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
xy00.511.52

Approximate ∫02 √(1 + x³) dx with the trapezoid rule using n = 4 strips. Give T.

f(x) = √(1 + x³)[a, b] = [0, 2]n = 4
T =
2.Numerical methodsHard
00.10.20.30.400.20.4ty(0, 0)

Use Heun's method (improved Euler) with h = 0.2 for y′ = 1 + sin(ty), y(0) = 0 to estimate y2 ≈ y(0.4).

y′ = 1 + sin(ty)y(0) = 0h = 0.2
y2 =
3.Sequences & seriesMedium
kSn​12345678910S = ?

Find the sum S = Σk=1∞ 4/(k(k + 4)).

ak = 4/(k(k + 4))
S =
4.Sequences & seriesHard

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

f(x) = ln xa = 2k = 2
c₂ =
5.Matrices & determinantsMedium
A =50−314−4−1−4−5

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

A = [5 0 −3; 1 4 −4; −1 −4 −5]
det A =
6.Matrices & determinantsHard
A =3−4−52b =9−2

Solve the system 3x − 4y = 9 and −5x + 2y = −2. Find x.

3x − 4y = 9−5x + 2y = −2
x =
7.Differential equationsMedium
tyy(0) = 5y(1) = ?

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

y″ + 6y′ + 9y = 0y(0) = 5y′(0) = -1
y(1) =
8.Differential equationsHard
tyy(0) = 5y(0.25) = ?

Solve y″ + 2y′ + 10y = 0 with y(0) = 5 and y′(0) = 0 (released from rest). Find y(0.25); angles in radians.

y″ + 2y′ + 10y = 0y(0) = 5y′(0) = 0
y(0.25) =

3Solutions

1.Numerical methods

Answer: T = 3.283

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)weight001110.51.060662211.414214231.52.0916524231

Nodes and values

h = 0.5
Σ wᵢfᵢ = 1 + 2·1.06066 + 2·1.414214 + 2·2.09165 + 3 = 13.133048

Trapezoid rule

T = (0.5/2)·13.133048 = 3.283

Answer

T = 3.283

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

2.Numerical methods

Answer: y2 = 0.4245

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₊₁00011.0399890.20399910.20.2039991.0407881.1641170.424489

Heun steps

y₁ = 0 + (0.2/2)(1 + 1.039989) = 0.203999
y₂ = 0.203999 + (0.2/2)(1.040788 + 1.164117) = 0.424489
00.10.20.30.400.20.4ty

Graph

y2 = 0.4245

Answer

y2 = 0.4245
3.Sequences & series

Answer: S = 2.083

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.

Partial fractions

4/(k(k + 4)) = (4/4)·(1/k − 1/(k + 4))

Cancel

S = (4/4)·(1 + 1/2 + 1/3 + 1/4) = 2.083

Answer

S = 2.083
4.Sequences & series

Answer: c₂ = −0.125

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) = −1/x²

Coefficient

c₂ = f⁽²⁾(2)/2! = −1/(2²·2) = −0.125
c₂ = −0.125

Answer

c₂ = −0.125
5.Matrices & determinants

Answer: det A = −180

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 = −36, M₁₂ = −9, M₁₃ = 0

Cofactor expansion

det A = 5·(−36) − 0·(−9) + (−3)·0 = −180

Answer

det A = −180
6.Matrices & determinants

Answer: x = −0.7143

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 = 3·2 − (−4)·(−5) = −14

Replace the x column

det Aₓ = 9·2 − (−4)·(−2) = 10
x = 10/(−14) = −0.7143

Answer

x = −0.7143
y = −2.786
7.Differential equations

Answer: y(1) = 0.946

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(1) = (5 + 14·1)·e−3 = 0.946

Answer

y(1) = 0.946
8.Differential equations

Answer: y(0.25) = 3.734

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.

Complex roots

r = −1 ± 3j

Constants

A = 5, B = 1·5/3 = 1.667

Evaluate

y(0.25) = e−0.25(5·cos 0.75 + 1.667·sin 0.75) = 3.734

Answer

y(0.25) = 3.734

ExamProblems · examproblems.com/courses/numerical-methods