(a) Locate the poles and their order for the function f(z) = 1/[z(sin πz)(z + 1/2)]. Also, find the residue of f(z) at these…

(a) The poles are zeros of the denominator z(sin πz)(z + 1/2).

• At z = 0: z has a simple zero and sin πz has a simple zero, while z + 1/2 ≠ 0. Hence the denominator has a zero of order 2, so f has a pole of order 2 at z = 0.
• At z = n, where n is an integer and n ≠ 0: sin πz has a simple zero, while n ≠ 0 and n + 1/2 ≠ 0. Hence f has a simple pole at each z = n, n ∈ ℤ, n ≠ 0.
• At z = −1/2: z + 1/2 has a simple zero, while z = −1/2 ≠ 0 and sin(−π/2) = −1 ≠ 0. Hence f has a simple pole at z = −1/2.

Residues:

For a simple pole at z = n, n ∈ ℤ, n ≠ 0, Res(f, n) = lim(z→n) (z − n) f(z) = 1/[n(n + 1/2)] · lim(z→n) (z − n)/sin πz. Since sin πz ≈ (−1)ⁿ π(z − n) near z = n, lim(z→n) (z − n)/sin πz = (−1)ⁿ/π. Thus Res(f, n) = (−1)ⁿ/[π n(n + 1/2)] = 2(−1)ⁿ/[π n(2n + 1)], n ∈ ℤ, n ≠ 0.

At z = −1/2, Res(f, −1/2) = lim(z→−1/2) (z + 1/2) f(z) = 1/[z sin πz] at z = −1/2 = 1/[(-1/2)(−1)] = 2.

At z = 0, let h(z) = (z + 1/2) sin πz. Near z = 0, sin πz = πz − π³z³/6 + …, so h(z) = (z + 1/2)(πz − π³z³/6 + …) = (π/2)z + πz² + … . Therefore 1/h(z) = 2/(πz) − 4/π + …, and f(z) = 1/[z h(z)] = 2/(π z²) − 4/(π z) + … . Hence Res(f, 0) = −4/π.

(b) Let Sₙ(x) = 1/(2n²) log(1 + n⁴x²), x ∈ [0, 1]. Then U₁ = S₁ and for n ≥ 2, Uₙ = Sₙ − Sₙ₋₁, with S₀ = 0.

First find the sum of the given series. For fixed x, if x = 0, Sₙ(x) = 0. If x > 0, Sₙ(x) = 1/(2n²) log(1 + n⁴x²) = 1/(2n²)[4 log n + 2 log x + o(1)] → 0. Also for 0 ≤ x ≤ 1, 0 ≤ Sₙ(x) ≤ log(1 + n⁴)/(2n²) → 0. Thus the series converges uniformly to f(x) = 0 on [0, 1].

Now differentiate Sₙ: Sₙ′(x) = 1/(2n²) · (2n⁴x)/(1 + n⁴x²) = n²x/(1 + n⁴x²).

Therefore U₁′(x) = S₁′(x) = x/(1 + x²), and for n ≥ 2, Uₙ′(x) = Sₙ′(x) − Sₙ₋₁′(x).

The partial sum of the derivative series is Σ(k=1 to N) Uₖ′(x) = S_N′(x) = N²x/(1 + N⁴x²).

For fixed x ≥ 0, if x = 0, S_N′(0) = 0; if x > 0, S_N′(x) ∼ 1/(N²x) → 0. Hence Σ(n=1 to ∞) Uₙ′(x) = 0 = f′(x). So the given series can be differentiated term-by-term on [0, 1].

However, the convergence of Σ Uₙ′(x) is not uniform. Its N-th remainder is 0 − S_N′(x) = − N²x/(1 + N⁴x²). The maximum of N²x/(1 + N⁴x²) on [0, 1] occurs at x = 1/N², and equals (N² · 1/N²)/(1 + 1) = 1/2. Thus sup(0 ≤ x ≤ 1) |remainder| = 1/2, which does not tend to 0 as N → ∞. Therefore Σ Uₙ′(x) does not converge uniformly on [0, 1], although term-by-term differentiation still holds here.

(c) Primal problem: Minimize z = 4x₁ + 3x₂ + x₃ subject to x₁ + 2x₂ + 4x₃ ≥ 12, 3x₁ + 2x₂ + x₃ ≥ 8, x₁, x₂, x₃ ≥ 0.

Using duality, the dual is: Maximize w = 12y₁ + 8y₂ subject to y₁ + 3y₂ ≤ 4, 2y₁ + 2y₂ ≤ 3, 4y₁ + y₂ ≤ 1, y₁, y₂ ≥ 0.

From 4y₁ + y₂ ≤ 1 and y₁ ≥ 0, w = 12y₁ + 8y₂ = 8(4y₁ + y₂) − 20y₁ ≤ 8 · 1 − 0 = 8. Equality holds when y₁ = 0 and 4y₁ + y₂ = 1, so y₂ = 1. This point satisfies all dual constraints. Hence the dual optimum is **w* = 8 at (y₁, y₂) = (0, 1)**.

By the duality principle, the primal optimum is also **z* = 8**.

Use complementary slackness. Since y₂ = 1 > 0, the second primal constraint is tight: 3x₁ + 2x₂ + x₃ = 8. At (y₁, y₂) = (0, 1), the dual constraints for x₁ and x₂ are y₁ + 3y₂ = 3 < 4, 2y₁ + 2y₂ = 2 < 3. Hence x₁ = 0 and x₂ = 0. Then from the tight second constraint, x₃ = 8. The first primal constraint gives 4x₃ = 32 ≥ 12, so it is satisfied.

Therefore the optimal primal solution is (x₁, x₂, x₃) = (0, 0, 8), and the minimum value is z_min = 8.