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Topic: Math professors of Univ Michigan endorsing Kepler Packing proof to arxiv
Replies: 22   Last Post: Jun 22, 2014 4:54 PM

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 plutonium.archimedes@gmail.com Posts: 18,572 Registered: 3/31/08
Math professors of Univ Michigan endorsing Kepler Packing proof to arxiv
Posted: Jun 17, 2014 2:55 PM

Alright, the reason Kepler Packing Problem KPP, was never proven in mathematics, is not because it is a difficult proof, but because there are two definitions involved that are very much cloudy, foggy and ill-defined. When you have one definition ill-defined is enough to prevent there ever being a math proof, but in the case of KPP, there are two ill-defined definitions:
(1) finite versus infinity with a borderline between them
(2) density definition

If we define density to be surface area contact by equal spheres, then the maximum kissing points is 12 and there is a theorem in mathematics that proves 12 is maximum. The theorem is called the Euler regular polyhedra formula of V-E+F = 2. In this definition of density, the surface area is the "Container involved". So if Kepler had stated his KPP as kissing point density, he probably could not have proven his own conjecture, because the Euler formula was not available, although Kepler could have reasoned that it is impossible for 13 unit spheres to surround and kiss a central sphere due to exhaustion of angles.

KEPLER PACKING PROVED:

Now to make the proof that 12 is the maximum kissing points allowed in sphere packing of Kepler and that no central sphere can have 13 kissing points uses the Euler regular-polyhedra formula of Vertices - Edges + Faces = 2. So that if a central sphere has surrounding it 13 spheres with kissing points violates this Euler formula.

But what I have neglected to mention over the course of the commentary on Kepler Packing is that the proof of the Moebius theorem used in 5 Color Mapping is proven also by the Euler formula. The Moebius theorem, if you remember states that the maximum allowed mutual adjacencies of countries is 4, since the fourth country surrounds at least one of the other three and encloses it to any further adjacency.

So here is a fine example of where a theorem-- Euler regular-polyhedra formula is crucial to Mapping and crucial to Packing. So why would the Euler formula be so versatile? The answer is simple and easy for each regular-polyhedra fits inside a sphere and the regular polyhedra is all about allowable angles for faces. A 13 kissing points of a sphere is not allowed by that formula of V-E+F=2, and that same formula does not allow 5 mutual adjacencies in a Moebius theorem.

The proof that a unit sphere has at most 12 surrounding unit spheres of kissing points and never 13 kissing points comes from that Euler formula. We model the dodecahedron where its 12 faces become the vertices of 12 kissing points on the sphere surface.

Perhaps an even easier proof is to realize that the 12 surrounding unit spheres are all in turn 6 spheres with 3 left and 3 right, or 3 up and 3 down but all the 12 spheres in turn are a ring of 6 around the central sphere. Now if 13 can be kissing points means that 7 would be in that ring instead of 6.  And we know from plane geometry that a regular heptagon is not constructible with compass and straightedge but that a regular hexagon is constructible.

128.57.. degrees internal angle for heptagon

120 degrees internal angle for hexagon

So we have 8.57.. degrees too much to support a 7th unit sphere kissing point.

QED

Now in modern times there is a silly objection to the above proof in that some coneheads of logic think that a higher dimension that kissing points start to vary is a objection. But such coneheads have never proven to math that a dimension higher than 3rd ever or even exists. In this textbook of "Correcting Math" it is proven that no 4th or higher dimension exists, but rubbish make belief. The proof that 3rd is the last dimension is proven that a 4th dimension causes the planes of 2nd dimension to be bent and have a 90 degree twist in the plane. So for anyone to object to a proof of kissing points because of higher dimensions is like someone objecting to something in "real life" because they staunchly believe in ghosts, witches and the fire breathing dragons.

However, if you do not define density as surface contact intersection, then density is defined as to Space as a container. And since Kepler and all his later day followers never defined "infinity" with precision, that there would never be a proof of KPP, so long as infinity is ill-defined.

If you take the unit Tractrix area and unit circle area and follow them out to Floor-pi*10^603, you discover there is a crossover in amount of area. That the area of circle is always greater except when it reaches floor-pi*10^603 and then the tractrix area becomes momentarily greater than the circle. In that crossover is the Infinity borderline.

So, to prove KPP, with a precision definition of what it means to be finite and infinite, that the 1*10^603 by 1*10^603 by 1*10^603 is a container, and the KPP question becomes, is the Hexagonal Closed Pack the most dense packing of that container (and all the smaller sized cube containers)?

Well, the answer surprises most people, in that a Hybrid form of Hexagonal Closed Pack is more dense than the pure Hexagonal Closed Pack, because near the walls, edges and corners, we can rearrange some of the Hexagonal Closed Pack to fit more unit spheres and make a denser packing.

I wrote awhile ago:

Alright, well, we could never have asked Kepler to define precisely infinity with a borderline, for he would never have had the ability to know the Tractrix and Circle area are equal at Floor-pi*10^603 and thus infinity border is 1*10^603. And then Kepler could have said for his Conjecture, that the hexagonal closed pack is the most dense packing ***for the majority of the packing*** of equal unit spheres in a container that is 1*10^603 by 1*10^603 by 1*10^603. Then, mathematicians would have looked to see if there was a more dense packing. And it turns out, that there is a more dense packing as we manipulate the last plane and last edge and last corners to fit more spheres in an otherwise pure Hexagonal Closed Pack. The most dense packing depends on the size of the container.

