What new set of numbers introduce tetration?

This thread is speculative, so if you find something too vague, is because it is not clearly defined. Please, help to make corrections and fill in the blanks.

Here I aim to "guess" what are the numbers introduced by tetration, in hope of getting a better grasp of the concept of tetration to any real exponent.

As understanding complex numbers helps with the calculation of exponentials, if we guess what is the field of numbers introduced by tetration, it may hep to understand tetration.

We do not actually know if those numbers strictly respect the definition of field, so, we should call them fracfield

__________________________________________________

Addition introduces negative numbers, product introduce rational numbers, which can be generalized to the real line, and exponentiation introduces complex numbers, which can be generalized to quaternions, octonions, sedenions, etc.

A pattern here is that integer numbers have zero dimension, real numbers are one dimensional, and complex numbers, with reals, quaternions, octonions, etc have integer dimension.

So maybe tetration introduces a set of numbers with real dimension. "Real dimension" means fractional dimension, not the real, unidimensional set of numbers. For example numbers with 1.5 dimension, which would be between reals and complex. One number is not enough to represent them, but a pair of numbers are redundant.

Yet, we do not have a way to represent more than one number, but less than two. What we can do to represent a number with dimension 1.5?

One thing we can do is to use redundancy: We can represent a fracfield with dimension r (\( \\[15pt]

{1 \leq r\leq 2 \)) with a pair of real numbers; one number may be the real part, and be used whole, like any real, but the remaining number can only grow up to a limit, and over that limit, it represents again a small number. For example, if reachs the value 1.5, it is zero.

This idea is not new. Polar representation of complex numbers already does it. If the angle of a complex number grows over \( \\[15pt]

{2\pi} \), it is the same as substracting \( \\[15pt]

{2\pi} \), i.e. \( \\[20pt]

{r.e^{i(2n\pi+\theta) }} \) is the same number for any integer n.

We can borrow this property, and, for a dimension \( \\[10pt]

{\omega } \) define a number representation with a period scaled by \( \\[15pt]

{\Omega \in \mathbb{R}} \), so, for any integer n, this number is the same:

\( \\[25pt]

{r.e^{\omega.(\Omega n\pi+\theta) } \,=\, r \, \left ( \cos{[\frac {2}{\Omega}(\Omega n\pi+\theta)]} \,+\, \omega . \sin{[\frac {2}{\Omega}(\Omega n\pi+\theta)] }\right) } \)

Is necessary to multiply \( \\[15pt]

{\theta } \) on the trigonometric functions to get the same number with period \( \\[15pt]

{\Omega\pi} \).

For example, for \( \\[15pt]

{\omega \,=\, i} \), which is the complex numbers, \( \\[10pt]

{\Omega \,=\, 2} \). Here \( \\[10pt]

{\omega } \) is a symbol and \( \\[15pt]

{\Omega \in \mathbb{R}} \) is a number associated to the dimension of that symbol.

This do not means that \( \\[15pt]

{\Omega \in \mathbb{R}} \) is a number equal to the dimensions of the symbol \( \\[10pt]

{\omega } \). \( \\[15pt]

{\Omega \in \mathbb{R}} \) is a function of the numbers of dimensions r, but his exact value is not trivial to define.

The definition for what is the same number tells us that at a radius r, we have a circumference with perimeter \( \\[12pt]

{r.\Omega \pi} \)

______________________________________________________

Another pattern which may be useful is that for a space with n integer dimensions, the space volume grows with the power of n.

For example, the length of the real line at a radius r from the origin is \( \\[15pt]

{L=r^1} \). For the plane of complex numbers \( \\[15pt]

{L^2=Area=\pi.r^2} \). In 3D space, a sphere with radius r has a volume \( \\[15pt]

{L^3=Vol=\pi \frac{4}{3}r^3} \).

So, a space with dimension t should grow with \( \\[15pt]

{L^t\propto r^t} \).

__________________________________________________

Geometrical interpretation

What is a space with fractional dimension? I conjecture, that it is a space curved on a higher dimension.

