(a) Find the maximum and minimum values of f(x) = x³ - 9x² + 26x - 24 for 0 ≤ x ≤ 1. (15 marks) (b) Let F be a field and f(x) ∈…

(a) Extreme values of f(x) = x³ − 9x² + 26x − 24 on [0, 1]

f′(x) = 3x² − 18x + 26. Its discriminant is 324 − 312 = 12, so f′ = 0 at x = 3 ± √3/3, about 2.42 and 3.58, both outside [0, 1]. Since f′(0) = 26 > 0 and f′ does not vanish on [0, 1], f′ > 0 there and f is strictly increasing.

Hence the minimum is f(0) = −24 at x = 0, and the maximum is f(1) = 1 − 9 + 26 − 24 = −6 at x = 1.

(b) Extension field and imbedding (Kronecker's theorem)

Since F[x] is a unique factorisation domain, f has an irreducible factor p(x) of positive degree. Put F′ = F[x]/(p(x)). As p is irreducible, (p) is a maximal ideal, so F′ is a field.

Define q: F → F′ by q(a) = a + (p). It is a ring homomorphism. If q(a) = 0 then a ∈ (p); a non-zero constant cannot be a multiple of p, which has positive degree, so a = 0. Thus q is injective, an imbedding.

Let α = x + (p) ∈ F′. If p(x) = a₀ + a₁x + … + aₙxⁿ, then

pᵠ(α) = q(a₀) + q(a₁)α + … + q(aₙ)αⁿ = (a₀ + a₁x + … + aₙxⁿ) + (p) = p(x) + (p) = 0.

Write f = p·h with h ∈ F[x]. Applying q to the coefficients, fᵠ = pᵠhᵠ, so fᵠ(α) = pᵠ(α)hᵠ(α) = 0.

Hence fᵠ has the root α in the field F′. ∎

(c) Laurent expansion of f(z) = (z² − z + 1)/[z(z² − 3z + 2)] in |z + 1| > 3

Since z² − 3z + 2 = (z − 1)(z − 2), partial fractions give

f(z) = 1/(2z) − 1/(z − 1) + 3/[2(z − 2)].

Put w = z + 1, so the poles z = 0, 1, 2 correspond to w = 1, 2, 3, and |w| > 3 lies outside all of them. Then 1/(2z) = 1/[2(w − 1)], −1/(z − 1) = −1/(w − 2), 3/[2(z − 2)] = 3/[2(w − 3)]. Expanding each in powers of 1/w (for |w| > 3):

1/[2(w − 1)] = Σₙ≥0 1/(2wⁿ⁺¹); −1/(w − 2) = −Σₙ≥0 2ⁿ/wⁿ⁺¹; 3/[2(w − 3)] = Σₙ≥0 (3/2)3ⁿ/wⁿ⁺¹.

Adding, with k = n + 1:

f(z) = Σₖ≥1 [ (1 + 3ᵏ)/2 − 2ᵏ⁻¹ ] / (z + 1)ᵏ, for |z + 1| > 3.

(The first terms are 1/(z + 1) + 3/(z + 1)² + 10/(z + 1)³ + ….)