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Course: MATH 5615, Fall 2011
School: Minnesota
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5615H: Math Introduction to Analysis I. Homework #10. Problems and Solutions. #1. The sequence {an } is defined by a1 = 0, and an+1 = 2an /2 Fall 2011 for n = 1, 2, 3, . . . . Prove that {an } is convergent and find its limit. You can use without proof the fact that the function y = f (x) = ax , where a &gt; 0, a = 1, is strictly convex, i.e. f (tx1 + (1 - t)x2 ) &lt; t f (x1 ) + (1 - t) f (x2 ) for all...

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5615H: Math Introduction to Analysis I. Homework #10. Problems and Solutions. #1. The sequence {an } is defined by a1 = 0, and an+1 = 2an /2 Fall 2011 for n = 1, 2, 3, . . . . Prove that {an } is convergent and find its limit. You can use without proof the fact that the function y = f (x) = ax , where a > 0, a = 1, is strictly convex, i.e. f (tx1 + (1 - t)x2 ) < t f (x1 ) + (1 - t) f (x2 ) for all x1 < x2 and 0 < t < 1. Solution. By induction, a1 = 0 < a2 = 1 < a3 < < an < < 2 for all n = 1, 2, 3, . . .. By Theorem 3.14 and its proof, there exists L = lim an = sup{an } 2. Moreover, an+1 = 2an /2 = = an+1 2L/2 , L 2an /2 for n = 1, 2, 3, . . . L 2L/2 , L 2L/2 = L = 2L/2 2. We claim that L = 2, i.e. the case L = 2L/2 < 2 is impossible. Indeed, if this is the case, then the equation x = 2x/2 has at least 3 distinct solutions x1 = L < x0 = 2 < x2 = 4. In other words, the graph of a strictly convex function y = f (x) := 2x/2 intersects the line y = x at 3 distinct points, which is impossible. Formally, we can write x0 = tx1 + (1 - t)x2 , where t := x2 - x0 x2 - x1 if x0 (x1 , x2 ). By the property of strict convexity, the desired contradiction follows: x0 = f (x0 ) = f (tx1 + (1 - t)x2 ) < t f (x1 ) + (1 - t) f (x2 ) = tx1 + (1 - t)x2 = x0 . #2. Investigate the behavior (convergence or divergence) of an if n+1- n n (a) an = n + 1 - n; (b) an = ; (c) an = n n - 1 . n (a). Solution. Since the partial sums n n sn := k=1 ak = k=1 k+1- k = n + 1 - 1 +, the series an diverges. (b). Note that 0 < an := By Theorem 3.28, the series n+1- n n = n 1 1 < cn := 3/2 . n n+1+ n an also converges. cn converges. Then by Theorem 3.25(a), 1 (c). We know that n n - 1 0 (Theorem 3.20(c)). Therefore, there exists a natural N0 such that 0 < n n - 1 < 1/2 for all n N0 . Then n 0 < an := n n - 1 2-n for all n N0 , and an converges by Theorems 3.20(e) and 3.25(a). #3. Using the fact that n=1 1 2 = , n2 6 find n=1 (-1)n-1 . n2 Solution. We have n=1 (-1)n-1 1 1 1 1 = 2 - 2 + 2 - 2 + 2 n 1 2 3 4 = = = 1 1 1 1 1 1 + 2 + 2 + 2 + - 2 + 2 + 2 2 1 2 3 4 2 4 1 1 1 1 1 1 1 + 2 + 2 + - + 2 + 2 + 12 2 3 2 12 2 3 1 1 1 1 2 1- + 2 + 2 + = . 2 12 2 3 12 #4. Find the radius of convergence of each of the following power series: (a) n2 z n , (b) 2n n z , n! (c) 2n n z , n2 (d) n3 n z . 3n Solution. For each of these series, cn z n , there is a limit := lim(cn+1 /cn ) 0. We can use Theorem 3.34 with an := cn z n , which satisfy n lim an+1 cn+1 z n+1 = lim = |z| . n an cn z n This implies that the radius of convergence of cn z n is R = 1/. Alternatively, by Theorem 3.37, := lim(cn+1 /cn ) = lim n cn , and the same equality R = 1/ follows from Theorem 3.39. In our particular cases, we have (a) = 1, R = 1; (b) = 0, R = +; 1 (c) = 2, R = ; 2 1 (d) = , R = 3. 3 2
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Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Homework #11. Problems and Solutions. #1. (12 points). Find n S := . 2n n=1Fall 2011Solution. Substituting n = m + 1, we get n m+1 ( 1 )m 2S = = =S+ = S + 2, 2n-1 m=0 2m 2 n=1 m=0so that S = 2. #2. (10 points).
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Homework #12. Problems and Solutions.Fall 2011#1. Let aj and bj,k be real numbers defined for all j, k = 0, 1, 2, . . ., such that|aj | A = const &lt; ,j=0|bj,n | B = const &lt; for all j, n;and lim bj,n = 0 for eac
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Homework #13. Problems and Solutions.Fall 2011#1. If f is a continuous mapping of a metric space X into a metric space Y , prove that f (E) f (E) for every set E X. (E denotes the closure of E). Show, by an exampl
Minnesota - MATH - 5615
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Minnesota - MATH - 5615
Math 5615H. November 16, 2011. Midterm Exam 2. Problems and Solutions. Problem 1. (10 points). Let A1 , A2 , A3 , . . . be subsets of a metric space (X, d). If B :=k=1Ak ,prove that B k=1Ak ,where B and Ak denote the closures of B and Ak . Show, b
Minnesota - MATH - 5615
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Minnesota - MATH - 8601
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Minnesota - MATH - 8601
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Minnesota - MATH - 8601
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Minnesota - MATH - 8601
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Minnesota - MATH - 8601
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Minnesota - MATH - 8601
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Minnesota - MATH - 8601
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Minnesota - MATH - 8601
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Minnesota - MATH - 8601
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Minnesota - MATH - 8601
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Minnesota - MATH - 8601
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Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS.Fall 2010Homework #3. Problems and Solutions. #1. Let f be a real function on R1 . The image and the inverse image of a subset A R1 under f are correspondingly f (A) = cfw_y : y = f (x) for some x A, f -1 (A) = cfw_x : f (x) A.
Minnesota - MATH - 8601
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Minnesota - MATH - 8601
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Minnesota - MATH - 8601
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Minnesota - MATH - 8601
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Minnesota - MATH - 8601
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