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Math 521, Lecture 2 - Spring 2009
Advanced Calculus
First Midterm Exam
Exercise 1: Course exercise (1) 20 points. Consider the set A := { 1 − (^1) n , n ∈ N∗}. Prove that sup A exists and is equal to 1. (2) 5 points. State the Bolzano-Weierstrass theorem.
Exercise 2 Let (xn) be the sequence defined by xn = (−1)n^ + (^1) n for every n ∈ N∗. (1) 10 points. Is (xn) convergent? Explain. (2) 5 points. Prove that − 1 and 1 are points of accumulation of (xn).
Exercise 3 Let (xn), (yn), (zn) be a three sequences satisfying yn ≤ xn ≤ zn for every n ∈ N. (1) 15 points. Assume that (yn) and (zn) converge to the same limit l ∈ R. Using the definition of the convergence, prove that (xn) converges to l (hint: remember that |xn − l| < ε is the same as l − ε < xn < l + ε). (2) 10 points. If (yn) and (zn) are convergent, prove or disprove by an example that (xn) is convergent. (3) 5 points. If (yn) and (zn) are convergent, prove that (xn) has a point of accumulation.
Exercise 4: (A fix point theorem) We define I := [0, 1]. And let f : I → I be an increasing function (it means that if x ≤ y then f (x) ≤ f (y)). Consider the set A := {x ∈ I, f (x) ≤ x}.
(1) 5 points. Prove that 1 ∈ A. (2) 5 points. Find a lower bound for A. (3) 5 points. Prove that inf A exists and that 0 ≤ inf A. Let us denote α := inf A. (4) 5 points. Prove that if x ∈ A then f (x) ∈ A. (5) 10 points. Prove that f (α) ≤ α (hint: assume by contradiction that we have α < f (α)). (6) 5 points. Prove that f (α) = α (hint: use (3) and (4)).
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Exercise 1: Course exercise (1) The proof of this is in your lecture notes; but here is a little bit different proof. The set A := { 1 − (^1) n , n ∈ N∗} is nonempty and since 1 − (^1) n < 1 for every positive integer n, it is bounded above by 1. So sup A exists and by definition sup A ≤ 1. To prove that sup A = 1, let ε > 0 , and let us find and element a ∈ A such that 1 − ε < a ≤ 1. But if this is wrong, then for any n ∈ N∗, we have 1 − (^) n^1 ≤ 1 − ε, and so (^1) n ≥ ε for any n ∈ N∗; this is a contradiction. (2) In your lecture notes.
Exercise 2 (1) You can use the second question to prove (1), as a convergent sequence has only one point of accumulation. But the easiest way to solve (1), was to use a result of Homework 3. (xn) is the sum of a divergent sequence and a convergent sequence; according to Homework 3, (xn) is not convergent. (2) (x 2 n) is a subsequence of (xn) satisfying x 2 n = 1 + (^21) n and thus converges to 1. And (x 2 n+1) is a subsequence of (xn) satisfying x 2 n = −1 + (^2) n^1 +1 and thus converges to − 1.
Exercise 3 Let (xn), (yn), (zn) be a three sequences satisfying yn ≤ xn ≤ zn for every n ∈ N. (1) Fix ε > 0. Since (yn) converge to l ∈ R, there exists n 1 ∈ N, such that for every n ≥ n 1 , |yn − l| < ε. And since (zn) converge to l ∈ R, there exists n 2 ∈ N, such that for every n ≥ n 2 , |zn − l| < ε. So taking no = max{n 1 , n 2 }, if n ≥ n 0 then |yn − l| < ε and |zn − l| < ε; wich is equivalent to l − ε < yn < l + ε and l − ε < zn < l + ε. Then we have the following inequalities: l − ε < yn ≤ xn ≤ zn < l + ε if n ≥ n 0. Thus l − ε < xn < l + ε whenever n ≥ n 0. (2) If (yn) and (zn) are convergent then (xn) can diverge. Take for instance yn = − 1 and zn = 1 and xn = (−1)n. (3) If (yn) and (zn) are convergent, then they are bounded, so is (xn). Then by Bolzano-Weierstras theorem, (xn) has a point of accumulation.
Exercise 4: (A fix point theorem) In this exercise, the most common errors were to say that inf A ∈ A without any proof and that I ⊆ A.
(1) Since f : I → I, we have f (1) ≤ 1 and thus 1 ∈ A. (2) 0 is a lower bound for A. (3) Then inf A exists because A is nonempty by (1), and bounded by below by 0. Moreover by defini- tion, 0 ≤ inf A. (4) Assume that x ∈ A. It means that f (x) ≤ x; and using the fact that f is increasing, we have f (f (x)) ≤ f (x). This proves f (x) ∈ A. (5) Assume by contradiction that α < f (α). Then, there exists an elements x ∈ A such that α ≤ x < f (α); then since f is increasing, we have f (α) ≤ f (x) < f (f (α)). So x < f (α) ≤ f (x) wich contradicts the fact that x ∈ A (f (x) ≤ x). (6) According to (5), α ∈ A and by (4) this implies f (α) ∈ A. Accoding to this and to the fact that α is the greatest lower bound, we have f (α) = α.
