(a) Prove that cosh ( z + 1/z ) = a₀ + Σₙ₌₁^∞ aₙ ( zⁿ + 1/zⁿ ) , where aₙ = (1 / 2π) ∫₀²π cos nθ cosh (2 cos θ) dθ, for n = 0…

(a) Laurent expansion of cosh(z + 1/z) Let f(z) = cosh(z + 1/z). The function f is analytic in the entire annulus 0 < |z| < ∞; it has an essential singularity at z = 0, so a Taylor expansion about 0 is impossible and only a Laurent series exists. Hence f admits a Laurent expansion f(z) = Σₙ₌₋∞^∞ cₙ zⁿ valid throughout that annulus, with coefficients given by the standard contour formula cₙ = (1/2πi) ∮_|z|=1 f(z) z⁻⁽ⁿ⁺¹⁾ dz, where the contour is the unit circle.

Symmetry of the coefficients. Observe that f(1/z) = cosh(1/z + z) = f(z). Substituting the Laurent series, f(1/z) = Σ cₙ z⁻ⁿ = Σ c₋ₙ zⁿ. Comparing term-by-term with f(z) = Σ cₙ zⁿ gives c₋ₙ = cₙ for every integer n. This symmetry is precisely what allows the series to be regrouped as f(z) = c₀ + Σₙ₌₁^∞ cₙ (zⁿ + z⁻ⁿ).

Evaluation of cₙ for n ≥ 0. Parametrise the unit circle by z = eⁱθ, so dz = i eⁱθ dθ and z⁻⁽ⁿ⁺¹⁾ = e⁻i(n+1)θ. On |z| = 1, z + 1/z = eⁱθ + e⁻iθ = 2 cos θ. Substituting into the coefficient integral: cₙ = (1/2πi) ∫₀²π cosh(2 cos θ) e⁻i(n+1)θ (i eⁱθ) dθ = (1/2π) ∫₀²π cosh(2 cos θ) e⁻inθ dθ. Now expand e⁻inθ = cos(nθ) − i sin(nθ). The factor cosh(2 cos θ) is unchanged under θ → 2π − θ, while sin(nθ) changes sign; hence the integral of cosh(2 cos θ) sin(nθ) over [0, 2π] vanishes. Therefore cₙ = (1/2π) ∫₀²π cosh(2 cos θ) cos(nθ) dθ. Setting aₙ = cₙ (with a₀ = c₀, the same formula with cos 0 = 1) and using c₋ₙ = cₙ = aₙ, the Laurent series becomes cosh(z + 1/z) = a₀ + Σₙ₌₁^∞ aₙ (zⁿ + 1/zⁿ), where aₙ = (1/2π) ∫₀²π cos(nθ) cosh(2 cos θ) dθ. Q.E.D.

(b) Upper and lower integrals The function f is defined on [0, 1] by f(x) = x² when x is rational and f(x) = x³ when x is irrational. To compute the lower and upper Darboux integrals, take an arbitrary partition P = {0 = x₀ < x₁ < … < xₙ = 1} of [0, 1].

Supremum and infimum on each subinterval. In every subinterval [xᵢ₋₁, xᵢ], both the rationals and the irrationals are dense. On [0, 1] we have 0 ≤ x ≤ 1, so x³ ≤ x². Consequently, the largest values of f on [xᵢ₋₁, xᵢ] are attained at the rational points (where f = x²), and the smallest at the irrational points (where f = x³). Because x² is increasing on [0, 1], the supremum is the value of x² at the right endpoint: sup f = xᵢ². Because x³ is increasing, the infimum is the value of x³ at the left endpoint: inf f = xᵢ₋₁³.

Upper and lower sums. The upper Darboux sum is U(f, P) = Σᵢ₌₁ⁿ xᵢ² Δxᵢ and the lower sum is L(f, P) = Σᵢ₌₁ⁿ xᵢ₋₁³ Δxᵢ. As the mesh ‖P‖ → 0, these Riemann sums converge to the definite integrals of the bounding functions x² and x³ respectively. Hence the upper integral is ∫₀¹ x² dx = [x³/3]₀¹ = 1/3, and the lower integral is ∫₀¹ x³ dx = [x⁴/4]₀¹ = 1/4.

Existence of the Riemann integral. By the Darboux criterion, the Riemann integral exists if and only if the upper and lower integrals are equal. Here 1/3 ≠ 1/4, so the Riemann integral ∫₀¹ f does not exist. This is consistent with the Lebesgue criterion: f is discontinuous at every point of (0, 1) (the rationals and irrationals are dense, and x² ≠ x³ there), a set of positive measure, so f is not Riemann integrable. (The Lebesgue integral does exist and equals 1/4, since f = x³ almost everywhere, but that is a different question.)

(c) Assignment problem There are 3 jobs and 4 machines, so the problem is unbalanced: add a dummy job J₄ with zero costs (the machine it receives stays idle). Cost matrix (rows J₁–J₄, columns M₁–M₄): J₁: 8 24 28 32 J₂: 8 13 17 19 J₃: 8 15 19 22 J₄: 0 0 0 0 Row reduction (row minima 8, 8, 8, 0); every column then already has a zero, so column reduction changes nothing: J₁: 0 16 20 24 J₂: 0 5 9 11 J₃: 0 7 11 14 J₄: 0 0 0 0 Iteration 1: all zeros are covered by 2 lines (column M₁, row J₄) < 4. The smallest uncovered entry is 5; subtract it from the uncovered entries and add it at the intersection (J₄, M₁): J₁: 0 11 15 19 J₂: 0 0 4 6 J₃: 0 2 6 9 J₄: 5 0 0 0 Iteration 2: minimum cover is 3 lines (column M₁, rows J₂ and J₄) < 4. Smallest uncovered entry 2; subtract from uncovered entries, add at (J₂, M₁) and (J₄, M₁): J₁: 0 9 13 17 J₂: 2 0 4 6 J₃: 0 0 4 7 J₄: 7 0 0 0 Iteration 3: minimum cover is 3 lines (columns M₁, M₂ and row J₄) < 4. Smallest uncovered entry 4; subtract from uncovered entries, add at (J₄, M₁) and (J₄, M₂): J₁: 0 9 9 13 J₂: 2 0 0 2 J₃: 0 0 0 3 J₄: 11 4 0 0 Now 4 independent zeros exist, so the assignment is optimal. J₁ has its only zero in M₁, so J₁ → M₁. Then J₄ must take M₄ (M₃ would leave M₄ with no zero among J₂, J₃). J₂ and J₃ share the zeros in M₂ and M₃: J₁ → M₁ (8), J₂ → M₂ (13), J₃ → M₃ (19), M₄ idle: total 8 + 13 + 19 = 40. Uniqueness: the zeros at (J₂, M₂), (J₂, M₃), (J₃, M₂), (J₃, M₃) form two independent choices, so the optimum is not unique. The alternate optimal assignment is J₁ → M₁ (8), J₂ → M₃ (17), J₃ → M₂ (15), M₄ idle: total 8 + 17 + 15 = 40. (Check by enumeration of all 24 assignments: 40 is the minimum and is attained only by these two; the next best is 42.)