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2.  Brownian friezes

As it was mentioned in the Introduction, the sequence    a0, a1a2 a3 , ...    containing the coefficients of the continued fraction expansion  x  = [a0 a1, a2 a3 , ... ] of a real number x Î can be used as a sequence of  instructions for creating a picture - the portrait of x . We will use a simple algorithm that allows us to get a kind of Brownian motion trajectory. The algorithm used here is as follows:

Algorithm A.

    1. Take the continued fraction expansion of  x as above. Cut the sequence {ak}   up to the (m+1)-th term so to obtain a0, a1 , ... , am ;

    2. Replace ai by a'i =  ai (mod 8);  In this way, the number x  is represented by a'0 a'1 ,  ... , a'm  where a'iÎ{0,1, ...,7}  We will write  a(x) = (a'0 a'1 ,  ... , a'm );

    3. Starting by the point  (x0 , y0) =  (0, 0) create a sequence of points in R2,  {(xk , yk), k = 1,2, ...,m+1} following the rule xk+1 = xk+ p,  yk+1 = yk+ q, where p and q depend on  a'i  as it is shown in the table:
 

 a'  1  2  3  4  5  6  7   0
  p  1  1  0   -1  -1  -1  0   1
 q  0  1  1  1  0  -1  -1  -1

    4. Draw the polygonal line  b (x) = {(x0 , y0), (x1, y1), (x2, y2), ... , (xm+1, ym+1)}.

Note that the the table encodes the direction of the trajectory of a moving particle according to the compass rose : E, NE, N, NW, W, SW, S and SE. This pseudorandom Brownian walk was used by the authors of [1] to investigate the quantity of information being hidden into the DNA molecule chain.

Let us try this algorithm on a simple example. Take the fraction x = 61 / 117. Its continued fraction expansion is 61/117 = [0, 1, 1, 11, 5], so, m = 4. Applying the second stage on the sequence 0, 1, 1, 11, 5 will result in "mod 8" reduction to   0, 1, 1, 3, 5. The  polygonal line defined by 6 vertices  (0, 0), (1,-1), (2,-1), (3,-1), (3, 0), (2, 0) is a result of applying the third stage. In this way we obtain the following "portrait" of the number x:
 
 


Quadratic irrationalities

The Euler-Lagrange theorem claims that a continued fraction expansion of x  is periodic, if and only if  x  is irrational of quadratic type. This means that  x  is a solution of  the quadratic equation x2 + p x + q = 0, with p, q ÎR.  The next figure shows the graphical outputs obtained by Algorithm A, for the numbers x1 = (35 + 3 Ö205)/62,  x2(831+31Ö957)/626 and x3 = (7 + Ö2117)/44.

The corresponding continued fractions are x1 =[á1,3,1,7ñ], x2 = [ á2,1, 6, 8,1,4ñ] and x3 = [á1, 4, 1, 7,1ñ].
 

Metallic means family

Maybe the best known continued fraction expansion is  f =  [1, 1, 1, 1, ...] = (1+Ö5)/2 - the famous Golden Mean number, alias the  positive solutions of quadratic equations  x2 - x - 1 = 0. It is easy to see that Algorithm A, applied to f  will produce a horizontal line. But there exist an interesting family of numbers that generalizes the Golden Mean number. This is so called Metallic Means Family (MMF)  defined as the  positive solution of quadratic equation x2 - px - q = 0 (p, q are natural numbers), i.e.

s(p, q) = (p+ (p2+4q)1/2)/2.

In special cases, we have [3, 4] the Golden Mean s(1, 1),  the Silver Mean s(2, 1), the Bronze Mean s(3, 1), etc. But these "noble" metal means are not very creative from the point of view of Brownian graphs, because all of them are of the form [p,ápñ] and therefore yield straight lines. But many of them are real "artists". For example, s(3,3)=(3+Ö21)/2,s(1, 8) = (1+ Ö33)/2, s(1, 32) = (1+ Ö129)/2  and  s(1, 578) = (1+ 3Ö257)/2 creates nice Brownian friezes that are shown below.


