Roots and Chaos by Means

Roots and Chaos by Means

         

          First define ‘harmonic addition’ <+> this way:

                   

          x <+> y   = 1 / ( 1/x + 1/y )   =    (xy) / (x+y)

          Therefore (x+y)(x<+>y)   =   xy.

          Now consider the arithmetic and harmonic means:

          AM    =    (x+y)/2         =     (x+y) <+> (x+y)    =   (x<+>x)+(y<+>y)

          HM    =    2(x <+> y)   =     (x<+>y) + (x<+>y)  =    (x+x)<+>(y+y)

          Recall that (x+y)(x <+> y) = xy:

          this implies that AM*HM = xy.

          Now consider this iteration:

                    (x₀, y₀)               =        (a, b)

                    (x_N+1, y_N+1)      =        ( 2(x_N<+>y_N) , (x_N+y_N)/2 )

          The recursion preserves the product of the components;

          therefore, by recurrence, y_N  = ab/x_N

          and therefore x_N+1  =  (x_N + ab/x_N)/2 =  (x_N² + ab)/(2x_N)

          - which is the Newton’s-method formula for the square root.

          If ab>0 then let z_N  = x_N/root(ab) ; this implies:

                    z_N+1 =  (z_N² + 1)/(2z_N)

          Compare that to the hyperbolic-cotangent identity:

                    coth(2A) =  (coth(A)² + 1)/(2coth(A))

          By recurrence we derive, for some A₀:

                    z_N+1 = coth(A₀*2^N)

          So therefore

                    x_N+1 = root(ab) coth(A₀*2^N)

          This converges to root(ab) if A₀ > 0, which happens if a>0;  and it converges to -root(ab) if A₀ < 0, which happens if a<0.

          If ab<0 then let z_N  = x_N/(root(-ab)) ; this implies:

                    z_N+1 =  (z_N² - 1)/(2z_N)

          Compare that to the cotangent identity:

                    cot(2A) =  (cot(A)² - 1)/(2cot(A))

          By recurrence we derive, for some A₀:

                    z_N+1 = cot(A₀*2^N)

          So therefore

                    x_N+1 = (root(-ab)) cot(A₀*2^N)

          This undergoes angle-doubling chaos. It is sensitive to initial conditions, and it has orbits that go arbitrarily close to any given countable set of values. It’s as if, when ab<0, the x’s are ‘trying to converge’ to root(ab), an imaginary number, even though the x’s are real.

          Experiments on a hand calculator (TI-83) show that if a and b have small imaginary parts, then the recursion quickly converges to one of the complex roots. The real line is a chaotic boundary between two basins of attraction.

          Now for the cubic case.

          Consider the arithmetic and harmonic means for three numbers:

                    AM = (x+y+z)/3    ;     
                    HM = 3(x<+>y<+>z)  =  3 / ( 1/x + 1/y + 1/z )

          Define the mediant as:

                    X @ Y @ Z  =    X*Y*Z  /  ((X+Y+Z)*(X<+>Y<+>Z))

                                        =     (XY+YZ+ZX) / (X+Y+Z)

          Therefore:

                    AM * mediant * HM    =    xyz

          Now consider this recursion:

                    (x₀,y₀,z₀)                =   ( a, b, c )

                    (x_N+1, y_N+1, z_N+1)   =  
                      ( (x_N+y_N+z_N)/3, x_N@y_N@z_N, 3(x_N<+>y_N<+>z_N) )

          This preserves product, and therefore by recurrence:

                    x_Ny_Nz_N = abc

          Conjectures:

          The number of positive entries in the triple (x_N, y_N, z_N) is constant. That is, all-positive remains all-positive, two-positive-one-negative remains two-positive-one-negative, one-positive-two-negative remains one-positive-two-negative, and all-negative remains all-negative.

          If a, b and c are all positive or all negative, then the above recursion converges rapidly to ( cuberoot(abc),  cuberoot(abc),  cuberoot(abc) ).

          If a, b and c do not all have the same sign, then the above recursion is chaotic.

          If a, b and c do not all have the same sign, and to each we add a small imaginary part, and iterate from there, then the above  recursion converges rapidly to a complex cube root of the new abc.

          These claims are supported by numerical experimentation on a hand calculator (TI-83). But support is not proof. How to prove this?

         

number limerick

A dozen, a gross and a score

plus three times the square root of four

divided by seven

plus nine times eleven

is five cubed plus zero, no more.

Exceptions to Fermat's Last Theorem

            Exceptions to Fermat's Last Theorem

Recall Fermat's Last Theorem:

For no integer n > 2  is there an integer solution to   aⁿ + bⁿ  =  cⁿ

This theorem, though proven true by Wiles, has loopholes; negative exponents, fractional exponents, and mod N.

