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
xy00.33330.666711.3331.6672

Approximate ∫02 sin(x²) dx with the trapezoid rule using n = 6 strips. Give T.

f(x) = sin(x²)[a, b] = [0, 2]n = 6
T =
2.Numerical methodsHard
00.050.11.91.952ty(0, 2)

Take one step of the classical Runge–Kutta method (RK4) with h = 0.1 for y′ = cos t − y, y(0) = 2. Find y₁ ≈ y(0.1).

y′ = cos t − yy(0) = 2h = 0.1
y1 =
3.Sequences & seriesMedium
kSn​12345678910S = ?

Find the sum S = Σk=1∞ 6/(k(k + 4)).

ak = 6/(k(k + 4))
S =
4.Sequences & seriesHard

f(x) = eˣ is approximated by its Maclaurin polynomial of degree 5 at x = 1.5. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = e1.5 = 4.482 to find the largest possible error |R₅|.

f(x) = eˣn = 5x = 1.5M = 4.482
|R₅| =
5.Matrices & determinantsMedium
A =−22−22−5−2444

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

A = [−2 2 −2; 2 −5 −2; 4 4 4]
det A =
6.Matrices & determinantsHard
A =6−1−5−1b =−4−7

Solve the system 6x − y = −4 and −5x − y = −7. Find x.

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

Solve y″ + 6y′ + 8y = 0 with y(0) = 5 and y′(0) = 6. Find y(1).

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

Solve the initial value problem y′ + 3y/x = 2 for x > 0, with y(1) = 5. Find y(3).

y′ + 3y/x = 2y(1) = 5
y(3) =

3Solutions

1.Numerical methods

Answer: T = 0.7795

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)weight000110.3333330.110883220.6666670.4299562310.841471241.3333330.978656251.6666670.355842262−0.7568021

Nodes and values

h = 0.3333
Σ wᵢfᵢ = 0 + 2·0.110883 + 2·0.429956 + 2·0.841471 + 2·0.978656 + 2·0.355842 − 0.756802 = 4.676813

Trapezoid rule

T = (0.3333/2)·4.676813 = 0.7795

Answer

T = 0.7795

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

2.Numerical methods

Answer: y1 = 1.905

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.

Four slopes

k₁ = f(0, 2) = −1
k₂ = f(0.05, 1.95) = −0.95125
k₃ = f(0.05, 1.952438) = −0.953687
k₄ = f(0.1, 1.904631) = −0.909627

Combine

y₁ = 2 + (0.1/6)(−1 + 2·−0.95125 + 2·−0.953687 − 0.909627) = 1.905

Answer

y1 = 1.905
3.Sequences & series

Answer: S = 3.125

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

6/(k(k + 4)) = (6/4)·(1/k − 1/(k + 4))

Cancel

S = (6/4)·(1 + 1/2 + 1/3 + 1/4) = 3.125

Answer

S = 3.125
4.Sequences & series

Answer: |R₅| = 0.0709

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.

Bound on the derivative

M = e1.5 = 4.482

Lagrange bound

|R₅| ≤ 4.482·1.5⁶/6! = 0.0709

Answer

|R₅| = 0.0709

The true error is smaller: |f(1.5) − P₅(1.5)| = 0.01997.

5.Matrices & determinants

Answer: det A = −64

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 = −12, M₁₂ = 16, M₁₃ = 28

Cofactor expansion

det A = (−2)·(−12) − 2·16 + (−2)·28 = −64

Answer

det A = −64
6.Matrices & determinants

Answer: x = 0.2727

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·(−1) − (−1)·(−5) = −11

Replace the x column

det Aₓ = (−4)·(−1) − (−1)·(−7) = −3
x = −3/(−11) = 0.2727

Answer

x = 0.2727
y = 5.636
7.Differential equations

Answer: y(1) = 1.613

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 = 5, −2A − 4B = 6
A = 13, B = −8

Evaluate

y(1) = 13·e−2 + (−8)·e−4 = 1.613

Answer

y(1) = 1.613
8.Differential equations

Answer: y(3) = 1.667

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)′ = 2x3 → x3y = 2x4/4 + C

Initial condition

5 = 2/4 + C → C = 4.5

Evaluate

y(3) = (2·34/4 + 4.5)/33 = 1.667

Answer

y(3) = 1.667

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