A weirdly mottled red sky over Crewe this morning.

I went out specially to catch the image before it faded.

A weirdly mottled red sky over Crewe this morning.

I went out specially to catch the image before it faded.

On Sacred Geometry

I've been looking at various YouTube sites that have videos on "Sacred Geometry". Some are completely vacuous waffle to me. However a few do contain genuine arithmetic and geometric results that seem of interest from a mathematical viewpoint. In particular there is a series from the "Jain Academy" fronted by an affable Australian lecturer who calls himself "Jain108". His mathematics is mostly correct, except where he obsesses about the "true" value of pi being 3.144... The Jain Academy deals in what I can only call "New Age Eclecticism". In other words it takes bits from everywhere, Hinduism, Kabbala, Christianity, Islam, Buddhism, Astrology, and so on, regardless of dogma.

The number 108 is apparently ubiquitous in Hindu mysticism. It is said in several sources to be the ratio of distance to diameter for both the Sun and the Moon. For any celestial body this ratio would be near enough the cotangent or cosecant of its apparent angular diameter. The angle whose cotangent or cosecant is 108 turns out to be 31 minutes and 50 seconds to the nearest second. In Patrick Moore's "Atlas of the Universe" (1994), which I happen to have to hand, the mean apparent diameter of the Sun is given as 32' 1" and the mean apparent diameter of the Moon as 31' 6". So the number 108 seems to be a reasonably good estimate. The closeness of the apparent diameters of Moon and Sun is of course why Solar Eclipses can be so spectacular.

Another obsession of the Sacred Geometers is, as might be expected, the dimensions of the Great Pyramid of Khufu at Giza. The Jain Academy makes much of the triangle formed by the pyramid as seen at a distance from a side. Taking its base width to be 2 units it is claimed that its slope length is the golden ratio 1.618 and its height is the square root of the golden ratio. According to Wikipedia the pyramid's original height was 280 cubits and its base 440 cubits. The slope length is thus the square root of (280^2 + 220^2) = root 126800 = 356.09 cubits. The ratio of slope to half base is thus 356.09/220 = 1.61859, and the golden ratio is 1.6180339. So again a plausible approximation. The angle of slope is 51 degrees 5 minutes so the visible angle at the summit would be 77 degrees 50 minutes.

When people make a model of a pyramid they nearly always take the faces to be exact equilateral triangles, but this gives a pyramid whose slope height is half of root three and whose height is half of root two. If the slope height of the pyramid is the golden ratio then this must be the altitude of the triangular face shape. The base angle then works out at 58 degrees 17 minutes and the apex angle as 63 degrees 23 minutes Differing from the equilateral by 3.6 degrees at the apex.

The symbol used for the golden ratio in Sacred Geometry is phi (though mathematical texts often use tau). The claim about the new value for pi is that is should be 4/(root phi) = 3.144605...as opposed to 3.14159... Or equivalently that 4/pi = 1.2732395... should be root phi = 1.2720196... Whether this means that the Laws of Nature themselves are going to change when the New Age dawns, or only human consciousness of them is not at all clear.

I've been looking through my books on number theory and combinatorics, but none of them seem to mention anywhere the following simple relationship of Fibonacci sequence to combinations. In the following I use the notation nCr = n!/(n-r)!r! for the number of ways of choosing r from n.

An explicit sum for F(n) can be found by summing the upward diagonals of the combination table that lists n against r and is sometimes presented in the form of Pascal's Triangle.

It can be expressed in the general form:

F(n) = (n-1)C0 + (n-2)C1 + (n-3)C2 + (n-4)C3 + ... + (n-k)C(k-1)

= Summation (r=1 to k) (n-r)C(r-1).

Where k = n/2 or (n+1)/ when n is even or odd respectively.

In the symmetric form of Pascal's Triangle the upward diagonals are transformed into knight-lines.

Instead of the above elementary formula for F(n) that uses only addition, subtraction, multiplication and division of whole numbers, the mathematical texts focus on the exotic Binet formula that expresses F(n) in terms of root five, or root five and the golden ratios, which are irrational numbers that all cancel each other out.

*A First Course of Combinatorial Mathematics* by Ian Anderson (Oxford University Clarendon Press (1979) page 43 derives from the Binet formula a more complicated relationship involving the summation of expressions 5^r.(n+1)Cr divided by 2^n, but does not simplify it to the above form. This is the nearest I have found in my limited sources.

Of course there is also a simple relationship of 2^n to the combinations:

2^n = nC0 + nC1 + nC2 + ... + nCn.

Does anyone have other sources they could direct me to? (see my Knight's Tour pages for my email)

Does anyone have a concrete combinatorial explanation for the formula?

CORRECTION: I have since noticed that Anderson does mention the above expression briefly, without further explanation, in Exercises 4.2 (4) on page 44.

FURTHER: I have found another reference in *Principles of Combinatorics* by Claude Berge (Academic Press 1971) page 31, though it has n+1 instead of n. It implies that (n-k)Ck counts the number of ways of choosing k items from a row of n, but with no two items being adjacent.

FURTHER: I have modified the equation, since there seems to be some divergence in the way that the Fibonacci numbers are labelled. I take F(0) = 0 and F(1) = 1, so that F(2) = 1 and F(3) = 2 and so on, whereas other sources start from F(0) = 1, F(1) = 1, F(2) = 2, etc. I hope it is now correct!

