Numerical Methods: cram pack

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

New problemsBack 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
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)051015202530v (m/s)3332397

A vehicle's speed is measured every 5 s (table). Estimate the distance s travelled from t = 0 to t = 30 s with the trapezoid rule.

h = 5 sn = 6
s = m
2.Numerical methodsHard
0120123ty(0, 0.5)

Use Euler's method with h = 0.5 for y′ = 1 + sin(ty), y(0) = 0.5 to estimate y5 ≈ y(2.5).

y′ = 1 + sin(ty)y(0) = 0.5h = 0.5
y5 =
3.Sequences & seriesMedium
AΣ 1/(2ᵏ + k)k = 1 … ∞BΣ (5/3)ᵏk = 1 … ∞CΣ 5/k⁴k = 1 … ∞DΣ (3/5)ᵏk = 1 … ∞

Exactly one of the four series diverges. Click it.

A: Σ 1/(2ᵏ + k)B: Σ (5/3)ᵏC: Σ 5/k⁴D: Σ (3/5)ᵏ
  1. 1A: Σ 1/(2ᵏ + k)
  2. 2B: Σ (5/3)ᵏ
  3. 3C: Σ 5/k⁴
  4. 4D: Σ (3/5)ᵏ
4.Sequences & seriesHard

Find the coefficient c₂ of (x − 2)² in the Taylor series of f(x) = e3x about a = 2.

f(x) = e3xa = 2k = 2
c₂ =
5.Matrices & determinantsMedium
A =3−6−25B =3−332

Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).

A = [3 −6; −2 5]B = [3 −3; 3 2]
c22 =
6.Matrices & determinantsHard
A =24−39

Find the larger eigenvalue λ of the matrix A shown.

A = [2 4; −3 9]
λ =
7.Differential equationsMedium
tyy(0) = 1y(0.5) = ?

Solve y″ + 7y′ + 12y = 0 with y(0) = 1 and y′(0) = 6. Find y(0.5).

y″ + 7y′ + 12y = 0y(0) = 1y′(0) = 6
y(0.5) =
8.Differential equationsHard
xy13(1, 5)y(3) = ?

Solve the initial value problem y′ + 3y/x = x2 for x > 0, with y(1) = 5. Find y(3).

y′ + 3y/x = x2y(1) = 5
y(3) =

3Solutions

1.Numerical methods

Answer: s = 125 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

3 + 2·3 + 2·3 + 2·2 + 2·3 + 2·9 + 7 = 50

Trapezoid rule

s = (5/2)·50 = 125 m

Answer

s = 125 m
2.Numerical methods

Answer: y5 = 3.027

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.

ntₙyₙf(tₙ, yₙ)yₙ₊₁000.51110.511.4794261.739713211.7397131.9857682.73259731.52.7325970.1823592.823776422.8237760.4063133.026932

Euler steps

y₁ = 0.5 + 0.5·(1) = 1
y₂ = 1 + 0.5·(1.479426) = 1.739713
y₃ = 1.739713 + 0.5·(1.985768) = 2.732597
y₄ = 2.732597 + 0.5·(0.182359) = 2.823776
y₅ = 2.823776 + 0.5·(0.406313) = 3.026932
0120123ty

Graph

y5 = 3.027

Answer

y5 = 3.027
3.Sequences & series

Answer: B: Σ (5/3)ᵏ

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/(2ᵏ + k) → Smaller than (1/2)ᵏ, a convergent geometric series: converges by comparison.
B: Σ (5/3)ᵏ → Geometric with r = 5/3 > 1: the terms grow, so it diverges.
C: Σ 5/k⁴ → p-series with p = 4 > 1: converges.
D: Σ (3/5)ᵏ → Geometric with r = 3/5, |r| < 1: converges.

The answer

B: Σ (5/3)ᵏ

Answer

B: Σ (5/3)ᵏ
4.Sequences & series

Answer: c₂ = 1815

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) = 3²·e3x

Coefficient

c₂ = 3²·e6/2! = 1815
c₂ = 1815

Answer

c₂ = 1815
5.Matrices & determinants

Answer: c22 = 16

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

c22 = (−2)·(−3) + 5·2 = 16

Answer

c22 = 16
6.Matrices & determinants

Answer: λ = 6

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 − (11)λ + (30) = 0

Solve

λ = (11 ± √(1))/2 → λ = 5, 6
λ = 6

Answer

λ = 6
7.Differential equations

Answer: y(0.5) = 1.013

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.

Characteristic roots

(r + 3)(r + 4) = 0 → r = −3, −4

Constants from the initial values

A + B = 1, −3A − 4B = 6
A = 10, B = −9

Evaluate

y(0.5) = 10·e−1.5 + (−9)·e−2 = 1.013

Answer

y(0.5) = 1.013
8.Differential equations

Answer: y(3) = 4.679

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∫3/x dx = x3

Integrate

(x3y)′ = x5 → x3y = 1x6/6 + C

Initial condition

5 = 1/6 + C → C = 4.833

Evaluate

y(3) = (1·36/6 + 4.833)/33 = 4.679

Answer

y(3) = 4.679

ExamProblems · examproblems.com/courses/numerical-methods