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 Fractionally dimensioned numbers marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 02/29/2016, 12:46 AM (This post was last modified: 02/29/2016, 12:57 AM by marraco.) 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 ($\vspace{15}{1 \leq r\leq 2$) 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 $\vspace{15}{2\pi}$, it is the same as substracting $\vspace{15}{2\pi}$, i.e. $\vspace{20}{r.e^{i(2n\pi+\theta) }}$ is the same number for any integer n. We can borrow this property, and, for a dimension $\vspace{10}{\omega }$ define a number representation with a period scaled by $\vspace{15}{\Omega \in \mathbb{R}}$, so, for any integer n, this number is the same: $\vspace{25}{r.e^{\omega.(\Omega n\pi+\theta) } \,=\, r \, \left ( \cos{[\frac {2}{\Omega}(\Omega n\pi+\theta)]} \,+\, \omega . \sin{[\frac {2}{\Omega}(\Omega n\pi+\theta)] }\right) }$ Is necessary to multiply $\vspace{15}{\theta }$ on the trigonometric functions to get the same number with period $\vspace{15}{\Omega\pi}$. For example, for $\vspace{15}{\omega \,=\, i}$, which is the complex numbers, $\vspace{10}{\Omega \,=\, 2}$. Here $\vspace{10}{\omega }$ is a symbol and $\vspace{15}{\Omega \in \mathbb{R}}$ is a number associated to the dimension of that symbol. This do not means that $\vspace{15}{\Omega \in \mathbb{R}}$ is a number equal to the dimensions of the symbol $\vspace{10}{\omega }$. $\vspace{15}{\Omega \in \mathbb{R}}$ 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 $\vspace{12}{r.\Omega \pi}$ ______________________________________________________ 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 $\vspace{15}{L=r^1}$. For the plane of complex numbers $\vspace{15}{L^2=Area=\pi.r^2}$. In 3D space, a sphere with radius r has a volume $\vspace{15}{L^3=Vol=\pi \frac{4}{3}r^3}$. So, a space with dimension t should grow with $\vspace{15}{L^t\propto r^t}$. __________________________________________________ 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 $\vspace{15}\epsilon^r$ when the area is scaled up by a factor of $\vspace{15}\epsilon^r$ 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: $\vspace{20}A_{(x)}=c_1 x^{(1-r)}$ I think that is the only solution for [I], but do not have proof. At least is the only solution for [II] The function $\vspace{15}P_{(x)}=c_1(1-r).x^{-r}$ has an area under the curve $\vspace{30} A_{(x)}\,=\,\int P_{(x)}dx$. (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 $\vspace{15}(x,\theta)$, with x in the real line, and $\vspace{15}\theta$ an angle with period than $\vspace{15}\Omega \pi$ If we fold this area, so each vertical line is turned into a circle, we get a surface with perimeter $\vspace{15}P_{(x)}$ (in direction normal to the real axis, over the circles). After the folding, we would have a radius equal to $\vspace{25}R_{(x)}=\frac{P_{(x)}}{2 \pi}$ (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). $\vspace{15}A_{(x)}=2 \pi \int R_{(x)}\sqrt{1+R'_{(x)}^2}$ 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, $\vspace{15}\wp$ and that it may have a closed solution for $\vspace{15} b \in \mathbb{Z}$. I found this paper, about using the $\vspace{15} \wp$ function for solving differential equations, but I can't figure how to use it. We also know that $\vspace{25}R(x) \approx \frac {P(x)}{2\pi}$ Any idea? I have the result, but I do not yet know how to get it. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 02/29/2016, 10:31 PM For simplicity and useful properties , let us say we want Sums and products to be Well defined. Sums and products to be commutative and associative. No zero-divisors. Algebraicly closed. Not isomorphic to known numbers. There are 3 types of thought for such number systems. Although they are all similar. 1) analytic This leads to stuff like double periodic meromorphic functions. So f(a+b i) has a lot of cancellation / copies. We could even work with polar notation and have 3 invariants. But in essense we end Up with complex numbers and functions. Or the reals. This tends to lead to functions alone. Notice that higher dimensions tend to zero-divisors or loose the Desired properties. 