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+--- Thread: Infinite tetration and superroot of infinitesimal (/showthread.php?tid=102)
RE: Infinite tetration and superroot of infinitesimal - Ivars - 02/27/2008
bo198214 Wrote:dx is used in the context of differentiation and integration. If you use it out of context, nobody knows what you mean. And if you additionally mix it into the theory of hyperreals, that doesnt simplify things. Limits however are well established in mathematics and
\( \lim_{x\downarrow 0} x^{1/x} = \lim_{x\downarrow 0} e^{\ln(x)/x} = 0 \) because \( \ln(x)/x \to -\infty \) for \( x\downarrow 0 \) (\( x\downarrow 0 \) means \( x\to 0 \) while \( x>0 \)).
There is nothing special about it.
In this thread, I was intereseted in infinitesimals in Leubniz/Euler sense, not limits. dx is an infinitesimal. What is dx^(1/dx)?
What is h(dx)? h(1/dx)? h(dx^2), h(1/dx^2)? etc.
If You use x>0, then You are of course right in evaluation of limit, but that is not considered as true limit, if values are different approaching from both sides, right?
Quote:What? The limit is a number and not a circle, how can a limit be a geometrical shape?
Undefined limits can be studied as limit cycles? To me this case (x^(1/x) looks similar to limit cycles in polar coordinates (even if we do not use complex values) , as even if it appraoches unit circle , it is not possible to tell where on Unit circle the end is as x-> infinity.
These spirals are perhaps similar to Gottfrieds lassoing with tetraseries, but I am not sure.
There has been suggestions to plot f(y,x) = y^(1/x) in 3D to see even better how it behaves close to y->0., x->0 . Or in Spherical. I just do not have the software/skills yet.
Ivars
RE: Infinite tetration and superroot of infinitesimal - bo198214 - 02/27/2008
Ivars Wrote:In this thread, I was intereseted in infinitesimals in Leubniz/Euler sense, not limits. dx is an infinitesimal. What is dx^(1/dx)?
What is h(dx)? h(1/dx)? h(dx^2), h(1/dx^2)? etc.
If you use something that is not properly defined, the answer will also not be properly defined. For my consideration the limit was the (properly defined mathematical) concept that comes closest to what you ask. Sometimes it seems to me that you intendedly try to avoid proper definitions because then the truth of your assertions could be verified.
Quote:If You use x>0, then You are of course right in evaluation of limit, but that is not considered as true limit, if values are different approaching from both sides, right?And infinitesimal (at least in the sense of hyperreals) is also positive or negative, isnt it? So you dont consider a both side limit when asking for dx^(1/dx).
Quote:Undefined limits can be studied as limit cycles?
I dont think so. Limit cycle is concept from dynamical systems and probably wrongly applied here.
Quote: To me this case (x^(1/x) looks similar to limit cycles in polar coordinates (even if we do not use complex values) , as even if it appraoches unit circle , it is not possible to tell where on Unit circle the end is as x-> infinity.
These spirals are perhaps similar to Gottfrieds lassoing with tetraseries, but I am not sure.
The limit \( x^{1/x} \) for \( x\to\infty \) is defined and so there is no need to complicate the matter.
Quote:There has been suggestions to plot f(y,x) = y^(1/x) in 3D to see even better how it behaves close to y->0., x->0 . Or in Spherical. I just do not have the software/skills yet.
You dont get more information by putting it into spirals (at least not in the way you do it), everything that is needed to see for the limit, you can see already from the usual cartesian function graph.
RE: Infinite tetration and superroot of infinitesimal - Ivars - 02/27/2008
Ok, that is then clear:
Thanks.
dx^(1/dx)= 0 where dx is infinitesimal as hyperreal if there is only one level of hyperreals.
But if there are still a smaller infinitesimal hyperreals of other level, smallest level? Because we are talking about selfroot, and self roots tend to be not 0.
I am interested in the type of spiral x^(1/x) creates in polar coordinates, anyway.
Ivars
RE: Infinite tetration and superroot of infinitesimal - bo198214 - 03/13/2008
Ivars Wrote:dx^(1/dx)= 0 where dx is infinitesimal as hyperreal if there is only one level of hyperreals.
dx^(1/dx) is not 0. It is another infinitesimal.
In the same way (dx)^2 is also another infinitesimal
and \( \lim_{x\to 0} x^2=0 \).