For example, dimensions between 1 and 2 are curved surfaces in 3D space.

Because we don't know if the fractional fracfield over this space satisfies the definition of vectorial space, we can call this a fractional vectorial space, or fracvectorial space, and the numbers over it are fracvectors.

As a space, we need a surface with fractional dimension. If we use the concept of Hausdorff dimension, then we need an area whose surface grows by a factor of \( \\[15pt]

\epsilon^r \) when the area is scaled up by a factor of \( \\[15pt]

\epsilon^r \)

Mathematically, this means that the area A(x) of the fractional space is defined as:

\( \epsilon A_{(\frac{x}{\epsilon})} = \epsilon^r A_{(x)} \) [I]

The derivative of A(x) is

\( A'_{(x)}=\lim_{\epsilon \to 1}\frac{A_{(x.\epsilon)}-A_{(x)}}{x.\epsilon-x} \) [II]

Because [I]

\( A_{(x.\epsilon)}=\epsilon^{1-r} A_{(x)} \)

So [II] is

\( A'_{(x)}=A_{(x)} \lim_{\epsilon \to 1}\frac{\epsilon^{1-r}-1}{x.\epsilon-x} = A_{(x)} \frac{1-r}{x} \)

Solving it: \( \\[20pt]

A_{(x)}=c_1 x^{(1-r)} \)

I think that is the only solution for [I], but do not have proof.

At least is the only solution for [II]

The function \( \\[15pt]

P_{(x)}=c_1(1-r).x^{-r} \) has an area under the curve \( \\[30pt]

A_{(x)}\,=\,\int P_{(x)}dx \).

(note that I left the integration limits undefined)

So, we have a continuous space with fractional dimension, and need a fracfield over it.

This figure shows the area A(x) under the curve P(x). Over this area, we can apply a fracfield made of pair of numbers \( \\[15pt]

(x,\theta) \), with x in the real line, and \( \\[15pt]

\theta \) an angle with period than \( \\[15pt]

\Omega \pi \)

If we fold this area, so each vertical line is turned into a circle, we get a surface with perimeter \( \\[15pt]

P_{(x)} \) (in direction normal to the real axis, over the circles).

After the folding, we would have a radius equal to \( \\[25pt]

R_{(x)}=\frac{P_{(x)}}{2 \pi} \) (I mean the radius of the circles). But that's only if folding the area does not distorts his surface, and, sadly, this is not true in general.

To be sure that the area preserves his Hausdorff dimension, we should define the radius R(x) as the one whose surface of revolution is equal to A(x).

\( \\[15pt]

A_{(x)}=2 \pi \int R_{(x)}\sqrt{1+R'_{(x)}^2} \)

We now have a 2D surface curved in 3D space with area A(x), but also need to check that this area to continuously transform from the real line (dimension r=1) to the complex plane (r=2).

For that, we need to find the function radius R(x) whose surface of revolution is A(x), and verify that it goes to zero when r=1 and r=2.

Deriving the last equation and squaring both sides:

\( \left (\frac {c_1 (1-r)}{2 \pi} \right)^2 x^{-2r} = R_{(x)} ^2 + \left ( R_{(x)} R'_{(x)} \right )^2 \)

Replacing

\( R_{(x)}^2 =2y \)

\( R_{(x)}R'_{(x)} =y' \)

\( a= - \left (\frac {c_1 (1-r)}{2 \pi} \right)^2 \)

\( b= - 2r \)

We get this differential equation:

\( 2y + y'^2 + a x^b=0 \)

Unfortunately, it is extremely hard to solve.

I had been vaguely told that should look on Weierstrass p function, \( \\[15pt]

\wp \) and that it may have a closed solution for \( \\[15pt]

b \in \mathbb{Z} \).

I found this paper, about using the \( \\[15pt]

\wp \) function for solving differential equations, but I can't figure how to use it.

We also know that \( \\[25pt]

R(x) \approx \frac {P(x)}{2\pi} \)

Any idea?

