(a) Let R be a field of real numbers and S, the field of all those polynomials f(x) ∈ R[x] such that f(0) = 0 = f(1). Prove that…

(a) Ideal S and the quotient ring R[x]/S

Let R denote the field of real numbers and S = {f ∈ R[x] : f(0)=0=f(1)}. First, S is nonempty since 0 ∈ S. If f,g ∈ S and a ∈ R, then (af+g)(0)=a f(0)+g(0)=0 and (af+g)(1)=a f(1)+g(1)=0, so af+g ∈ S. If h ∈ R[x] and f ∈ S, then (hf)(0)=h(0)f(0)=0 and (hf)(1)=h(1)f(1)=0, so hf ∈ S. Hence S is an ideal of R[x].

Define φ:R[x] → R×R by φ(f)=(f(0),f(1)). This is a ring homomorphism with kernel S. It is surjective: for any (a,b) ∈ R×R, the polynomial f(x)=a+(b−a)x satisfies f(0)=a, f(1)=b. Therefore, by the first isomorphism theorem, R[x]/S ≅ R×R. But R×R is not an integral domain, since (1,0)(0,1)=(0,0) while neither factor is zero. Hence R[x]/S is not an integral domain.

(b) Convergence of Σ nⁿxⁿ/n!

The general term is uₙ = nⁿ xⁿ/n!, n≥1, x>0. By the ratio test, uₙ₊₁/uₙ = [ (n+1)ⁿ⁺¹ xⁿ⁺¹/(n+1)! ] / [ nⁿ xⁿ/n! ] = x(1+1/n)ⁿ → xe. Thus the series converges if xe<1, i.e. 0<x<1/e, and diverges if xe>1, i.e. x>1/e.

At x=1/e the ratio test fails. Use Stirling’s formula: n! ∼ √(2πn)(n/e)ⁿ. Then uₙ = nⁿ(1/e)ⁿ/n! ∼ 1/√(2πn). Since ∑1/√n diverges (p=1/2≤1), the series diverges at x=1/e. Therefore the series converges for 0<x<1/e and diverges for x≥1/e.

(c) Transportation problem

Total supply = 11+13+19 = 43, total demand = 6+10+12+15 = 43, so the problem is balanced.

(c)(i) Initial BFS by Vogel’s method

By Vogel’s approximation method, initial penalties are: Rows: S₁=16−13=3, S₂=17−14=3, S₃=27−18=9. Columns: A=21−17=4, B=18−16=2, C=18−14=4, D=23−13=10. Largest penalty is D=10. Minimum in D is S₁D=13. Allocate min(11,15)=11 to S₁D. S₁ is exhausted; D demand left 4.

Remaining penalties: rows S₂=3, S₃=9; columns A=32−17=15, B=27−18=9, C=18−14=4, D=41−23=18. Largest is D=18. Minimum in D is S₂D=23. Allocate min(13,4)=4 to S₂D. D is satisfied; S₂ supply left 9.

Remaining rows S₂(9), S₃(19); columns A(6), B(10), C(12). Penalties: S₂=3, S₃=9; A=15, B=9, C=4. Largest A=15. Minimum in A is S₂A=17. Allocate min(9,6)=6 to S₂A. A is satisfied; S₂ supply left 3.

Remaining rows S₂(3), S₃(19); columns B(10), C(12). Penalties: S₂=4, S₃=9; B=9, C=4. Largest 9 (tie; choose column B). Minimum in B is S₂B=18. Allocate min(3,10)=3 to S₂B. S₂ is exhausted; B demand left 7.

Only S₃ remains. Allocate to cheaper C: min(19,12)=12 to S₃C, then remaining 7 to S₃B. Initial BFS: S₁D=11, S₂D=4, S₂A=6, S₂B=3, S₃C=12, S₃B=7; all other entries zero. Number of basic cells = 6 = 3+4−1, so it is nondegenerate. Initial cost = 11×13+4×23+6×17+3×18+12×18+7×27 = 796.

(c)(ii) Optimality test by MODI

Use MODI method on the basic cells S₁D, S₂D, S₂A, S₂B, S₃B, S₃C. Let u₁=0. Then v_D=13, u₂=10, v_A=7, v_B=8, u₃=19, v_C=−1. Reduced costs cᵢⱼ−(uᵢ+vⱼ) for nonbasic cells are: S₁A: 21−(0+7)=14 S₁B: 16−(0+8)=8 S₁C: 25−(0−1)=26 S₂C: 14−(10−1)=5 S₃A: 32−(19+7)=6 S₃D: 41−(19+13)=9 All are nonnegative, so the solution is optimal.

Optimal allocation: S₁D=11, S₂A=6, S₂B=3, S₂D=4, S₃B=7, S₃C=12. Optimal transportation cost = 11×13+6×17+3×18+4×23+7×27+12×18 = 143+102+54+92+189+216 = 796.