http://www.theguardian.com/science/alexs-adventures-in-numberland/2014/aug/1... Explanations of the work of the medallists
On 14 Aug 2014, at 12:32 am, math-fun-request@mailman.xmission.com wrote:
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Today's Topics:
1. Re: Alternatoids? (Dan Asimov) 2. Re: Dumping water (Warren D Smith) 3. Re: Steiner chain constraints (Fred Lunnon) 4. junk DNA - the case of the Antarctic Midge (Warren D Smith) 5. Maryam Mirzakhani first female Fields Medal winner (Tehran-Iran born prof. at Stanford) (Warren D Smith) 6. Re: Dumping water (Eugene Salamin) 7. Re: Steiner chain constraints (Fred Lunnon)
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Message: 1 Date: Tue, 12 Aug 2014 17:15:35 -0700 From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Alternatoids? Message-ID: <F9A6EADB-5E46-4604-9EA2-227D5EE3E7B5@earthlink.net> Content-Type: text/plain; charset=us-ascii
Actually, a special case of Moufang loops are groups.
The statement below refers to only the Moufang loops of size <= 31 that are *not* groups -- i.e., that are not associative.
--Dan
On Aug 12, 2014, at 5:05 PM, Dan Asimov <dasimov@earthlink.net> wrote:
A paper states: "There are 13 [Moufang loops of size <= 31]: one of order 12, five of order 16, one of order 20, five of order 24, and one of order 28.
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Message: 2 Date: Tue, 12 Aug 2014 20:16:35 -0400 From: Warren D Smith <warren.wds@gmail.com> To: math-fun@mailman.xmission.com Subject: Re: [math-fun] Dumping water Message-ID: <CAAJP7Y14zigtfWo9uSrSD_odnk-Sqp=N2xPuQMqe-6+pzf_2YQ@mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1
I had ignored pressure, i.e. was regarding the water in the bottle as both shallow and far wider than the hole.
Yes, Eugene is correct, if the bottle were very tall there would be high pressure and this eventually should overcome any surface tension attempt to hold it in. (You could easily modify my maths to incorporate pressure...)
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Message: 3 Date: Wed, 13 Aug 2014 01:51:13 +0100 From: Fred Lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Steiner chain constraints Message-ID: <CAN57YqtsNfdj5KPnuMPrndjy1JEsP2_Q1YbNgVrwvz1Sr6w85g@mail.gmail.com> Content-Type: text/plain; charset=UTF-8
Aaarghhh! Lost the mods --- should have read <<
So the constraints for n = 5 should have read: 5 cyclic shifts (fixing k0,k6 ) of k1^2 + k3*k5 + k2*k4 + k2*k3 = k3^2 + k5*k2 + k4*k1 + k1*k2 ; and 5 shifts of 4*k5^2 - 6*k4*k1 + 2*k2*k3 + k0*k5 + k6*k5 = constant .
WFL
On 8/12/14, Fred Lunnon <fred.lunnon@gmail.com> wrote: Ahah! RWG's second set of Steiner 5-ring constraints failed my numerical tests because of a wrong sign for k0 --- I assume curvatures have sign built-in, allowing for circles to evert under transformation. I didn't initially suspect this problem, since by Murphy's law the dodgy sign passed when tested on the canonical concentric ring.
So the constraints for n = 5 should have read: 5 cyclic shifts (fixing k0,k6 ) of k1^2 + k3*k5 + k2*k4 + k2*k3 = k3^2 + k5*k2 + k4*k1 + k1*k2 ; and 5 shifts of 4*k5^2 - 6*k4*k1 + 2*k2*k3 - k0*k5 + k6*k5 = constant .
[My updated versions --- I can't be bothered to correct all the signs in his original! Could the first set also be pruned to some equivalent LHS = const form perhaps?]
Unfortunately, this makes no difference to the rank, which remains 5 rather than 4; I remain baffled by this discrepancy.
Fred Lunnon
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Message: 4 Date: Tue, 12 Aug 2014 20:53:27 -0400 From: Warren D Smith <warren.wds@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] junk DNA - the case of the Antarctic Midge Message-ID: <CAAJP7Y2OmVDFeE=4bicmK6yQznnTjRhb2foa+e5q9YPX1O8-iA@mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1
The Antarctic Midge, Belgica antarctica, a small insect native to antarctica, just had its genome sequenced, result 99 million base pairs and about 13500 genes (7333 BPs per gene).
The Anopheles mosquito has 278 million base pairs and approximately 13683 genes. (20317 BPs/gene.)
The Mountain Grasshopper Podisma pedestris has 16600 million base pairs and probably about the same number of genes as mosquito. (1213184 BP/gene.)
Humans: 3200 million BP, about 23000 genes. (139130 BP/gene.)
Some of the largest animal genomes at present are the small arctic crustacean, Ampelisca macrocephala with 63200 million base pairs, and the marbled lungfish, Protopterus aethiopicus, with 133000 million.
If crustacean does not have as many genes as human(?) then at least 2747826 BPs/gene.