This thread is speculative, so if you find something too vague, is because it is not clearly defined. Please, help to make corrections and fill in the blanks.

Here I aim to "guess" what are the numbers introduced by tetration, in hope of getting a better grasp of the concept of tetration to any real exponent.

As understanding complex numbers helps with the calculation of exponentials, if we guess what is the field of numbers introduced by tetration, it may hep to understand tetration.

We do not actually know if those numbers strictly respect the definition of field, so, we should call them fracfield

__________________________________________________

Addition introduces negative numbers, product introduce rational numbers, which can be generalized to the real line, and exponentiation introduces complex numbers, which can be generalized to quaternions, octonions, sedenions, etc.

A pattern here is that integer numbers have zero dimension, real numbers are one dimensional, and complex numbers, with reals, quaternions, octonions, etc have integer dimension.

So maybe tetration introduces a set of numbers with real dimension. "Real dimension" means fractional dimension, not the real, unidimensional set of numbers. For example numbers with 1.5 dimension, which would be between reals and complex. One number is not enough to represent them, but a pair of numbers are redundant.

Yet, we do not have a way to represent more than one number, but less than two. What we can do to represent a number with dimension 1.5?

One thing we can do is to use redundancy: We can represent a fracfield with dimension r (\( \\[15pt]

{1 \leq r\leq 2 \)) with a pair of real numbers; one number may be the real part, and be used whole, like any real, but the remaining number can only grow up to a limit, and over that limit, it represents again a small number. For example, if reachs the value 1.5, it is zero.

This idea is not new. Polar representation of complex numbers already does it. If the angle of a complex number grows over \( \\[15pt]

{2\pi} \), it is the same as substracting \( \\[15pt]

{2\pi} \), i.e. \( \\[20pt]

{r.e^{i(2n\pi+\theta) }} \) is the same number for any integer n.

We can borrow this property, and, for a dimension \( \\[10pt]

{\omega } \) define a number representation with a period scaled by \( \\[15pt]

{\Omega \in \mathbb{R}} \), so, for any integer n, this number is the same:

\( \\[25pt]

{r.e^{\omega.(\Omega n\pi+\theta) } \,=\, r \, \left ( \cos{[\frac {2}{\Omega}(\Omega n\pi+\theta)]} \,+\, \omega . \sin{[\frac {2}{\Omega}(\Omega n\pi+\theta)] }\right) } \)

Is necessary to multiply \( \\[15pt]

{\theta } \) on the trigonometric functions to get the same number with period \( \\[15pt]

{\Omega\pi} \).

For example, for \( \\[15pt]

{\omega \,=\, i} \), which is the complex numbers, \( \\[10pt]

{\Omega \,=\, 2} \). Here \( \\[10pt]

{\omega } \) is a symbol and \( \\[15pt]

{\Omega \in \mathbb{R}} \) is a number associated to the dimension of that symbol.

This do not means that \( \\[15pt]

{\Omega \in \mathbb{R}} \) is a number equal to the dimensions of the symbol \( \\[10pt]

{\omega } \). \( \\[15pt]

{\Omega \in \mathbb{R}} \) is a function of the numbers of dimensions r, but his exact value is not trivial to define.

The definition for what is the same number tells us that at a radius r, we have a circumference with perimeter \( \\[12pt]

{r.\Omega \pi} \)

______________________________________________________

Another pattern which may be useful is that for a space with n integer dimensions, the space volume grows with the power of n.

For example, the length of the real line at a radius r from the origin is \( \\[15pt]

{L=r^1} \). For the plane of complex numbers \( \\[15pt]

{L^2=Area=\pi.r^2} \). In 3D space, a sphere with radius r has a volume \( \\[15pt]

{L^3=Vol=\pi \frac{4}{3}r^3} \).

So, a space with dimension t should grow with \( \\[15pt]

{L^t\propto r^t} \).

__________________________________________________

Geometrical interpretation

What is a space with fractional dimension? I conjecture, that it is a space curved on a higher dimension.

