Timeline for How do we know that analytic continuation agrees with UV regulators?
Current License: CC BY-SA 4.0
17 events
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| Jun 22, 2022 at 8:14 | history | edited | CommunityBot |
replaced http://www.mat.univie.ac.at/~neum with http://arnold-neumaier.at
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| Jul 16, 2019 at 17:44 | history | edited | Arnold Neumaier | CC BY-SA 4.0 |
added a comment on nonuniqueness
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| Jul 16, 2019 at 16:27 | comment | added | user76284 | @Arnold Thanks. To me, the idea of extending summation to divergent series seems analogous to the idea of extending powers to zero, negative, fractional, irrational, and then complex exponents. That is, “multiplying something by itself pi times” and “multiplying something by itself i times”, may seem like nonsense and (at one time) be considered meaningless, but they turn out to have natural, meaningful answers. | |
| Jul 16, 2019 at 10:53 | history | edited | Arnold Neumaier | CC BY-SA 4.0 |
added comment on rigorous summation methods
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| Jul 16, 2019 at 10:52 | comment | added | Arnold Neumaier | @user76284: see the addition at the end on my answer. | |
| Jul 15, 2019 at 18:12 | comment | added | user76284 | @ArnoldNeumaier But why do different summation methods yield the same answer for many of these divergent series? Doesn't that indicate there's something meaningful about those answers? | |
| Jul 15, 2019 at 10:57 | comment | added | Arnold Neumaier | @user76284: yes, for series that are convergent but not absolutely convergent. But for divergent series you don't even need to reorder, just bracket in different ways.... | |
| Jul 1, 2019 at 20:24 | comment | added | user76284 | @ArnoldNeumaier Doesn’t reordering change the value of convergent series too, by the Riemann series theorem? | |
| Aug 22, 2016 at 2:04 | vote | accept | knzhou | ||
| Jul 18, 2016 at 21:45 | history | bounty ended | CommunityBot | ||
| Jul 15, 2016 at 8:11 | history | edited | Arnold Neumaier | CC BY-SA 3.0 |
added remark about the relevance of the context
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| Jul 15, 2016 at 8:07 | comment | added | Arnold Neumaier | Detached from physical content, a divergent series is just that - divergent and hence meaningless. The series $\sum_{i=0}^\infty (-1)^i$ can be given any integer value by appropriately reordering and taking partial sums, and all these are different from the value you get by interpreting it as a limit of a geometric series. It is only the context that makes the value of a divergent series possibly well-defined. | |
| Jul 15, 2016 at 0:58 | comment | added | knzhou | This doesn't give me much faith in the method though... how do I know when it will work, if I see a divergent series "in the wild", detached from physical context? | |
| Jul 15, 2016 at 0:54 | comment | added | knzhou | So the conclusion from my contradictory result trying to subtract $1+1+...$ from $\int 1 dx$ is that such a series/counterterm would never arise in a physical problem? | |
| Jul 12, 2016 at 9:00 | comment | added | Arnold Neumaier | @knzhou: just taking a series or an integral, renormalization is ill-defined. To know what the result should be one must start with a well-defined expression whose limit is soughtr and then transform the parameters in this expression in a way that after the transformation the limit exists. Then the result should be regulator independent. But if one doesn't have a finite context to start with there is no way to know what one should do. | |
| Jul 11, 2016 at 20:27 | comment | added | knzhou | Thanks for the response! However, I attempted to regularize the vacuum energy contribution and got the wrong answer: in analogy with the series, it is $\int_0^\infty (1-r)^x dx = -1/\log(1-r) = 1/r-1/2 + O(r)$. The difference in energies is $-1/2$, rather than $+1/2$ with zeta regularization. Do you have an idea what went wrong, if there are extra terms I'm forgetting? | |
| Jul 11, 2016 at 12:48 | history | answered | Arnold Neumaier | CC BY-SA 3.0 |