I will prove here a theorem I have not seen before:

The sum of an arithmetic progression with common difference k

is the sum of k triangular numbers of two sizes.

Consider the arithmetic progression h, h + k, h + 2k, h + 3k, ..., h+(n-1)k which has n terms.

The summation of this is the series: hn + k(1 + 2 + 3 + ...+ (n-1))

= hn + kn(n-1)/2 = [kn^2 + (2h - k)n)]/2

This formula can be expressed as (k-h)n(n-1)/2 + hn(n+1)/2

= (k-h)T(n-1) + hT(n).

That is as the sum of k triangles, h of side n and (k-h) of side (n-1).

In the case when h= 0 or h = k we get k triangles all of the same size.

Figurate representation of the numbers

When h = 1 and k = 3 we get the sequence 1, 4, 7, 10, 13, 16, 19, 22, 25, ...

its summation is the series: 1, 5, 12, 22, 35, 51, 70, 92, 117, ...

which is traditionally known as the 'pentagonal numbers'.

When h = 2 and k = 3 we get the sequence 2, 5, 8, 11, 14, 17, 20, 23, 26, ...

its summation is the series: 2, 7, 15, 26, 40, 57, 77, 100, 126, ...

which are known as the 'pentagonal numbers of the second kind'.

When h = 0 (or 3) we get the sequence 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, ...

its summation is the series 0, 3, 9, 18, 30, 45, 63, 84, 108, 135, ...

which are triple triangle numbers.

It was reported that a couple of members of Crewe Chess Club had tested positive for covid19, and one was in hospital. So I tested myself last night using a Lateral Flow Test that I obtained from a local pharmacy last week. Fortunately the result was negative.

I tried reporting the result to the NHS as is required but the website said it needed to know a mobile phone number, and I don't have a mobile phone. So I phoned 119 this morning. It required answering an awful lot of multiple choice questions before I could get through to the right place.

It seems both chess club members are recovering, though one is still in hospital for other health reasons. The club is closed this week, which is probably the right decision. I was due to take part in a team match.

I've also put off going to a meeting of a Writer's group at the local Library for the present. It seems best to wait until the news about the new omicron variant becomes more definite. At least I now know how to carry out the test when necessary.

UPDATE. Regrettably the club member who was in hospital, Les Hall, died on 8th December, though his cause of death may not have been covid19. He had diabetes and other conditions. Les Hall will be well known to chess players who have attended the Crewe Congress in former years. He was a larger than life personality, and one of the stalwarts behind the day-to-day running of the club. We will miss him.

Snowy scene from my back window yesterday.

Put my winter boots on this morning to trek to the supermarket and make sure I have enough coffee in stock until the new year.

Back in *The Games and Puzzles Journal*, #13 p.223 (May 1996), [the first issue of volume 2 which came out 25 years ago] the first puzzle was on "A Question of Proportion". It was about rectangles of various shapes including the standard paper sizes and the golden rectangle.

The 'Numberphile' videos on YouTube, some by Ben Sparks, have recently been featuring questions of this nature. I also borrowed from our local library the book *One to Nine* by Andrew Hodges which includes similar material, including the "Plastic" number.

Towards the end of the book Hodges implies that "Computable" numbers as defined in Alan Turing's famous paper are denumerable, that is they can be matched with the set of natural numbers.

I hadn't realised this before!

This also raises the philosophical question of whether any other "real" numbers "really" exist, if they cannot be calculated by a finite set of rules. Such numbers are those that appear in Cantor's diagonal proof of the nondenumerability of the continuum.

All the usual suspects like root 2, root 3, the golden section, e and pi and Euler's constant gamma are all computable numbers. This just means that they can be computed to any number of decimal places and that the calculation by whatever method always gives the same digit in the nth place for any n.

They really are real real numbers!

Going back to my puzzle in the *G&PJoutrnal* the last question asked was: What shape of rectangle gives a smaller rectangle of the same proportions when a rectangle of m squares by n squares is removed from one end?

The answer was given in G&PJournal #14 p.245 (December 1996). By applying the same methods as for the golden ratio I found a formula for such a generalised golden ratio, in terms of m and n.

It now occurs to me that we can denote the larger and smaller golden ratios by capital and lower case Phi as before, but with a suffix 1, and the generalised golden ratios by the same letters with a suffix m/n. This also indicates that these golden ratios all form a denumerable set since they can be placed in one-to-one correspondence with the ratios m/n, which are known to be denumerable.

The formulas are: Phi suffix m/n = [root (m^2 + 4.n^2) +/- m]/2.n

For example Large Phi suffix 1/2 = (root 17 + 1)/4 = 1.2807764...

Small phi suffix 1/2 = (root 17 - 1)/4 = 0.7807764...

Their difference is 1/2 and their product is 1 (or 0.99999...).

Some of these generalised ratios work out to be rational. The simplest case is when m=3, n=2 which gives the two Phi with suffix 3/2 as 2 and 1/2, but most of the ratios will be irrational.

ADDENDUM: It now occurs to me that the ratio m/n of the rectangle removed can in fact be any number not just a rational fraction. And conversely the shape of the whole rectangle can be given by any number x, the rectangle removed being x - 1/x.

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