Timeline for Analytic continuation for disjoint domains
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 8, 2023 at 18:20 | comment | added | Christian Remling | @RichardDiagram: Glad to hear this was of some value for you, that's more than one can usually hope for when posting something on MO. | |
| Mar 8, 2023 at 0:37 | comment | added | Richard Diagram | @ChristianRemling Thanks again for this answer. This gave me a good reality check. I had a feeling I was doing something wrong, but I wasn't sure what I was doing wrong. Deepest thanks! | |
| Mar 8, 2023 at 0:32 | comment | added | Richard Diagram | @ChristianRemling You have pointed out an obvious error. But it didn't affect the work I am trying to do in the grand scheme. This actually made my other work make more sense. The integral $\int_C f_2(w)/(w-z)\,dw = 3 \pi i$, and is constant in $z$. The work I was trying to do still works out though. I was trying to write the reflection formula $f_1(z) = f_1(0) + \frac{1}{2\pi i} \int_C f_2(w)/(w-1/z) \, dw$, which comes out much much cleaner now. I think I just got ahead of myself when I wrote this question, and I was confused why the numbers were working but it contradicted Cauchy's theorem. | |
| Mar 1, 2023 at 18:01 | comment | added | Christian Remling | The calculation Conrad outlined shows that $\int_C f_2(w)/(w-z)\, dw$ is constant on $|z|<2$. This should also follow from the fact that $f_2$ is holomorphic at $\infty$. (If $g\in H^p(D)$, then the Cauchy integral is zero for $z$ outside $\overline{D}$.) | |
| Mar 1, 2023 at 4:04 | comment | added | Richard Diagram | Oh! Okay, this makes sense. I definitely made a typo somewhere. And in my Cauchy formula there's some $\delta$ factor which is added in once the contour passes the wall of singularities. I think the fact you noticed $3/2 = 1/2 + 1$ makes a lot of sense as I think about it. I was confused as to how this could possibly be an analytic continuation. I had a suspicion I was doing something wrong! Thanks a lot! | |
| Mar 1, 2023 at 4:00 | vote | accept | Richard Diagram | ||
| Mar 1, 2023 at 2:20 | comment | added | Conrad | one can easily compute the integral in general since one can write everything as geometric series; eg above $\frac{\zeta^{n-1}}{\zeta^n+1}=\sum_{k \ge 0} (-1)^k\zeta^{-kn-1}$ for $n \ge 1$ and that integrates to zero unless $k=0$ and in general same thing except that one has a series for $1/(\zeta-z)=\sum z^k/\zeta^{k+1}$ and one for $\zeta^n/(1+\zeta^n)=\sum (-1)^k\zeta^{-kn}$ unless $n=0$ etc | |
| Mar 1, 2023 at 1:29 | history | answered | Christian Remling | CC BY-SA 4.0 |