MTH631 ASSIGNMENT 1 SOLUTION FALL 2023 | MTH631 ASSIGNMENT 1 SOLUTION 2023 | MTH631 ASSIGNMENT 1 2023 | REAL ANALYSIS - II | VuTech

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Question: 1

Consider the sequence`{f_n}` of functions `f_n:R \to R` defined by `f_n (x) = \frac{nx}{\sqrt {1 + n^2x^2} }`

Find the pointwise limit of this sequence as `n \to \infty.`

Does the sequence converge uniformly on R? Justify it.

Solution:

To find the pointwise limit as n approaches infinity for the sequence of functions `f_n (x) = \frac{nx}{\sqrt {1 + n^2x^2} }` defined on the real numbers R, we'll examine the behavior of the function for each x in R as n tends to infinity.

As n approaches infinity, consider the function values for different values of x:

1. For x=0:

`f_n (0) = \frac{n.0}{\sqrt {1 + n^2 .0 }}`

Irrespective of the value of n,`f_n (0) = 0` for all n, indicating the pointwise limit at x = 0 is 0.

2. For x ≠ 0:

`f_n (x) = \frac{nx}{\sqrt {1 + n^2 x^2} } = \frac{x}{\frac{1}{n}\sqrt {1 + n^2 x^2} }`

 As n approaches infinity, the denominator `\frac{1}{n}\sqrt {1 + n^2 x^2}`  approaches ∣x∣, irrespective of the value of x. Thus, `f_n (x)` tends towards `\frac{x}{|x|} = sgn (x)`, where sgn(x) is the signum function that yields -1 for negative x, 0 for  x=0, and 1 for positive x.

Therefore, the pointwise limit of the sequence of functions `\f_n (x)` as n tends to infinity is a function that is 0 at x=0 and sgn(x) for x ≠ 0.

Now, to determine if the sequence converges uniformly on R, we must check whether the pointwise limit is also the limiting function for `f_n (x)` uniformly.

For uniform convergence, we would need to show that the supremum of the difference between `f_n (x)`.  and the limiting function tends to zero as n approaches infinity.

However, as the limiting function for x ≠ 0 is different from the limiting function at x=0 (0), the sequence of functions does not converge uniformly on R.

The pointwise limit is sgn(x) for x ≠ 0 and 0 at x=0, but this limit function does not represent the uniform convergence of the sequence.

Question: 2

State Cauchy’s criterion and Weierstrass’s test for uniform convergence for series. Test the uniform convergence of the series `\sum u_n ; u_n = \frac{e^ -nx}{4n^2 - 1}` by using Weierstrass’s test.

Solution:

Cauchy's Criterion for Uniform Convergence:

For a series of functions `\sum u_n (x)` to converge uniformly on a domain D, it is necessary that for every ϵ >0, there exists an N such that for all m,n ≥ N and for all x ∈ D, the inequality `\|u_n (x) +  u_{n + 1} (x) +  \ldots  + u_m (x)| < \in` holds.

Weierstrass’s M-Test for Uniform Convergence:

Suppose `\sum u_n (x)` is a series of functions defined on a set D and `\sum M_n` is a series of positive real numbers such that `| u_n (x)| \le M_n` for all x ∈ D and all n ≥ 1. If the series `\sum M_n` converges, then the series of functions `\sum u_n (x)` converges uniformly on D.

Now, let's test the uniform convergence of the series `\sum u_n` where `u_n = \frac{e^{-nx}}{4n^2 - 1}` using Weierstrass's M-Test.

We need to find Mn such that `|u_n (x)| \le M_n` for all x in a given domain.

Given `u_n = \frac{e^{-nx}}{4n^2 - 1}` , it's necessary to find an upper bound `M_n` that satisfies `| u_n (x) | \le M_n` for all x in the domain.

Notice that the exponential term `e^{-nx}` can be maximized by minimizing the exponent −nx. This occurs when x is at its maximum value. Considering x ≥ 0, as x increases, the value of `e^- nx` decreases.

So, let's find the maximum value of `\frac{e^-nx}{4n^2 - 1}` for x ≥ 0.

As x increases, the term `\frac{e^-nx}{4n^2 - 1}` decreases, as the denominator is increasing and the exponential term is decreasing.

To find an upper bound for `M_n`, consider the term at the point where it's minimized, at x = 0:

`M_n = \frac{e^0}{4n^2 - 1} = \frac{1}{4n^2 - 1}` 

Now, let's check if the series `\sum \frac{1}{4n^2 - 1}` converges. If it does, according to Weierstrass's M-Test, the series `\sum u_n` will uniformly converge.

To test the convergence of the series `\sum \frac{1}{4n^2 - 1}`, we need to use some convergence test methods such as the comparison test or the limit comparison test.

If the series `\sum \frac{1}{4n^2 - 1}` converges, then by Weierstrass's M-Test, the series  `\sum u_n`.  will uniformly converge on the specified domain.

This procedure doesn't directly apply Cauchy's Criterion but instead uses Weierstrass's M-Test to establish uniform convergence.

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