Now, if density is defined as how many other spheres can be packed onto a single sphere-- the kissing points, then in that density definition, the proof of Kepler Packing is a one paragraph proof that 12 kissing points is maximum. The surface of a single sphere would be considered "the container".

But Kepler had in mind something different for density as a container. Now if one does not define finite with infinity properly as a borderline, then empty space that is infinite empty space is meaningless, because the Space cannot tell you whether you are in "finite space" or infinite space. In total empty space, means there is only finite space, because no border was reached to say, now we are in infinite space.

So if we define Infinite Space as a borderline that is crossed and that forms a container such as the 1*10^603 by 1*10^603 by 1*10^603 cube.

Now we can get an induction notion of this container of Space by smaller cubes such as 10 by 10 by 10 or the 100 by 100 by 100 cube and ask whether the most dense packing is the same as we go higher in cubes?

It turns out that a Hybrid form of Hexagonal Closed Pack is the most dense packing possible for it allows more unit spheres into a container than the pure Hexagonal Closed Pack.

In the below old post of mine, I illustrate this Hybrid Hexagonal Closed Pack.

I wrote awhile ago:

Alright, in my prior post, the illustration was messed up in format, so here I hope to have cleared up view of the illustration.

Now I think that the maximum hybrid of hexagonal closed pack is going to add spheres in the corners and edges, but I fail to see or recollect in my memory how an entire plane of new spheres can be added
to a stock of hexagonal closed pack. If the body of the packing is hexagonal closed pack, I fail to see how an entire plane can be added in a hybrid. I think I came to that conclusion many years back when I first did these hybrid packings, but want to make sure of that memory.

Here is the illustration:

OOOO
?OOOOO

when the headroom of the ceiling face is large enough to scoot ?that upper layer to the rightward side then we can add a new sphere ?shown by the X

XOOOO
?OOOOO

And if the upper layer is full then we do the scooting on the second ?to top ?layer as such:
OOOOO ?
XOOOO

And here is the illustration of Oblong Hex when the square or cube has ?a 0.99 after the integer

O  9
?_O  8
?O   7
?_O  6
?O    5
?_O   4

The underline indicates a distance of 0.99 to the wall. Those ?are either spheres in 3D or circles in 2D and they are numbered ?for easy reference.
In the Oblong technique we scoot 5 downwards leaving a gap ?large enough to pack a new sphere or circle. We leave 6 undisturbed ?but move 7 downwards leaving a gap which we add a new circle or ?sphere. And we do that entire face of both the length face and width ?face, depending on whether we are in 3D or 2D.

Newsgroups: sci.math, sci.physics, sci.logic
From: Archimedes Plutonium <plutonium.archime...@gmail.com>
Date: Mon, 18 Jul 2011 09:16:34 -0700 (PDT)
Local: Mon, Jul 18 2011 11:16 am
Subject: Chapt2 seeing if Kepler Packing can match the Omega function for Infinity #xxx Correcting Math 3rd ed

Just minutes ago I posted a reply to Carl for his
finding the 10^603 of the 1.000..01 form in the Omega function
(Lambert W function)
What I want to see is whether Kepler style Packing is another form of ?the Omega function, perhaps a hybrid of a trigonometric function along ?with the logarithmic omega function.
See whether the most dense packing uniquely occurs at 10^603.
So I start a table below using Unit Diameter and where the trough ?height is 0.133974 from that of ?0.8660254
sq size,  max pack,  max rows, height of max pack,  #circles, circl/ ?square density
1x1           pole        1 ?1                            1           0.7853
2x2           pole        2 ?2                            4         0.7853
3x3           pole        3 ?3                            9        0.7853
4x4           pole        4                4 ?16
5x5          pole         5                5 ?25
6x6          pole         6                6 ?36
7x7          pole         7                7 ?49
8x8          hex         9              7.9282 ?68     0.8344
9x9    oblong-hex    10            8.7942 + 0.1339       86     0.8338
10x10 oblong-hex   11            9.66025 + 0.1339     105
11x11 oblong-hex   12           10.5262 + 0.1399
12x12 oblong-hex   13          11.3923 + 0.1399
13x13 oblong-hex   14          12.2583 + 0.1399
14x14    hex           16          13.99
15x15 oblong-hex    17          14.8564 +0.1399
16x16
Tables are good for seeing general trends. If I can see a trend with ?the small ?numbers, I could find out what happens at 10^603
The Logic would be that at Infinity, there is a maximum of empty space ?or a ?maximum of occupied space. One or the other, making Infinity special.
Archimedes Plutonium ?http://www.iw.net/~a_plutonium ?whole entire Universe is just one big atom ?where dots of the electron-dot-cloud are galaxies

--- end of old 2011 post of mine on the depression number in Kepler Packing ---

--

Recently I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of front-page-hogs since they crowd out everyone else reading "mobile" instead of "desktop". Many posters in sci.physics and sci.math are not doing science but seeing whether they can dominate the front page of the newsgroup, and PAU is free of hogs, mockers and hatemongers.