Timeline for Does this method analytically continue gap series series?
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7 events
| when toggle format | what | by | license | comment | |
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| Jul 18, 2021 at 0:34 | comment | added | Joseph Van Name | Then $F$ is a generalized analytic continuation of $f$ to a holomorphic function with domain $\mathbb{C}\setminus S^{1}$, but $F$ is the continuation of $f$ defined by the functional equation $F(z)=F(1/z)$. This means that without being careful, one can generalized analytically continue one function to any other function. | |
| Jul 18, 2021 at 0:34 | comment | added | Joseph Van Name | Here is a cautionary tale about what could possibly go wrong with generalized analytic continuation (I came up with this example when trying to produce a generalized analytic continuation to answer this question). Suppose $f$ is holomorphic on the unit disk with $f(0)=0$. Then let $f(z)=\sum_{k=1}c(k)x^{k}$ for some $c$. Define $F(z)=\sum_{k=1}^{\infty}(c*T(\mu))(k)\frac{z^{k}}{(1-z^{k})^{2}}$ where $*$ is the Dirichlet convolution, $\mu$ is the Mobius-Mu function, and $T$ is the transform defined by letting $T(b)(k)=k\cdot b(k)$. | |
| Jul 17, 2021 at 22:48 | comment | added | Joseph Van Name | There are notions of generalized analytic continuation where for example, if $A$ is a countable subset of $S^{1}$, $\sum_{a\in A}|c_{a}|<\infty$, and $f(z)=\sum_{a\in A}\frac{c_{a}}{x-a}$, and $g$ is an entire function, then one can coherent extend $f+g$ from inside $S^{1}$ to outside $S^{1}$ since one can recover $\{(a,c_{a})\mid a\in A\}$ from the restriction of $f+g$ to the inside of $S^{1}$ since $c_{a}=\lim_{r\rightarrow 1^{-}}(1-r)(f+g)(ra)$ for each $a\in A$. See this Master's thesis for more details. scholarspace.manoa.hawaii.edu/bitstream/10125/29513/1/… | |
| Jun 17, 2021 at 13:29 | comment | added | Alexandre Eremenko | @Caleb Briggs: by definition of an analytic continuation. The functions must agree on an open set. | |
| Jun 16, 2021 at 17:36 | comment | added | Wojowu | @CalebBriggs The functions have to agree on some connected open set. Without this condition the idea of analytic continuation is useless, because pretty much any function can be analytically continued to any other | |
| Jun 16, 2021 at 17:27 | comment | added | Caleb Briggs | For what reason does a natural boundary stop an analytical continuation? Since the function and the proposed continuation agree at angles which are a rational multiples of $\pi$, then they agree in a dense set. Isn't this enough-- or do the functions have to agree on a larger set to be a continuation? | |
| Jun 16, 2021 at 16:03 | history | answered | Alexandre Eremenko | CC BY-SA 4.0 |