(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) Let M = sup f(x) and m = inf f(x) for x∈[a,b]. Since f is real-valued and bounded, M and m are finite and M≥m. The oscillation is defined as M-m. Put S = {|f(x₁)-f(x₂)| : x₁, x₂∈[a,b]}. For any x₁, x₂, m≤f(x₁), f(x₂)≤M, so |f(x₁)-f(x₂)|≤M-m. Thus sup S≤M-m. Conversely, let ε>0. By the definitions of supremum and infimum, choose x₁, x₂∈[a,b] with f(x₁)>M-ε/2 and f(x₂)<m+ε/2. Then |f(x₁)-f(x₂)|>M-m-ε, so sup S≥M-m-ε. Since ε>0 is arbitrary, sup S≥M-m. Therefore sup S=M-m. The oscillation is the supremum of S.

(b) Use the Maclaurin series, valid for all 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 is 1 at z=0, so by continuity it is nonzero in a neighbourhood of 0; hence z-sin z has a zero of exact order 3 at 0. Since eᶻ is analytic and e⁰=1≠0, z=0 is a pole of order 3. For the principal part, write R(z)=(1-z²/20+z⁴/840-...)⁻¹. Let R(z)=1+c z²+O(z⁴). Multiplying by the bracket and equating the z² coefficient gives c=1/20, so R(z)=1+z²/20+O(z⁴). Therefore, in 0<|z|<r, where r is the distance to the nearest other zero of z-sin z, or ∞ if none, eᶻ/(z-sin z)=6z⁻³(1+z+z²/2+O(z³))(1+z²/20+O(z⁴)) =6z⁻³(1+z+(1/2+1/20)z²+O(z³)) =6z⁻³+6z⁻²+(66/20)z⁻¹+O(1) =6z⁻³+6z⁻²+(33/10)z⁻¹+O(1). Principal part: 6z⁻³+6z⁻²+(33/10)z⁻¹.

(c) Use the Hungarian method. Row, column, and Hungarian adjustments preserve optimality, since each complete assignment changes by a constant. Costs are in hours. Row minima are 4, 5, 7, 9, 6. After row reduction:

• 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 Column minima are all 0. The zeros are I-A, I-C, I-E, II-D, III-B, IV-A, IV-D, V-A. No five independent zeros exist. A maximum matching of size 4 is I-C, II-D, III-B, IV-A. From unmatched row V, the alternating tree reaches rows V, IV, II and columns A, D; hence the minimum cover is rows I, III and columns A, D. Since 4<5, let Δ be the smallest uncovered entry. Uncovered entries are in rows II, IV, V and columns B, C, E; their minimum is Δ=2 (IV,C). Subtract Δ from uncovered entries and add Δ to intersections of the covering lines. New reduced matrix:
• 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 choose five independent zeros: I-E, II-D, III-B, IV-C, V-A. Since all reduced entries are nonnegative, a zero reduced total is optimal. Original cost: 4+5+7+11+6=33. Minimum assignment: I-E, II-D, III-B, IV-C, V-A; total minimum time = 33 hours.

For the restriction, forbid IV-C by replacing 11 with M=100 hours. Every feasible assignment without this cell has cost at most 5×20=100 hours, while any assignment using IV-C has cost at least 100+4+5+7+6=122 hours, so the modified problem has the same required optimum. Row reduction gives:

• I: 0, 5, 0, 8, 0
• II: 10, 6, 15, 0, 3
• III: 10, 0, 8, 5, 11
• IV: 0, 4, 91, 0, 5
• V: 0, 5, 6, 3, 8 Column minima are 0. The same minimum cover, rows I, III and columns A, D, covers all zeros. The smallest uncovered entry is Δ=3 (II,E). Subtract Δ from uncovered entries and add Δ to intersections. New reduced matrix:
• I: 3, 5, 0, 11, 0
• II: 10, 3, 12, 0, 0
• III: 13, 0, 8, 8, 11
• IV: 0, 1, 88, 0, 2
• V: 0, 2, 3, 3, 5 Five independent zeros are I-C, II-E, III-B, IV-D, V-A. The forbidden cell IV-C is not used, so the restriction is satisfied. Original restricted cost: 4+8+7+9+6=34. With IV not assigned C, minimum assignment: I-C, II-E, III-B, IV-D, V-A; total minimum time = 34 hours.