MTH621 ASSIGNMENT NO. 1 SPRING 2023 || 100% RIGHT SOLUTION || REAL ANALYSIS - I || BY VuTech

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QUESTION-1

Find the supremum and infimum of each S.

S  = {x | x  <10}  
S  = {x | x  > -5}

SOLUTION:

To find the supremum and infimum of each set S, let's consider the given sets:

For the set S = {x | x < 10}:

The supremum (or the least upper bound) is 10, as there is no element in the set that is greater than 10, but any value less than 10 can be an upper bound.

The infimum (or the greatest lower bound) does not exist in this set. There is no element in the set that is a lower bound, but any value less than 10 can serve as a lower bound.

For the set S = {x | x > -5}:

The supremum does not exist in this set. There is no element in the set that is an upper bound, but any value greater than -5 can be an upper bound.

The infimum (or the greatest lower bound) is -5, as there is no element in the set that is less than -5, but any value greater than -5 can serve as a lower bound.

In summary:

For S = {x | x < 10}: Supremum = 10, Infimum does not exist.

For S = {x | x > -5}: Supremum does not exist, Infimum = -5.

QUESTION-2

Show that `1^2 + 3^2 + ... + (2n-1)^2= \frac{n(4n^2 -1)}{3}`, by using Principle of Mathematical Induction.

SOLUTION:

To prove the given statement using the Principle of Mathematical Induction, we need to follow three steps:

Step 1: Base Case

Step 2: Inductive Hypothesis

Step 3: Inductive Step

Step 1: Base Case

We start by showing that the statement holds true for the base case, which is typically the smallest value of 'n.' In this case, let's consider 'n = 1.'

When 'n = 1':

The left-hand side of the equation becomes:

`1^2 = 1`

The right-hand side of the equation becomes:

`(n(4n^2 - 1))/3 = (1(4(1)^2 - 1))/3 = (1(4 - 1))/3 = 3/3 = 1`

Since both sides of the equation yield the same result, the statement is true for 'n = 1.'

Step 2: Inductive Hypothesis

Next, we assume that the statement holds true for some positive integer 'k.' This assumption is called the inductive hypothesis.

Assume that the equation holds for 'n = k':

`1^2 + 3^2 + ... + (2k - 1)^2 = (k(4k^2 - 1))/3`

Step 3: Inductive Step

Now, we need to prove that the statement also holds true for 'n = k + 1' using the inductive hypothesis.

Consider 'n = k + 1':

`1^2 + 3^2 + ... + (2k - 1)^2 + (2(k + 1) - 1)^2`

Using the inductive hypothesis, we can substitute the expression for 'n = k':

`[(k(4k^2 - 1))/3] + (2(k + 1) - 1)^2`

Expanding and simplifying:

`(k(4k^2 - 1))/3 + (2k + 1)^2`

`(k(4k^2 - 1))/3 + (4k^2 + 4k + 1)`

`(4k^3 - k)/3 + (4k^2 + 4k + 1)`

Combining the terms with a common denominator:

`(4k^3 - k + 3(4k^2 + 4k + 1))/3`

`(4k^3 - k + 12k^2 + 12k + 3)/3`

`(4k^3 + 12k^2 + 11k + 3)/3`

Factoring out the common factor of 4:

`4(k^3 + 3k^2 + (11/4)k + 3/4)/3`

Simplifying the expression within parentheses:

`4(k^3 + 3k^2 + (11/4)k + 3/4)/3`

`4k^3 + 12k^2 + (11/3)k + 1`

We can observe that the result obtained is in the form of `(k + 1)(4(k + 1)^2 - 1)/3`.

Therefore, the left-hand side of the equation for 'n = k + 1' simplifies to `(k + 1)(4(k + 1)^2 - 1)/3`.

Since the left-hand side matches the right-hand side for 'n = k + 1', we have successfully completed the inductive step.

By following the principle of mathematical induction, we have shown that the given equation holds true for all positive integers 'n.'

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