# Pointwise Convergence .... Abbott, Example 6.2.2 (iii) .... ....

• MHB
• Math Amateur
In summary, the author is discussing how to find the limit of a function as n approaches infinity. He provides an example of how to do this for a function that is represented by a series with an odd power on the variable. When x<0, the power of the series becomes negative, and as n approaches infinity, the power of the series becomes 1.
Math Amateur
Gold Member
MHB
I am reading Stephen Abbott's book: "Understanding Analysis" (Second Edition) ...

I am focused on Chapter 6: Sequences and Series of Functions ... and in particular on pointwise convergence...

I need some help to understand the 'mechanics' of Example 6.2.2 (iii) ...

Example 6.2.2 reads as follows:View attachment 8580In the above text from Abbott, in Example 6.2.2 (iii), we read the following:

" ... ... $\displaystyle \lim_{n \to \infty }$ $$\displaystyle h_n (x) = x$$ $\displaystyle \lim_{n \to \infty }$ $$\displaystyle x^{ \frac{1}{ 2n -1 } } = \ \mid x \mid$$ ... ... "

Can someone please explain exactly how ...$$\displaystyle x$$ $\displaystyle \lim_{n \to \infty }$ $$\displaystyle x^{ \frac{1}{ 2n -1 } } = \ \mid x \mid$$ ... ... "
Peter

#### Attachments

• Abbott - Example 6.2.2 (iii) .png
25.5 KB · Views: 82
Peter said:
I am reading Stephen Abbott's book: "Understanding Analysis" (Second Edition) ...

I am focused on Chapter 6: Sequences and Series of Functions ... and in particular on pointwise convergence...

I need some help to understand the 'mechanics' of Example 6.2.2 (iii) ...

Example 6.2.2 reads as follows:In the above text from Abbott, in Example 6.2.2 (iii), we read the following:

" ... ... $\displaystyle \lim_{n \to \infty }$ $$\displaystyle h_n (x) = x$$ $\displaystyle \lim_{n \to \infty }$ $$\displaystyle x^{ \frac{1}{ 2n -1 } } = \ \mid x \mid$$ ... ... "

Can someone please explain exactly how ...$$\displaystyle x$$ $\displaystyle \lim_{n \to \infty }$ $$\displaystyle x^{ \frac{1}{ 2n -1 } } = \ \mid x \mid$$ ... ... "
Peter

Keep in mind that $x\in[-1,1]$. Note that
$\lim_{n\to\infty} x^{\frac{1}{2n-1}}=\begin{cases}-1, & x\in[-1,0),\\ 0, & x=0,\\ 1, & x\in(0,1].\end{cases}$
So in the end, we find that
$x\lim_{n\to\infty}x^{\frac{1}{2n-1}} = \begin{cases}x, & x\geq 0,\\ -x,& x<0\end{cases} = |x|.$
I hope this makes sense!

Chris L T521 said:
Keep in mind that $x\in[-1,1]$. Note that
$\lim_{n\to\infty} x^{\frac{1}{2n-1}}=\begin{cases}-1, & x\in[-1,0),\\ 0, & x=0,\\ 1, & x\in(0,1].\end{cases}$
So in the end, we find that
$x\lim_{n\to\infty}x^{\frac{1}{2n-1}} = \begin{cases}x, & x\geq 0,\\ -x,& x<0\end{cases} = |x|.$
I hope this makes sense!
Thanks Chris ...

... just a clarification ...

How did you determine that

$$\displaystyle \lim_{n\to\infty} x^{\frac{1}{2n-1}} = -1, \ \ \text{ for } \ \ x\in [-1,0)$$

and

$$\displaystyle \lim_{n\to\infty} x^{\frac{1}{2n-1}} = 1, \ \ \text{ for } \ \ x \in (0,1]$$ ...Do you just have to know these limits ... ?PeterPS I'm particularly curious about the case where x is negative ...

Peter said:
Thanks Chris ...

... just a clarification ...

How did you determine that

$$\displaystyle \lim_{n\to\infty} x^{\frac{1}{2n-1}} = -1, \ \ \text{ for } \ \ x\in [-1,0)$$

and

$$\displaystyle \lim_{n\to\infty} x^{\frac{1}{2n-1}} = 1, \ \ \text{ for } \ \ x \in (0,1]$$ ...Do you just have to know these limits ... ?PeterPS I'm particularly curious about the case where x is negative ...

