(a) Let m₁, m₂, …, mₖ be positive integers and d > 0 the greatest common divisor of m₁, m₂, …, mₖ. Show that there exist integers…

(a) d = x₁m₁ + … + xₖmₖ

Induct on k. For k = 1, d = m₁ = 1·m₁. For two integers a, b, Bézout's identity gives integers u, v with gcd(a, b) = ua + vb (take the least positive element of the set {ua + vb}; it divides a and b by the division algorithm, and any common divisor divides it).

Assume the result for k − 1 integers: g = gcd(m₁, …, mₖ₋₁) = y₁m₁ + … + yₖ₋₁mₖ₋₁. Then d = gcd(g, mₖ), so d = rg + smₖ for some integers r, s. Hence

d = (ry₁)m₁ + … + (ryₖ₋₁)mₖ₋₁ + smₖ,

so take xᵢ = ryᵢ (i < k) and xₖ = s. ∎

(b) Uniform convergence of x⁴ + x⁴/(1 + x⁴) + x⁴/(1 + x⁴)² + … on [0, 1]

For x > 0 this is a geometric series with first term x⁴ and ratio q = 1/(1 + x⁴). The partial sum is

Sₙ(x) = x⁴(1 − qⁿ⁺¹)/(1 − q) = (1 + x⁴)(1 − qⁿ⁺¹) = 1 + x⁴ − 1/(1 + x⁴)ⁿ.

So S(x) = 1 + x⁴ for 0 < x ≤ 1, while at x = 0 every term is 0 and S(0) = 0.

Then |Sₙ(x) − S(x)| = 1/(1 + x⁴)ⁿ for x > 0, which tends to 1 as x → 0⁺ for every fixed n. Hence sup|Sₙ − S| = 1 does not tend to 0, and the convergence is not uniform on [0, 1].

Equivalently, each Sₙ is continuous but the limit S is discontinuous at 0, which uniform convergence would forbid.

(c) Monotonic functions are Riemann integrable

Let f be increasing on [a, b] (for a decreasing f apply the argument to −f). Then f is bounded, with f(a) ≤ f ≤ f(b).

For a partition P: a = x₀ < … < xₙ = b, on [xᵢ₋₁, xᵢ] the infimum is f(xᵢ₋₁) and the supremum is f(xᵢ). With ‖P‖ the largest Δxᵢ,

U(P, f) − L(P, f) = Σ[f(xᵢ) − f(xᵢ₋₁)]Δxᵢ ≤ ‖P‖ Σ[f(xᵢ) − f(xᵢ₋₁)] = ‖P‖[f(b) − f(a)].

If f(b) = f(a), f is constant and integrable. Otherwise, given ε > 0, choose P with ‖P‖ < ε/[f(b) − f(a)]; then U − L < ε. By Riemann's criterion, f is integrable on [a, b]. ∎

(d) Contour integral

2z² − 5z + 2 = 2(z − 1/2)(z − 2), so the poles are z = 1/2 and z = 2.

c(t) = e^(4πit) is the unit circle traversed twice anticlockwise. The pole 1/2 lies inside (winding number 2) and the pole 2 lies outside (winding number 0).

Residue at 1/2: 1/[2(1/2 − 2)] = −1/3.

∫_c dz/(2z² − 5z + 2) = 2πi × 2 × (−1/3) = −4πi/3.

(e) Assignment problem (Hungarian method)

Row minima 5, 3, 2, 7, 4 subtracted give:

A: 5 0 8 10 11; B: 0 6 15 10 3; C: 8 5 0 0 0; D: 0 4 2 0 5; E: 3 5 6 0 8.

Column minima are all 0. The zeros can cover at most four independent positions (for example A–II, B–I, C–III, D–IV), and four lines (columns I, II, IV and row C) cover all zeros, so no complete assignment exists yet.

The smallest uncovered element is 2. Subtract it from uncovered elements and add it at the intersections of lines:

A: 5 0 6 10 9; B: 0 6 13 10 1; C: 10 7 0 2 0; D: 0 4 0 0 3; E: 3 5 4 0 6.

A complete zero assignment now exists: A → II, B → I, C → V, D → III, E → IV. Cost: 5 + 3 + 2 + 9 + 4 = 23.

Minimum total processing time = 23 hours.