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)01234v (m/s)0561214

A vehicle's speed is measured every 1 s (table). Estimate the distance s travelled from t = 0 to t = 4 s with the trapezoid rule.

h = 1 sn = 4
s = m
2.Numerical methodsHard
00.20.411.5ty(0, 1)

Use Heun's method (improved Euler) with h = 0.25 for y′ = 1 + sin(ty), y(0) = 1 to estimate y2 ≈ y(0.5).

y′ = 1 + sin(ty)y(0) = 1h = 0.25
y2 =
3.Sequences & seriesMedium
k|aₖ₊₁/aₖ|123456789101112

Apply the ratio test to Σk=1∞ (k!)²·5ᵏ/(2k)!. Find L = lim |ak+1/ak|.

ak = (k!)²·5ᵏ/(2k)!
L =
4.Sequences & seriesHard
xyfP₃2.5

Use the Maclaurin polynomial of degree 3 of f(x) = e0.5x to approximate f(2.5). Give P₃(2.5).

f(x) = e0.5xn = 3x = 2.5
P₃(2.5) =
5.Matrices & determinantsMedium
A =34−4−401−32−1

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

A = [3 4 −4; −4 0 1; −3 2 −1]
det A =
6.Matrices & determinantsHard
A =−3−197

Find the larger eigenvalue λ of the matrix A shown.

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

Solve y″ + 8y′ + 15y = 0 with y(0) = 3 and y′(0) = 0. Find y(1).

y″ + 8y′ + 15y = 0y(0) = 3y′(0) = 0
y(1) =
8.Differential equationsHard
ty1y(1) = ?

Solve y′ + 0.5y = 7e−3t with y(0) = 5. Find y(1).

y′ + 0.5y = 7e−3ty(0) = 5
y(1) =

3Solutions

1.Numerical methods

Answer: s = 30 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

0 + 2·5 + 2·6 + 2·12 + 14 = 60

Trapezoid rule

s = (1/2)·60 = 30 m

Answer

s = 30 m
2.Numerical methods

Answer: y2 = 1.668

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₊₁00111.3074391.2884310.251.288431.3165661.7234491.668432

Heun steps

y₁ = 1 + (0.25/2)(1 + 1.307439) = 1.28843
y₂ = 1.28843 + (0.25/2)(1.316566 + 1.723449) = 1.668432
00.20.411.5ty

Graph

y2 = 1.668

Answer

y2 = 1.668
3.Sequences & series

Answer: L = 1.25

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.

Ratio of consecutive terms

|ak+1/ak| = 5·(k + 1)²/((2k + 2)(2k + 1)) → 5/4

Conclusion

L = 1.25 > 1

Answer

L = 1.25

L > 1, so the series diverges.

4.Sequences & series

Answer: P₃(2.5) = 3.357

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 + 0.5x + 0.125x² + 0.02083x³

Substitute

P₃(2.5) = 1 + 1.25 + 0.78125 + 0.325521 = 3.357

Answer

P₃(2.5) = 3.357

The exact value is f(2.5) = 3.490343; the error 0.133572 shrinks when you take more terms or a point closer to 0.

5.Matrices & determinants

Answer: det A = −2

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 = −2, M₁₂ = 7, M₁₃ = −8

Cofactor expansion

det A = 3·(−2) − 4·7 + (−4)·(−8) = −2

Answer

det A = −2
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 − (4)λ + (−12) = 0

Solve

λ = (4 ± √(64))/2 → λ = −2, 6
λ = 6

Answer

λ = 6
7.Differential equations

Answer: y(1) = 0.3431

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 = 3, −3A − 5B = 0
A = 7.5, B = −4.5

Evaluate

y(1) = 7.5·e−3 + (−4.5)·e−5 = 0.3431

Answer

y(1) = 0.3431
8.Differential equations

Answer: y(1) = 4.592

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 = e0.5t
(e0.5ty)′ = 7e−2.5t

General solution

y = 7/(−2.5)·e−3t + C·e−0.5t = −2.8·e−3t + C·e−0.5t

Initial condition

5 = −2.8 + C → C = 7.8

Evaluate

y(1) = −2.8·0.04979 + 7.8·0.6065 = 4.592

Answer

y(1) = 4.592

ExamProblems · examproblems.com/courses/numerical-methods