RE: Infinite tetration and superroot of infinitesimal - Ivars - 04/05/2008
If we apply Mother Law of hyperoperations to infinite tetration (and with a=i^(1/i)=e^(pi/2) than:
i^(1/i)+(i^(1/i)*oo) = i^(1/i)*(oo+1)
(i^(1/i))*(i^(1/i)^oo) = i^(1/i)^(oo+1)
(i^(1/i))^(i^(1/i)#oo)) =i= i^(1/i)#(oo+1)
Tetration is the first operation where something qualitatively new happens.
so (i^(1/i))^i=i=(i^(1/i))#(oo+1)
This allows(?) to apply formula recursively, so that by denoting oo'=oo+1:
(i^(1/i))^(i^(1/i))#oo')) =i= (i^(1/i))#(oo'+1)
If this is continued infinitely, then
(i^(1/i))^((i^(1/i))#(oo+ oo-1) )) =i= (i^(1/i))#(2*oo)
This again can be repeated, so going via 3 oo, 4oo etc we get:
(i^(1/i))^((i^(1/i))#(oo*oo-1)) )) =i= (i^(1/i))#(oo*oo)=(i^(1/i))#(oo^2)
It does not end there as well, as by repeating (oo^2+1, oo^2+2...) we get to oo^2+oo, then oo^2+oo^2= 2*oo^2 then n*oo^2 then oo*oo^3= oo^3
(i^(1/i))^(i^(1/i)#(oo^3-1)) )) =i= (i^(1/i))#(oo^3) and continuing this:
(i^(1/i))^(i^(1/i)#(oo^oo-1)) )) =i= (i^(1/i))#(oo^oo) and it still goes on, so as a point to have a look at this and may be stop:
(i^(1/i))^(i^(1/i)#(oo[oo]oo-1)) )) =i= (i^(1/i))#(oo[oo]oo)
What could be done next? Is this a full circle now or is there a way to add 1 still to oo[oo]oo? or next operation is 0[0]0?
(i^(1/i))^(i^(1/i)#(oo[oo]oo)) )) =(i^(1/i))^((i^(1/i))#(0[0]oo)-1) =i= i^(1/i)#(0[0]0)?
What it shows is that starting from tetration, i^(1/i) = (e^pi/2) = 4,81...ignores all operations, returning the same i at infinite application of any hyperoperation. Except for (1/i)^i returning 1/i I do not think there are ANY (or not MANY) other numbers able to stay the same during so fast hyperoperations. The smallest deviation from i^(1/i) or (1/i)^i will lead to immedeate amplifying of it via hyperfast operations away from itself .
Ivars
RE: Infinite tetration and superroot of infinitesimal - Ivars - 05/10/2008
I have stumbled upon Time calculus while searching for substructures of infinitesimals and how to include time in mathematics.
I thought it might give a handle on infinitesimals, since, how I understand, it, structureless linear infinitesimal h->0 in derivative definition is replaced in general case by graininess \( \mu(t) ->0 \) to obtain normal continuous derivative.
That means that infinite tetration etc of infinitesimal can be written(?):
lim \( \mu(t)->0 \): \( \mu(t) [4]\infty \)
And superroot :lim \( \mu(t)->0 \): \( q(t)^{q(t)}= \mu(t) \)
And self root: lim \( \mu(t)->0 \): \( \mu(t)^{1/\mu(t)} \)
The limit may depend on how \( \mu \) depends on t since it is not obvious what will happen faster- tetration or changes in \( \mu(t) \).
Gut feeling is, these speeds may be comparable.
The next question is, what will happen with this tetration and time calculus if \( \mu(t) \) itself is hyperoperation?
for example, \( \mu(t) = b[4]t \) ? or \( \mu(t) = t[4]\infty \)? Or \( \mu(t) = t[x]y \)?
And generally, hyperoperations in time scales might be interesting.
Please point out obvious mistakes. Thank You in advance,
Ivars
RE: Infinite tetration and superroot of infinitesimal - Ivars - 06/27/2008
I was sure \( I \) is not so simple as it has been portrayed. Below is text from Wikipedia on quaternions, but my idea is related to the "point" at origin of pure imaginary 2-sphere, whose projection on unit sphere in that \( ikj \) space is equivalent to \( I \) of a complex plane, and its antipode to \( -I \). One unit sphere in imaginary 3D space \( ikj \) contain infinity of complex planes , each corresponding to axis going via 2 points on the surface of imaginary 2 sphere.
The question is about the nature of the point at origin, which is not normal 0, since it is present in pure imaginary space.
I still have and always have had a gut feeling that the infinity of such axis going via imaginary 0 and tetration of real numbers to imaginary values are linked. That is why I have been trying to use dI since change from one complex plane defined by one of the axis in purely imaginary 3D space to another defined by another infinitesimaly close axis would lead to ANOTHER \( I \) in that space, dI away from previous. dI would be than a vector on the surface of the Unit sphere in \( ikj \) space, probably orthogonal to \( I \), and dI would be possible to decompose into dj and dk, or have spherical imaginary angle representation.