NEGATIVE EXPONENTS

We have these Diophantine formulas:

a¹ + b¹                                                  =          (a+b)¹

[m² ‑ n²]² + [2mn]²                               =          [m² + n²]²

[a(a+b)] ^‑1 + [b(a+b)] ^‑1                        =          [ab] ^‑1

[m⁴ ‑ n⁴] ^‑2  +  [2mn(m² + n²)] ^‑2            =          [2mn(m² - n²)] ^‑2

The last two equations involve 'reciprocal addition':

x "1/+" y   =   1 / ( (1/x) + (1/y) )   =    xy /  ( x+ y )

This 'conjugate of addition' is involved in Kirkhoff's Law and flow‑rate algebra.

Evidently the Fermat/Wiles theorem should read "... for |n|>2..." Now the question is; what's so special about ‑2, ‑1, 1, 2?

FRACTIONAL EXPONENTS

            For practically any a,b,c you can find a suitable n, provided that it may be fractional ‑ and probably transcendental. For instance:

a=1, b=2, c=5  ‑‑‑‑>  n=0.5638955243...

a=2, b=3, c=4  ‑‑‑‑>  n=1.507126592...

You can find these n’s by solving a^x+b^x-c^x = 0 by Newton’s method.

MODULO N

If the function xⁿ is a one-to-one function on the integers modulo k, then we can find c in Fermat's equation for any value of a and b.

For instance, consider the cube mod 5:

x    mod 5        0   1   2   3   4

x³  mod 5         0   1   3   2   4

It swaps 2 and 3, leaving the others unchanged. Therefore the function (x³) is its own inverse; (x³)³ = x mod 5. Now consider addition conjugated by cubing:

            ( x³ + y³ )³   =   x  "3|/+"   y    ; 

This is the 'cube root of addition';

            x 3|/+  y  =  z    iff   x³ + y³ =  z³ .

3|/+    0   1   2   3   4

0       0   1   2   3   4       

1       1   3   4   2   0        this is a field under * mod 5;

2       2   4   1   0   3        * distributes over this

3       3   2   0   4   1

4       4   0   3   1   2

In this 'modular Fermat field', the Fibonacci sequence goes:

1,1,3,2,0,2,2,1,4,0,4,4,2,3,0,3,3,4,1,0,1,1,...

So mod 5, we have a whole conjugate field full of exceptions to Fermat's theorem! This trick doesn't work for every modulus; in some moduli, x³ is not 1 to 1.

This theorem applies:

x³ is one-to-one mod M  (and M has a cube root of addition)       if and only if  

M is a product of distinct primes, none of form 6k+1.

These moduli include 2, 3, 5, 6, 10, 11, 15, 17, 22, 23, 29, 30, 33, 34, 41, 43, 46, 47, 53, 55, 57, 58, 66, 69, 71, 82, 83, 86, 87, 89, 94... A motley crew!

This generalizes:  If p is an odd prime, then

x^p is one-to-one mod M           if and only if  

M is a product of distinct primes, none of form 2pk+1.

x^pq  = (x^p)^q  ;  therefore: If N is an odd composite number, then 

x^N is one-to-one mod M          if and only if

M is a product of distinct primes, none of form 2pk+1, for any prime factor p of N.

Any modulus equal to a Fermat prime (M = 2^{(2^n)}+1) will pass all these tests: therefore Fermat-prime moduli have one-to-one odd-power functions; and therefore all Fermat-prime moduli have all odd roots of addition. (So do moduli equalling distinct products of Fermat primes.)

There are infinitely many Fermat primes if and only if there are infinitely many moduli which have every odd root of addition.

Conjugates of Fields, 6 of 6; Other Conjugate Fields

6. Other Conjugate Fields

Here are some useful conjugate field operators: 2 arctan[+], arccos[*] , tanh[+], tan[+].

2 arctan[+]  ;  for subtended angles.

If a line segment of length L is held with midpoint at distance D away from an observer, at right angles to the observer's line of sight, then it will subtend an angle of  

θ   =   2 arctan ( L/(2D) ) .

Therefore two lengths, which subtend angles of (respectively) θ₁ and θ₂, when joined together will subtend an angle of

θ₁₂   =   θ₁  (2 arctan)[+]  θ₂

arccos[*]  ;  for spherical right triangles.

If a spherical right triangle has sides A, B and hypotenuse C, (as measured in radians), then we have the equation:

cos(A) * cos(B)   =  cos(C)

Ergo:

C       =     A  arccos[*]   B

tanh[+]  ;  for relativistic velocity addition.

The relativistic velocity addition law can be written with tanh[+]:

V₁₂   =    ( V₁ + V₂ ) / ( 1 + V₁V₂/c²)    =   c ( V₁/c  tanh[+]  V₂/c )

tan[+]  ;  for slope rotation.

If two lines have slopes x and y, and a third line is tilted from the horizontal at an angle equal to the sum of the other two line’s tilts, then the third line has this slope:

tan( arctan(x) + arctan(y))   =    ( x + y ) / ( 1 – x y )

Note that ( x tan[+] y )  =  i ( x/i  + y/i ) / ( 1 + (x/i)(y/i) ) = i ( (x/i) tanh[+] (y/i))

Therefore :      tan[+]  =  (i*tanh)[+]

And likewise:  tanh[+]  =  (i*tan)[+]

Slope rotation and relativistic velocity addition are imaginary conjugates of each other!