2) geometric Things like giving the same value for everything ( points ) on a sphere. Related to modular , norms , determinants and bicomplex , although those examples are of algebraic or algebraic-geo nature. 3) algebraic Defining sums and products is a typical way. Groups and rings. Group rings , isomorphisms. That kind of stuff. I got inspired by the riemann surface of the log and the hyperoperators to study my numbers. I consider them of this " 3) algebraic " type. Z1,Z2 are complex Numbers. R1,R2 are reals. Tommy's spiral Numbers --------------------------- (z1,r1) * (z2,r2) = (z1 z2, r1 + r2). (z1,r1) + (z2,r2) = (z1 + z2,[r1 + r2]\2). Notice they satisfy the properties AND the distributive property. Also the complex are a subset of them ( r1 and r2 = 0 ) , what explains the algebraic closure alot. Here are some links http://math.eretrandre.org/tetrationforu...p?tid=1037 http://math.eretrandre.org/tetrationforu...p?tid=1036 In general the number of variables is the dimension. Reductions can lower the dimension but normally with An integer amount. For instance the groupring R+(C_4) - iso to the couple (z,r). Or R+(C_3) iso to the complex Numbers. This is the closest i could find as An answer to your question for commutative Numbers with Nice properties ... How this is gonna relate to tetration ... ¯\_(ツ)_/¯ Maybe if you had a plan to relate numbers in Some way ... Anyway hope this enlightens you. Regards Tommy1729 The master marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 03/01/2016, 12:12 AM (This post was last modified: 03/01/2016, 01:54 AM by marraco.) (02/29/2016, 10:31 PM)tommy1729 Wrote: No zero-divisors.Sometimes I found myself thinking that tetration introduces zero divisors, and that some of the new numbers have a geometrical representation as lines. In particular, a real number divided by zero should produce a line of infinite length, which is somewhat related to the Dirac delta. For example the Dirac delta is his derivative and crazy ideas like that... So, the limit for $\vspace{15}^{-2}b$ is not just an asymptotic, but a number, and the square waves which are limit to $\vspace{15}lim_{x \to \infty}\,^{x}b$ for $\vspace{15}0 \leq b \leq e^{-e}$ should be continuous functions. (02/29/2016, 10:31 PM)tommy1729 Wrote: 2) geometricI was planning to sum along geodesics, and multiply in such a way that I get the complex numbers for dimension r=2 But for that is needed a clearly defined surface for dimensions between 1 and 2. (02/29/2016, 10:31 PM)tommy1729 Wrote: Tommy's spiral Numbers --------------------------- (z1,r1) * (z2,r2) = (z1 z2, r1 + r2). (z1,r1) + (z2,r2) = (z1 + z2,[r1 + r2]\2).Should aim to have a matrix representation with a trace equal to the dimension, because of this. Quote:The dimension of a vector space may alternatively be characterized as the trace of the identity operator. Note that the identity of the product is 1, and if the identity of adding angles is the period $\vspace{15}\Omega \pi$ then the identity operator with your definition of product is $\vspace{15} (1,\Omega \pi)$. That gives a definition for $\vspace{15}\Omega = \frac{r-1}{\pi}$, which look suspiciously like: (02/29/2016, 12:46 AM)marraco Wrote: $a= - \left (\frac {c_1 (1-r)}{2 \pi} \right)^2$ (02/29/2016, 10:31 PM)tommy1729 Wrote: How this is gonna relate to tetration ...Maybe this makes clear how to calculate tetration to real exponents, the same way that complex numbers facilitate calculation with polynomials, roots, and other equations. I have the result, but I do not yet know how to get it. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 03/01/2016, 09:45 PM Correction : the spiral Numbers do not have associative addition. Regards Tommy1729 « Next Oldest | Next Newest »

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