Heptagons, etc

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Heptagons, etc

Postby wendy » Sat Oct 06, 2012 8:57 am

At the moment, i have been going through the heptagons again, including their strange version of the fibonacci numbers (the sevenly flat).

One of the keys to an understanding of deep geometry is a grounding in number theory. Here's the sevenly flat. The eye (0,0) is based on the common modular point. Where a period is set, it passes through the eye. 5, for example, has a 31-place period, so it is cyclic over a 31-size animal. You can see a new start for it at (0,0), and (4,1) There's one at (8,9) too.

Code: Select all
        -1    0     1    2    3     4    5    6     7       8     
  8    157   283   510  919 1656 2984 5377  9689 17459   31460
  7     70   126   227  409  737 1328 2393  4312  7770   14001
  6     31    56   101  182  328  591 1065  1919  3458    6231
  5     14    25    45   81  146  263  474   854  1539    2773
  4      6    11    20   36   65  117  211   380   685    1234
  3      3     5     9   16   29   52   94   169   305     549
  2      1     2     4    7   13   23   42    75   136     244
  1      1     1     2    3    6   10   19    33    61     108
  0      0     0     1    1    3    4    9    14    28      47
-1      1     0     1    0    2    1    5     5    14      19


One playes with coins on the table, to replicate the abacus of the various bases, like phi, r2+1 by r2, and (r6+r2)/2 by r2, the calculators for the pentagon, octagon, and dodecagon. The ones given in 'B by C', means that the vertical rows go by powers of C, while the horizontal powers go by powers of B. Generally, the powers of C are kept minimal. (1, 2). I managed to calculate numbers as far as 120 in the duodecimal numbers (base 2+r3, with digits 0, 1, 2, R = 1+r3). In the main, the duodecimal numbers are the hardest to reckon.

The heptagon numbers lend themselves to an abacus, but not a base, since the system is a class 3 one, is not inclusive of all members equating to a 'finite number'. Still, calculating with the stone-board is faster than the algebra, and more extensive, since the same calculation can be moved around the board.
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Re: Heptagons, etc

Postby Klitzing » Sun Oct 07, 2012 9:27 am

wendy wrote:At the moment, i have been going through the heptagons again, including their strange version of the fibonacci numbers (the sevenly flat).[...]

Sorry, Wendy,
but except that you are hunting for kind of heptagonal Fibonaccis, I really got nothing from that post.
What is a sevenly flat?
What are the calculation rules for your heptagonal Fibonaccis?
What are your intended applications?
Most probably there are some broader contexts, where you embed that theory!?
E.g. cf. your parallel post of the other thread: http://teamikaria.com/hddb/forum/viewtopic.php?f=25&t=1726#p18152 - from which I've to admit that I likewise got nothing!
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Re: Heptagons, etc

Postby wendy » Sun Oct 07, 2012 10:04 am

The post actually gives the heptagonal fibonacci series (sevenly flat). This is a two-dimensional series, where there is a convergent region where the three values stand in the ratios of the chords of the heptagon. The thing is periodic over every prime, generally in a period of p²+p+1 (for primes mod 2,3,4,5 mod 7), or p-1 (for primes modulo 0,1,6 mod 7). You will probably recognise some of these numbers from your thesis (which deals with Z7 in a large part).

Code: Select all
     2.246 979 603 7                          5377         =  2.246 970 330
    1.000 000 000 0    1.801 937 736     eg  2393  4312   =  1.000 000 000   1.801 922 273


Where B is one vertical, and A is one step horizontal, the calculation rules are the same as what holds in multiplication: A² = 1+B, AB = A=B, B² = 1+AB. For example, where 4312 is at (8,4), then 14001 = 4312 + 9689, 17459 = 7770 + 9689, and 21771 = 4312 + 17459.

Using the same relations on an grid with counters, one can show from these three relations (alone), things like A solves A³+1 = A²+2A, B solves B³+1 = 2B²+B, that (-1+AB)³ = 7 AB, and that any one of the relations can be found from the other two. This is an 'abacus' or stone-board, for calculating heptagonal numbers.

Such convergent regions exist for all polygons, the region for the 13-gon is also been found by a general process.

One can do similar things with the pentagonal numbers (by the rule that F² = F+I, which is the same rule that creates the fibonacci series), with octagonal numbers, (where one uses a two-dimensional grid, vertical by Q = root-2, and horizontal by A = sqrt(2)+1, the two rules are A = Q+I, and AQ = A + I. These by themselves generate the two octagonal series. A = alpha, the unit of the octagonal numbers.

