Power Series .... Abbott, Theorem 6.5.1 .... ....

[Math Amateur]

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 power series ...

I need some help to understand Theorem 6.5.1 ... specifically, some remarks that Abbott makes after the proof of the theorem ...

Theorem 6.5.1 and Abbott's remarks read as follows:View attachment 8585In the above text by Abbott (after the proof ... ) we read the following:

" ... ... The main implication of Theorem 6.5.1 is that the set of points for which a given power series converges must necessarily be \(\displaystyle \{ 0 \} , \mathbb{R} \), or a bounded interval centred around \(\displaystyle x = 0\) ... ... "I was wondering why the above quote would be true ...

My thinking is that since \(\displaystyle \mid \frac{x}{x_0} \mid \lt 1\) we have that \(\displaystyle -x_0 \lt x \lt x_0\) ... ...

If \(\displaystyle x_0\) was simply \(\displaystyle 0\) ( ... and there were no other points where the power series converged) then \(\displaystyle \{ 0 \}\) would be the set of points for which the power series converged ...

If \(\displaystyle x_0 = b\), say, then the power series would converge in the bounded interval \(\displaystyle ( -b, b )\) ...

Is that correct so far?BUT ... how does Abbott arrive at the fact that an implication of the above theorem is that the power series may converge on all of \(\displaystyle \mathbb{R}\) ...Hope that someone can help ...

Peter

 

[Opalg]

In the above text by Abbott (after the proof ... ) we read the following:

" ... ... The main implication of Theorem 6.5.1 is that the set of points for which a given power series converges must necessarily be \(\displaystyle \{ 0 \} , \mathbb{R} \), or a bounded interval centred around \(\displaystyle x = 0\) ... ... "

I was wondering why the above quote would be true ...

My thinking is that since \(\displaystyle \mid \frac{x}{x_0} \mid \lt 1\) we have that \(\displaystyle -x_0 \lt x \lt x_0\) ... ...

If \(\displaystyle x_0\) was simply \(\displaystyle 0\) ( ... and there were no other points where the power series converged) then \(\displaystyle \{ 0 \}\) would be the set of points for which the power series converged ...

If \(\displaystyle x_0 = b\), say, then the power series would converge in the bounded interval \(\displaystyle ( -b, b )\) ...

Is that correct so far?

BUT ... how does Abbott arrive at the fact that an implication of the above theorem is that the power series may converge on all of \(\displaystyle \mathbb{R}\) ...

The point here is that $x_0$ is not unique (except in the case where $x_0=0$ is the only point where the power series converges). In fact, if $x_0=b$ is a possible value for $x_0$, then so is $x_0=a$, for any $a$ satisfying $0<a<b$. There may also be values $c>b$ that are also possible values for $x_0$.

Let $S$ be the set of possible positive values for $x_0$. If $S$ is bounded above then it has a supremum $r$. This has the property that that the series converges for $x\in (-r,r)$, but diverges whenever $|x|>r$. But if $S$ is not bounded then the series converges for all $x\in\Bbb{R}$.

 

[Math Amateur]

The point here is that $x_0$ is not unique (except in the case where $x_0=0$ is the only point where the power series converges). In fact, if $x_0=b$ is a possible value for $x_0$, then so is $x_0=a$, for any $a$ satisfying $0<a<b$. There may also be values $c>b$ that are also possible values for $x_0$.

Let $S$ be the set of possible positive values for $x_0$. If $S$ is bounded above then it has a supremum $r$. This has the property that that the series converges for $x\in (-r,r)$, but diverges whenever $|x|>r$. But if $S$ is not bounded then the series converges for all $x\in\Bbb{R}$.

Thanks for the help, Opalg ...

Peter