(a) Let f be an entire function whose Taylor series expansion with centre z = 0 has infinitely many terms. Show that z = 0 is an…

(a) z = 0 is an essential singularity of f(1/z)

Since f is entire, f(w) = Σₙ≥0 aₙwⁿ for all w, with aₙ ≠ 0 for infinitely many n. For z ≠ 0,

g(z) = f(1/z) = Σₙ≥0 aₙz⁻ⁿ,

which is the Laurent expansion of g about z = 0 (it converges for all z ≠ 0). Its principal part contains the infinitely many non-zero terms aₙz⁻ⁿ.

If z = 0 were removable, the principal part would vanish, so aₙ = 0 for all n ≥ 1. If z = 0 were a pole of order m, aₙ = 0 for all n > m. Both contradict the hypothesis. By uniqueness of the Laurent expansion, z = 0 is an essential singularity of f(1/z). ∎

(b) Stationary values of x² + y² + z² under ax² + by² + cz² = 1 and lx + my + nz = 0

Let S = x² + y² + z² and use L = S − λ(ax² + by² + cz² − 1) − 2ν(lx + my + nz). Setting the partial derivatives to zero:

(1 − λa)x = νl, (1 − λb)y = νm, (1 − λc)z = νn.

Multiplying by x, y, z and adding, S − λ(ax² + by² + cz²) = ν(lx + my + nz) = 0, so S = λ: the multiplier equals the stationary value.

Substituting x = νl/(1 − λa) etc. in lx + my + nz = 0 (ν ≠ 0 for a non-trivial point):

l²/(1 − λa) + m²/(1 − λb) + n²/(1 − λc) = 0.

Clearing denominators, with P = l² + m² + n², Q = l²(b + c) + m²(c + a) + n²(a + b), R = l²bc + m²ca + n²ab:

Rλ² − Qλ + P = 0, so λ = [Q ± √(Q² − 4PR)]/(2R).

Geometric meaning: the plane lx + my + nz = 0 through the origin cuts the quadric ax² + by² + cz² = 1 in a central conic. S is the squared distance of a point of the conic from its centre, so its two stationary values λ₁, λ₂ are the squares of the semi-axes of the conic; the semi-axes are √λ₁ and √λ₂. For a, b, c > 0 the section is an ellipse and both roots are positive.

(c) Dual of the linear programme

Primal: minimise Z = x₁ − 3x₂ − 2x₃ subject to 3x₁ − x₂ + 2x₃ ≤ 7, 2x₁ − 4x₂ ≥ 12, −4x₁ + 3x₂ + 8x₃ = 10; x₁, x₂ ≥ 0, x₃ unrestricted.

Write the first constraint as −3x₁ + x₂ − 2x₃ ≥ −7, so all inequalities are of the ≥ type. Attach y₁ ≥ 0, y₂ ≥ 0 to the two inequalities and y₃ (unrestricted) to the equation.

Dual: maximise W = −7y₁ + 12y₂ + 10y₃ subject to

−3y₁ + 2y₂ − 4y₃ ≤ 1 (for x₁ ≥ 0), y₁ − 4y₂ + 3y₃ ≤ −3 (for x₂ ≥ 0), −2y₁ + 8y₃ = −2 (x₃ is unrestricted, so an equation),

with y₁ ≥ 0, y₂ ≥ 0 and y₃ unrestricted.