The power on $x$ is $\frac{1}{2n-1}$ which implies these are odd radicals for $n\geq 2$ (when $n=1$, the expression is just $x$). So if $x>0$, $x^{\frac{1}{2n-1}}>0$ and since $\frac{1}{2n-1}\rightarrow 0$ as $n\rightarrow\infty$, we find that $x^{\frac{1}{2n-1}}\rightarrow 1$ (since $a^0=1$ for $a\neq 0$).

On the other hand, when $x<0$, we can factor negatives out of odd radicals, so we find that $x^{\frac{1}{2n-1}}=-(-x)^{\frac{1}{2n-1}}<0$. Since $x<0\implies -x>0$, we find that as $n\to\infty$, $x^{\frac{1}{2n-1}}=-(-x)^{\frac{1}{2n-1}}\rightarrow -1$ for $x<0$ (note that $(-x)^{\frac{1}{2n-1}}\rightarrow 1$ as $n\rightarrow \infty$ by similar reasoning in the previous paragraph).

I hope this clarifies things!

Last edited:
Chris L T521 said:
The power on $x$ is $\frac{1}{2n-1}$ which implies these are odd radicals for $n\geq 2$ (when $n=1$, the expression is just $x$). So if $x>0$, $x^{\frac{1}{2n-1}}>0$ and since $\frac{1}{2n-1}\rightarrow 0$ as $n\rightarrow\infty$, we find that $x^{\frac{1}{2n-1}}\rightarrow 1$ (since $a^0=1$ for $a\neq 0$).

On the other hand, when $x<0$, we can factor negatives out of odd radicals, so we find that $x^{\frac{1}{2n-1}}=-(-x)^{\frac{1}{2n-1}}<0$. Since $x<0\implies -x>0$, we find that as $n\to\infty$, $x^{\frac{1}{2n-1}}=-(-x)^{\frac{1}{2n-1}}\rightarrow -1$ for $x<0$ (note that $(-x)^{\frac{1}{2n-1}}\rightarrow 1$ as $n\rightarrow \infty$ by similar reasoning in the previous paragraph).

I hope this clarifies things!
Very clear thanks Chris ...

Your help is much appreciated ...

Peter

## 1. What is pointwise convergence?

Pointwise convergence is a type of convergence in which a sequence of functions approaches a limit function point by point. In other words, for each individual point in the domain of the functions, the sequence of function values approaches the value of the limit function at that point.

## 2. How is pointwise convergence different from uniform convergence?

Uniform convergence is a stronger form of convergence in which the entire sequence of functions approaches the limit function uniformly, meaning that the distance between each function and the limit function becomes smaller and smaller as you move away from the point of interest. In pointwise convergence, the distance between the functions and the limit function may vary at different points in the domain.

## 3. What does Example 6.2.2 (iii) in Abbott refer to?

Example 6.2.2 (iii) is an example in the textbook "Understanding Analysis" by Stephen Abbott that illustrates the concept of pointwise convergence. It specifically refers to a sequence of functions that converges pointwise but not uniformly.

## 4. How is pointwise convergence related to continuity?

In order for a sequence of functions to converge pointwise, the limit function must also be continuous at each point in the domain. This is because the limit function is the pointwise limit of the sequence of functions, and the definition of continuity requires that the limit of a function at a point is equal to the function value at that point.

## 5. Can a sequence of discontinuous functions converge pointwise?

Yes, a sequence of discontinuous functions can still converge pointwise as long as the limit function is continuous. This is because the pointwise convergence only requires that the functions converge at each individual point, regardless of the continuity of the functions themselves.

• Topology and Analysis
Replies
4
Views
2K
• Topology and Analysis
Replies
11
Views
1K
• Topology and Analysis
Replies
2
Views
676
• Topology and Analysis
Replies
9
Views
1K
• Topology and Analysis
Replies
2
Views
1K
• Topology and Analysis
Replies
2
Views
1K
• Topology and Analysis
Replies
3
Views
1K
• Topology and Analysis
Replies
21
Views
2K
• Topology and Analysis
Replies
2
Views
1K
• Topology and Analysis
Replies
2
Views
1K