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Real Analysis Assignment: Some of the Maths Problems (Math Problem Sample)

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these are some of the maths problems coming under the pure and abstract maths category. as the solutions are my own understanding, i haven't used any resources.

source..
Content:
Real analysis
solutions for the problems
1 Given that the sequence {an} n=1∞ is convergent. Let its limit be a. Hence, by the definition, there exists a N such that,
│ an - a│< ε for all n ≥ N.
Consequently, consider the sequence {an+k} n=1∞ where k is a fixed natural number. Case (i)
Now, since the nature of k is not given, let us assume that k > 0. Then all the terms of the sequence {an+k} n=1∞ will naturally be greater than the terms of {an} n=1∞.
This means the sequence {an+k} n=1∞ is an increasing sequence. Hence, this limit a will be, a = sup {an+k / n, k € N} € R (By the characterization of supremum). Now, we have, a – ε < aN+k ≤ an ≤ a < a + ε for all n ≥ N. (This claim is valid, because of the lemma, every increasing bounded sequence say {xn} converges and that xn ↑ sup {xn / n € N} . let a = sup {xn / n € N} . then, xn ≤ a for all n and for ε > 0, we have a N such that a – ε < xN).
Case (ii)
Conversely, let us assume that k < 0. Then all the terms of the sequence {an+k} n=1∞ will naturally be lesser than the terms of {an} n=1∞. This means the sequence {an+k} n=1∞ is a decreasing sequence. Hence, this limit a will be, a = inf{an+k / n, k € N} € R (By the characterization of infimum). Thus, the inequality mentioned above will be reversed and can be given by, a – ε > aN+k ≥ an ≥ a > a + ε for all n ≥ N. (This claim is valid, because of the lemma, every decreasing bounded sequence say {xn} converges and that xn ↓ inf{xn / n € N} . let a = inf{xn / n € N} . then, xn ≥ a for all n and for ε > 0, we have a N such that a – ε > xN).
Thus, in either case the sequence is bounded and hence converges and also converges to the same limit. Thus,
│ an - a│= │ an+k - a│< ε for all n ≥ N, proving the equality.
2 Given that {an} n=1∞ is a sequence of real numbers. Then, we need to now prove that the statements, for every n1 > n2, where n1, n2 € N, we have an1 > an2 ------------ (1) and for every n € N, an+1 > an---------------(2). By these two statements we mean that the sequence {an} n=1∞ is an increasing sequence.
Substituting n = 1, 2, 3,... consequently in (2), we get a2 > a1, a3 > a2, a4 > a3 and so on. Now, given that for every n1 > n2, where n1, n2 € N, we have an1 > an2. In this manner, we have n1 > n2 > n3 > n4 >…., so that the terms will also be an1 > an2 > an3 > an4 >…. where n1, n2,... € N If we substitute n4 = 1, n3 = 2, n2 = 3,… we can see that a1 < a2 < a3 < a4 <…. This is nothing but the expanded form of (2). Hence, both the statements (1) and (2) are equivalent.
3 Let us consider the sequence {an} n=1∞ defined by the recurrence relation.
a1 = 1 and an+1 = 1-an for n ≥ 1
To establish that the given argument is wrong, let us begin by expanding the given sequence. Given that a1 = 1, a2 will be, a2 = 1-a1 = 1-1 = 0. Now consequently a3, a4,… can be calculated as, a3 = 1-a2 = 1-0 = 1 and a4 = 1-a3 = 1-1 = 0. Continuing in this manner, the whole sequence can be written as {1, 0, 1, 0,….}. This sequence doesn’t approach anywhere and hence has no limit. Also, this sequence does...
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