 ∫₀² √(1 + x³) dx with Simpson's rule using n = 2 strips. Give S.

f(x) = √(1 + x³)[a, b] = [0, 2]n = 2

S =

2.Numerical methodsHard

Use Euler's method with h = 0.2 for y′ = 2y, y(0) = 2 to estimate y₃ ≈ y(0.6).

y′ = 2yy(0) = 2h = 0.2

y₃ =

3.Sequences & seriesMedium

Exactly one of the four series diverges. Click it.

A: Σ (3/4)ᵏB: Σ (−1)ᵏ/√kC: Σ (4k + 1)/(k + 2)D: Σ 5/k³

1A: Σ (3/4)ᵏ
2B: Σ (−1)ᵏ/√k
3C: Σ (4k + 1)/(k + 2)
4D: Σ 5/k³

4.Sequences & seriesHard

Find the coefficient c₄ of (x − 1)⁴ in the Taylor series of f(x) = ∛x about a = 1.

f(x) = ∛xa = 1k = 4

c₄ =

5.Matrices & determinantsMedium

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

A = [4 2 2; 0 −1 −2; −5 −2 3]

det A =

6.Matrices & determinantsHard

Find the larger eigenvalue λ of the matrix A shown.

A = [3 6; −5 −8]

λ =

7.Differential equationsMedium

Solve y″ + 5y′ + 4y = 0 with y(0) = 3 and y′(0) = 0. Find y(0.5).

y″ + 5y′ + 4y = 0y(0) = 3y′(0) = 0

y(0.5) =

8.Differential equationsHard

Solve y″ + y′ + 4.25y = 0 with y(0) = 5 and y′(0) = 0 (released from rest). Find y(1.5); angles in radians.

y″ + y′ + 4.25y = 0y(0) = 5y′(0) = 0

y(1.5) =

3Solutions

1.Numerical methods

Answer: S = 3.219

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.

Nodes and values

h = 1

Σ wᵢfᵢ = 1 + 4·1.414214 + 3 = 9.656854

Simpson's rule

S = (1/3)·9.656854 = 3.219

Answer

S = 3.219

This integral has no elementary antiderivative, which is exactly when a numerical rule is needed.

2.Numerical methods

Answer: y₃ = 5.488

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 = 2 + 0.2·(4) = 2.8

y₂ = 2.8 + 0.2·(5.6) = 3.92

y₃ = 3.92 + 0.2·(7.84) = 5.488

Graph

y₃ = 5.488

Answer

y₃ = 5.488

The exact value is y(0.6) = 6.640234.

3.Sequences & series

Answer: C: Σ (4k + 1)/(k + 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 |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.

Test each series

A: Σ (3/4)ᵏ → Geometric with r = 3/4, |r| < 1: converges.

B: Σ (−1)ᵏ/√k → Alternating, with 1/√k decreasing to 0: converges (conditionally).

C: Σ (4k + 1)/(k + 2) → The terms tend to 4, not 0: diverges by the n-th term test.

D: Σ 5/k³ → p-series with p = 3 > 1: converges.

The answer

C: Σ (4k + 1)/(k + 2)

Answer

C: Σ (4k + 1)/(k + 2)

4.Sequences & series

Answer: c₄ = −0.04115

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) = −0.9877·x^{1/3 − 4}

Coefficient

c₄ = −0.9877·1^{1/3 − 4}/4! = −0.04115

c₄ = −0.04115

Answer

c₄ = −0.04115

5.Matrices & determinants

Answer: det A = −18

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 = −7, M₁₂ = −10, M₁₃ = −5

Cofactor expansion

det A = 4·(−7) − 2·(−10) + 2·(−5) = −18

Answer

det A = −18

6.Matrices & determinants

Answer: λ = −2

Characteristic equation

λ² − (−5)λ + (6) = 0

Solve

λ = (−5 ± √(1))/2 → λ = −3, −2

λ = −2

Answer

λ = −2

7.Differential equations

Answer: y(0.5) = 2.291

Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = e^rt: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·e^r₁t + B·e^r₂t (overdamped); a double root r gives y = (A + Bt)·e^rt (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″ + 2ζω_ny′ + ω_n²y = 0, ω_n 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 + 1)(r + 4) = 0 → r = −1, −4

Constants from the initial values

A + B = 3, −1A − 4B = 0

A = 4, B = −1

Evaluate

y(0.5) = 4·e^−0.5 + (−1)·e⁻² = 2.291

Answer

y(0.5) = 2.291

8.Differential equations

Answer: y(1.5) = −2.255

Complex roots

r = −0.5 ± 2j

Constants

A = 5, B = 0.5·5/2 = 1.25

Evaluate

y(1.5) = e^−0.75(5·cos 3 + 1.25·sin 3) = −2.255

Answer

y(1.5) = −2.255

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