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

Method

  1. 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.
  2. For a table value: interpolate linearly between the two neighbouring points.
  3. 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.
  4. 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

  1. Identify the type: arithmetic (constant difference), geometric (constant ratio), a standard series, a power series or a Taylor polynomial.
  2. 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.
  3. 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.
  4. 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
t (s)0510152025v (m/s)71115141314

A vehicle's speed is measured every 5 s (table). Estimate the distance s travelled from t = 0 to t = 25 s with the trapezoid rule.

h = 5 sn = 5
s = m
2.Numerical methodsHard
00.20.40.623ty(0, 2)

Use Heun's method (improved Euler) with h = 0.2 for y′ = y − t², y(0) = 2 to estimate y3 ≈ y(0.6).

y′ = y − t²y(0) = 2h = 0.2
y3 =
3.Sequences & seriesMedium
AΣ 1/√kk = 1 … ∞BΣ 2ᵏ/k²k = 1 … ∞CΣ ln k / kk = 1 … ∞DΣ k²/2ᵏk = 1 … ∞

Exactly one of the four series converges. Click it.

A: Σ 1/√kB: Σ 2ᵏ/k²C: Σ ln k / kD: Σ k²/2ᵏ
  1. 1A: Σ 1/√k
  2. 2B: Σ 2ᵏ/k²
  3. 3C: Σ ln k / k
  4. 4D: Σ k²/2ᵏ
4.Sequences & seriesHard

Recognise the series and find its sum: S = Σk=0∞ (−1)ᵏ/k!.

ak = (−1)ᵏ/k!
S =
5.Matrices & determinantsMedium
A =3−5−3−1B =−6−646

Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).

A = [3 −5; −3 −1]B = [−6 −6; 4 6]
c12 =
6.Matrices & determinantsHard
A =0−1−2−1

Find the larger eigenvalue λ of the matrix A shown.

A = [0 −1; −2 −1]
λ =
7.Differential equationsMedium
tyy(0) = 5y(2) = ?

Solve y″ + 6y′ + 9y = 0 with y(0) = 5 and y′(0) = -1. Find y(2).

y″ + 6y′ + 9y = 0y(0) = 5y′(0) = -1
y(2) =
8.Differential equationsHard
xy12(1, 2)y(2) = ?

Solve the initial value problem y′ + y/x = 1 for x > 0, with y(1) = 2. Find y(2).

y′ + y/x = 1y(1) = 2
y(2) =

3Solutions

1.Numerical methods

Answer: s = 317.5 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

7 + 2·11 + 2·15 + 2·14 + 2·13 + 14 = 127

Trapezoid rule

s = (5/2)·127 = 317.5 m

Answer

s = 317.5 m
2.Numerical methods

Answer: y3 = 3.545

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.

ntₙyₙk₁k₂yₙ₊₁00222.362.43610.22.4362.3962.75522.9511220.42.951122.791123.1493443.545166

Heun steps

y₁ = 2 + (0.2/2)(2 + 2.36) = 2.436
y₂ = 2.436 + (0.2/2)(2.396 + 2.7552) = 2.95112
y₃ = 2.95112 + (0.2/2)(2.79112 + 3.149344) = 3.545166
00.20.40.623ty

Graph

y3 = 3.545

Answer

y3 = 3.545

The exact value is y(0.6) = 3.56.

3.Sequences & series

Answer: D: Σ k²/2ᵏ

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.

Test each series

A: Σ 1/√k → p-series with p = 1/2 ≤ 1: diverges.
B: Σ 2ᵏ/k² → Ratio test: L = 2 > 1, diverges.
C: Σ ln k / k → Larger than 1/k for k ≥ 3: diverges by comparison with the harmonic series.
D: Σ k²/2ᵏ → Ratio test: L = 1/2 < 1, converges.

The answer

D: Σ k²/2ᵏ

Answer

D: Σ k²/2ᵏ
4.Sequences & series

Answer: S = 0.3679

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
S = e−1 = 0.3679

Answer

S = 0.3679
5.Matrices & determinants

Answer: c12 = −48

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 = 3·(−6) + (−5)·6 = −48

Answer

c12 = −48
6.Matrices & determinants

Answer: λ = 1

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.

Characteristic equation

λ² − (−1)λ + (−2) = 0

Solve

λ = (−1 ± √(9))/2 → λ = −2, 1
λ = 1

Answer

λ = 1
7.Differential equations

Answer: y(2) = 0.0818

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.

Double root

(r + 3)² = 0 → r = −3

Constants

A = y(0) = 5
y′(0) = B − 3A → B = -1 + 15 = 14

Evaluate

y(2) = (5 + 14·2)·e−6 = 0.0818

Answer

y(2) = 0.0818
8.Differential equations

Answer: y(2) = 1.75

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 = e∫1/x dx = x1

Integrate

(x1y)′ = x → x1y = 1x2/2 + C

Initial condition

2 = 1/2 + C → C = 1.5

Evaluate

y(2) = (1·22/2 + 1.5)/21 = 1.75

Answer

y(2) = 1.75

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