So when we get as a result of infinite tetration e.g. \( h(e^{\pi/2})=+- I \) we do not know do we speak about one of these complex planes or all of them. We do not know are there quaternions or even higher dimensional imaginary spaces involved or not; finally, if x^2+1=0 has infinitely many solutions in such space, what about x^x+1=0 in infinitedimensional imaginary space?
Of course, I am not sure if the point of origin in 3D imaginary space is not represented as some hyper surface in e.g. 7D pure octonion space which in turn is a hyper surface in 15D pure sedenion space etc.
Quote:H as a union of complex planes
Informal Introduction
There exists an intriguing way of understanding H that links its structure closely to the surface of an ordinary sphere of radius 1. In mathematics such a sphere is called a unit 2-sphere to emphasize that only its two-dimensional surface is being considered.
The first step is to translate the XYZ coordinates of the unit 2-sphere into the ijk coordinate system of quaternions, keeping the scalar (first) value of the quaternions set to zero. For example, the XYZ point <1,0,0> becomes the quaternion 0 + 1i + 0j + 0k. Since quaternion absolute lengths are calculated in the same way as XYZ radii, the resulting unit 2-sphere quaternions also all have absolute lengths (radii) of 1.
A less intuitive property of unit 2-sphere quaternions is that their squares all equal -1. This is true by definition for the three main axes of i, j, and k, but it can also be verified easily by trial for any arbitrary unit 2-sphere quaternion.
Since a length of 1 and a square of -1 are the defining properties of i, these unit 2-sphere quaternions look suspiciously like mathematical analogs to i. Furthermore, since each such quaternion has an "unused" scalar value associated with it, a fascinating conjecture becomes possible:
For any given ijk-only point on the quaternion unit 2-sphere, does the set of all quaternions that can be expressed as the sum of a real number and a multiple of that ijk-only point behave like a complex plane?
Somewhat unsurprisingly, the answer is yes.
That is, H can be partitioned in such a way that it looks like an infinite set of complex planes. Each such plane has its own unique version of i, although they all share the same real (scalar) axis. Furthermore, each unique i value corresponds to and is fully defined by a point on the surface of an ordinary unit-radius sphere, thus providing a strong connection between the geometry of ordinary spheres and the far less intuitive four-dimensional properties of H. Once a point on the unit 2-sphere has been selected, there is no mathematical difference in the behavior of the resulting subset of H and the more traditional concept of a single abstract complex plane.
Thus quaternions do not just extend the concept of i just to the two new axes j and k. They generalize i to an infinite set of points that happen to be the same ones found on the surface of an ordinary unit-radius sphere!
A more precise mathematical profile of how H can be interpreted as a union of complex planes is provided below.
Detailed Specification
Isomorphisms to the imaginary unit
The set of quaternions of absolute length (radius) 1 has the form of a 3-sphere or hypersphere, which is also called S³. Within this hypersphere there exists a subset of quaternions with the additional property that their squares are equal to −1. This subset has the geometric form of an ordinary sphere, or 2-sphere (S²). It can be understood as a three-dimensional "slice" of the larger hypersphere in much the same way that a circle is a two-dimensional "slice" of an ordinary sphere. For reasons explained below, this sphere-like subset of H is referred to here as Hi, where the i subscript refers to the imaginary unit.
This all can be found in Wikipedia under Quaternions.
Ivars
RE: Infinite tetration and superroot of infinitesimal - Ivars - 07/04/2008
I was thinking how infinitesimal Volume of any dimension ( including 3) may reach 0.
Given the dimensionality of space, at each dimension x there exists a smallest indivisible which can not be divided further. Further division can happen only by reducing its dimensionality, i.e. x.
Since we are used to work in 3D space, the smallest spatial indivisible could be found as a thing when dimensionality of space is x=3 which still has 3D shape and area but 0 volume in 3D- it would have volume in space with less than 3D , but not in 3D. As such, since its volume in 3D is 0, it is not further divisible spatially, but it can be divided further only by reducing space dimensionality.
The reasoning why this should be possible is based on fact that if You reduce small volume of space in 3D to infinitesimal, its volume ( since it is a cube of edge, roughly) reduces to 0 infinitely faster than its area, which in turn reduces to zero infinitely faster than edge. So at each dimensionality, there has to be an smallest indivisible just on border between previous and next smaller dimensionality.
The thought of the order in which infinitesimal volume disappears - first dimension, then edge, leads to a notion of continuos space dimensionalities (which in language of AN number of variables would mean non-integer number of variables, which should not be impossible, since there are non-itnteger number of terms in sums, non-integer iteration of functions, non-integer Euler characteristics (or measures) of polytopes etc.) where integer ones are just very important special cases. If there are continuos space dimensionalities y, then there can be also imaginary
and negative.
We may assume in the beginning these continuos dimensionalities to be real number, but very soon ( e.g. via derivatives or tetration) some special cases will start to pop out, perhaps in the end leading to some prime polytype structures in space being more important than others.
