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

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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:


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

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

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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:


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

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

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

and

$$\lim_{n\to\infty} x^{\frac{1}{2n-1}} = 1, \ \ \text{ for } \ \ x \in (0,1]$$ ...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

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

and

$$\lim_{n\to\infty} x^{\frac{1}{2n-1}} = 1, \ \ \text{ for } \ \ x \in (0,1]$$ ...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
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.

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