The dodecagonal series uses Q and W, being sqrt(2) and (sqrt(6)+sqrt(2))/2. One uses the relation that QQ = I + I, and WW = WQ + I. (W = omega, the symbol i use for the shortchord of the dodecagon). Its square is the unit of the numbers of the system Z6 (span of heptagonal chords).

Once i learnt enough from these systems, and the transport across branches on the dynkin symbol, one can successfully predict the sorts of numbers that is necessary to make the chords of any polygon. A slightly less advanced trick, based on the same system, gives the sorts of numbers that make the chords of rational angles, from which one can (by looking at the cosine of an angle), reject without further consideration, that certian angles are not rational. For example, the van Oss polygon of the {3,5/2,3} gives a shortchord of 1.070466 or sqrt(3/2.61803398875). However, the only rational angles that can have a weight of sqrt(3) belong to 120 deg ie (sqrt(3)), and that this particular value is not a valid rational angle for that weight. The van Oss polygon does not close, and therefore {3,5/2,3} is infinitely dense.

The use of transport across the dynkin branch helps to calculate the sides etc to give required edges. It also helps to determine the sorts of chords necessary for any composite polygon, eg {21} or {14}.

Many of the things learnt here were then used to develop hyperbolic geometry from first principles.
The general theory exists with the general polytope. It's one of the filters that i use to tell me what works and what does not.

It should be remembered, that polytopes are the integers of geometry, and one can as much understand the nature of the containing geometry by looking on the polytopes it contains, as one does find the reals from looking at the integers of various types, that rest on the reals (and complex, and quarterions etc).
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Re: Heptagons, etc

Postby Klitzing » Thu Nov 22, 2012 3:42 pm

Polygonal shortchords do occure as the edges of polytopal vertex figures (of any dimension). This is why those are most relevant.

For a regular polygon with unit edge-length and Schläfli symbol {n/d} I denote by x(n/d, m) the length of the m-th chord (secant), i.e. of the line-segment connecting one vertex to the m-th successor. Clearly we would have x(n/d, 0) = 0 and x(n/d, 1) = 1. Likewise we would get x(n/d, n) = 0 and x(n/d, n-1) = 1. More generally we would have x(n/d, n-k) = x(n/d, k). Relevant as edges of vertex figures esp. are the shordchords, i.e. the cases x(n/d, 2). This is why I use a more special function here: x(n/d) = x(n/d, 2).

Assume for a moment that the polygon would be scaled down to an unit circumradius. Therefrom it gets obvious that we would have x(n/d, k) / r = 2 sin(pi k d/n). Together with x(n/d, 1) = 1 we hence deduce generally that r = 1 / (2 sin(pi d/n)). Accordingly we finally result in x(n/d, k) = sin(pi k d/n) / sin(pi d/n). If we use the trigonometric formula for doubled angles, sin(2 phi) = 2 sin(phi) cos(phi), one furthermore derives x(n/d) = 2 cos(pi d/n).

With respect to the regular heptagons, i.e. {7} = {7/1}, {7/2}, {7/3} which Wendy emphasizes in this thread, we get the following characteristic numbers:

Set of chords of a single unit-edged {7}:
Code: Select all
x(7, 1) = sin(pi 1/7) / sin(pi 1/7) = 1
x(7, 2) = sin(pi 2/7) / sin(pi 1/7) = 1.8019377358048382524722046390149
x(7, 3) = sin(pi 3/7) / sin(pi 1/7) = 2.2469796037174670610500097680085


Only the shortchords of the set of unit-edged {7/k}:
Code: Select all
x(7/1) = 2 cos(pi 1/7) = 1.8019377358048382524722046390149
x(7/2) = 2 cos(pi 2/7) = 1.2469796037174670610500097680085
x(7/3) = 2 cos(pi 3/7) = 0.44504186791262880857780512899359


As one can see, those numbers are furthermore closely related within its module: x(7, 3) - x(7, 2) = x(7/3) resp. x(7, 3) - x(7, 1) = x(7/2).

Now comming back to Wendy's sevenly flat. As already mentioned in my former reply, that flat is kind a Fibonacci sequence; just that it is 2-dimensional. As the quotioent of 2 subsequent Fibonacci numbers approaches the golden ratio, we would be not too much surprised to revisite the above numbers here as well: And indeed, the quotient of 2 subsequent numbers of a single horizontal line, called A in her post, approaches x(7, 3), while the quotient of 2 subsequent numbers of a single vertical column, called B in her post, approaches x(7, 2).

