Mathematics Homework 9: Proofs about Continuous Functions and Geometric Constructions, Assignments of Mathematics

Download Mathematics Homework 9: Proofs about Continuous Functions and Geometric Constructions and more Assignments Mathematics in PDF only on Docsity! Math 3110 Spring ’09 Homework 9: Selected Solutions [12-3]: Claim: Let f : [−a, a] → R be continuous, with f(0) > f(−a), f(a). In this case there exist x1, x2 ∈ [−a, a] such that a horizontal chord of length exactly a joins (x1, f(x1)) to (x2, f(x2)). Proof Let g : [−a, 0] → R be defined by g(x) = f(x + a) − f(x). The function g(x) is clearly continuous on its domain and g(x) = 0 exactly when a horizontal chord of length a joins f(x) to f(x + a). Since g(−a) = f(0)− f(−a) > 0 and g(0) = f(a)− f(0) < 0, Bolzano’s Theorem tells us that g(x) = 0 for some x ∈ [−a, 0], or equivalently that there is a horizontal chord of length exactly a joining two points on the graph of f([−a, a]). [12.5]: Claim: An equilateral triangle can be inscribed in any smooth convex closed curve C in R2. Proof [See Figure 2 on last page.] Choose any point P on C. Because C is both smooth and convex, there is a tangent line TP to C at P whose only intersection with C occurs at P . Let L1,L2 : [0, 1] × ( 0, 2π3 ) → R2 describe line segments originating at P , making angles θ and( θ + π3 ) respectively with TP and ending at their points of first intersection Q(θ) = L1(1, θ) and R(θ + π3 ) = L2(1, θ) with C. Due to the fact that C is smooth closed and convex, Qθ = Q(θ) and R(θ+π 3 ) = R(θ + π3 ) are continuous functions of θ, as are `1(θ) = |PQθ| and `2(θ) = ∣∣∣PR(θ+π 3 ) ∣∣∣. Moreover ∠QθPR(θ+π 3 ) = π3 for all θ ∈ ( 0, 2π3 ) by construction, so ∆QθPR(θ+π 3 ) is equilateral as soon as `1(θ) = `2(θ). Let F (θ) = `1(θ)− `2(θ). F is continuous on ( 0, 2π3 ) since `1 and `2 are. The properties of C ensure that lim θ→0+ `1(θ) = 0, lim θ→0+ `2(θ) = c2 > 0 and lim θ→( 2π3 ) − `2(θ) = 0, lim θ→( 2π3 ) − `1(θ) = c1 > 0. So there is clearly a small ² > 0 for which F (²) < 0 and F (2π3 − ²) > 0. Bolzano’s Theorem now guarantees the existence of a θ̂ ∈ (0, 2π3 ) for which F (θ̂) = 0 and thus the existence of an equilateral triangle ∆Qθ̂PR(θ̂+π 3 ) inscribed in C. [13.2-1]: Claim: (a) If x(t), y(t) are continuous on I = [a, b], then {(x(t), y(t)) : t ∈ I} is contained in a large square centered at the origin, however (b) this statement need not hold if we consider the curve {(x(t), y(t)) : t ∈ I ′} for x(t), y(t) continuous on I ′ = (a, b). Proof (a) Since I is compact and x(t), y(t) are continuous on I, both x(t) and y(t) are bounded on I and there is a K such that |x(t)|, |y(t)| ≤ K for all t ∈ I. So for any ² > 0, a square of sidelength 2K + ² centered at the origin will contain the curve {(x(t), y(t)) : t ∈ I}. For noncompact I ′ = (a, b) this need not be the case. Counterexamples abound; an obvious one is a = 0, b = 1, x(t) = t, y(t) = 1 t . 1 [13.5-5]: Claim: f(x) = lnx is uniformly continuous on [1,∞). Proof We are given at the outset that f is continuous at x = 1, so for any ² > 0 there is δ1(²) such that |x− 1| < δ1(²) ⇒ |lnx− ln 1| = |ln x| < ². An equivalent way to say this is that for h > −1, if |h| < δ1(²) then |ln(1 + h)− ln 1| < ². Now use the same ² and take any α > 1. Again let h > −α be a positive or negative increment. Notice that |ln (α + h)− lnα| = ∣∣∣∣ln ( 1 + h α )∣∣∣∣ < ² when ∣∣∣∣ h α ∣∣∣∣ < δ1(²). Since α > 1, this gives |ln (α + h)− ln α| < ² whenever |h| < δ1(²) < αδ1(²) and proves the uniform continuity of lnx on [1,∞). [13-2]: Claim: Let I be a compact interval and f : I → R be a continuous function. The Maximum Theorem asserts that M = sup x∈I f(x) ∈ I. Applying the Boundedness Theorem to g(x) = 1 M − f(x) yields an indirect proof of Maximum Theorem. Proof Suppose that M /∈ I. Then g(x) = 1 M − f(x) is continuous on the compact interval I and the Boundedness Theorem therefore guarantees that g will be bounded on I. So there must be some ² > 0 such that M − f(x) > ² for all x ∈ I. But this means that M − ² < M is an upper bound for f on I violating our assumption that M is the least upper bound, ie. the supremum of f on I. [Extra Problem # 2]: Claim: Let x̂ = 2 3π . The function F (x) =    0 x = 0 x sin 1 x − x x ∈ (0, x̂) −2x x = x̂∣∣∣∣(x− x̂) sin ( 1 x− x̂)− (x− x̂) ∣∣∣∣− 2x̂ x ∈ (x̂, 2x̂) − ( 2x̂ 1− 2x̂ ) (x− 2x̂) x ∈ [2x̂, 1] is continuous on all of I = [0, 1], constant on no subinterval I ′ ⊂ I and achieves both its minimum and maximum values infinitely many times on I. Proof x sin 1 x − x and ∣∣∣∣(x− x̂) sin ( 1 x− x̂)− (x− x̂) ∣∣∣∣ − 2x̂ have removable discontinuities at x = 0 and x = x̂ respectively. This is easily corrected by setting F (0) = 0 and F (x̂) = −2x̂, rendering F continuous on all of I = [0, 1]. See Figure 1 below for evidence that F has infinitely many maxima and minima; this follows from the fact that ( sin 1 x (1− x) ) is non-positive and accumulates infinitely many zeros near x = 0. 2