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Transcript

Math 500 – Intermediate Analysis Homework 8 – Solutions Fall 2014 23.1 For each of the following power series, find the radius of convergence and determine the exact interval of convergence. P∞ n3 n (d) n=1 3n x . Solution: Setting an = n3 , 3n we have that 1/n |an | n1/n = 3 3 and hence limn→∞ |an |1/n = 13 . It follows that the radius of convergence is R = 3. To determine the exact interval of convergence, we must check P the3 convergence of the series at x = ±3. When x = 3, the series is ∞ n=1 n th 3 which diverges by the n term test (limn→∞ n 6= 0). Similarly, at x = −3, the series again diverges by the nth term test, and hence the interval of convergence is precisely (−3, 3). P∞ (−1)n n x . (h) n=1 n2 4n Solution: Set an = (−1)n n2 4n and notice that |an |1/n = 1 2 (n1/n ) 4 and hence limn→∞ |an |1/n = 41 . Hence, the radius of convergence is R = 4. Moreover, at the endpoints we have convergence at x = 4 (by the integral test) and convergence at x = −4 (by the alternating series test). n n . 23.4 For n = 0, 1, 2, . . ., let an = 4+2(−1) 5 an+1 (a) Find lim sup |an |1/n , lim inf |an |1/n , lim sup an+1 , and lim inf an . an Solution: Notice that 6 n , if n is even 5 n an = 2 , if n is odd 5 1 In particular, the subsequential limits of the sequence |an |1/n are 65 and 25 . Thus, lim sup |an |1/n = 56 and lim inf |an |1/n = 52 . Moreover, we have 1 n an+1 , if n is even 3 an = 3n , if n is odd an+1 an+1 and hence lim inf an = 0 and lim sup an = +∞. P P∞ n (b) Do the series ∞ n=0 an and n=0 (−1) an converge? Explain. Solution: The easiest way to do this is to note that the sequence (an ) does n not converge to zero as n → ∞. Indeed, when n is even we get an = 56 which clearly does not converge to zero as n → ∞. Thus, neither of the two series listed converges. P n (c) Now, consider the power series ∞ n=0 an x with the coefficients an as above. Find the radius of convergence and determine the exact interval of convergence of the series. Solution: From part (a), the radius of convergence of the given power series is R = 56 . To determine the exact interval of convergence, we must now test the endpoints. At x = 56 , we have the series n ∞ X 2 + (−1)n 3 n=0 which diverges by the nth term test (CHECK!). Similarly, at x = − 56 , we have n ∞ X 2 + (−1)n n (−1) 3 n=0 which again diverges by the nth term test (CHECK!). Thus, the interval of convergence for this power series is − 65 , 56 . 24.2 For x ∈ [0, ∞), let fn (x) = nx . (a) Find f (x) = limn→∞ fn (x). Solution: For a given x0 ∈ [0, ∞), the numerical sequence fn (x0 ) = xn0 clearly converges to zero as n → ∞. Thus, the sequence of functions (fn ) converges pointwise to the function f (x) = 0 on [0, ∞). (b) Determine whether fn → f uniformly on [0, 1]. Solution: Let ε > 0 be given and notice that for any x ∈ [0, 1], we have |fn (x) − f (x)| = |f (x)| = 2 |x| 1 ≤ . n n Thus choosing N = 1ε (which, notice, does NOT depend on x), it follows for all n > N we have |fn (x) − f (x)| ≤ 1 < ε. n Since ε > 0 was arbitrary, we conclude that fn → f uniformly on [0, 1]. (c) Determine whether fn → f uniformly on [0, ∞). Solution: The convergence here is not uniform. If it were, then there would exist and N ∈ N such that |fn (x)| < 1 for all x ∈ [0, 1] and all n > N . However, this is clearly can’t be the case since |fN +1 (x)| > 1 for all x > N + 1. Thus, the we conclude that (fn ) does not converge to the function f uniformly on the set [0, ∞]. 1 24.3 Repeat exercise 24.2 for fn (x) = 1+x n. Solution: In class, we argued that xn → 0 pointwise on [0, 1), and that xn → 1 at x = 1. Moreover, for x > 1 we have xn → +∞. Thus, we see that the sequence (fn ) converges pointwise to the function f : [0, ∞) → R defined by 1, if x ∈ [0, 1) 1 , if x = 1 f (x) = 2 0, if x > 1. In particular, f is not continuous on either of the sets [0, 1] nor [0, ∞) and hence the convergence can not be uniform on either of these sets by Theorem 24.3. 24.7 Let fn (x) = x − xn for x ∈ [0, 1]. (a) Does the sequence (fn ) converge pointwise on the set [0, 1]? If so, give the limit function. Solution: Given a fixed x0 ∈ [0, 1), we know from class that xn0 → 0 and that xn0 → 1 for x0 = 1. Thus, the sequence of functions (fn ) converges pointwise to the function x, if x ∈ [0, 1) f (x) = x − 1, if x = 1 on the set [0, 1]. 