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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=1âˆž is convergent. Let its limit be a. Hence, by the definition, there exists a N such that,

â”‚ an - aâ”‚< Îµ for all n â‰¥ N.

Consequently, consider the sequence {an+k} n=1âˆž 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=1âˆž will naturally be greater than the terms of {an} n=1âˆž.

This means the sequence {an+k} n=1âˆž is an increasing sequence. Hence, this limit a will be, a = sup {an+k / n, k â‚¬ N} â‚¬ R (By the characterization of supremum). Now, we have, a â€“ Îµ < aN+k â‰¤ an â‰¤ a < a + Îµ for all n â‰¥ N. (This claim is valid, because of the lemma, every increasing bounded sequence say {xn} converges and that xn â†‘ sup {xn / n â‚¬ N} . let a = sup {xn / n â‚¬ N} . then, xn â‰¤ a for all n and for Îµ > 0, we have a N such that a â€“ Îµ < xN).

Case (ii)

Conversely, let us assume that k < 0. Then all the terms of the sequence {an+k} n=1âˆž will naturally be lesser than the terms of {an} n=1âˆž. This means the sequence {an+k} n=1âˆž is a decreasing sequence. Hence, this limit a will be, a = inf{an+k / n, k â‚¬ N} â‚¬ R (By the characterization of infimum). Thus, the inequality mentioned above will be reversed and can be given by, a â€“ Îµ > aN+k â‰¥ an â‰¥ a > a + Îµ for all n â‰¥ N. (This claim is valid, because of the lemma, every decreasing bounded sequence say {xn} converges and that xn â†“ inf{xn / n â‚¬ N} . let a = inf{xn / n â‚¬ N} . then, xn â‰¥ a for all n and for Îµ > 0, we have a N such that a â€“ Îµ > xN).

Thus, in either case the sequence is bounded and hence converges and also converges to the same limit. Thus,

â”‚ an - aâ”‚= â”‚ an+k - aâ”‚< Îµ for all n â‰¥ N, proving the equality.

2 Given that {an} n=1âˆž is a sequence of real numbers. Then, we need to now prove that the statements, for every n1 > n2, where n1, n2 â‚¬ N, we have an1 > an2 ------------ (1) and for every n â‚¬ N, an+1 > an---------------(2). By these two statements we mean that the sequence {an} n=1âˆž 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 â‚¬ N, we have an1 > an2. In this manner, we have n1 > n2 > n3 > n4 >â€¦., so that the terms will also be an1 > an2 > an3 > an4 >â€¦. where n1, n2,... â‚¬ N If we substitute n4 = 1, n3 = 2, n2 = 3,â€¦ we can see that a1 < a2 < a3 < a4 <â€¦. 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=1âˆž defined by the recurrence relation.

a1 = 1 and an+1 = 1-an for n â‰¥ 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,â€¦ 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,â€¦.}. This sequence doesnâ€™t approach anywhere and hence has no limit. Also, this sequence doesnâ€™t have a generalized pattern and is nothing but the sequence of the partial sum of the Grandi Series, n=0âˆž(-1)n . Splitting this sequence into its subsequences namely {1,1,1,â€¦} and {0,0,0,â€¦} we see that both the subsequences have the limits 1 and 0 respectively and hence are convergent. Now, the given limit calculation is untenable because, the limit is calculated for the term an+1, denoted by a relation which considers the terms right from a2 and the first term, namely a1 is perceived to be a constant. Obviously, the term an+1 defined by the given relation, doesnâ€™t represent the whole sequence as a general pattern. Instead, it is just a recurrence relation used for calculating the terms occurring after a1. Had it been the original pattern, then the term a1 would fit in into it and would not have mentioned separately. Also, by the problem 1, it is well established that limn â†’âˆžan+1 = limn â†’âˆžan and this holds good only for the convergent sequences. Hence, the usage of this tenet itself is a flaw. Thus, the limit obtained through this manner will be wrong, giving the misconception that the given sequence is indeed a convergent sequence, but is not originally so.

4 Given that a1 = 1 and an+1 = 2+an for n â‰¥ 1.

* The term a2 will be, a2 = âˆš(2+a1) = âˆš(2+1) = âˆš3 = 1.7320

The term a3 will be, a3 = âˆš(2+a2) = âˆš(2+âˆš3) = 1.9318

The term a4 will be, a4 = âˆš(2+a3) = âˆš(2+1.9318) = 1.9828

* Let us consider the procedure of mathematical induction

Given that a1 = 1 and 0 < 1 < 2. Now, according to the subsection a, all the terms a2, a3, a4 satisfy the inequality that 0 < an < 2, n = 2,3,4. In order to prove that the general term an also satisfy the inequality, let us assume that the relation holds good for an-1. Hence, 0 < an-1 < 2. Now, an according to the recurrence realtion will be âˆš(2+âˆšan-1). But, an-1 is less than 2, which means

âˆšan-1 is also less than 2.

Now, consider the inequality, âˆšan-1 < 2. Adding 2 to both sides, we have 2+âˆšan-1 < 4. Taking square root on both sides, will yield âˆš(2+âˆšan-1) < 2, proving that an is less than 2, establishing the given inequality.

* Obviously, from the previous subsections it is clear that a1 < a2 < a3 < a4 <â€¦.

Let us now assume that the result is true for an-1. Then the relation will be a1 < a2 < a3 < a4 <â€¦.< an-2 < an-1, so that in particular an-2 < an-1 which means âˆš an-2 < âˆš an-1 (as according to our assumption that the sequence is increasing holds good and hence, the lemma, If a sequence {an} n=1âˆž is an increasing sequence of real numbers, such that a1 < a2 < a3 < a4 <â€¦., then a1 < a2 ...

solutions for the problems

1 Given that the sequence {an} n=1âˆž is convergent. Let its limit be a. Hence, by the definition, there exists a N such that,

â”‚ an - aâ”‚< Îµ for all n â‰¥ N.