Furthermore Wendy already did deduce from the building rules of her sevenly flat the minimal polynoms of those 2 numbers:
Code: Select all
A = x(7, 3) solves A^3 = A^2 + 2A - 1 and
B = x(7, 2) solves B^3 = 2B^2 + B - 1.


As is known for the Fibonacci numbers, the building rule alone provides the limiting ratio, but not the actual numbers. Those depend on the chosen starting numbers. In case of Fibonacci those are 2 (subsequent ones). In case of Wendy's sevenly flat the seed numbers are 3: one number at the left-bottom corner, and both numbers directly to its right respectively directly atop. (The remainder then can be calculated by the building rules.) - Sure the building rules could be reverted so that one deduces building rules to proceed into the reverse direction(s), i.e. getting Fibonacci numbers with negative indices. Here too Wendy could proceed to negative numbers as well (as she already did, providing numbers for index -1 each). - But for sure, the convergence realm still is for very high positive indices only...

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Re: Heptagons, etc

Postby wendy » Fri Nov 23, 2012 11:24 am

The actual convergence is a region, bounded by two rays. It converges in a region that includes b but not a. The thing that governs it is that the various isomorphs must be less than one, ie N + N' + N"" converges when N' and N"" are both less than 1, and prehaps N.N'.N"" = 1. (It generally will converge, because there is always a region where N' and N" is less than one, regardless of the product.

The actual convergence can occur at quite small values, eg the values like 5:9:11 is a convergence. It must always converge. The region of convergence for {13} is also known. The convergence happens regardless of the starting animal, like the fibbonacci series. It just takes a bit larger for the convergence to become apparent. The direction of convergence is always in the direction of a^5 b^14.

There is a general direction of maximum convergence, where N' is approximately N"". You start with b^8 is nearly 8^11, and thence find the direction of convergence is in the direction a^5 b^14 = 170921,307989,384056 = 1588864. The two isomorphs are about 1/sqrt(15888864). Subsequent powers of the matrix of this value also converge fastest.

The actual convergent relation for x,y is a^(x).b^y /(3+a+2b), which replicates F(n) = f^(n)/(2f-1).
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Re: Heptagons, etc

Postby Klitzing » Sat Nov 24, 2012 10:19 am

Hy Wendy,

the thing you are dealing in the first paragraph, most probably is what is called a Pisot-Vijayaraghavan number, right? I.e. an algebraic number greater than 1, the other solutions of its minimal polynom have magnitude less than 1. In fact right that property will serve for the convergence. Those numbers too play a significant part within my thesis. Esp. I showed that substitution rules for pattern scalings (and a Fibonacci sequence and your flats clearly are essentially such integer substitution things) would lend for high-dimensional embedding into some grid-projection scenario with inflation symmetry, provided the scaling factor is such a PV number. - But clearly, I there discussed also the effect to the acceptance region of such projection scenarios, if the order of such a number is greater than 2 ...

The remainder of your reply is a bit nubilous. Part of it obviously handles the seed numbers. Part of it deals about convergence-directions and -velocities, I suppose. For Fibo numbers one usually considers the ratio F(n+1)/F(n) for increasing n, getting that this ratio reaches the golden number. This is what I would get from convergence region. What were you after here?

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Re: Heptagons, etc

Postby wendy » Sun Nov 25, 2012 7:16 am

It's a convergent region, not a number.

When you construct a series like this, which can be extended in integers in every direction, then the steps are units in some system, and all of the possible seeds represent the same number multiplied by some unit. In the present case, the seed generates from X, every XU, where U is a unit of the system.

We also have X. F(X) . FF(X) is an element of Z, where FF(X) = F(F(X)), and FFF(X) = F(F(F(X))) = X. Because of this, we see that among other things, that every heptagonal number divides some z in Z, and that there must be a region where XU is large enough that XU >1, F(XU) >1, FF(XU) > 1. The value then converges on XU + F(XU) + FF(XU) ~= XU, because the other two terms go to zero. This happens when one supplies successive powers of A.B³, and fastest for A^5 B^14.

Since this scale is generally a logrithmic scale, there are lines that cross the region where XU = 1, F(XU) = 1, and FF(XU) = 1. All three can't be less than one, so the region inside the finite triangle has all three greater than one. The region of convergence is where XU>1, F() and FF() <1. It converges slower when nearer the edge, when one of these F() or FF() are near one, and quickest when these are equal and small. For the heptagonal flat, the ray of convergence is in the general direction of ab³ or a^5 b^14. The second relies on the extraordinary fact that ln(a):ln(b)::8:11, very nearly ie 11 ln a / ln b = 8.000 979 925,
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