For example, dimensions between 1 and 2 are curved surfaces in 3D space.

Because we don't know if the fractional fracfield over this space satisfies the definition of vectorial space, we can call this a fractional vectorial space, or fracvectorial space, and the numbers over it are fracvectors.

As a space, we need a surface with fractional dimension. If we use the concept of Hausdorff dimension, then we need an area whose surface grows by a factor of \( \\[15pt]

\epsilon^r \) when the area is scaled up by a factor of \( \\[15pt]

\epsilon^r \)

Mathematically, this means that the area A(x) of the fractional space is defined as:

\( \epsilon A_{(\frac{x}{\epsilon})} = \epsilon^r A_{(x)} \) [I]

The derivative of A(x) is

\( A'_{(x)}=\lim_{\epsilon \to 1}\frac{A_{(x.\epsilon)}-A_{(x)}}{x.\epsilon-x} \) [II]

Because [I]

\( A_{(x.\epsilon)}=\epsilon^{1-r} A_{(x)} \)

So [II] is

\( A'_{(x)}=A_{(x)} \lim_{\epsilon \to 1}\frac{\epsilon^{1-r}-1}{x.\epsilon-x} = A_{(x)} \frac{1-r}{x} \)

Solving it: \( \\[20pt]

A_{(x)}=c_1 x^{(1-r)} \)

I think that is the only solution for [I], but do not have proof.

At least is the only solution for [II]

The function \( \\[15pt]

P_{(x)}=c_1(1-r).x^{-r} \) has an area under the curve \( \\[30pt]

A_{(x)}\,=\,\int P_{(x)}dx \).

(note that I left the integration limits undefined)

So, we have a continuous space with fractional dimension, and need a fracfield over it.

This figure shows the area A(x) under the curve P(x). Over this area, we can apply a fracfield made of pair of numbers \( \\[15pt]

(x,\theta) \), with x in the real line, and \( \\[15pt]

\theta \) an angle with period than \( \\[15pt]

\Omega \pi \)

If we fold this area, so each vertical line is turned into a circle, we get a surface with perimeter \( \\[15pt]

P_{(x)} \) (in direction normal to the real axis, over the circles).

After the folding, we would have a radius equal to \( \\[25pt]

R_{(x)}=\frac{P_{(x)}}{2 \pi} \) (I mean the radius of the circles). But that's only if folding the area does not distorts his surface, and, sadly, this is not true in general.

To be sure that the area preserves his Hausdorff dimension, we should define the radius R(x) as the one whose surface of revolution is equal to A(x).

\( \\[15pt]

A_{(x)}=2 \pi \int R_{(x)}\sqrt{1+R'_{(x)}^2} \)

We now have a 2D surface curved in 3D space with area A(x), but also need to check that this area to continuously transform from the real line (dimension r=1) to the complex plane (r=2).

For that, we need to find the function radius R(x) whose surface of revolution is A(x), and verify that it goes to zero when r=1 and r=2.

Deriving the last equation and squaring both sides:

\( \left (\frac {c_1 (1-r)}{2 \pi} \right)^2 x^{-2r} = R_{(x)} ^2 + \left ( R_{(x)} R'_{(x)} \right )^2 \)

Replacing

\( R_{(x)}^2 =2y \)

\( R_{(x)}R'_{(x)} =y' \)

\( a= - \left (\frac {c_1 (1-r)}{2 \pi} \right)^2 \)

\( b= - 2r \)

We get this differential equation:

\( 2y + y'^2 + a x^b=0 \)

Unfortunately, it is extremely hard to solve.

I had been vaguely told that should look on Weierstrass p function, \( \\[15pt]

\wp \) and that it may have a closed solution for \( \\[15pt]

b \in \mathbb{Z} \).

I found this paper, about using the \( \\[15pt]

\wp \) function for solving differential equations, but I can't figure how to use it.

We also know that \( \\[25pt]

R(x) \approx \frac {P(x)}{2\pi} \)

Any idea?

I have the result, but I do not yet know how to get it.