Timeline for How does Mathematica understand branchcuts of the complex logarithm?
Current License: CC BY-SA 3.0
21 events
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| Aug 8, 2015 at 20:20 | history | edited | chris | CC BY-SA 3.0 | deleted 14 characters in body |
| Jul 5, 2015 at 5:36 | comment | added | Jens | If I understand you correctly, this is already included in my definition. You just have to change the angle $\sigma$ by $2\pi$. E.g., $\sigma=0$ and $\sigma=-2\pi$ are what I think you call continuous from above and below (I just don't think one should call this above and below because that's ambiguous for non-horizontal branch cuts). | |
| Jul 4, 2015 at 0:32 | comment | added | J. M.'s missing motivation | Currently. Log[] is "continuous from above", because evaluating it at the branch cut gives a result with positive imaginary part. To get the corresponding function that is "continuous from below", we use -Log[1/z], as whuber demonstrated in another answer, where evaluating at the cut gives a result with negative imaginary part. (Sometimes you'll see the notation $\log(-x\pm i0)=\log\, x \pm i\pi$ denoting this choice of continuity.) I'm thinking your arg and log can use this additional flexibility. :) | |
| Jul 3, 2015 at 21:23 | comment | added | Jens | @Guesswhoitis. not sure what you mean. Do you want a multivalued function? | |
| Jul 3, 2015 at 17:16 | comment | added | J. M.'s missing motivation | For full flexibility, you might consider generalizing arg[] and log[] so that one can choose them to be continuous either from above or below the chosen branch cut. | |
| Apr 4, 2014 at 0:00 | history | bounty awarded | user6818 | ||
| Apr 2, 2014 at 14:16 | vote | accept | user6818 | ||
| Apr 1, 2014 at 2:51 | comment | added | Jens | @user6818 No, that's not quite right. The choice of contour in complex integration is actually the way in which you, the user, define the quantity you want Mathematica to compute. This is becoming more of a math discussion, so I don't think it's fruitful to continue in the context of the original question. | |
| Apr 1, 2014 at 0:29 | comment | added | user6818 | If Mathematica is not choosing a contour then how is it computing the contour integral? Is there any valid meaning to this integral without a choice of branch avoiding contour and an associated limiting prescription? Thats is why I asked if Mathematica's answer is something meaningful? Because if it is not choosing a proper branch avoiding contour and a limiting process, it means it doesn't understand what a branch-cut is - right? Because technically one cannot just integrate across a branch-cut - right? | |
| Mar 30, 2014 at 23:13 | comment | added | Jens | Mathematica cannot know along which contour you want to integrate. You have to tell it. Your question is about Mathematica's understanding of branch cuts of Log, and I explained that they arise from the phase angle convention. But Log is only one of infinitely many choices for the inverse of the Exp function, and all others can be generated by fixing a different $\phi$ interval for the phase angles. The quantity $\sigma$ is the lower bound of that interval of length $2\pi$, as seen in the definition of log above. It's pretty standard textbook stuff, but looks nice in your example. | |
| Mar 30, 2014 at 23:01 | comment | added | user6818 | And how do I understand your $\sigma$ - what is the graphical/physical significance of that number? How do I see that physically/graphically? Is there a geometrical way to see this $\sigma$? Why did you think of it? | |
| Mar 30, 2014 at 22:52 | comment | added | user6818 | Thanks for the efforts! So if I let the branchcut be the default - as in upwards along the positive imaginary axis starting from point $ai$ then doing this contour integration on the UHP semicircle would have entailed taking a key-hole contour with the hole about the branchpoint and a semicircular bump about each pole on the positive imaginary axis on both sides of the key etc - so is Mathematica's answer when asked to integrate $f$ (your notation) along the semicircle the same? I mean when Mathematica is integrating $f$ is it taking into account the branch point at $ai$ and poles of Tanh? | |
| Mar 30, 2014 at 21:25 | history | edited | Jens | CC BY-SA 3.0 | Added gif animation |
| Mar 30, 2014 at 21:16 | history | edited | Jens | CC BY-SA 3.0 | Uploaded wrong image |
| Mar 30, 2014 at 18:02 | comment | added | Jens | @user6818 I added more information to my answer, although I had to guess what it is that you really want to know. | |
| Mar 30, 2014 at 17:55 | history | edited | Jens | CC BY-SA 3.0 | Added comments on the more mathematical issues of the question |
| Mar 26, 2014 at 22:04 | comment | added | Jens | I may have time to get back to this later, but in the meantime you may want to look at this post: How to calculate contour integrals with Mathematica | |
| Mar 26, 2014 at 6:00 | comment | added | user6818 | In your last plot all the red lines are branch cuts -right? Then you have moved the branchcut along the x-axis? That will complicate matters - because I want to understand via complex analysis what is the behaviour of the integral of the function along the real line - thats the main objective... | |
| Mar 26, 2014 at 5:56 | comment | added | user6818 | Thanks or this amazing answer - I simply don't know so much Mathematica! Can you say if my conclusions were right that I got using Mathematica about the $A$ dependence and the vanishing of the integral about the branchpoint? | |
| Mar 26, 2014 at 4:17 | history | edited | Jens | CC BY-SA 3.0 | Moved branch cuts |
| Mar 26, 2014 at 4:06 | history | answered | Jens | CC BY-SA 3.0 |