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
xy00.16670.33330.50.66670.83331

Approximate ∫01 1/(1 + x³) dx with Simpson's rule using n = 6 strips. Give S.

f(x) = 1/(1 + x³)[a, b] = [0, 1]n = 6
S =
2.Numerical methodsHard
00.20.40.60.8123ty(0, 1)

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
xx₀ = 0??R

Find the radius of convergence R of the power series Σk=1∞ 4ᵏx²ᵏ/k.

Σ 4ᵏx²ᵏ/kx₀ = 0
R =
4.Sequences & seriesHard
xyfP₂0.3

Use the Maclaurin polynomial of degree 2 of f(x) = ln(1 − x) to approximate f(0.3). Give P₂(0.3).

f(x) = ln(1 − x)n = 2x = 0.3
P₂(0.3) =
5.Matrices & determinantsMedium
A =6−4−3−1B =−634−6

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

A = [6 −4; −3 −1]B = [−6 3; 4 −6]
c12 =
6.Matrices & determinantsHard
A =6−5−4−4b =83

Solve the system 6x − 5y = 8 and −4x − 4y = 3. Find x.

6x − 5y = 8−4x − 4y = 3
x =
7.Differential equationsMedium
tyy(0) = 4y(1) = ?

Solve y″ + 8y′ + 15y = 0 with y(0) = 4 and y′(0) = 6. Find y(1).

y″ + 8y′ + 15y = 0y(0) = 4y′(0) = 6
y(1) =
8.Differential equationsHard
ty2y(2) = ?

Solve y′ + y = 3e−3t with y(0) = 5. Find y(2).

y′ + y = 3e−3ty(0) = 5
y(2) =

3Solutions

1.Numerical methods

Answer: S = 0.8357

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.

ixᵢf(xᵢ)weight001110.1666670.995392420.3333330.964286230.50.888889440.6666670.771429250.8333330.6334314610.51

Nodes and values

h = 0.1667
Σ wᵢfᵢ = 1 + 4·0.995392 + 2·0.964286 + 4·0.888889 + 2·0.771429 + 4·0.633431 + 0.5 = 15.042275

Simpson's rule

S = (0.1667/3)·15.042275 = 0.8357

Answer

S = 0.8357

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

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: 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₊₁00111.210.21.21.41.4820.41.481.881.85630.61.8562.4562.3472

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
00.20.40.60.8123ty

Graph

y4 = 2.347

Answer

y4 = 2.347

The exact value is y(0.8) = 2.651082.

3.Sequences & series

Answer: R = 0.5

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

The series is in x²: it converges for 4·x² < 1, so |x| < 1/2.

Radius

R = 0.5

Answer

R = 0.5

The ratio test says nothing at the two end points; they have to be checked separately.

4.Sequences & series

Answer: P₂(0.3) = −0.345

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.

Maclaurin polynomial

P₂(x) = −x − 0.5x²

Substitute

P₂(0.3) = −0.3 − 0.045 = −0.345

Answer

P₂(0.3) = −0.345

The exact value is f(0.3) = −0.356675; the error −0.0116749 shrinks when you take more terms or a point closer to 0.

5.Matrices & determinants

Answer: c12 = 42

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 = 6·3 + (−4)·(−6) = 42

Answer

c12 = 42
6.Matrices & determinants

Answer: x = 0.3864

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·(−4) − (−5)·(−4) = −44

Replace the x column

det Aₓ = 8·(−4) − (−5)·3 = −17
x = −17/(−44) = 0.3864

Answer

x = 0.3864
y = −1.136
7.Differential equations

Answer: y(1) = 0.5866

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 + 5) = 0 → r = −3, −5

Constants from the initial values

A + B = 4, −3A − 5B = 6
A = 13, B = −9

Evaluate

y(1) = 13·e−3 + (−9)·e−5 = 0.5866

Answer

y(1) = 0.5866
8.Differential equations

Answer: y(2) = 0.876

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 = et
(ety)′ = 3e−2t

General solution

y = 3/(−2)·e−3t + C·e−t = −1.5·e−3t + C·e−t

Initial condition

5 = −1.5 + C → C = 6.5

Evaluate

y(2) = −1.5·0.002479 + 6.5·0.1353 = 0.876

Answer

y(2) = 0.876

ExamProblems · examproblems.com/courses/numerical-methods