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
xy00.751.5

Approximate ∫01.5 e−x² dx with Simpson's rule using n = 2 strips. Give S.

f(x) = e−x²[a, b] = [0, 1.5]n = 2
S =
2.Numerical methodsHard
00.10.20.30.411.5ty(0, 1)

Use Heun's method (improved Euler) with h = 0.2 for y′ = t² + y², y(0) = 1 to estimate y2 ≈ y(0.4).

y′ = t² + y²y(0) = 1h = 0.2
y2 =
3.Sequences & seriesMedium
k|aₖ₊₁/aₖ|123456789101112

Apply the ratio test to Σk=1∞ k³·5ᵏ/3ᵏ. Find L = lim |ak+1/ak|.

ak = k³·5ᵏ/3ᵏ
L =
4.Sequences & seriesHard
xyfP₂0.4

Use the Maclaurin polynomial of degree 2 of f(x) = e2x to approximate f(0.4). Give P₂(0.4).

f(x) = e2xn = 2x = 0.4
P₂(0.4) =
5.Matrices & determinantsMedium
A =33110−35−53

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

A = [3 3 1; 1 0 −3; 5 −5 3]
det A =
6.Matrices & determinantsHard
A =241−1

Find the larger eigenvalue λ of the matrix A shown.

A = [2 4; 1 −1]
λ =
7.Differential equationsMedium
tyy(0) = 1y(1.5) = ?

Solve y″ + 8y′ + 15y = 0 with y(0) = 1 and y′(0) = 4. Find y(1.5).

y″ + 8y′ + 15y = 0y(0) = 1y′(0) = 4
y(1.5) =
8.Differential equationsHard
tyy(0) = 5y(0.5) = ?

Solve y″ + 4y′ + 20y = 0 with y(0) = 5 and y′(0) = 0 (released from rest). Find y(0.5); angles in radians.

y″ + 4y′ + 20y = 0y(0) = 5y′(0) = 0
y(0.5) =

3Solutions

1.Numerical methods

Answer: S = 0.8461

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)weight001110.750.569783421.50.1053991

Nodes and values

h = 0.75
Σ wᵢfᵢ = 1 + 4·0.569783 + 0.105399 = 3.384531

Simpson's rule

S = (0.75/3)·3.384531 = 0.8461

Answer

S = 0.8461

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

2.Numerical methods

Answer: y2 = 1.669

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₊₁00111.481.24810.21.2481.5975042.6170591.669456

Heun steps

y₁ = 1 + (0.2/2)(1 + 1.48) = 1.248
y₂ = 1.248 + (0.2/2)(1.597504 + 2.617059) = 1.669456
00.10.20.30.411.5ty

Graph

y2 = 1.669

Answer

y2 = 1.669
3.Sequences & series

Answer: L = 1.667

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| = ((k + 1)/k)³·5/3 → 5/3

Conclusion

L = 1.667 > 1

Answer

L = 1.667

L > 1, so the series diverges.

4.Sequences & series

Answer: P₂(0.4) = 2.12

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) = 1 + 2x + 2x²

Substitute

P₂(0.4) = 1 + 0.8 + 0.32 = 2.12

Answer

P₂(0.4) = 2.12

The exact value is f(0.4) = 2.225541; the error 0.105541 shrinks when you take more terms or a point closer to 0.

5.Matrices & determinants

Answer: det A = −104

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 = 18, M₁₃ = −5

Cofactor expansion

det A = 3·(−15) − 3·18 + 1·(−5) = −104

Answer

det A = −104
6.Matrices & determinants

Answer: λ = 3

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 − (1)λ + (−6) = 0

Solve

λ = (1 ± √(25))/2 → λ = −2, 3
λ = 3

Answer

λ = 3
7.Differential equations

Answer: y(1.5) = 0.04805

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 = 1, −3A − 5B = 4
A = 4.5, B = −3.5

Evaluate

y(1.5) = 4.5·e−4.5 + (−3.5)·e−7.5 = 0.04805

Answer

y(1.5) = 0.04805
8.Differential equations

Answer: y(0.5) = 0.07082

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 = −2 ± 4j

Constants

A = 5, B = 2·5/4 = 2.5

Evaluate

y(0.5) = e−1(5·cos 2 + 2.5·sin 2) = 0.07082

Answer

y(0.5) = 0.07082

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