Math 317 HW #6 Solutions

Solutions to Take-Home Part of Math 317 Exam #1

where the first inequality is an application of the triangle inequality, the second follows from (1) and (2), and the third from the choice of m. Therefore, since our choice of > 0 was arbitrary, we conclude that the subsequence (xnm)→ b. A similar argument using the sequence (zm) given by zm = inf{xm, xm+1, . . .} and the version of Lemma 1.3.7 suitable for infima (see Exercise 1.3.2, which yo...

Math 317 HW #11 Solutions

Proof. Suppose f is differentiable and f ′ is continuous on [a, b]. Then, since [a, b] is compact, f ′ is bounded. In other words, there exists M > 0 such that |f ′(x)| ≤ M for all x ∈ [a, b]. Then, for any x, y ∈ [a, b], the Mean Value Theorem guarantees that there exists c between x and y such that ∣∣∣∣f(x)− f(y) x− y ∣∣∣∣ = |f ′(c)| ≤M. Since the choice of x and y was arbitrary, we see that ...

Math 317 HW #2 Solutions

Proof. If b ∈ B, then b is a lower bound for A, meaning that b ≤ a for all a ∈ A. Therefore, every element of A is an upper bound for B, so B is bounded above and thus, by the Axiom of Completeness, has a least upper bound supB. We want to see that supB satisfies the two conditions for being the greatest lower bound of A. i. If supB is not a lower bound for A, then there exists a ∈ A such that ...

Math 317 HW #5 Solutions

Proof. I intend to use the Monotone Convergence Theorem, so my goal is to show that (xn) is decreasing and bounded. To do so, I will prove by induction that, for any n ∈ {1, 2, 3, . . .}, 0 < xn+1 < xn < 4. Base Case When n = 1, we have that x1 = 3 and x2 = 1 4−3 = 1, so 0 < x2 < x1. Inductive Step Suppose 0 < xk+1 < xk < 4; our goal is to show that this implies that 0 < xk+2 < xk+1 < 4. Since ...

Math 215 HW #4 Solutions

Math 215 HW #2 Solutions