Timeline for What is the relationship between $\sum_{n=0}^\infty f(n) x^n$ and $-\sum_{n=1}^\infty f(-n) x^{-n}$?
Current License: CC BY-SA 4.0
9 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
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 |
added 185 characters in body
|
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 |
added 693 characters in body
|
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 |