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Dec 1, 2022 at 23:12 comment added Caleb Briggs I checked out the specific case of $\sum_{n=0}^\infty x^{n^k}$ in a previous question: mathoverflow.net/questions/395443/…. As a result, I'm more interested in functions like $\sum_{n=0}^\infty \frac{1}{n!} x^{n^k}$ because 1), these functions have nicer functional equations and 2) the function extends continuously to the boundary, so there seems to be a better chance a nice connection between the function inside and outside its natural boundary can be found.
Dec 1, 2022 at 22:32 history edited Sidharth Ghoshal CC BY-SA 4.0
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Dec 1, 2022 at 22:20 comment added Sidharth Ghoshal See my addition to the answer^
Dec 1, 2022 at 22:19 history edited Sidharth Ghoshal CC BY-SA 4.0
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Dec 1, 2022 at 20:27 comment added Caleb Briggs Hmm that is a very interesting idea, I will surely think about that (note that one has to multiply by an extra $\frac{1}{2 \sqrt{z}}$ to cancel out the residues). Interestingly, this gives the same (unappealing) value as my method along the real line $w>1$, which is a constant $\frac{1}{2}$, and I plan to look in detail later about the relationship when $w$ is a complex number
Dec 1, 2022 at 19:53 comment added Sidharth Ghoshal Actually we can get even smarter than that... we can use $\csc(\pi \sqrt{x})$ to filter terms and then your residue theorem trick. So I imagine something like $$ \frac{1}{2i} \int_{C} \csc(\pi \sqrt{z}) w^z dx = \sum_{n=0}^{\infty} (-1)^n w^{n^2}$$ (note we have a branch cut here so that makes things a little complex)
Dec 1, 2022 at 18:45 comment added Sidharth Ghoshal Interesting, these are theta function like objects. What happens if we drop the taylor term and just look at $\sum_{n=0}^{\infty} w^{n^2}$ does the residue trick yield anything well behaved function "reciprocal"-series?
Nov 30, 2022 at 22:14 comment added Caleb Briggs Actually, part of my initial interest in this question comes from trying to analyze lacunary functions. In particular, series like $\sum_{n=0}^\infty \frac{(-1)^n x^n}{n!} w^{n^k}$ can be continued (not analytically) along the real line by taking a contour starting from near the origin with rays in the directions $\sqrt[k]{i}$ and $\sqrt[k]{-i}$. The contour here continues to converge when $w$ is real and $|w|>1$, and this extension continuously extends the derivatives beyond the boundary. Thus, some of my motivation is understanding ill-conditioned applications of the residue theorem.
Nov 29, 2022 at 21:56 history answered Sidharth Ghoshal CC BY-SA 4.0