The Great Pyramid of Gizah was built for Pharaon Chufu …



Phidias numbers

Adam Marlewski1), Govert Werther2)

1) Institute of Mathematics, Poznań University of Technology, ul.Piotrowo 3a, 60-965 Poznań, Poland; e-mail: amarlew@math.put.poznan.pl; 2) EuroSchool, Nassauplein 8, 1815 GM Alkmaar, the Netherlands;

Abstract. First, we recall the definition and basic properties of the Golden Number (, known also as the Phidias number. Next, we generalise this notion to have Phidias numbers of arbitrary order and of various degree in wealth (from the poor ones up to the rich ones). We prove theorems on the convergence of the sequences of these numbers.

1. Definition and basic formulae for the Golden Number (

The Great Pyramid of Giza was built for Pharaoh Khufu (known also by his Greek name Cheops) about 2560 BC. It is made of 2.3 million limestone blocks of average weight 2.5 tons, the biggest ones weight 15 tons. But granite blocks covering the King's Cell weight more than 50 tons each one! Thus the Great Pyramid weights more than 6 million tons.

Since centuries the Great Pyramid (and its neighbours: the Sphinx statue and pyramids named after Khefren and Mykerinos, both pharaohs of the Fourth Dynasty) fascinates and still keeps its mysteries. Some of them are already revealed. As an example let's say about the position of its basis. This is a square with diagonals which are almost collinear with the directions north-south and east-west. As far as we can follow the migration of the Earth magnetic pole, the word "almost" would be cancelled in the time the pyramid was in construction.

The magnificent Tomb of Khufu has been erected during 20 years only. It still dominates the region which is covered by the desert sands today, but 4500 years ago hosted the capital of Lower and Upper Egypt. The passage of centuries made that it is hard to know exact dimensions at the time Khufu Pyramid was in construction. Today's measurements [AP], [Me] say that the apex is 146.6 meters over the square basis having sides of 230.4 meters long. It gives that the angle between a side of the pyramid and its basis is equal to

atan(146.6/230.4) (  51.84˚ ( 51˚50'.

This number is very close to the angle 51.82˚ ( 51˚48'.

In the right-angle triangle, where one of shortest sides is of length 1 and the adjacent angle equal to the 51.6272˚, the length of the hypotenuse is the reciprocal of the value

cos(51.6272˚) ( 0.618035256.. ( 

(1.1) ((5 - 1)/2 =: (.

Thus the hypotenuse is of length

(1.2) ( := 1/( = ((5 + 1)/2 

= 2 cos((/5)

( 1.618033988.

Ahmes Papyrus (written about 1650 BC, in 1858 brought to Britain by Henry Rhind) is one of the oldest mathematical works reporting the problems studied and methods applied by Egyptians. In this paper it is mentioned that the number (, then named the sacred ratio, was used in the construction of the Pyramid of Cheops (although no details are given). 13 centuries later Euclid, in his 13-tome "Elements" (c.310 BC), divides a linear segment of length 1 into two parts observing the proportion

(1.3) ( : (1 - (),

i.e. 1/( : (1 - 1/() = 1/( : ((( - 1)/() = 1/(( - 1) = ( = 

(1.4) ( : 1.

Fig.1. Golden section of the line segment (on the left) and the golden triangle in the pyramid (below)

A point marked after this rule is called by him the golden point, and the relation between the longer part and shorter one - the golden mean (or: golden ratio, mean ratio). It seems that the same problem and name were treated by earlier Greek mathematicians, especially Phytagoras (580-500 BC). The Greeks felt that rectangles with sides of lengths observing the ratio ( : 1 are the most harmonious, aesthetically most pleasing among all rectangles (see Fig.1). It is believed that Phidias (living in fifth century BC), the greatest sculptor of antic Athens, consciously used this ratio in his constructions, in particular in the statue of Olympic Zeus which is considered as one of seven wonders of the world (the Great Pyramid is another one). In honour to Phidias' mastery the American mathematician Mark Barr proposed to use the capital letter Phi, (, to denote the ratio at hand, and its reciprocal 1/( by the small letter phi, ( (note: ( = ( - 1). For this reason these both numbers are called sometimes (great) Phidias number and small Phidias number, respectively. Other denotation for (, applied a.o. in [JC], is the letter ( (tau) - the initial letter of Greek word tome meaning cut.