This leads to natural (?) interpretation of exponentiation as hypercube of
edge x and dimension y= x^y.
Particularly interesting seems case where y<1 in which case it seems to me space
will become disconnected, or ? I have not read anything about this yet.
Which in turn allows to consider case x^x but here both x have
DIFFERENT geometrical meaning- one is edge, another SPACE dimension.
The speed of growth or decline of hypervolume x^x depends DIFFERENTLY on each x, so the derivative of the function makes them distinquishable. For some
reason, if the edge of such hypercube is e, (so e^x) all its
derivatives are e^x.
If we take partial derivatives of x^y, then they are:
y*x^y-1 - this is derivative by length of edge x, dimensionality y=constant
and
x^y*ln(y) - this is derivative by dimension y, edge x= constant.
If we now replace y with x, these derivatives are equal with oppossite signs at x= 1/e. The value of hypervolume is (1/e)^(1/e)=0,69... At x=e,at hypervolume e^e=15,.. the derivative by edge length gets smaller than derivative by dimension, so if edge is smaller than e AND dimension is smaller than e, then volume disapppear first by reducing edge in dimension x, than dimension x itself. If edge is bigger than e AND dimension is bigger than e, then increase of hypervolume happens first by increase in DIMENSION x , and then by increase of EDGE to x in THAT dimension x.
Understanding case x^y and exponentiation in general in dimensional
terms should allow to interpret tetration x^x^x^ ........
geometrically, and what infinite tetration of e^pi/2 = +-i might mean
geometrically and physically.
Of interest are also other geometrical figures in X dimensional space,
like sphere, and its volume/area values as function of radius.
Here are separated derivatives in picture:
Ivars
RE: Infinite tetration and superroot of infinitesimal - Ivars - 07/04/2008
When mentioning continuos dimensions of space above, I do no mean fractal dimension
which should be used in this context, but topological dimension or some equivalent measure that is possible to extend to rational and real numbers. So it is more linked to:
Fractional Calculus
which is actually continuous real number iterates of differentiation/integration operations and involves gamma function ( factorial!).
Since integer differentation reduces dimension in exponent by 1, while integration increases, e.g.
d/dx (x^n) = n*x^(n-1)
it is reasonable to assume that any noninteger topological dimension (or what could be its equivalent) perhaps including negative and imaginary would be accesable via fractional differentiation/integration.
Ivars
RE: Infinite tetration and superroot of infinitesimal - Ivars - 07/05/2008
When looking at x^y as hypercube in dimension y with edge x, I have made implicit assumption that the ANGLE in which each edge of such hypercube creates with another is 90 degrees in any dimension y, just by analogy, or Pi/2, as in general definitions of hypercube.
However, this angle may be another,may be negative, may be imaginary.
This leads to interpretation of real etc. extensions of hyperoperation number z in x[z]y as angle between such edges x of a hypercube in dimension y either directly, or via some exponential relation like log (pi/2)^n/log (pi/2) = n for integer z, and z*log pi/2/log (pi/2) =z for all other. , or n= log (e^n*I*pi/2)/log I , z= log (e^z*I*pi/2)/log I.
As for imaginary, negative angles and other imaginary things like points, edges etc. it seems obvious that such interpretation of hyperoperations and its extension for real/imaginary/negative number of operations leads to/is connected with projective geometry in its original complex form, as developed by Lambert, Gauss, Stoudt, Cayley, Klein. It seems to me that real numbers as foundations of geometry fail here(?) (in interpretation of hyperoperations to real , imaginary, negative numbers) , which is of no big surprise since their foundation are based on physical convenience but not abstract proofs (since it is unprovable so far) , so alternative models are possible as well.
But since I only conjectured this yesterday, there are many things to read, particularly about origins of non-euclidian geometry . Just to summarize:
In expression :
x[z]y
x- edge of hypervolume in y dimensions
y- dimensions of hypervolume with edge x
z- is related to varying angle between edges x of hypervolume in y dimensions, perhaps via logarithm or some trigonometric function. A meaningful definition should recover usual angles pi/2 in case of hypercube in integer dimensions.
The geometric interpretation of hyperoperation in general and its real, complex extensions is related to projective geometry in imaginary form extended to non-integer, negative and imaginary dimensions.
To give a more concrete test example:
I[I]I is a hypervolume with edge I in I dimensions with angle between imaginary edges (e^(I*I*pi/2) = e^(-pi/2)
I[-I]I is a hypervolume with edge I in I dimensions with angle between imaginary edges e^pi/2. In I dimensional space, such angle might have a meaning.
It also seems that his hyperangle conveyed via z is a composite one, so related to twisting of hypervolume x[z]y of dimension y. This is probably possible to check if some volumes in ordinary integer spaces of twisted hypercubes are known.
Ivars