3 (b) Does (fn ) converge uniformly on [0, 1]? Prove your assertion. Solution: No, the convergence is not uniform. Indeed, notice that although each function fn is continuous on [0, 1], the pointwise limit f is not continuous on [0, 1]. Thus, the convergence can not be uniform by Theorem 24.3. 24.12 Let (fn ) be a sequence of functions defined on a set S ⊂ R. Then (fn ) converges uniformly to a function f : S → R if and only if lim sup |f (x) − fn (x)| = 0. n→∞ x∈S Solution: First, suppose that limn→∞ supx∈S |f (x) − fn (x)| = 0 and let ε > 0 be given. Then by definition there exists an N ∈ R such that 0 ≤ sup |f (x) − fn (x)| < ε x∈S for all n > N . By the definition of the supremum then, it follows that we have |f (x) − fn (x)| < ε for all n > N and for all x ∈ S. That is, the sequence (fn ) converges to f uniformly on S. Conversely, suppose now that fn → f uniformly on a set S and let ε > 0 be given. Then by definition, given any ε > 0 there exists a N ∈ R such that |f (x) − fn (x)| < ε for all n > N and for all x ∈ S. Thus, for all n > N we have sup |f (x) − fn (x)| ≤ ε x∈S and hence the numerical sequence (supx∈S |f (x) − fn (x)|)∞ n=0 must converge to zero as n → +∞, i.e. we must have lim sup |f (x) − fn (x)| = 0 n→∞ x∈S as claimed. 24.15 Let fn (x) = nx 1+nx for x ∈ [0, ∞). 4 (a) Find f (x) = limn→∞ f (x). Solution: First, notice that fn (0) = 0 for all n ∈ N, and hence fn (0) → 0 as n → ∞. Moreover, for x > 0 we have by the limit theorems that x nx lim = lim 1 =1 n→∞ 1 + nx n→∞ +x n and hence the sequence of functions (fn ) converges pointwise to the function 0, if x = 0 f (x) = 1, if x > 0 on the set [0, ∞). (b) Does fn → f uniformly on [0, 1]? Explain. Solution: Notice that although each function fn is continuous on [0, 1], the pointwise limit f is not continuous on [0, 1]. Thus, the convergence can not be uniform by Theorem 24.3. (c) does fn → f uniformly on [1, ∞)? Explain. Solution: Notice that on [1, ∞) the pointwise limit f is continuous, so the argument above does not hold. Instead, we use the result from exercise 24.12. To this end, we use calculus to find the global extrema of the 1 functions fn . First, notice that fn0 (x) = (1+nx) 2 > 0 for all x ∈ [1, ∞) and hence the function fn is strictly increasing on the set [1, ∞). Moreover, n fn (1) = 1+n and limx→+∞ fn (x) = 1. Thus, given any n ∈ N and x ∈ [1, ∞) we have n . |fn (x) − f (x)| = |fn (x) − 1| ≤ 1 − n+1 n Since limn→∞ n+1 = 1 by the limit theorems, it follows that given any n < ε provided that ε > 0 there exists an N >∈ R such that 0 < 1 − 1+n n > N and hence |fn (x) − f (x)| < ε for all x ∈ [1, ∞). This directly verifies that the convergence is indeed uniform on [1, ∞). P P∞ n 25.6 (a) Show that if ∞ n=1 |an | < ∞, then the series n=1 an x converges uniformly on [−1, 1] to a continuous function. n Solution: Notice P∞that given any x ∈ [−1, 1], we have |an x | < |an |. Since the series n=1 |an | converges by hypothesis, it follows by the WeierP∞ strass M-Test that the series of functions n=1 an xn converges uniformly P on [−1, 1]. Since each partial sum kn=1 an xn is aP continuous on [−1, 1], it n follows by Theorem 24.3 that the uniform limit ∞ n=1 an x must also be continuous on [−1, 1] as claimed. 5 P xn (b) Does ∞ n=1 n2 represent a continuous P∞ 1 function on [−1, 1]? Solution: Since the series P∞ xn n=1 n2 is convergent by the integral test, the series of functions n=1 n2 converges uniformly to a continuous function on [−1, 1] by part (a). P xn 25.8 Show that ∞ n=1 n2 2n has radius of convergence 2, and that the series converges uniformly to a continuous function on [−2, 2]. Solution: First, notice that if an = n212n , then |an |1/n = 1 2 (n1/n ) 2 and hence limn→∞ |an |1/n = 21 . It follows therefore that the radius of convergence of this series is R = 2. Secondly, notice that the series is convergent at the endpoints x = ±2 and hence the interval of convergence is [−2, 2]. Now, to see this represents a continuous P function on [−2, 2] notice that |an xn | ≤ n12 for 1 all x ∈ [−2, 2]. Since the series ∞ is convergent by the integral test, the n=1P n2 n Weierstrass M-Test implies the series ∞ n=1 an x converges uniformly on [−2, 2] and hence the infinite series must represent a continuous function on [−2, 2] by Theorem 24.3. 6