Timeline for Is there a decision procedure for analytic continuation?
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Sep 5, 2021 at 12:54 | comment | added | LSpice | Please use TeX, not triple backquotes, to delimit math. I have edited accordingly. | |
| Sep 5, 2021 at 12:53 | history | edited | LSpice | CC BY-SA 4.0 |
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| Sep 5, 2021 at 12:45 | answer | added | Gerald Edgar | timeline score: 3 | |
| Sep 5, 2021 at 2:03 | comment | added | zhtprog | @Arno, see quantifier elimination for example. One can reduce one representation to another. Deciding the truthfulness of a formula like: $ \exists x\in \mathbb {R} .(a\neq 0\wedge ax^{2}+bx+c=0)\ \ \Longleftrightarrow \ \ a\neq 0\wedge b^{2}-4ac\geq 0 $ is not entirely vacuous and is considered arithmetic in my book. | |
| Sep 4, 2021 at 23:00 | comment | added | Arno | @zhtprog Ok, so what logical formalism do you want to use? If you put arithmetic in, is all very undecidable straight away. If you don't put arithmetic in, how can you make sense of powerseries? You might get away with a two-sorted setting using a very weak arithmetic on the natural numbers, and treat the complex numbers separately. But I still think the question you ought to be asking is about $\Sigma^1_1$-completeness. | |
| Sep 4, 2021 at 21:32 | comment | added | zhtprog | @Arno, I am thinking of decidability as showing whether two logic formulae using defined primitives and logical connectives can be transformed into each other through finite amount of deductions. | |
| Sep 4, 2021 at 21:19 | comment | added | zhtprog | @Igor Thanks for the comment. I want to be sure I am not missing already discovered algorithms (even if it only works in some highly restricted situations). | |
| Sep 4, 2021 at 21:18 | comment | added | Arno | @zhtprog Your edit makes less sense than the previous version. Do you want to restrict your powerseries to having rational (or maybe algebraic coefficients) to make your argument for polynomials actually go through? Or maybe taking a step back, are you looking for an algorithm in a formal sense at all? | |
| Sep 4, 2021 at 21:12 | comment | added | zhtprog | @Arno, I am not familiar with the complexity question you raised. I am interested in formalizing complex analysis/Riemann surface using power series and continuation and what kind of proof automation can be done in such a framework. | |
| Sep 4, 2021 at 21:04 | history | edited | zhtprog | CC BY-SA 4.0 |
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| Sep 4, 2021 at 21:03 | comment | added | Igor Khavkine | The practical answer is to look for an alternative representation for the analytic elements, such that they would have overlapping domains of convergence where they could be compared. E.g., the some integral representations converge in a half-plane, other (non-power) series approximations converge on non-disc domains, etc. I doubt that this approach can be made into an algorithm, but it does sometimes work in practice. | |
| Sep 4, 2021 at 20:27 | review | Close votes | |||
| Sep 12, 2021 at 3:04 | |||||
| Sep 4, 2021 at 20:24 | comment | added | Arno | @zhtprog "How can we tell" is either incredibly vague, or the answer is: "We can't." If you want a meaningful question, you could ask "What is the complexity (in the sense of descriptive set theory) of having a joint analytic continuation?". The definition itself is a $\Sigma^1_1$, so if the problem is indeed $\Sigma^1_1$-complete, there is no simpler way to express this. | |
| Sep 4, 2021 at 20:17 | comment | added | zhtprog | @Arno,@Alex is right. I will add his comment to the question to clarify. Obviously if the A.E.s are polynomials of degrees at most n, we can just pick any n+1 points on the plane to see if they are equivalent. If such questions can not be answered for general A.E.s what restrictions do we need? For example what if the power series f: N->C is inductively defined? | |
| Sep 4, 2021 at 19:44 | comment | added | CommunityBot | Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. | |
| Sep 4, 2021 at 19:44 | comment | added | Alex M. | @Arno: Maybe the OP does not have computability in mind. Maybe he thinks of: if $(U,f)$ and $(V,g)$ are analytic elements, and $U \cap V = \emptyset$, how can we tell whether there exists $(U \cup V, F)$ such that $F = f$ on $U$ and $F = g$ on $V$? | |
| Sep 4, 2021 at 19:32 | comment | added | Arno | Not even equality on reals is decidable, how could this be? | |
| Sep 4, 2021 at 19:31 | history | edited | zhtprog | CC BY-SA 4.0 |
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| S Sep 4, 2021 at 19:12 | review | First questions | |||
| Sep 4, 2021 at 19:44 | |||||
| S Sep 4, 2021 at 19:12 | history | asked | zhtprog | CC BY-SA 4.0 |