MTH631 ASSIGNMENT NO. 1 SPRING 2023 || 100% RIGHT SOLUTION || REAL ANALYSIS - II || BY VuTech

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QUESTION-1

Consider the sequence of functions `(x) = 1/{1+x^n}` for `x \in  [0,1]`. Check whether the sequence of functions `f_n (x)` converges uniformly on [0,1]. 

SOLUTION:

To check whether the sequence of functions `f_n(x) = 1/(1+x^n)` converges uniformly on the interval [0,1], we need to analyze the behavior of the sequence as n approaches infinity.

For a sequence of functions to converge uniformly, the limit function must satisfy the following condition: For any ε > 0, there exists an integer N such that for all x in the interval [0,1] and for all n ≥ N, the difference between `f_n(x)` and the limit function, denoted as f(x), must be less than ε.

Let's examine the behavior of the sequence. For any fixed x in the interval [0,1], as n increases, the term `x^n` becomes smaller. Consequently, the denominator `1 + x^n` becomes larger, leading to the fraction `1/(1+x^n)` approaching zero.

If we take the limit as n approaches infinity, we can evaluate the limit function f(x) by substituting n = ∞ into the expression:

`f(x) = 1/(1+x^∞) = 1/(1+0) = 1/1 = 1`

Therefore, the limit function is f(x) = 1 for all x in the interval [0,1].

Now, let's examine the uniform convergence. For a given ε > 0, we need to find an integer N such that for all n ≥ N and for all x in the interval [0,1], `|f_n(x) - f(x)| < ε`.

Let's choose an arbitrary x in the interval [0,1] and evaluate `|f_n(x) - f(x)|`:

`|f_n(x) - f(x)| = |1/(1+x^n) - 1|`

To simplify further, we can find the maximum value of `|f_n(x) - f(x)|` within the interval [0,1].

If we differentiate `|f_n(x) - f(x)|`with respect to x and equate it to zero, we find the critical points of the expression `1/(1+x^n)`. However, since x^n is strictly increasing for x ∈ [0,1], the maximum value occurs at either x = 0 or x = 1.

Evaluating `|f_n(0) - f(0)| and |f_n(1) - f(1)|`:

`|f_n(0) - f(0)| = |1/(1+0^n) - 1| = |1/(1+0) - 1| = |1 - 1| = 0`

`|f_n(1) - f(1)| = |1/(1+1^n) - 1| = |1/(1+1) - 1| = |1/2 - 1| = 1/2`

We can observe that `|f_n(0) - f(0)| = 0` for all n, indicating that the sequence of functions converges pointwise at x = 0. However, `|f_n(1) - f(1)| = 1/2`, which does not approach zero as n tends to infinity. Therefore, the sequence of functions `{f_n(x)}` does not converge uniformly on the interval [0,1].

In conclusion, the sequence of functions `f_n(x) = 1/(1+x^n)` does not exhibit uniform convergence on the interval [0,1].

QUESTION-2

Using Weierstrass M-test determine the set on which the series `\sum 1/n^{3/2} (x/{1+x})^n` converges absolutely uniformly.

SOLUTION:

To determine the set on which the series `\sum 1/n^{3/2} (x/{1+x})^n` converges absolutely uniformly, we can utilize the Weierstrass M-test. This test allows us to establish uniform convergence by comparing the series to a convergent series with known bounds.

Let's analyze the given series: `\sum 1/n^{3/2} (x/{1+x})^n`.

First, consider the sequence of functions `f_n(x) = 1/n^{3/2} (x/{1+x})^n`. To apply the M-test, we need to find a convergent series `\sum M_n` such that `|f_n(x)| ≤ M_n` for all x.

For the given series, note that `x/{1+x}` is always less than or equal to 1 on the interval [0,1]. Therefore, we can simplify the inequality as follows:

`|f_n(x)| = 1/n^{3/2} (x/{1+x})^n ≤ 1/n^{3/2}`

Now, let's consider the series `\sum 1/n^{3/2}`. This series is a convergent p-series with p = 3/2, which means it converges. Therefore, we can choose `M_n = 1/n^{3/2}` as our convergent series.

Now, we need to show that `|f_n(x)| ≤ M_n = 1/n^{3/2}` for all x in the interval [0,1]. Since we already established that `x/{1+x} ≤ 1`, we can rewrite the inequality as:

`1/n^{3/2} (x/{1+x})^n ≤ 1/n^{3/2}`

This inequality holds for all x in the interval [0,1] and for all n. Hence, the series `\sum 1/n^{3/2} (x/{1+x})^n` satisfies the conditions of the Weierstrass M-test.

By the M-test, if a series of functions `\sum f_n(x)` converges absolutely and uniformly on a set S, then it also converges uniformly on S. Since the series `\sum 1/n^{3/2} (x/{1+x})^n` is absolutely convergent and satisfies the conditions of the M-test on the interval [0,1], we can conclude that the series converges absolutely and uniformly on the interval [0,1].

In summary, the series `\sum 1/n^{3/2} (x/{1+x})^n` converges absolutely and uniformly on the interval [0,1].

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