Conjugates of Fields, 5 of 6; Logarithmic and Exponential Fields

5. Logarithmic and Exponential Fields

Logarithmic Addition

Logarithmic addition and subtraction (respectively, "accretion" and "erosion") are logarithmic conjugates of plus and minus:

X  ln[+] Y    =    X # Y   =   ln( e^X  +  e^Y )  =  "X along Y"

X  ln[-] Y    =     X $ Y   =   ln( e^X   -  e^Y )  =  "X beyond Y"

Logarithmic multiplication is addition:

X ln[*] Y   =   ln ( e^X * e^Y )   =   X+Y

Therefore logarithmic addition is a field under +, with identity   - ∞:

X + ( Y # Z )  =  ( X + Y ) # ( X + Z )

X + ( Y $ Z )  =  ( X + Y ) $ ( X + Z )

- ∞  #  X         =        X

X  $  X           =       - ∞

Defining logarithmic negation require using Euler's formula e^πi  =  -1.

Define $X  =  - ∞ $ X  =  ln(-e^X)  =  "beyond X";  then

$X  =  X  + $0

$0  =  πi

"Beyond zero equals pi times i." This formula refers, in five symbols, to the five major branches of mathematics; analysis, arithmetic, logic, geometry, and algebra.

We can define log addition to other bases. Let

X  #_C  Y  =   log_C( C^X + C^Y )  = ( (X lnC)#(Y lnC) ) / lnC  =  X  [#/lnC] Y

X  $_C  Y  =   log_C( C^X - C^Y )  = ( (X lnC)$(Y lnC) ) / lnC  =  X  [$/lnC] Y

If   P₁ = A b^X   and  P₂ = A b^Y  then   (P₁+P₂)  =  A b ^Z  , where Z  =  X #_b Y.

When exponentials add, exponents log-add. If  b = e^r,  then

Z  =   X [#/r] Y  =  ( rX # rY ) / r   =   X   ln ^r\/+  Y  ;  log-fermat addition!

In the decibel scale, sound energy is proportional to 10^D/10 , where D is decibels; therefore if two sound energies add linearly, then the decibels add logarithmically;

D₁₂   =   D₁   #_10^0.1  D₂

In the Richter scale, seismic energy is proportional to 10^R. If energies add, then;

R₁₂   =   R₁   #₁₀  R₂

In the pH scale, concentration is proportional to  10 ^{- pH} ; therefore, if we mix equal volumes of two solutions and no buffering occurs, so that the concentrations average, then;

pH₁₂   =   ( pH₁   #_0.1  pH₂ )  +  ln2

Exponential Multiplication

Exponential multiplication (or "eruption") is * conjugated by exp:

X  e^*  Y    =    X  @  Y   =   e ^{( lnX * lnY )}  =  X ^lnY  =  Y ^lnX

Its inverse operation is exponential division, or "subduction":

X e^¸ Y     =    X ^(1/lnY)

Unlike exponentiation, eruption is commutative and associative:

X @ Y  =  Y @ X  

X @ ( Y @ Z )   =  ( X @ Y ) @ Z

* is a field under @, with identities 1 and e:

X @ ( Y * Z )   =   ( X @ Y ) * ( X @ Z )

X @ 1              =           1

X @ e              =           X

X @ e^N            =           X^N

X @ Y^N           =       (X@Y)^N

We can exponentiate eruption:    X  e^@  Y   =   e^{(lnX @ ln Y)}. This is a field over @, with identities e and e^e. And so on; we get an "Exponential Tower" of operations:

#  ,  +  ,  *  ,  @  ,  e^@ ,  ...

- each one of which is a field over the preceding operation, with identities

-∞  ,  0  ,  1 ,  e  ,  e^e  ,  ...

Conjecture: Continuous Exponential Tower: there exists a continuous, invertible, analytic "super-exponential" operator E on the positive reals such that E(x+1) = e^E(x).  Given such an E, we can define the continuous iterate of exponentiation:

exp_R(X)   =   E( E^-1(X) + R )

So exp_-1(X) =  lnX  ;   exp₀(X)  =  X  ;  exp₁(X)   =  e^X  ;  exp₂(X)  =  e^{(e^X)} ; ...