There are other names used to the ratio ( : 1, namely the divine proportion (introduced by Fra Luca Pacioli in his book De divina proportione edited in 1509) and the golden section (first used in Latin sectio aurea probably by Leonardo da Vinci, 1452-1519). The name Golden Number to the value ( gained popularity since Kepler's admiration.

The proportions (1.3) and (1.4) say that

(1.5) 1/( : (1 - 1/() = ( : 1,

hence

(1.6) (2 = ( + 1.

This relation can be written as

(1.7) ( = ((1+ ()

and

(1.8)   ( =  1 + 1/(.

Both these equalities can be treated as generators of following infinite representations of the golden number (:

(1.9) ( = ((1 + ((1 + ((1 + ((1+ ...)))),

(1.10) ( = 1+ 1/(1 + 1/(1 + 1/(1 + ...))).

Let us pay attention to the formula (1.10). Its right side is systematically built of 1's, summations and divisions only, it is an example of the continued fraction. The Golden Number ( is the essentially continued fraction, so it is not a fraction. In other words: in spite of its name the mean ratio ( is not a rational number (for the proof see [JS], [RK]).

2. Fibonacci numbers and Kepler limit

It is interesting to count successive finite fractions produced by taking into account initial 1's appearing in the continued fraction (1.10). We have (so-called convergents)

1  =   1/1,

1 + 1/1  =   2/1,

1 + 1/(1 +1)  =   3/2,

1 + 1/(1 +1/(1 + 1))  =   5/3,

1 + 1/(1 +1/(1 + 1/(1 + 1)))  =   8/5,

1 + 1/(1 +1/(1 + 1/(1 + 1/(1 + 1))))  = 13/8,

...

One can notice that the successive denominators are simply the sums of two previous denominators (starting with 1, 1), and the same rule governs the nominators (starting with 1, 2). So the sequence of (de)nominators is produced by the formula

(2.1) F0 := 0, F1 := 1,  

(2.2) Fn+1 := Fn + Fn-1, n = 1, 2, 3, 4, ...

It was Fibonacci, a.k.a Leonardo of Pisa, or Leonardo Pisano (c.1180-1240), who, in his book Liner abaci (Book of numbers) edited in 1202, first defined this sequence of numbers Fn, which are worldwide recognised as Fibonacci numbers. This name was proposed by French mathematician François Édouard Anatole Lucas (1842-91), the author of famous Towers of Hanoi puzzle (see e.g. [GKP]). He found Fibonacci numbers in the Pascal triangle and he generalised them (to numbers named by his name today).

Johannes Kepler (1571-1630), the author of three empirical laws describing the movement of planets around the Sun, was the first man who observed (in 1608) that the ratio Fn+1/Fn of two consecutive Fibonacci numbers tends to Golden Number (,

(2.3) lim Fn+1/Fn  (  ( as n  ( (.

By the way let's mention Kepler's admiration to Golden Number (: two great jewels of the geometry are Theorem of Phytagoras and Golden Number (.

By (2.2) we have

(2.4) Fn+1/Fn = 1 + Fn-1/Fn.

Thus, under the assumption that the sequence of quotients Fn+1/Fn has a finite limit (denoted here by) g as n  ( (, the limiting of both sides in (2.4) gives

g = 1 + 1/g.

This is nothing else than the square equation

(2.5) g2 ( g ( 1 = 0.

It is solved by two values:

g = (( and g = (.

All Fibonacci numbers are positive, the first root is negative, so it has to be rejected. That's why the limit g = (, as it is said by (2.3).

Fully complete proof of the Kepler relation (2.3) goes very easy if we refer to the formula

(2.6)   Fn = {(n ( ((()n }/(5

which expresses the n-th Fibonacci number by Golden Number (. This relation, first discovered by Leonhard Euler (1707-83) in 1765 and left behind, was independently found 78 years later by Jacques Binet (1786-1856), and today it is referred to as the Binet formula. It is not hard to derive it if the generating function is applied (see e.g. [GPK]).

