(a) Let m₁, m₂, …, mₖ be positive integers and d > 0 the greatest common divisor of m₁, m₂, …, mₖ. Show that there exist integers…

(a) Let d = gcd(m₁,m₂,…,mₖ). Use Bézout’s identity for two integers: if g = gcd(a,b), then there exist integers u,v such that g = ua + vb.

Prove by induction on k. For k=1, d = m₁, so d = 1·m₁. Assume the result for k−1 integers. Let g = gcd(m₁,m₂,…,mₖ₋₁). By induction, there exist integers y₁,y₂,…,yₖ₋₁ such that g = y₁m₁ + y₂m₂ + … + yₖ₋₁mₖ₋₁.

Now d = gcd(m₁,m₂,…,mₖ) = gcd(g,mₖ). By Bézout’s identity applied to g and mₖ, there exist integers r,s such that d = rg + smₖ.

Substituting, d = r(y₁m₁ + y₂m₂ + … + yₖ₋₁mₖ₋₁) + smₖ = (ry₁)m₁ + (ry₂)m₂ + … + (ryₖ₋₁)mₖ₋₁ + smₖ.

Take x₁ = ry₁, x₂ = ry₂, …, xₖ₋₁ = ryₖ₋₁, xₖ = s. Then d = x₁m₁ + x₂m₂ + … + xₖmₖ.

(b) The series is x⁴ + x⁴/(1+x⁴) + x⁴/(1+x⁴)² + x⁴/(1+x⁴)³ + …

For x > 0, put q = 1/(1+x⁴). Then the terms are x⁴qʳ, r = 0,1,2,… The n-th partial sum is Sₙ = x⁴(1 + q + q² + … + qⁿ) = x⁴(1 − qⁿ⁺¹)/(1 − q).

Since 1 − q = 1 − 1/(1+x⁴) = x⁴/(1+x⁴), we get Sₙ = (1+x⁴)(1 − qⁿ⁺¹) = 1 + x⁴ − (1+x⁴)qⁿ⁺¹ = 1 + x⁴ − 1/(1+x⁴)ⁿ.

Hence, for x > 0, lim Sₙ(x) = 1 + x⁴. At x = 0, every term is 0, so the sum is 0. Thus the pointwise limit is S(x) = 0 for x = 0, and S(x) = 1 + x⁴ for x > 0.

Now for x > 0, |Sₙ(x) − S(x)| = 1/(1+x⁴)ⁿ. At x = 0, this difference is 0. But as x → 0⁺, 1/(1+x⁴)ⁿ → 1. Therefore sup over [0,1] of |Sₙ(x) − S(x)| = 1 for every n. This does not tend to 0 as n → ∞. Hence the convergence is not uniform on [0,1].

Equivalently, the partial sums are continuous on [0,1], but the pointwise limit S(x) is discontinuous at x = 0, so uniform convergence is impossible.

(c) Assume first that f is monotonically increasing on [a,b]. Since f is monotone, it is bounded on [a,b], with f(a) ≤ f(x) ≤ f(b).

Let P = {a = x₀ < x₁ < x₂ < … < xₙ = b} be any partition of [a,b]. On each subinterval [xᵢ₋₁,xᵢ], because f is increasing, mᵢ = inf f = f(xᵢ₋₁), Mᵢ = sup f = f(xᵢ). Therefore U(P,f) − L(P,f) = sum over i=1 to n of (Mᵢ − mᵢ)Δxᵢ = sum over i=1 to n of (f(xᵢ) − f(xᵢ₋₁))Δxᵢ.

Let ‖P‖ = max Δxᵢ. Then U(P,f) − L(P,f) ≤ ‖P‖ · sum over i=1 to n of (f(xᵢ) − f(xᵢ₋₁)) = ‖P‖(f(b) − f(a)).

If f(b) = f(a), then f is constant, hence Riemann integrable. Otherwise f(b) − f(a) > 0. Given ε > 0, choose a partition P with ‖P‖ < ε/(f(b) − f(a)). Then U(P,f) − L(P,f) < ε. By Riemann’s criterion, f is Riemann integrable on [a,b].

If f is monotonically decreasing, apply the same argument to −f, which is increasing. Hence a monotonic function on [a,b] is always Riemann integrable. Therefore f is Riemann integrable in [a,b].

(d) Factor the denominator: 2z² − 5z + 2 = (2z − 1)(z − 2) = 2(z − 1/2)(z − 2). So the poles are z = 1/2 and z = 2.

The curve is c(t) = e⁴πit, 0 ≤ t ≤ 1. This is the unit circle traversed twice counterclockwise. The point z = 1/2 lies inside the unit circle, so its winding number is 2. The point z = 2 lies outside the unit circle, so its winding number is 0.

Compute the residue at z = 1/2: Res(1/(2z² − 5z + 2), 1/2) = lim as z → 1/2 of (z − 1/2)/(2(z − 1/2)(z − 2)) = 1/(2(1/2 − 2)) = 1/(2(−3/2)) = −1/3.

By the residue theorem, integral along c of dz/(2z² − 5z + 2) = 2πi · 2 · (−1/3) = −4πi/3.

Therefore the value of the contour integral is −4πi/3.

(e) Use the Hungarian method. The cost matrix is:

A: 10 5 13 15 16 B: 3 9 18 13 6 C: 10 7 2 2 2 D: 7 11 9 7 12 E: 7 9 10 4 12

Row minima are 5,3,2,7,4. Subtract them from the respective rows. The reduced matrix is:

A: 5 0 8 10 11 B: 0 6 15 10 3 C: 8 5 0 0 0 D: 0 4 2 0 5 E: 3 5 6 0 8

The column minima are all 0, so the matrix remains unchanged. A maximum matching of zeros has size 4, for example A-II, B-I, C-III, D-IV. Thus no perfect zero assignment exists.

A minimum cover of all zeros is given by columns I, II, IV and row C. The uncovered entries lie in rows A, B, D, E and columns III, V. The smallest uncovered entry is 2, at D-III. Subtract 2 from all uncovered entries and add 2 at intersections of covered rows and covered columns. The new matrix is:

A: 5 0 6 10 9 B: 0 6 13 10 1 C: 10 7 0 2 0 D: 0 4 0 0 3 E: 3 5 4 0 6

Now a perfect zero assignment exists: A → II, B → I, C → V, D → III, E → IV.

Using the original matrix, the costs are: A-II = 5, B-I = 3, C-V = 2, D-III = 9, E-IV = 4.

Hence the minimum total processing time is 5 + 3 + 2 + 9 + 4 = 23 hours.

Optimal assignment: A → II, B → I, C → V, D → III, E → IV; minimum total processing time = 23 hours.