Therefore we can make the exponential tower continuous, thus:

X  _R+  Y   =    exp_R( exp_-R(X) + exp_-R(Y) )

                =    E (  E^-1{ E[E^-1(X) - R]  +  E[E^-1(Y) - R] }+ R )

            So   _-1+   =  #  ,    ₀+   =  +  ,    ₁+   =  *   ,    ₂+   =  @   ,   ... 

Each  _R+ distributes over _R-1+ , and distributes under _R+1+ .

Open questions remain. For instance, how does  _1/2+  relate to  +  and  * ?

Conjugates of Fields, 4 of 6; Modular Fermat Fields

4. Modular Fermat Fields

If the function x^N is one-to-one on the integers modulo M, then for any value of a and b, we can solve Fermat's equation, modulo M:

 a^N+b^N   =   c^N    mod M

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. It is therefore self‑inverse;   ((x³)³) =  x  mod 5.

Now consider the cube root of addition, mod 5;

x   ³\/+  y       =        ((x³)+(y³))^1/3      =        ((x³)+(y³))³    mod 5   

x   ³\/+ y      =   z      if and only if      x³+y³   =   z³    mod 5

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 as well as +

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,...

Now consider cubing modulo 11:

x      0  1  2  3  4  5  6  7  8  9  10

x³     0  1  8  5  9  4  7  2  6  3  10

Here's the cube root of addition mod 11:

³\/+     0   1    2    3    4    5    6    7    8    9   10

0        0    1    2    3    4    5    6    7    8    9   10                     

1        1    7    4    8   10   3    2    9    6    5    0    

2        2    4    3    7    8    1    5   10   9    0    6           

3        3    8    7   10   9    4    1    6    0    2    5

4        4   10   8    9    6    7    3    0    5    1    2

5        5    3    1    4    7    2    0    8   10   6    9

6        6    2    5    1    3    0    9    4    7   10   8           

7        7    9   10   6    0    8    4    5    2    3    1

8        8    6    9    0    5   10   7    2    1    4    3

9        9    5    0    2    1    6   10   3    4    8    7

10     10    0   6    5    2    9    8    1    3    7    4

This has the Fibonacci sequence 1, 1, 7, 9, 3, 2, 7, 10, 1, 0, 1, 1, ...

In the integers we solve Fermat's equation for N=2 via the rational‑right‑triangle identity:

(m² ‑ n²)² + (2mn)²  =  (m² + n²)²

This equation still applies in cube-root addition mod 11, but with a change in the second term; for   1 ³\/+  1  =   7  mod 11. Therefore, in this context, the correct equation is:

(m^{2  3}\/‑ n²)^{2  3}\/+ (7mn)²    =    (m^{2  3}\/+ n²)²          mod 11

Cubing yields: (m^{2  3}\/‑ n²)⁶  +  (7mn)⁶      =    (m^{2  3}\/+ n²)⁶          mod 11

We get these triples:

m      n             (m^{2  3}\/‑ n²)⁶  +  (7mn)⁶     =    (m^{2  3}\/+ n²)⁶     mod 11

2      1                              2 ⁶  +  3 ⁶          =          10 ⁶        mod 11

3      2                              3 ⁶  +  9 ⁶          =          1 ⁶          mod 11

3      1                              7 ⁶  +  10 ⁶        =          5 ⁶          mod 11

4      1                              9 ⁶  +  6 ⁶          =          3 ⁶          mod 11

4      3                              1 ⁶  +  7 ⁶          =          6 ⁶          mod 11

These in turn imply other sums:          

1 ⁶  +  7 ⁶    =   5 ⁶          mod 11

1 ⁶  +  3 ⁶    =   4 ⁶          mod 11

3 ⁶  +  2 ⁶    =   1 ⁶          mod 11

Note the colliding triplets (1,7;6), (1,7;5). Mod 11, 6 =  ‑ 5  and  6² = 5²

This method obtains 2*3 power Fermat equations, but not 4*3 power, nor 8*3, nor higher orders of evenness.

So modulo 5 and 11, we have conjugate fields full of loopholes in 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.