3. Natural powers of the Golden Number

The relation (1.6) can be regarded as the formula to get the square (2 simply by the increasing the number ( by 1. Higher powers of ( may be also calculated in analogical way. In fact,

(3 = (((2 = ((( ( + 1) =  (2 +   ( =  (( + 1) +  ( = 2( + 1 = F3(( + F2,

(4 = (((3 = (((2( + 1) = 2(2 +  ( = 2(( + 1) +  ( = 3( + 2 = F4(( + F3,

(5 = (((4 = (((3( + 2) = 3(2 + 2( = 3(( + 1) + 2( = 5( + 3 = F5(( + F4,

etc. If we take into account that ( = 1(( + 0 =  F1(( + F0 and (2 =  1(( + 1 = F2(( + F1, then by the mathematical induction it is easy to state that

(3.1) (n = Fn(( + Fn-1, n = 1, 2, 3, ...,

where Fn's are Fibonacci numbers defined by (2.1)-(2.2).

There can be found analogous formulae for numbers which are generated in analogous way the Golden Number ( is. In next section we will define such numbers and we will denote them by (k. The Golden Number is the first of them. In aim to systemise the notions we denote it by (2, i.e. (2 := (, and the index 2 is saying that solves the quadratic equation (2.5). Now the identity (3.1) can be written in the form

(3.2)   (2 n = Fn((2 + Fn-1,

and the equation (2.5) may be considered as that defining the number (2.

In aim to systemise our consideration we will treat Fibonacci numbers in analogous way we deal with Golden Number, i.e. we enrich them with the index 2. So we set

F2,n :=  Fn.

Then (2.1), (2.2) and (3.2) take form:

(3.3) F2,0 := 0, F2,1 := 1,  

(3.4)  F2,n+1 := F2,n + F2,n-1, n = 1, 2, 3, 4, ...

(3.5)   (2 n = F2,n((2 + F2,n-1.,  n = 1, 2, 3, 4, ...

Elements of the sequence

(F2,n)n = 0, 1, 2, ... = ( 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, ...),

i.e. Fibonacci numbers can be called 2-Fibonacci numbers. Analogously, Golden Number (2 can be called the 2-Phidias number. (or: Phidias number of order 2) In next section we define 3-Fibonacci numbers F3,n and 3-Phidias ratio (3.

4. Silver Number

Analogously to the formulae (3.3)-(3.4) we define 3-Fibonacci numbers (a.k.a. Fibonacci numbers of order 3, Tribonacci numbers) F3,n as follows:

(4.1) F3,0 := F3,1 := 0, F3,2 := 1,  

(4.2)   F3,n := F3,n-1 + F3,n-2 + F3,n-3, n = 3, 4, 5, ...

The above recurrence produces the sequence

(F3,n)n = 0, 1, 2, ... = ( 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, ...).

Analogously as in case of 2-Fibonacci numbers F2,n we can observe that the sequence of the quotients F3,n+1/F3,n converges, and for instance

F3,10/F3,9 =  149/81  = 1.83950617..

F3,15/F3,14 =  3136/1705  = 1.83929618.. 

F3,20/F3,19 = 35890/19513 = 1.83928662.. 

Reasoning as in Section 2 we find that a finite

g = lim F3,n+1/F3,n as n ( (

has to satisfy the equation

3)   g3 ( g2 ( g ( 1 = 0.

This polynomial equation of the third degree has two complex zeroes and the real zero equal to

(4.4)   {(19 + 3(33)1/3 + (19 - 3(33)1/3}/3 = 1.839286795..

According to our convention we denote this value by (3 and we call it the 3-Phidias number In [GH] and [RR] this number is called the silver number (so next, (4, by the relation to Olympic medals, should be called the bronze number!). Taking into account the extensions are to be presented below we refer to (3 as the rich Phidias number of order 3.

We can write the formula analogous to (3.5), namely

4) (3 n+1 = F3,n((32 + (F3,n-1 + F3,n-2)((3 + F3,n-1, n = 3, 4, 5, 6, 7, ...