E - friezes

An important part of the set of irrational numbers is the set of  transcendental numbers. These numbers fail to be roots of any algebraic polynomial equation with real coefficients and their continued fraction expansion is non periodic.

We will start with  the number  e = 2.71828182845905... - the basis of natural logarithms. It has the continued fraction expansion  e = [2,1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,1,1,14,1,1,...] .  Note that, starting from the third term, the pattern  {2n, 1, 1}, n = 1, 2, 3, ... repeats endlessly. Obviously, the "mod 8" operation will transform this pattern into {{2n mod 8}, 1, 1}, n = 1, 2, 3, .... In this way, a non periodic expansion  becomes periodic, i.e. we will have

        b(e) = (2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 0, 1, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 0, ...) ,

where the number "1" appears twice as frequently as the other numbers. This will make the "portrait"  b(e) a horizontal frieze. What is peculiar is that the same property, to produce horizontal friezes, remains if we perform some simple operations over e. The next figure shows  b(e),  b(e1/3)  (with m = 60),  b(e2) (with m = 100) and  b(e/3) (with m = 130).



Other transcendental numbers

 Finally we want to examine some other transcendental numbers, like   p, Log(p),  or Euler gamma g » 0.577216...,  for example. The figure below shows  b(p), b(Log(p)) and b(g) for m = 80 points. Note that no rules or patterns are visible on these graphs and therefore they really looks like Brownian walk paths.

The following figure looks similar as the previous one although  the numbers shown are not transcendental but irrational. The graph on the top corresponds to the number  t = Ö2 + Ö3 which is the greatest solution of the biquadratic equation  t4 - 10 t2 + 1  =  0. The middle graph is  b( 5 1/3 ) while the last one is  b( 7 1/5 ).

Hyperbolic functions are composed of exponential ones. Therefore, they also can produce periodic sequences a(x) that causes periodic graphical outputs that sometimes may look like simple geometric shapes. Displayed below are "portraits" of   Tanh(0.5),  Coth(1) and  Coth(-1).

Trigonometric functions are supposed to be a reach source of transcendental numbers. The next figure shows "Brownian paths" made by tangent function. It is composed out of three rows. Each row is made by drawing graphs  b(Tan(1/2)), b(Tan(1/3)) and  b( Tan(1)) put side by side.



The inverse problem

Can we find a number that corresponds to a certain plane figure being drawn as a union of eight types of segments:  {(0,0),(1,0)} - having  code "1",  {(0,0),(1,1)} - code "2",  {(0,0),(0,1)} - "3", {(0,0),(-1,1)} - "4",  {(0,0),(-1,0)}- "5", {(0,0),(-1,-1)} - "6", {(0,0),(0,-1)} - "7", {(0,0),(1,-1)} - having code "0" or "8"? The answer is "yes" if the figure is finite (bounded) or infinite (unbounded) but periodic. If these requirements are not fulfilled, then it is a composition of two or more finite or periodic graphs. If the figure is finite, the solution is trivial. If it is periodic, given by repeating the period  a0 , a1, a2, ..., ak, then the corresponding number  x  is the biggest solution of quadratic equation x = [á a0, a1, a2, ..., ak, xñ]. If the figure is composed of one finite and one periodic part, where the first is given by a0 , a1, ..., an  while the periodic part  is given by a0, a1, ..., ak,  then the corresponding number is y = [a0, a1, ..., an, x], where x is given by the same equation as above.

Example. The "3D staircase" figure



can be drawn by repeating the motif

that is encoded by the sequence  7, 3, 6, 7, 3, 5, 7, 1, 2, 1. The number that will cause an infinite repeating of this motif  is given by the infinite continued fraction  x = [á7, 3, 6, 7, 3, 5, 7, 1, 2, 1 ñ]. It is therefore the biggest solution of the quadratic equation x = [á7, 3, 6, 7, 3, 5, 7, 1, 2, 1, x ñ], i.e. the number as strange as x=(474047+Ö337648155621)/144218.


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