MATH 210 Exam 2 Solution, Fall 2016

Math 210, Introduction to Analysis- Fall 2016-2017 Quiz2 Solutions Question 1. a) Let p0 = (x0, y0) ∈ R2, we need to prove f is continuous at any such point. i.e show: ∀ϵ > 0, ∃δ > 0 s.t d1((x, y), (x0, y0)) < δ =⇒ |f(x, y) − f(x0, y0)| < ϵ. However |f(x, y) − f(x0, y0)| = | cos x2 − cos x2 0 − ( � |y| − � |y0|)| ≤ | cos x2 − cos x2 0| + | � |y| − � |y0|| ≤ |x2 − x2 0| + ����� � |y| − � |y0| ���� 2 | ≤ |x2 − x2 0| + �����( � |y|)2 − ( � |y0|)2 ���� since | � |y| − � |y0|| ≤ � |y| + � |y0|. Then |f(x, y) − f(x0, y0)| ≤ |x − x0||x + x0| + �����|y| − |y0| ���� ≤ |x − x0||x + x0| + � |y − y0| Now if both |x− x0| and |y − y0| are < δ then |x| < |x0| +δ < |x0| +1 if we take δ < 1. Thus |f(x, y) − f(x0, y0)| ≤ δ(|x0| + 1|) + √ δ ≤ √ δ(|x0 + 1|) + √ δ = √ δ(|x0| + 2) since δ < 1. Choose δ = 1 2 min{1, ( ϵ |x0|+2)2} to get the needed result: ∀ϵ > 0, ∃δ = 1 2 min{1, ( ϵ |x0|+2)2} > 0 s.t d1((x, y), (x0, y0)) < δ =⇒ |f(x, y) − f(x0, y0)| < ϵ. b) i. A = f −1({0}) and B = f −1((−∞, −1]) and both {0} and (−∞, −1] are closed in R and f is continuous thus A and B are closed in (R2, d1) . ii. U = f −1((−4, 3)), f is continuous and (−4, 3) is open in R then U is open in (R2, d1). Question 2. a) This is an alternating series �(−1)nbn where bn = √n + 1−√n = 1 √n+1+√n is a decreasing sequence to zero. Thus, by the alternating series test this is a convergent series. It is not however absolutely convergent because � |(−1)nbn| = �(√n + 1 − √n) = � 1 √n+1+√n ≥ � 1 n+1+n ≥ � 1 3n which is a divergent series by the p-series test. b)i. ���� cn+1 cn ���� = ���� (3n+3)! ((n+1)!)3 (n!)3 (3n)! ���� = (3n+3)(3n+2)(3n+1) (n+1)3 −→ 27 as n −→ ∞ thus R = 1 27. ii. lim n−→∞ ���� cn+1 cn ���� = lim ���� �(2 − 3i)n+1 √3n + 3 − (n + 1)2 � 1 (2−3i)n √ 3n − (n)2 ���� = lim ���� (2 − 3i)n+1 (2 − 3i)n ���� = |2−3i| = √ 13 as |2−3i|n = √ 13 n will dominate other terms since √ 13 > 1 Thus R = 1 √ 13. Question 3. a) Refer to lecture notes. b) i. cos is differentiable on (a, b) thus ∃c ∈ (a, b) s.t cos a − cos b = (cos)′(c)(b − a) = sin(c)(b − a) by the MVT. Hence | cos a − cos b| = | sin(c)||b − a| ≤ |b − a| since | sin x| ≤ 1 ∀x ∈ R. ii. Let f(x) = cos(πx2) − x3 − x2 − 8x then f is continuous on [0, 1], and f(0) = 1 > 0, f(1) = −11 < 0 and 0 ∈ (f(1), f(0)) thus by the IVT ∃c ∈ (0, 1) s.t f(c) = 0 i.e c is a solution for the equation in question. Let’s show uniqueness now: Suppose there are c1 ̸= c2 ∈ (0, 1) s.t f(c1) = 0 = f(c2), then by the MVT (or Rolle’s 1 2 which is a special case of MVT) ∃d ∈ (c1, c2) s.t f ′(d) = f(c1)−f(c2) c1−c2 = 0. However, f ′(x) = −2πx sin(πx2) − 3x2 − 2x − 8 ≤ 2π − 8 < 0 ∀x ∈ (0, 1). Hence f ′ cannot be zero. This is a contradiction. Question 4. a) We need to show that ∀ϵ > 0, ∃δ > 0, s.t ∀f, g ∈ C(X), d(f, g) < δ =⇒ |T(f) − T(g)| < ϵ. However, ∀ϵ > 0, ∃δ = ϵ > 0, s.t ∀f, g ∈ C(X), d(f, g) < δ =⇒ sup x∈X |f(x) − g(x)| < δ = ϵ just by the definition of the distance d, Hence |f(x0) − g(x0)| ≤ sup x∈X |f(x) − g(x)| < ϵ =⇒ |T(f) − T(g)| < ϵ just by the definition of T.Thus T is uniformly continuous on C(X). b) A = T −1 � (0, ∞) � , and since T is continuous(uniform continuity =⇒ continuity), and (0, ∞) is open in R then A is open in � C(X), d � .