4
$\begingroup$

Let an analytic element be a power series associated with an open disc of the complex plane over which the series is convergent. W.l.o.g. assume the series is a Taylor expansion about the center of the disc. It is easy to show that analytic continuation is an equivalence relationship between analytic elements.

Given two analytic elements $(U, f)$ and $(V, g)$, how can we tell whether they are in the same class? Obviously if $f$ and $g$ are both polynomials of degrees at most $n$ we could see if the two are in the same class by evaluating at $n + 1$ points. On the other extreme: even comparing coefficients of two general series may take infinite steps. So let me bifurcate the question:

  • Without concern for computability, what can we say in general on how to decide if two analytic elements are in the same class?

  • What restrictions do we need to restore computability to this question?

Improve this question
$\endgroup$
13
  • 5
    $\begingroup$ Not even equality on reals is decidable, how could this be? $\endgroup$
    – Arno
    Sep 4, 2021 at 19:32
  • 4
    $\begingroup$ @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 $V$? $\endgroup$
    – Alex M.
    Sep 4, 2021 at 19:44
  • $\begingroup$ 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. $\endgroup$
    – Community Bot
    Sep 4, 2021 at 19:44
  • $\begingroup$ @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? $\endgroup$
    – zhtprog
    Sep 4, 2021 at 20:17
  • 1
    $\begingroup$ @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. $\endgroup$
    – Arno
    Sep 4, 2021 at 23:00

1 Answer 1

Reset to default
3
$\begingroup$

!!Without regard for computability!!
One classic criterion is this. $(U,f)$ analytically continues to $(V,g)$ iff there is a finite sequence $(U_k,f_k), k=0,\dots,n$ of analytic elements, with $(U_0,f_0) = (U,f)$ and $(U_n,g_n) = (V,g)$ where, for $k=1,\dots,n$, we have $U_{k-1}\cap U_k \ne \varnothing$ and $f_{k-1}(z) = f_k(z)$ for all $z \in U_{k-1}\cap U_k$.

A simple example.
Take $(U,f)$ where $f(z) = \log z$ where $f(1) = 0$ and $U$ is the disk $|z-1|<1$.
Take $(V,g)$ where $g(z) = \log(-z)+C$ where $g(-1) = C$, $C$ a fixed constant, and $V$ is the disk $|z+1|<1$.
Question: Is $(U,f)$ continuable to $(V,g)$?
Answer: If and only if $C$ is an odd integer multiple of $i\pi$.

Improve this answer
$\endgroup$
2
  • 6
    $\begingroup$ Is not this just the definition of analytic continuation? $\endgroup$
    – Alexandre Eremenko
    Sep 5, 2021 at 12:53
  • 2
    $\begingroup$ @AlexandreEremenko Of course. At least is some texts. $\endgroup$
    – Gerald Edgar
    Sep 5, 2021 at 13:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.