I was wondering if anybody here has anything to say about this. I'd love to know if there are related topics.

Consider the following definition:

r:log(x) is a superfunction of log(x); where r is the iteration count.

Therefore

2:log(x) = log(log(x))

etc etc...

r:log(y:log(x)) = r+y:log(x)

and therefore:

-1:log(x) = b ^ x

-2:log(x) = b ^ (b ^ x)

so on and so forth. r:f(x) is taken to be the superfunction of f(x) aswell.

0:f(x) = x

We must first observe tetration and its connections to the superfunction of log(x).

if b {0} x = b + x and b {1} x = b * x and b {2} x = b ^ x

b {3} x will denote tetration. (Do not be fooled by the number 3).

by definition (logs base b)

log(b {3} x) = b {3} (x-1)

and therefore, connected to superfunctions:

r:log(b {3} x) = b {3} (x-r)

this holds for b > 0, b =/= 1; r < x; x > -1; b, r, x E R

As you can see rational iterations of the logarithm are defined by rational tetration. There is still no clear concensus on the evaluation of rational tetration, however, I hope to further argue the model which states over domain [-1, 0];

-1 <= f <= 0

b {3} f = f + 1

Now comes the area of my paper where one must open their minds. m,n > 0 E R

Consider:

log(m* n) = log(m) + log(n)

and

log(m ^ n) = log(m) * n

and

2:log(m ^ n) = 2:log(m) + log(n)

One can see that logarithms work to lower the operator magnitude across any operator lower than {2}

The assertion I make is that taking rational iterations of logarithms gives us rational operators.

Or:

0 <= q <= 1

q:log(m {1} n) = q:log(m) {1-q} q:log(n)

q:log(m {2} n) = q:log(m) {2-q} n

These operators would have the following property; if S(q) returns the identity of any operator:

m {q} S(q) = m

q:log(m) + q:log(S(q)) = q:log(m)

therefore

q:log(S(q)) = 0

which is true regardless of logarithm base.

m {1+q} S(1+q) = m

q:log(m) * S(1+q) = q:log(m)

therefore:

S(1+q) = 1

and which in general becomes all operators greater than or equal to one have identity one.

Operators less than or equal to one are commutative:

m {q} n = n {q} m

q:log(m) + q:log(n) = q:log(n) + q:log(m)

m {1+q} n =/= n {1+q} m

q:log(m) * n =/= q:log(n) * m

Operators less than or equal to one are associative:

m {q} (l {q} n) = l {q} (m {q} n)

Rational operators preserve the law of recursion found in natural operators.

m {1 + q} 2 = m {q} m

q:log(m) * 2 = q:log(m) + q:log(m)

therefore:

m {1+q} n = m {q} m {q} m ... {q} m n amount of times

m {2+q} n is therefore defined recursively.

if k;log(x) is the inverse of any function b {k} x:

0;log(x) = x - b

1;log(x) = x/b

2;log(x) = log_b(x)

1+q;log(b {2+q} x) = b {2+q} (x-1)

or more generally:

r: (1+q);log(b {2+q} x) = b {2+q} (x-r)

r:q;log(b {1+q} x) = b {1+q} (x-r)

Rational operators are not distributive over addition, however, as multiplication is to exponentiation and as exponentiation is to multiplication [i]{q} is to {1+q} as {1+q} is to {q}

therefore:

(m {q} n) {1 + q} l = (m {1+q} l) {q} (n {1+q} l)

(m {1+q} n) {q} m = m {1+q} (n+1)

m {1+q} 0 = S(q)

since

q:log(m) * 0 = q:log(S(q)) = 0

Now comes the difficult task of evaluating these new found operators. With our knowledge that:

r:log(b {3} x) = b {3} (x-r)

r:log(b {3} x) = b {3} (slog(b {3} x) - r)

and therefore:

r:log(m) = b {3} (slog(m) - r)

Where slog(x) is the inverse function of tetration.

now, since:

-r:log(r:log(x)) = x

m {q} n = -q:log(q:log(m) + q:log(n))

m {q} n = b {3} (slog( (b {3} (slog(m) - q)) + (b {3} (slog(n) - q))) + q)

m {1+q} n = -q:log(q:log(m) * n)

m {1+q} n = b {3} (slog((b {3} (slog(m)-q)) * n) + q)

Now, further observing the identity function:

since:

q:log(S(q)) = 0

b {3} (slog(S(q)) -q) = 0

slog(S(q)) - q = -1

slog(S(q)) = q - 1

S(q) = b {3} (q-1)

And now if the critical strip of tetration is defined as:

-1 <= f <= 0

b {3} f = f + 1

S(q) = q

and therefore:

m {q} q = m

Further notes:

Consider the function

A(x) = m {x} n

Which is a generalization of the Ackerman function, extending it to domain real.