The proof goes immediately by the induction.

5. Rich Phidias numbers

Let us recall the equations (2.5) and (4.3) which define the 2-Phidias and 3-Phidias numbers (2 and (3, respectively. Both these equations are special cases of the equation

(5.1)     gk ( gk(1 ( gk(2 ... ( g ( 1 = 0

where k is arbitrary natural number.

This equation, called the rich k-Phidias equation (or: rich Phidias equation of degree k), may be derived in the analogous way as that presented in Sections 2 and 4. To do it we define n-th k-Fibonacci number by the recurrence

(5.2)   Fk,n := Fk,n-1 + Fk,n-2 + ... + Fk,n-k, n = k, k+1, k+2, ...

applied with k initial values

(5.3) Fk,0 = Fk,1 = ... = Fk,k-2 = 0,  Fk,k-1 = 1.

Now we investigate the quotients

qn := Fk,n+1/Fk,n = { Fk,n + Fk,n-1 +  Fk,n-2 + ... + Fk,n-k+1}/ Fk,n =

= 1 

+ 1/{ Fk,n / Fk,n-1 } 

+ 1/{ Fk,n / Fk,n-1 }(1/{ Fk,n-1 / Fk,n-2 } 

+ ...

+ 1/{ Fk,n / Fk,n-1 }(1/{ Fk,n-1 / Fk,n-2 }(...(1/{ Fk,n-k+1 / Fk,n-k }(

If we assume that there exists the finite limit g of the quotients qn, the above equality turns into the relation (5.1). This limit is denoted by (k and called the rich k-Phidias number, or: rich Phidias number of order k (the motive to call this number a rich one will justify already in the next section). Exact values of are easily obtained for k = 1, 2, 3 and 4 only, the next ones may be determined numerically. Some of them are listed in Table 1.

Fig.2. Graphs of the functions g ( gk (k = 2, 3, 4, 5),

g ( 1/(2 - g) and of the line g = 2

In [PC] this equation (5.1) is rewritten in the form

1/g = gk(1 ( g k(2 ( ... ( g ( 1.

But we can take use of the formula on the sum of k initial elements of the geometric progression:

gk-1 + gk-2 + ... + g + 1 = (gk ( 1)/(g ( 1) for any g ( 1.

Then (5.1) takes form

gk ( (gk ( 1)/(g ( 1) = 0.

Multiplying both sides by the difference (g - 1) turns it into the equation gk((g ( 2) + 1 = 0, so finally we come to the form

(5.4)   gk = 1/(2 ( g).

This relation, which can be called the rich Phidias equation of order k, or: rich k-th Phidias equation, immediately shows (see Fig.2) that there holds

Theorem 1. The sequence ((k)k = 1, 2, 3, ... of rich Phidias numbers (k converges to 2.

The circular relation (5.4) may be used to fast calculation of arbitrary integer power of (k. For instance, (kk+1 = (k/(2 ( (k) with any natural k.

6. Poor Phidias numbers

Both Fibonacci numbers Fn and Phidias number ( may be generalised in various ways and there is a very huge literature on them. k-Fibonacci numbers Fk,n and k-Phidias numbers (k presented in previous section are one of possible results. Investigating them we deal with regular polynomials denoted in the left side of the equation (5.1). This regularity appears in the presence of all consecutive powers of the variable g, as well as all elements identified by lower indexes in (5.2). Now let us work with the case when some terms are absent. So we consider the recursive definition

(6.1)   L0 = L1 = ... = Lk-2 = 0,  Lk-1 = 1,

(6.2)   Ln+k := Ln+ms + ... + Ln+m2 + Ln+m1, n = 0, 1, 2, ...

with k > ms, and its corresponding equation (easily produced by the changing Ln+r for gr)

(6.3)   gk ( gms ( ... ( gm2 ( gm1 ( 1 = 0

where k > ms > ... > m2 > m1 ( 0.