Consequently, consider the sequence {an+k} n=1âˆž 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=1âˆž will naturally be greater than the terms of {an} n=1âˆž.

This means the sequence {an+k} n=1âˆž is an increasing sequence. Hence, this limit a will be, a = sup {an+k / n, k â‚¬ N} â‚¬ R (By the characterization of supremum). Now, we have, a â€“ Îµ < aN+k â‰¤ an â‰¤ a < a + Îµ for all n â‰¥ N. (This claim is valid, because of the lemma, every increasing bounded sequence say {xn} converges and that xn â†‘ sup {xn / n â‚¬ N} . let a = sup {xn / n â‚¬ N} . then, xn â‰¤ a for all n and for Îµ > 0, we have a N such that a â€“ Îµ < xN).

Case (ii)

Conversely, let us assume that k < 0. Then all the terms of the sequence {an+k} n=1âˆž will naturally be lesser than the terms of {an} n=1âˆž. This means the sequence {an+k} n=1âˆž is a decreasing sequence. Hence, this limit a will be, a = inf{an+k / n, k â‚¬ N} â‚¬ R (By the characterization of infimum). Thus, the inequality mentioned above will be reversed and can be given by, a â€“ Îµ > aN+k â‰¥ an â‰¥ a > a + Îµ for all n â‰¥ N. (This claim is valid, because of the lemma, every decreasing bounded sequence say {xn} converges and that xn â†“ inf{xn / n â‚¬ N} . let a = inf{xn / n â‚¬ N} . then, xn â‰¥ a for all n and for Îµ > 0, we have a N such that a â€“ Îµ > xN).

Thus, in either case the sequence is bounded and hence converges and also converges to the same limit. Thus,

â”‚ an - aâ”‚= â”‚ an+k - aâ”‚< Îµ for all n â‰¥ N, proving the equality.

2 Given that {an} n=1âˆž is a sequence of real numbers. Then, we need to now prove that the statements, for every n1 > n2, where n1, n2 â‚¬ N, we have an1 > an2 ------------ (1) and for every n â‚¬ N, an+1 > an---------------(2). By these two statements we mean that the sequence {an} n=1âˆž 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 â‚¬ N, we have an1 > an2. In this manner, we have n1 > n2 > n3 > n4 >â€¦., so that the terms will also be an1 > an2 > an3 > an4 >â€¦. where n1, n2,... â‚¬ N If we substitute n4 = 1, n3 = 2, n2 = 3,â€¦ we can see that a1 < a2 < a3 < a4 <â€¦. 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=1âˆž defined by the recurrence relation.

a1 = 1 and an+1 = 1-an for n â‰¥ 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,â€¦ 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,â€¦.}. This sequence doesnâ€™t approach anywhere and hence has no limit. Also, this sequence doesnâ€™t have a generalized pattern and is nothing but the sequence of the partial sum of the Grandi Series, n=0âˆž(-1)n . Splitting this sequence into its subsequences namely {1,1,1,â€¦} and {0,0,0,â€¦} we see that both the subsequences have the limits 1 and 0 respectively and hence are convergent. Now, the given limit calculation is untenable because, the limit is calculated for the term an+1, denoted by a relation which considers the terms right from a2 and the first term, namely a1 is perceived to be a constant. Obviously, the term an+1 defined by the given relation, doesnâ€™t represent the whole sequence as a general pattern. Instead, it is just a recurrence relation used for calculating the terms occurring after a1. Had it been the original pattern, then the term a1 would fit in into it and would not have mentioned separately. Also, by the problem 1, it is well established that limn â†’âˆžan+1 = limn â†’âˆžan and this holds good only for the convergent sequences. Hence, the usage of this tenet itself is a flaw. Thus, the limit obtained through this manner will be wrong, giving the misconception that the given sequence is indeed a convergent sequence, but is not originally so.

4 Given that a1 = 1 and an+1 = 2+an for n â‰¥ 1.

* The term a2 will be, a2 = âˆš(2+a1) = âˆš(2+1) = âˆš3 = 1.7320

The term a3 will be, a3 = âˆš(2+a2) = âˆš(2+âˆš3) = 1.9318

The term a4 will be, a4 = âˆš(2+a3) = âˆš(2+1.9318) = 1.9828

* Let us consider the procedure of mathematical induction

Given that a1 = 1 and 0 < 1 < 2. Now, according to the subsection a, all the terms a2, a3, a4 satisfy the inequality that 0 < an < 2, n = 2,3,4. In order to prove that the general term an also satisfy the inequality, let us assume that the relation holds good for an-1. Hence, 0 < an-1 < 2. Now, an according to the recurrence realtion will be âˆš(2+âˆšan-1). But, an-1 is less than 2, which means

âˆšan-1 is also less than 2.

Now, consider the inequality, âˆšan-1 < 2. Adding 2 to both sides, we have 2+âˆšan-1 < 4. Taking square root on both sides, will yield âˆš(2+âˆšan-1) < 2, proving that an is less than 2, establishing the given inequality.

* Obviously, from the previous subsections it is clear that a1 < a2 < a3 < a4 <â€¦.

Let us now assume that the result is true for an-1. Then the relation will be a1 < a2 < a3 < a4 <â€¦.< an-2 < an-1, so that in particular an-2 < an-1 which means âˆš an-2 < âˆš an-1 (as according to our assumption that the sequence is increasing holds good and hence, the lemma, If a sequence {an} n=1âˆž is an increasing sequence of real numbers, such that a1 < a2 < a3 < a4 <â€¦., then a1 < a2 ...

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