Interesting results are:

A(x) = 2 {x} 2 peaks at A(1 - 1/ln(2)) = 4.248828844

A(x) = 2 {x} 2 is periodic with period one, and therefore has a fourier series.

Results found using the following derivatives:

(b {3} x)' = ln(b)^floor(x) * [E(k=0, floor(x)) b {3} (x - k)]

(slog(x))' = (ln(b)^floor(slog(x)) * [E(k=0, floor(slog(x))) k:log(x)])^(-1)

Where [E(k=0, n) f(k)] is an Euler product.

Edit:

Also if

(x {q} y) }q{ y = x

or if }q{ is rational division and subtraction.

x {1+q} -1 = q }q{ x

Which is a special case of a more general formula

x {1+q} e^ji = q (e^ji){q} x

if (-1){q} = }q{

(x (e^ji){q} y) (e^ji)}q{ y = x

(e^ji)}q{ = (e^(j+pi)i){q}

Not much is really known about artificial operators. They are created by multiplying any natural operator with a complex coefficient of magnitude 1. }2{ is roots. }3{ is super roots

EDIT 2:

Also, one should note that

0.5:log(0.5:log(x)) = b {3} (slog(b {3} (slog(x) - 0.5)) - 0.5) = log(x)

Which is probably my main argument for the extension of tetration that I use.

Also, if one doesn't like this extension: rational operators are an independent discovery consistent with any rational tetration. However, if domain [-1, 0] is not universal for each base rational operators become dependent on a logarithm base.

Edit 3:

Actually, I see now that there is another method of evaluating rational iterations of the logarithm function.

as long as:

-q:log(q:log(x)) = x; this should maintain consistency.

Actually nvm this last part, my rational iteration model is symmetric to the other method.

Edit 4:

Here is a graph of x {0} 3 transforming into x {1.8} 3, counting up by .2

window screen is (xmin = 0, xmax=50, ymin=0, ymax=50)

The fact that it's squiggly bewilders me and leaves me in awe.

I believe it has something to do with the extension of tetration I use...

Consider the following definition:

r:log(x) is a superfunction of log(x); where r is the iteration count.

Therefore

2:log(x) = log(log(x))

etc etc...

r:log(y:log(x)) = r+y:log(x)

and therefore:

-1:log(x) = b ^ x

-2:log(x) = b ^ (b ^ x)

so on and so forth. r:f(x) is taken to be the superfunction of f(x) aswell.

0:f(x) = x

We must first observe tetration and its connections to the superfunction of log(x).

if b {0} x = b + x and b {1} x = b * x and b {2} x = b ^ x

b {3} x will denote tetration. (Do not be fooled by the number 3).

by definition (logs base b)

log(b {3} x) = b {3} (x-1)

and therefore, connected to superfunctions:

r:log(b {3} x) = b {3} (x-r)

this holds for b > 0, b =/= 1; r < x; x > -1; b, r, x E R

As you can see rational iterations of the logarithm are defined by rational tetration. There is still no clear concensus on the evaluation of rational tetration, however, I hope to further argue the model which states over domain [-1, 0];

-1 <= f <= 0

b {3} f = f + 1

Now comes the area of my paper where one must open their minds. m,n > 0 E R

Consider:

log(m* n) = log(m) + log(n)

and

log(m ^ n) = log(m) * n

and

2:log(m ^ n) = 2:log(m) + log(n)

One can see that logarithms work to lower the operator magnitude across any operator lower than {2}

The assertion I make is that taking rational iterations of logarithms gives us rational operators.

Or:

0 <= q <= 1

q:log(m {1} n) = q:log(m) {1-q} q:log(n)

q:log(m {2} n) = q:log(m) {2-q} n

These operators would have the following property; if S(q) returns the identity of any operator:

m {q} S(q) = m

q:log(m) + q:log(S(q)) = q:log(m)

therefore

q:log(S(q)) = 0

which is true regardless of logarithm base.

m {1+q} S(1+q) = m

q:log(m) * S(1+q) = q:log(m)

therefore:

S(1+q) = 1

and which in general becomes all operators greater than or equal to one have identity one.

