Timeline for nth-order generalizations of the arithmetic-geometric mean
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Dec 27, 2022 at 10:20 | comment | added | Martin Sleziak | I have added at least a Wayback Machine link instead of the broken link - but it seems that the paper can be found in various places. J. M. Borwein and P. B. Borwein: A Cubic Counterpart of Jacobi's Identity and the AGM; DOI: 10.1090/S0002-9947-1991-1010408-0, DOI: 10.2307/2001551; MR1010408, Zbl 0725.33014. | |
| Dec 27, 2022 at 10:15 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added a Wayback Machine link for the dead link
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| Nov 28, 2011 at 15:41 | comment | added | J. M. isn't a mathematician | @Will: IIRC Bille Carlson derived a "nice" integral expression for the surface area of an ellipsoid; I'll post it when I find it. | |
| Nov 15, 2011 at 5:15 | answer | added | Noam D. Elkies | timeline score: 15 | |
| Nov 15, 2011 at 4:28 | comment | added | Will Sawin | Or I could just find it myself. Since people upvoted this comment, presumably they care: en.wikipedia.org/wiki/Landen%27s_transformation. One just repeatedly applies this and then uses the formula for the circumference of a circle. If there is some integral representation for this mean, it would presumably come from a similar type of transformation. Has anyone checked that it's not the surface area of an ellipsoid in $n$ dimensions? | |
| Nov 14, 2011 at 21:39 | comment | added | Will Sawin | Does anyone have a reference for Gauss's proof (or a modern version) of the statement about elliptic integrals? | |
| Nov 14, 2011 at 17:47 | answer | added | M-Maesumi | timeline score: 2 | |
| Sep 3, 2010 at 16:05 | answer | added | KConrad | timeline score: 22 | |
| Sep 3, 2010 at 12:36 | comment | added | J. M. isn't a mathematician | For Gerry: It seems Plouffe's is unable to recognize even outputs from the classical AGM; I am thus not surprised it does not recognize outputs for the nth-order generalizations. | |
| Sep 3, 2010 at 12:06 | history | edited | J. M. isn't a mathematician | CC BY-SA 2.5 |
Edited question, taking Darsh's comments into account
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| Sep 3, 2010 at 7:49 | comment | added | Darsh Ranjan | J. M. - not at all. I'm guessing there are lots of interesting generalizations of the AGM (not that I know anything about them). My point was just that the particular "generalizations" you chose don't seem very relevant. Actually, I think your second set of sequences becomes more interesting if you just change the denominators from (4,3,2) to (4,6,4). (Note that if the sequences have a common limit x, then the next iterate of (x,x,x,x) had better be (x,x,x,x).) | |
| Sep 3, 2010 at 7:03 | comment | added | J. M. isn't a mathematician | Hans, thanks for the note; I always had the impression that Gauss did this in his teens (in connection with rectifying the lemniscate of Bernoulli) before anyone else cared. | |
| Sep 3, 2010 at 6:45 | comment | added | Hans Lundmark | A slightly off-topic remark: According to Stillwell's Mathematics and its history (p. 160 in the 1st ed.), Lagrange actually discovered (and published) the relation between elliptic integrals and the AGM long before Gauss. | |
| Sep 3, 2010 at 6:30 | comment | added | J. M. isn't a mathematician | Darsh, thanks for the notes, I'll retry my experiments later and see if I can confirm what you said. So your point is that AGM is "special" in that the n>2 cases cannot exhibit interesting behavior? | |
| Sep 3, 2010 at 6:21 | comment | added | Darsh Ranjan | Your second attempt falls to a similar elementary counterargument: the four sequences cannot have a nonzero common limit. You can make it more interesting by defining your sequences so that when all four numbers start out the same, you get stationary sequences. :-) | |
| Sep 3, 2010 at 6:18 | comment | added | Darsh Ranjan | It's easy to see that your first example can have no common limit but zero when n isn't 2: (a, b) = ((a+b)/n, (ab)^(1/n)) immediately implies a = (n-1)a, so a is 0 unless n-1 = 1 (i. e., n = 2). Then of course b = 0 from the second component. | |
| Sep 3, 2010 at 6:15 | comment | added | J. M. isn't a mathematician | Hmm, now that you mention it, I haven't gotten around to feeding the stuff I got to Plouffe's beastie. I'll try later when I see my notebooks again. Thanks for reminding me Gerry! | |
| Sep 3, 2010 at 5:45 | comment | added | Gerry Myerson | Have you worked through any examples? If you get a numerical limit, there are places where you can look it up to see whether it's a "recognizable" number. | |
| Sep 3, 2010 at 4:01 | history | asked | J. M. isn't a mathematician | CC BY-SA 2.5 |