So anyhow, some animals have 400 times as much DNA per gene, as other animals.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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Message: 5 Date: Tue, 12 Aug 2014 21:36:46 -0400 From: Warren D Smith <warren.wds@gmail.com> To: math-fun@mailman.xmission.com Subject: [math-fun] Maryam Mirzakhani first female Fields Medal winner (Tehran-Iran born prof. at Stanford) Message-ID: <CAAJP7Y0wBha-g-_c2b3QtzB7=tidE80e-eAk6kcr=vQK4kkt1A@mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1
and what did she actually do?
http://en.wikipedia.org/wiki/Maryam_Mirzakhani says some stuff, but basically I did not understand a word of this.
http://www.mathunion.org/fileadmin/IMU/Prizes/2014/news_release_mirzakhani.p... contains a few actually comprehensible claims.
Despite Iran culture and education system allegedly being anti-female, unlike the enlightened West, you'll notice the first female Fields medalist came from Iran.
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Message: 6 Date: Tue, 12 Aug 2014 20:06:02 -0700 From: Eugene Salamin <gene_salamin@yahoo.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Dumping water Message-ID: <1407899162.58290.YahooMailNeo@web162105.mail.bf1.yahoo.com> Content-Type: text/plain; charset=utf-8
The pressure difference P across a surface of radius of curvature r with surface tension ?is
P = 2?/r.
The hydrostatic pressure of a fluid of density ? with column height h in gravity g is
P = ?gh.
Suppose there is a circular hole of radius R in the bottom of the jar.? A bead of fluid extends beneath the hole, and if R << h, we may neglect the variation of hydrostatic pressure within the bead, so that the bead is a spherical cap of radius of curvature r.
Setting the pressures equal,
rh = 2?/?g.
For the bead to be stable, r >= R, so there is a maximum column height H with
RH = 2?/?g.
On Earth g = 9.8 m/s^2, and for water, ? = 1000 kg/m^3, ? = 0.073 N/m.
Thus RH = 1.5e-5 m^2 = 15 mm^2.
For example, a 1 mm radius hole should support a 15 mm column height.? Note that one must use a jar, not a tube, in order to avoid the attractive capillary force.
? --? Gene
________________________________ From: James Propp <jamespropp@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Monday, August 11, 2014 9:14 PM Subject: [math-fun] Dumping water
Say I have a jar full of water covered by a thin plate. Then I turn the jar upside down, holding the plate firmly against the lip of the jar to prevent water from spilling. Then I whisk away the plate so that instead of the plate pushing against the water, only the air beneath the jar is pushing upward.
Of course, the water will leave the jar. But what will the geometry of the process be? The water can't leave as a cylindrical slug; intuitively, it seems that the process has instability, so that spontaneous fingering in the air-water interface will break the initial cylindrical symmetry.
Maybe Doc Edgerton made high-speed pictures of this?
Jim Propp
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Message: 7 Date: Wed, 13 Aug 2014 15:31:42 +0100 From: Fred Lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Steiner chain constraints Message-ID: <CAN57YqsAsqYBJo5DpC6gx_hxugKaEOSdUESTaCT0a-H8h1G5kA@mail.gmail.com> Content-Type: text/plain; charset=UTF-8
And so to hunt missing relations when n = 4 . Via brute-force equation-solving I discovered that
(k0 + k5) - (k1 + k3) , (k0 + k5) - (k2 + k4) , [RWG]
(k0 - k5)^2 + 4*(k3 + k4)*(k0 + k5) + 4*(k3^2 + k4^2) , (k0 - k5)^2 + 4*(k2 + k3)*(k0 + k5) + 4*(k2^2 + k3^2) , (k0 - k5)^2 + 4*(k1 + k2)*(k0 + k5) + 4*(k1^2 + k2^2) , (k0 - k5)^2 + 4*(k4 + k1)*(k0 + k5) + 4*(k4^2 + k1^2) ,
all vanish (verified numerically).
But the Jacobian (again!) has rank 5, instead of 3. Flubber, flubber ...
Fred Lunnon
On 8/13/14, Fred Lunnon <fred.lunnon@gmail.com> wrote: Aaarghhh! Lost the mods --- should have read <<
So the constraints for n = 5 should have read: 5 cyclic shifts (fixing k0,k6 ) of k1^2 + k3*k5 + k2*k4 + k2*k3 = k3^2 + k5*k2 + k4*k1 + k1*k2 ; and 5 shifts of 4*k5^2 - 6*k4*k1 + 2*k2*k3 + k0*k5 + k6*k5 = constant .
WFL
On 8/12/14, Fred Lunnon <fred.lunnon@gmail.com> wrote: Ahah! RWG's second set of Steiner 5-ring constraints failed my numerical tests because of a wrong sign for k0 --- I assume curvatures have sign built-in, allowing for circles to evert under transformation. I didn't initially suspect this problem, since by Murphy's law the dodgy sign passed when tested on the canonical concentric ring.
So the constraints for n = 5 should have read: 5 cyclic shifts (fixing k0,k6 ) of k1^2 + k3*k5 + k2*k4 + k2*k3 = k3^2 + k5*k2 + k4*k1 + k1*k2 ; and 5 shifts of 4*k5^2 - 6*k4*k1 + 2*k2*k3 - k0*k5 + k6*k5 = constant .
[My updated versions --- I can't be bothered to correct all the signs in his original! Could the first set also be pruned to some equivalent LHS = const form perhaps?]
Unfortunately, this makes no difference to the rank, which remains 5 rather than 4; I remain baffled by this discrepancy.
Fred Lunnon
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