Operators less than or equal to one are commutative:

m {q} n = n {q} m

q:log(m) + q:log(n) = q:log(n) + q:log(m)

m {1+q} n =/= n {1+q} m

q:log(m) * n =/= q:log(n) * m

Operators less than or equal to one are associative:

m {q} (l {q} n) = l {q} (m {q} n)

Rational operators preserve the law of recursion found in natural operators.

m {1 + q} 2 = m {q} m

q:log(m) * 2 = q:log(m) + q:log(m)

therefore:

m {1+q} n = m {q} m {q} m ... {q} m n amount of times

m {2+q} n is therefore defined recursively.

if k;log(x) is the inverse of any function b {k} x:

0;log(x) = x - b

1;log(x) = x/b

2;log(x) = log_b(x)

1+q;log(b {2+q} x) = b {2+q} (x-1)

or more generally:

r: (1+q);log(b {2+q} x) = b {2+q} (x-r)

r:q;log(b {1+q} x) = b {1+q} (x-r)

Rational operators are not distributive over addition, however, as multiplication is to exponentiation and as exponentiation is to multiplication [i]{q} is to {1+q} as {1+q} is to {q}

therefore:

(m {q} n) {1 + q} l = (m {1+q} l) {q} (n {1+q} l)

(m {1+q} n) {q} m = m {1+q} (n+1)

m {1+q} 0 = S(q)

since

q:log(m) * 0 = q:log(S(q)) = 0

Now comes the difficult task of evaluating these new found operators. With our knowledge that:

r:log(b {3} x) = b {3} (x-r)

r:log(b {3} x) = b {3} (slog(b {3} x) - r)

and therefore:

r:log(m) = b {3} (slog(m) - r)

Where slog(x) is the inverse function of tetration.

now, since:

-r:log(r:log(x)) = x

m {q} n = -q:log(q:log(m) + q:log(n))

m {q} n = b {3} (slog( (b {3} (slog(m) - q)) + (b {3} (slog(n) - q))) + q)

m {1+q} n = -q:log(q:log(m) * n)

m {1+q} n = b {3} (slog((b {3} (slog(m)-q)) * n) + q)

Now, further observing the identity function:

since:

q:log(S(q)) = 0

b {3} (slog(S(q)) -q) = 0

slog(S(q)) - q = -1

slog(S(q)) = q - 1

S(q) = b {3} (q-1)

And now if the critical strip of tetration is defined as:

-1 <= f <= 0

b {3} f = f + 1

S(q) = q

and therefore:

m {q} q = m

Further notes:

Consider the function

A(x) = m {x} n

Which is a generalization of the Ackerman function, extending it to domain real.

Interesting results are:

A(x) = 2 {x} 2 peaks at A(1 - 1/ln(2)) = 4.248828844

A(x) = 2 {x} 2 is periodic with period one, and therefore has a fourier series.

Results found using the following derivatives:

(b {3} x)' = ln(b)^floor(x) * [E(k=0, floor(x)) b {3} (x - k)]

(slog(x))' = (ln(b)^floor(slog(x)) * [E(k=0, floor(slog(x))) k:log(x)])^(-1)

Where [E(k=0, n) f(k)] is an Euler product.

Edit:

Also if

(x {q} y) }q{ y = x

or if }q{ is rational division and subtraction.

x {1+q} -1 = q }q{ x

Which is a special case of a more general formula

x {1+q} e^ji = q (e^ji){q} x

if (-1){q} = }q{

(x (e^ji){q} y) (e^ji)}q{ y = x

(e^ji)}q{ = (e^(j+pi)i){q}

Not much is really known about artificial operators. They are created by multiplying any natural operator with a complex coefficient of magnitude 1. }2{ is roots. }3{ is super roots

EDIT 2:

Also, one should note that

0.5:log(0.5:log(x)) = b {3} (slog(b {3} (slog(x) - 0.5)) - 0.5) = log(x)

Which is probably my main argument for the extension of tetration that I use.

Also, if one doesn't like this extension: rational operators are an independent discovery consistent with any rational tetration. However, if domain [-1, 0] is not universal for each base rational operators become dependent on a logarithm base.

Edit 3:

Actually, I see now that there is another method of evaluating rational iterations of the logarithm function.

as long as:

-q:log(q:log(x)) = x; this should maintain consistency.

Actually nvm this last part, my rational iteration model is symmetric to the other method.

Edit 4:

Here is a graph of x {0} 3 transforming into x {1.8} 3, counting up by .2

window screen is (xmin = 0, xmax=50, ymin=0, ymax=50)

The fact that it's squiggly bewilders me and leaves me in awe.

I believe it has something to do with the extension of tetration I use...