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Timeline for Divergent series summation beyond natural boundaries

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Jul 17, 2023 at 0:11 comment added Sidharth Ghoshal Yes! I should look into the concept of parity for generating functions but the point you are making is understood in that they are both $1-1+1-1...$. Whats interesting is the taylor series of $\frac{1}{1-x}$ centered at $-1$ has radius of convergence (2) I believe. But the same "taylor" series for the theta function has infinite radius of convergence (even though it is only equal at a single point). I just thought that it was interesting you found a totally different representation of the theta function that also suggests this $\frac{1}{2}$ everywhere behavior.
Jul 17, 2023 at 0:06 comment added Jorge Zuniga @Sidharth Ghoshal $\sum_{n=0}^{\infty} x^{n^2} = \sum_{n=0}^{\infty} x^{n} = \frac{1}{2}$ for $x = -1$ since $n^2$ has the same parity as $n$
Jul 15, 2023 at 18:08 comment added Sidharth Ghoshal Interesting, I’ll dig more into this, it’s also worth noting that $\sum_{n=0}^{\infty} x^{n^2}$ supports a (this is no longer rigorous) “Taylor expansion” with infinite radius of convergence at $x = -1$ this expansion is $f(x) = \frac{1}{2} + 0 + 0 ….$ by evaluating the derivatives of the theta function using the alternating zeta function . The result you have found of $s(1,q) = \frac{1}{2}$ might be related to that unrigorous calculation somehow as it feels like an unusual coincidence to me atleast
Jul 9, 2023 at 13:26 comment added Jorge Zuniga @Sidharth_Ghoshal. Nope, I did not verify because strongly divergent series beyond factorial divergence have generating functions represented by integral transforms that are NOT unique (the classic Stieltjes' moment problem of distributions) and the same divergent series can represent infinite functions beyond the natural boundary. At the end of Grecchi-Maioli's paper (see above) there is an example. I think, other methods are needed. Caleb_Briggs have been publishing some in this line here in MO.
Jul 8, 2023 at 22:07 comment added Sidharth Ghoshal Did the two of you find any time to verify the Jacobi theta results? How many of those relationships do still survive?
Nov 8, 2021 at 3:34 history edited Jorge Zuniga CC BY-SA 4.0
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Oct 6, 2021 at 2:38 comment added Caleb Briggs Wow, the new addenda is incredible. Thank you so much for taking the time to write such a marvelous answer!
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Sep 21, 2021 at 4:53 comment added Caleb Briggs I think I can see the s(1,x) argument pretty clearly, but I am more curious about the right way to extend this into the complex plane. I may have done something wrong, but allowing x to be complex doesn't quite work. Another consideration would be splitting the sum into parts, with each part being of the same form as the $s(1,x)$. So for the angle $\frac{1}{6} \tau$, I've split into $$\sum_{n=0}^\infty (-1)^n x^{(3n)^2} +e^{\frac{1}{6} \tau} \sum_{n=0}^\infty (-1)^n x^{(3n+1)^2} + e^{\frac{4}{6} \tau} \sum_{n=0}^\infty (-1)^n x^{(3n+2)^2}$$
Sep 21, 2021 at 4:45 comment added Jorge Zuniga @CelebBriggs. That's correct $s(1,x) = 1/2$. I will extend my answer to prove this.
Sep 21, 2021 at 3:04 comment added Caleb Briggs Am I interpreting your equation correctly? When I compute $$\sum_{n=0}^\infty (-1)^n x^{n^2}=\frac{1}{\sqrt{4\pi\ln(x)}}\int_{0}^{\infty}\frac{e^{\left(-\frac{\ln^{2}(t)}{4\ln(x)}\right)}}{t(1+t)}dt$$ I get 1/2 everywhere when x>1. Similarly, there are lots of angles in the complex plane where the function becomes a constant 1/2 when |z|>1.
Sep 21, 2021 at 2:55 history edited Jorge Zuniga CC BY-SA 4.0
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Sep 21, 2021 at 2:39 history edited Jorge Zuniga CC BY-SA 4.0
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Sep 21, 2021 at 2:04 history edited Jorge Zuniga CC BY-SA 4.0
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Sep 21, 2021 at 1:49 history edited Jorge Zuniga CC BY-SA 4.0
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Sep 18, 2021 at 23:37 vote accept Caleb Briggs
Sep 17, 2021 at 22:16 comment added Caleb Briggs This is an amazing answer! I'll take a more thorough look at it as soon as I have time
S Sep 17, 2021 at 22:13 history suggested Caleb Briggs CC BY-SA 4.0
If you like, I've rewritten this answer in latex. You can choose to keep the edits or not, depending on your preference
Sep 17, 2021 at 22:08 review Suggested edits
S Sep 17, 2021 at 22:13
Sep 17, 2021 at 20:39 history edited Jorge Zuniga CC BY-SA 4.0
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Sep 17, 2021 at 20:19 history answered Jorge Zuniga CC BY-SA 4.0
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