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
xy12.54

Approximate ∫14 1/x dx with the trapezoid rule, n = 2, and compare it with the exact value. Find the absolute error |E| = |T − I|.

f(x) = 1/x[a, b] = [1, 4]n = 2
|E| =
2.Numerical methodsHard
00.20.40.600.5ty(0, 0)

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

y′ = e−tyy(0) = 0h = 0.2
y3 =
3.Sequences & seriesMedium
xx₀ = 0??R

Find the radius of convergence R of the power series Σk=1∞ (2k)!·xᵏ/(k!)².

Σ (2k)!·xᵏ/(k!)²x₀ = 0
R =
4.Sequences & seriesHard
xyfP₃−1.5

Use the Maclaurin polynomial of degree 3 of f(x) = e−x to approximate f(−1.5). Give P₃(−1.5).

f(x) = e−xn = 3x = −1.5
P₃(−1.5) =
5.Matrices & determinantsMedium
A =−51−5−5−45−530

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

A = [−5 1 −5; −5 −4 5; −5 3 0]
det A =
6.Matrices & determinantsHard
A =−1−6−6−4b =23

Solve the system −x − 6y = 2 and −6x − 4y = 3. Find x.

−x − 6y = 2−6x − 4y = 3
x =
7.Differential equationsMedium
tyy(0) = 3y(1) = ?

Solve y″ + 6y′ + 8y = 0 with y(0) = 3 and y′(0) = 1. Find y(1).

y″ + 6y′ + 8y = 0y(0) = 3y′(0) = 1
y(1) =
8.Differential equationsHard
t (min)m (kg)90 minm = ?

A tank holds 1000 L of water in which 10 kg of salt is dissolved. Brine with 0.25 kg of salt per litre flows in at 20 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 90 min?

V = 1000 Lm0 = 10 kgc = 0.25 kg/Lr = 20 L/min
m = kg

3Solutions

1.Numerical methods

Answer: |E| = 0.1512

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.

Exact value

I = 1.386294

Trapezoid rule

T = (1.5/2)·(1 + 2·0.4 + 0.25) = 1.5375

Error

|E| = |1.5375 − 1.386294| = 0.1512

Answer

|E| = 0.1512

Relative error: 10.91%.

2.Numerical methods

Answer: y3 = 0.5357

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₊₁00010.9607890.19607910.20.1960790.9615430.8561110.37784420.40.3778440.8597290.7190140.535719

Heun steps

y₁ = 0 + (0.2/2)(1 + 0.960789) = 0.196079
y₂ = 0.196079 + (0.2/2)(0.961543 + 0.856111) = 0.377844
y₃ = 0.377844 + (0.2/2)(0.859729 + 0.719014) = 0.535719
00.20.40.600.5ty

Graph

y3 = 0.5357

Answer

y3 = 0.5357
3.Sequences & series

Answer: R = 0.25

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

ck+1/ck = (2k + 2)(2k + 1)/(k + 1)² → 4, so R = 1/4

Radius

R = 0.25

Answer

R = 0.25

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

4.Sequences & series

Answer: P₃(−1.5) = 4.188

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

P₃(x) = 1 − x + 0.5x² − 0.1667x³

Substitute

P₃(−1.5) = 1 + 1.5 + 1.125 + 0.5625 = 4.188

Answer

P₃(−1.5) = 4.188

The exact value is f(−1.5) = 4.481689; the error 0.294189 shrinks when you take more terms or a point closer to 0.

5.Matrices & determinants

Answer: det A = 225

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 = −15, M₁₂ = 25, M₁₃ = −35

Cofactor expansion

det A = (−5)·(−15) − 1·25 + (−5)·(−35) = 225

Answer

det A = 225
6.Matrices & determinants

Answer: x = −0.3125

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 = (−1)·(−4) − (−6)·(−6) = −32

Replace the x column

det Aₓ = 2·(−4) − (−6)·3 = 10
x = 10/(−32) = −0.3125

Answer

x = −0.3125
y = −0.2813
7.Differential equations

Answer: y(1) = 0.8156

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 + 2)(r + 4) = 0 → r = −2, −4

Constants from the initial values

A + B = 3, −2A − 4B = 1
A = 6.5, B = −3.5

Evaluate

y(1) = 6.5·e−2 + (−3.5)·e−4 = 0.8156

Answer

y(1) = 0.8156
8.Differential equations

Answer: m = 210.3 kg

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.

Set up the equation

dm/dt = 20·0.25 − (20/1000)·m = 5 − 0.02·m

Solve

The integrating factor is ert/V; the solution approaches the equilibrium c·V.

m(t) = 250 + (−240)·e−t/50

Substitute t

m(90) = 250 + (−240)·0.1653 = 210.3 kg

Answer

m = 210.3 kg

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