|Fig.3. Convergence of the quotients Ln/Ln-1 |Fig.4. Graph of the polynomial |

|of numbers Ln defined by formulae (6.4)-(6.5) |g ( g4 ( g2 ( g ( 1 |

For example, with k = 4 > m2 = 2 > m1 = 1 we have

(6.4)  L0 = L1 = L2 = 0,  L3 = 1,

(6.5)  Ln+4 := Ln+2  + Ln+1 + Ln, n = k, k+1, k+2, ...,

(6.6)  g4 ( g2 ( g ( 1 = 0,

so the numbers form the sequence (Ln)n = 0, 1, 2, ... = (0, 0, 0, 1, 0, 1, 1, 2, 2, 4, 5, 8, 11, 17, 24, 36, 52, 77,112, ...). For n = 10, 20, 30, 40, 50 and 100 their quotients Ln/Ln-1 assumes values

5/4  = 1.25,

241/165  = 1.46606060..,

11032/7528  = 1.46546227..,

504355/344136  = 1.46556884..,

23057166/15732546  = 1.46557130.. , 6747757084228261/4604182272046428  = 1.46557123..

The sequence of these quotients converges (comp. Fig.3) to the positive root of the corresponding Phidias equation (6.6), i.e. to the value 1.4655712309999489.. (see Fig.4).

In the classical Fibonacci recursion (2.2) the right side is the sum of only two terms. Just the particular case in (6.2) holds if on the right side we skip all terms except to the most extreme ones. Then we have

(7.1)   L0 = L1 = ... = Lk-2 = 0,  Lk-1 = 1,

(7.2)  Ln := Ln-1 + Ln-k, n = k, k+1, k+2, ...

3) gk = gk-1 + 1.

For any k > 2 the right side in (7.2) is scantier than that in (6.2), and it validates that the corresponding equation (7.3) is called the poor k-th Phidias equation, or: poor Phidias equation of degree k. The same name embraces its other form:

(7.4)  gk-1 = 1/(1 ( g).

It is easy to verify that this equation has always a single positive zero. Following adopted system of names we call it the poor k-th Phidias number, or: poor Phidias numbers of order k Identically as in Section 6, this number is the limit of quotients Ln+1/Ln of successive numbers produced by formulae (7.1)-(7.2) with arbitrarily fixed gap k. The equation (7.4) immediately reveals

Theorem 2. The sequence of poor Phidias numbers of order k converges to 1 as k approaches the infinity.

8. In-between Phidias numbers

Theorems 1 and 2 say that rich k-th Phidias numbers tends 2, poor ones - to 1 as k increases to the infinity. Let us recall that these numbers are defined by the number of terms present in the recursive formula (6.2). If in its right side there are involved in all the terms, we have rich Phidias numbers - see (5.2). If there are taken only two terms, with indices distant at the value k, we have poor Phidias numbers of order k - see (7.2). Obviously, only in case k = 2 these two formulae (and, in consequence, rich and poor Phidias numbers) coincide. If k > 2, there may be defined so-called the (in-between) Phidias numbers of order k, and there are so many as many are various are the recurrences (6.2). There are 2k-1 such recurrences, because there is the number of words composed of k characters such that every words ends with 1 and any of other k-1 characters is 1 or 0. For instance, with k = 4 we have 23 = 8 configurations: 1111, 0111, 1011, 1101, 0011, 0101, 1001, 0001. The first of them refers to the rich fourth Phidias number, the last one - to the poor fourth Phidias number. The word 0111 corresponds to the recurrence (6.5) and generates the incomplete Phidias equation (6.6). The graph of the polynomial staying on the left side of this equation crosses the abscissa axis at the point 1.4655712309999489.. . This observation may be generalised as follows:

Theorem 3. Let k, ms, ..., m2, m1 be natural numbers such that k > ms > ... > m2 > m1 > 0. Then the equation (6.3) has at least one zero in the interval (1, 2).

Proof.

Denote the left side in (6.3) by p(g), i.e.

p(g) := gk ( gms ( ... ( gm2 ( gm1 ( 1.

There is p(1) = 1k ( (1ms + ... + 1m2 + 1m1 + 1) = 1 ( (s + 1) = (s  ................
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