mafa11e1s
2
Modern Analysis EXAM 1. F rida y , Septe mber 23, 2011 INSTRUCTIONS • You do not have to copy the statements of the exercis es. But make sure you clearly indica te where your answer to any given exercise begins, and where it ends. • The ma in thing I’d like to verif y with thi s exa m is whether you are able to wri te a proo f correc tly . So, please, write proofs correctly. Don’t improvise, don’t invent strange things, don’t fly off tangents. Don’t use undefi ned terms. What you write should be clear and coherent. 1. Let A, B be sets of real numbers. Assume they are bounded below and non-empty. Prove inf(A + B) = inf A + inf B. Do not waste time proving separately that the set A + B is non- empt y or bounded below. Y ou may accept that it is. The set A + B is defined by A + B = { a + b : a ∈ A, b ∈ B}. Solution. Let x ∈ A + B ; then x = a + b for some a ∈ A, b ∈ B. Bec ause infi ma are lower bounds, a ≥ inf A, b ≥ infB, thus x = a + b ≥ inf A + inf B. Sinc e x was an arbitrry element of A + B , this proves that inf A + inf B is a lower bound of A + B ; in particular, inf( A + B ) ≥ inf A + inf B. For the converse inequality, let > 0 be a positive number. Then inf A < inf A + 2 thus there exists a ∈ A such that a < inf A + 2 . Similarly, there exists inf A < inf A + 2 thus there exists b ∈ B such that b < inf B + 2 . Then inf(A + B) ≤ a + b < inf A +inf B + ; since > 0 is arbitrary, this implies inf( A + B) ≤ inf A +inf B. Having proved inf(A + B) ≥ inf A + inf B and inf(A + B) ≤ inf A + inf B, we are done. 2. Let {a n } be a sequence of real numbers. Assume {a n } is increasing and bounded above. Prove: lim n→∞ a n = sup{a n : n ∈ N}. Solution. Because the sequ ence is boun ded above, the non empty set {a n : n ∈ N} is bounded above; let β be its supremum. Let > 0 be given. Then β − < β , thus β − is not an upper bound of the set of which β is the least upper bound, hence there is x ∈ {a n : n ∈ N} such that β − < x. Given the definition of the set, x = a N for some N ∈ N, thus there exists N ∈ N with a N > β − . If n > N , because the sequence is increasing, we’ll have a n ≥ x N > β − ; by the definition of β , a n ≤ β . Thus β − < a n ≤ β < β + ; that is, |a n − β | < whenever n > N . 3. Prove, using only the definition of limit, that the follo wing sequen ces {a n } converge to the given limit L. (a) {a n } = { n 2 + 5 3n 2 − 2 }, L = 1 3 . Solution. (b) {a n } = { (−1) n + 2 √ n √ n }, L = 2. (c) a 1 = 1, a 2 = 4/2, a 3 = 5/3, a 4 = 9/4, and in general a n = n-th digit after the decimal point of the expansion of π n ; L = 0. 4. Let {a n }, {b n } be sequences of real num bers. Assume {a n }, {b n } both converge and a n ≤ b n for all n ∈ N. Prove: lim n→∞ a n ≤ lim n→∞ b n . Hint: You could give names to the limits, say a = lim n→∞ a n , b = lim n→∞ b n . You have to prove a ≤ b; the alternative is a > b . What happens if you take = (a − b)/2?
-
Upload
eddiekimbowa -
Category
Documents
-
view
219 -
download
0
Transcript of mafa11e1s
8/13/2019 mafa11e1s
http://slidepdf.com/reader/full/mafa11e1s 1/1
Modern Analysis
EXAM 1. Friday, September 23, 2011
INSTRUCTIONS
• You do not have to copy the statements of the exercises. But make sure you clearly indicate where youranswer to any given exercise begins, and where it ends.
• The main thing I’d like to verify with this exam is whether you are able to write a proof correctly. So,please, write proofs correctly. Don’t improvise, don’t invent strange things, don’t fly off tangents. Don’t useundefined terms. What you write should be clear and coherent.
1. Let A, B be sets of real numbers. Assume they are bounded below and non-empty. Prove
inf(A + B) = inf A + inf B.
Do not waste time proving separately that the set A + B is non-empty or bounded below. You may acceptthat it is. The set A + B is defined by A + B = {a + b : a ∈ A, b ∈ B}.
Solution. Let x ∈ A + B; then x = a + b for some a ∈ A, b ∈ B. Because infima are lower bounds,a ≥ inf A, b ≥ infB, thus x = a + b ≥ inf A + inf B. Since x was an arbitrry element of A + B , thisproves that inf A + inf B is a lower bound of A + B ; in particular, inf(A + B) ≥ inf A + inf B. For theconverse inequality, let > 0 be a positive number. Then inf A < inf A +
2 thus there exists a ∈ A such that
a < inf A +
2. Similarly, there exists inf A < inf A +
2 thus there exists b ∈ B such that b < inf B +
2. Then
inf(A + B) ≤ a + b < inf A +inf B + ; since > 0 is arbitrary, this implies inf(A + B) ≤ inf A +inf B. Havingproved inf(A + B) ≥ inf A + inf B and inf(A + B) ≤ inf A + inf B, we are done.
2. Let {an} be a sequence of real numbers. Assume {an} is increasing and bounded above. Prove:
limn→∞
an = sup{
an : n ∈
N
}.
Solution. Because the sequence is bounded above, the non empty set {an : n ∈ N} is bounded above; letβ be its supremum. Let > 0 be given. Then β − < β , thus β − is not an upper bound of the set of whichβ is the least upper bound, hence there is x ∈ {an : n ∈ N} such that β − < x. Given the definition of theset, x = aN for some N ∈ N, thus there exists N ∈ N with aN > β − . If n > N , because the sequence isincreasing, we’ll have an ≥ xN > β − ; by the definition of β , an ≤ β . Thus β − < an ≤ β < β + ; that is,|an − β | < whenever n > N .
3. Prove, using only the definition of limit, that the following sequences {an} converge to the given limit L.
(a) {an} = { n2 + 5
3n2 − 2}, L =
1
3.
Solution.
(b) {an} = {(−1)n
+ 2√ n√ n
}, L = 2.
(c) a1 = 1, a2 = 4/2, a3 = 5/3, a4 = 9/4, and in general
an = n-th digit after the decimal point of the expansion of π
n ; L = 0.
4. Let {an}, {bn} be sequences of real numbers. Assume {an}, {bn} both converge and an ≤ bn for all n ∈ N.Prove: limn→∞an ≤ limn→∞ bn.
Hint: You could give names to the limits, say a = limn→∞an, b = limn→∞ bn. You have to prove a ≤ b; thealternative is a > b. What happens if you take = (a − b)/2?
