MHB 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 $$\{ 0 \} , \mathbb{R} $$, or a bounded interval centred around $$x = 0$$ ... ... "I was wondering why the above quote would be true ...

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

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

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

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 $$\mathbb{R}$$ ...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 $$\{ 0 \} , \mathbb{R} $$, or a bounded interval centred around $$x = 0$$ ... ... "

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

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

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

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

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 $$\mathbb{R}$$ ...

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