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Stream: learning: questions

Topic: is convergence an open condition?

John Baez (Jul 09 2024 at 12:33):

Right now I don't have a great place to ask math questions just as they pop into my head - questions that may seem trivial or even stupid moments later. (Maybe I shouldn't post such questions? But sometimes that helps me figure out that answer, and sometimes they turn out to be interesting.)

For category theory I don't mind asking such questions here. But these two are actually analysis. The first one is more standard, but the second is the one I really care about.

Question 1. Suppose we have a series

$f(x) = \sum_{i = 0}^\infty a_i x^i$

with each $a_i \ge 0$. Is the following true?

Either the series diverges for all $x > 0$, or there exists some $R$ such that the series converges for $0 \le x < R$ and diverges for $x \ge R$.

My guess is yes. This would be a sense in which "convergence is an open condition". I'm mainly trying to rule out the possibility that it converges for $0 \le x \le R$ but diverges for $x > R$. There are also, a priori, other possibilities, but I think these are easier to rule out.

I also guess that $f(x) \to +\infty$ as $x$ approaches $R$ from below, and I'd like to know this too if true.

Question 2. Suppose we have a series

$f(x) = \sum_{i = 0}^\infty \exp(-a_i x)$

where $a_i \in \mathbb{R}$. Is the following true?

Either the series diverges for all $x \ge 0$, or there exists some $R > 0$ such that the series converges for $x > R$ and diverges for $0 \le x \le R.$

Note that in this second one, if we assume the series converges for some $x \ge 0$, it must be true that the $a_i$ are bounded below, and there must be only finitely many $\le A$ for any $A \in \mathbb{R}$. Thus we can assume without loss of generality that $a_1 \le a_2 \le a_3 \le \cdots$, since we're free to rearrange terms in a sequence whose terms are all positive.

Todd Trimble (Jul 09 2024 at 12:59):

For question 1, what about $x + x^2/4 + x^3/9 + \ldots$ which converges at $x = 1$ but not for $x > 1$?

John Baez (Jul 09 2024 at 13:20):

Oh, duh, thanks. I knew about that - I just wasn't engaging my history of teaching calculus. :grimacing: That's exactly what I meant by firing off a question just as it came into my head.

So now maybe I can come up with a counterexample to the one I'm really interested in, question 2.

John Baez (Jul 09 2024 at 13:29):

For a counterexample I want to find a series

$\sum_{i = 0}^\infty \exp(-a_i x)$

and an $x > 0$ that makes it converge, but for which no smaller $x$ makes it converge.

In case anyone is curious about why I care: it's because I'm writing about the Hagedorn temperature in my little book Tweets on Entropy. This is a temperature above which the partition function of some system diverges so the statistical mechanics of that system makes no sense. Basically it gets so hot that all hell breaks loose. :fire: :fire: :fire:

John Baez (Jul 09 2024 at 13:32):

My intuition is that when a Hagedorn temperature exists, the partition function must diverge at the Hagedorn temperature. If the answer to question 2 is "yes, your claim is true", then my intuition is right. Todd has just shaken that intuition. But it's possible question 2 works differently from question 1.

John Baez (Jul 09 2024 at 13:35):

(By the way: for anyone who cares, Wikipedia defines Hagedorn temperature in a way that assumes my intuition is right! They don't consider the possibility that the partition function is well-defined that Hagedorn temperature but diverges at all higher temperatures. They only consider the possibility where the partition function approaches infinity as you approach the Hagedorn temperature and is thus (one can show) divergent at that temperature.)

John Baez (Jul 09 2024 at 13:35):

Anyway, my questions deliberately stripped off all this physics, because you don't need to know physics to answer question 2!

Oscar Cunningham (Jul 09 2024 at 14:29):

The quantity $\mathrm{exp}(-a_ix)$ is the same as $\mathrm{exp}(-a_i)^x$. Pick $a_i$ so that $\mathrm{exp}(-a_i) = \frac{1}{i\mathrm{log}(i)^2}$. Then when $x = 1$ the sum of $\frac{1}{i\mathrm{log}(i)^2}$ converges, but when $x < 1$ the sum of $\frac{1}{i^x\mathrm{log}(i)^{2x}}$ diverges.

John Baez (Jul 09 2024 at 14:38):

Nice! Thanks so much, @Oscar Cunningham. I was trying something less clever, like $\exp(-a_i) = 1/i^2$. Somehow I failed to use all my wonderful knowledge, like that the sum of $1/n \log n$ diverges but $1/n (\log n)^2$ converges, and so on.

John Baez (Jul 09 2024 at 14:39):

Okay, so the answer to question 2 is "false": the borderline case can go either way.

Morgan Rogers (he/him) (Jul 09 2024 at 19:42):

Didn't you say $a_i \in \R$? Does Oscar's answer still apply?

John Baez (Jul 09 2024 at 19:46):

Yes, I foolishly used $i$ to mean $1, 2, 3, \dots$ and Oscar kindly went along with me. So all the numbers here are real, even though they look $i$-maginary.

Say, @Morgan Rogers (he/him), could you please delete the comments above where I marked this topic as resolved and then thought better of it?