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### mt2_sol

Course: MAT 320, Fall 2008
School: SUNY Stony Brook
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Word Count: 967

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Two Midterm - Solutions November 20, 2006 (1) (a) Which of the following series converge? 1 (i) n4 1n3 converges by comparison with n4 . Indeed, for large n we have n4 n n=1 n=1 1 2 2 3 &gt; 1 n4 , so n4 1n3 &lt; n4 . Because n=1 n4 , and so n4 converges, n4 1n3 will also n=1 n=1 2 converge. (You would need more eort or a limit comparison argument to formalize this, but this reasoning is good...

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Two Midterm - Solutions November 20, 2006 (1) (a) Which of the following series converge? 1 (i) n4 1n3 converges by comparison with n4 . Indeed, for large n we have n4 n n=1 n=1 1 2 2 3 > 1 n4 , so n4 1n3 < n4 . Because n=1 n4 , and so n4 converges, n4 1n3 will also n=1 n=1 2 converge. (You would need more eort or a limit comparison argument to formalize this, but this reasoning is good enoght to give you an answer. Note that all series here consist of positive terms.) n n (ii) n2 +1 diverges because the n-th term n2 +1 converges to 1. (We know that lim an = 0 is a n=1 necessary condition for convergence of the series an . n=1 1 n=1 3n +n 1 n=1 3n . 2 2 (iii) converges by comparison with More specically, 1 1 < n. +n 3 3n < 3n + n = 3n (1) 1 n=1 3n +n which converges. Since both series are made up of positive terms, this implies that the converges, because 31 converges (the latter is a geometric series). n=1 n (iv) 1 n=1 3 n diverges. Indeed, we know that 1 n=1 n . 1 n=1 np diverges for p 1. We could also go by comparison with We have n 3 1 1 1 1 n = = 3 3 n n n n=1 n n=1 1 (2) Since the lower series diverges, both series diverge. (b) Which of the following statements imply that (xn ) is a Cauchy sequence? (i) limn |xn xn+1 | = 0 This statement does not imply that (xn ) is Cauchy. Otherwise the n-th term test for series would be enough to guarantee convergence. For example, let xn = xn diverges. n 1 k=1 k . Then |xn xn+1 | |xn | + |xn+1 | < 2/n 0. However, we know Another example is provided by one of the practice questions: the sequence (xn ) with xn = n satises the condition limn |xn xn+1 | = 0, but the sequence is not Cauchy because it diverges to innity. (ii) For every > 0, there exists an N N such that for all m, p > N we have that |xm xp | < . This is the denition of Cauchy sequence. Clearly, then, it implies that (xn ) is Cauchy. (iii) |xn xn+1 | = Since 1 n=1 2n 1 2n . = 1, for any > 0, there must be some N N for which n=N 1 < 2n (3) Hence, m > n > N = m1 m1 m1 k =N |xm xn | = | k =n (xk+1 xk )| k =n |xk+1 xk | = k =n 1 2k 1 < 2k (4) Thus (xn ) is Cauchy. (iv) If (xn ) converges, then (xn ) is Cauchy. In fact, we had a theorem saying that a sequence is Cachy if and only if it is convergent. (This is one of the corollaries of completeness of R). (c) Which of the following functions are uniformly continuous on the given intervals? 2 (i) f (x) = x2 is uniformly continuous on [0, 1] because it is continuous to begin with, and [0, 1] is a closed bounded interval. (ii) f (x) = x2 is not uniformly continuous on [0, ). To show this, it is sucient to nd sequences (xn ), (yn ) in [0, ) such that 2 |xn yn | < 1/n but |x2 yn | > 1. n We can do this because the slope of x2 (in other words, the derivative) grows innity. to Of course, we dont know about derivatives yet. More formally, we can assume that xn yn and write 2 2 |x2 yn | = yn x2 = (yn + xn )(yn xn ) 2xn (yn xn ). n n (5) Thus, let us take xn = n and yn = n + 1 2n . Then clearly |xn yn | = 1 2n 1 < n , while 2 |x2 yn | 1. n (6) Thus x2 cannot be uniformly continuous. (iii) f (x) = x21 is uniformly continuous on R. This is because it is continuous, and its limits at +1 are zero. A very similar question was discussed in Homework 9, Problem 4; see solutions for HW 9 for more details. (iv) Consider the function dened on [1, 1] by f (0) = 0 and f (x) = x sin(1/x) otherwise. This function is continuous; a very similar question with cosine instead of sine was on one of he homeworks. Intuitively, you can assume continuity from the graph. Since f (x) is continuous on a closed bounded interval, it must be uniformly continuous. (2) Let (xn ) be a sequence in [1, 1]. Then the sequence yn = max{x1 , x2 , ...., xn } converges. Proof. The sequence is increasing (because yn = max{yn1 , xn } and bounded, so it converges by the Monotone Convergence Theorem. 3 (3) Let f, g be two functions dened for all x, except perhaps at x = 2. Suppose limx2 f (x) = 0 ( x) and limx2 g (x) = +. Then limx2 f (x) = 0. g Proof. Choose any > 0. Then for some 1 > 0, 0 < |x 2| < 1 = g (x) > 1 = |g (x)| > 1 Next, for some 2 > 0, we have (7) 0 < |x 2| < 2 = |f (x)| < Let = min{1 , 2 }. Then |f (x)| f (x) 0 = < < g (x) |g (x)| 1 (8) 0 < |x 2| < = (9) Hence indeed limx2 f (x) g ( x) exists and is equal to 0. (4) Recall the Bolzano-Weierstrass Theorem. Every bounded sequence has a convergent subsequence. Let f, g : [0, 1] R be two continuous functions. Suppose that for some sequence (xn ) in [0, 1] 1 we have f (xn ) = g (xn ) + n . Then f (y ) = g (y ) for some y [0, 1]. Proof. By Bolzano-Weierstrass, (xn ) must have a convergent subsequence xnk . We will write its limit as y . Note that 0 y 1 because 0 xnk 1. By continuity, 1 = g (y ) nk f (y ) = lim f (xnk ) = lim g (xnk ) + lim k k k (10) ( Since nk k goes to , it follows that 1 nk must go to 0.) (5) Let f : [0, +) (0, ) be a continuous function such that limx f (x) = . 4 Then f > M for some M > 0. (The statement may be false if we drop the requirement that limx f (x) = .) Proof. Since limx f (x) = , we can nd a number R > 0 such that x > R = f (x) 1 Let M = inf {f (x) | x [0, R]}, and let M = 1 2 (11) min{1, M }. Then M > 0 if M > 0. Indeed, M > 0 because f must attain its inmum M on the closed interval [0, R], and f is strictly positive. Thus, f > M > 0, as desired. For a counterexample in Part 2, any function with limx f (x) = 0 would do (it is then impossible to nd M > 0 such that f (x) remains greater than M for all x). 5
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