(a) Prove that the oscillation of a real-valued bounded function f defined on [a, b] is the supremum of the set {|f(x₁)-f(x₂)| …

(a) Oscillation as a supremum

Let M = sup f(x) and m = inf f(x) over [a, b]. Since f is bounded, M and m are finite, and the oscillation of f on [a, b] is defined as ω = M − m. Let S = {|f(x₁) − f(x₂)| : x₁, x₂ ∈ [a, b]}.

Upper bound: for any x₁, x₂ ∈ [a, b], m ≤ f(x₁) ≤ M and m ≤ f(x₂) ≤ M, so |f(x₁) − f(x₂)| ≤ M − m. Hence S is bounded above by M − m, and sup S ≤ M − m.

Lower bound: let ε > 0. By the definitions of supremum and infimum there are x₁, x₂ ∈ [a, b] with f(x₁) > M − ε/2 and f(x₂) < m + ε/2. Then

|f(x₁) − f(x₂)| ≥ f(x₁) − f(x₂) > M − m − ε, so sup S > M − m − ε. As ε > 0 is arbitrary, sup S ≥ M − m.

Combining the two, sup S = M − m = ω. The oscillation of f is the supremum of S.

(b) Nature of the singularity at z = 0 and the principal part

Use the Maclaurin series of eᶻ and sin z: eᶻ = 1 + z + z²/2 + z³/6 + … sin z = z − z³/6 + z⁵/120 − z⁷/5040 + …

Then z − sin z = z³/6 − z⁵/120 + z⁷/5040 − … = (z³/6)(1 − z²/20 + z⁴/840 − …).

The bracket equals 1 at z = 0, so z − sin z has a zero of order exactly 3 at 0. Since eᶻ is analytic with e⁰ = 1 ≠ 0, z = 0 is a pole of order 3.

For the principal part, write 1/(1 − z²/20 + z⁴/840 − …) = 1 + z²/20 + O(z⁴). Then eᶻ/(z − sin z) = (6/z³)(1 + z + z²/2 + O(z³))(1 + z²/20 + O(z⁴)) = (6/z³)(1 + z + (1/2 + 1/20)z² + O(z³)) = 6/z³ + 6/z² + (33/10)/z + O(1), since 6 × 11/20 = 33/10.

Principal part: 6/z³ + 6/z² + 33/(10z). (The residue at 0 is 33/10.)

(c) Assignment problem by the Hungarian method

Cost matrix (hours), rows I to V, columns A to E: I: 4 9 4 12 4 II: 15 11 20 5 8 III: 17 7 15 12 18 IV: 9 13 11 9 14 V: 6 11 12 9 14

Row reduction: the row minima are 4, 5, 7, 9, 6. Subtracting them gives I: 0 5 0 8 0 II: 10 6 15 0 3 III: 10 0 8 5 11 IV: 0 4 2 0 5 V: 0 5 6 3 8

Every column already has a zero. The zeros can be covered by four lines only: rows I and III and columns A and D. Four lines for a 5 × 5 matrix means no complete assignment yet. The smallest uncovered entry is 2 (row IV, column C). Subtract 2 from uncovered entries and add 2 where two lines cross: I: 2 5 0 10 0 II: 10 4 13 0 1 III: 12 0 8 7 11 IV: 0 2 0 0 3 V: 0 3 4 3 6

Now five independent zeros exist: II-D, III-B and V-A are forced, then IV-C and I-E. The assignment is I-E, II-D, III-B, IV-C, V-A, with total time 4 + 5 + 7 + 11 + 6 = 33 hours.

Restriction (IV cannot do C): put a very large cost M in cell IV-C and repeat. Row reduction gives the same matrix as before except IV: 0, 4, M − 9, 0, 5. The same four lines cover all zeros, and now the smallest uncovered entry is 3 (row II, column E). Subtract 3 from uncovered entries and add 3 at the crossings: I: 3 5 0 11 0 II: 10 3 12 0 0 III: 13 0 8 8 11 IV: 0 1 M−12 0 2 V: 0 2 3 3 5

Five independent zeros: III-B, V-A, IV-D, II-E and I-C. The assignment is I-C, II-E, III-B, IV-D, V-A with total time 4 + 8 + 7 + 9 + 6 = 34 hours, one hour more than the unrestricted minimum.

Answer: unrestricted, I-E, II-D, III-B, IV-C, V-A in 33 hours; with IV barred from C, I-C, II-E, III-B, IV-D, V-A in 34 hours.