10/31/2016, 02:57 PM (This post was last modified: 10/31/2016, 03:26 PM by Stanislav.)
In previous time I supposed to make a paper in English explaining my thoughts about zeration, but now I nly am able to present here an article in Russian. It is not provided with a literature overview but I've read some works by G.F. Romerio & Rubtsov, and some other works named in my paper. Finally I have to note that there's a lot of questionable points in my suggestions about zeration and I am ready to discuss them.
UPD. Some questionable points.
1) Our resulting solution (an algorithm for calculating collection operation) is depending on what requirements we want to satisfy. E. g., I wanted to achieve an implicit algorithm based on hyperboloidous approximations, plus some 'common sense' suppositions - so then I've got my algorithm which is in this paper. But nevertheless I've also got another, explicit formula with exponents, satisfying all the constarints too, but in another manner. Furthermore, I can put some new constraints built on a sand, in a manner 'what if also ...', in a 'common sense' manner too, which are making collection operation more similar to summ and multiplication but are not allowing to use my algorithm from the paper.
So the question is what do we want? What is the only set of requirements necessary for an operation, such that we could call it pre-addition?
My first algorithm presented in the paper is based on polynoimal equations and hence leads to everywhere we want in mathematics.
My second algorithm not presented here is based on exponential functions and allows to connect this operation to complex numbers, Fourier transformation and so on.
A way chosen by Romerio & Rubtsov, max(a, b) + 1, tends to give non-smooth functions and thus is harder to extend I guess.
Is it probable to connect these and some other ways together in order to find a non-controversary image of a pre-addition?
We all would like to finally get an infinite array of numerical operations or, probably, an infinite tree or multidimensional set of them, but we should list all the requirements we want to satisfy, and issue all the features we are able to discover in the operations which are already known to us.
06/13/2022, 10:43 PM (This post was last modified: 11/04/2022, 08:35 AM by Catullus.)
(03/20/2015, 09:44 AM)marraco Wrote: This definition for zeration has similar properties to the common zeration:
\( a \circ b \,=\, ln(e^a+e^b) \)
\( a \circ -\infty \,=\, a \,+\, ln(1),=\, a \)
\( a \circ a \,=\, a \,+\, ln(2) \)
\( (a_n \circ \,...\,(a_3 \circ (a_2 \circ a_1))) \,=\, a \,+\, ln(n) \)
if b>>a
\( a \circ b \,=\, b \,+\ ln(1+e^{a-b}) \,\approx \, b \)
if b>a
\( a \circ b \,=\, b \,+\ ln(1+e^{a-b}) \,<\, b \,+\, ln(2) \)
\( a \circ -\infty \,=\, -\infty \circ a \,=\, a \)
\( -\infty \circ -\infty \,=\,-\infty \)
\( x \,\circ\, -x \,=\, ln(cosh(x)) \,+\, ln(2) \)
Under that definition of zeration, a°-∞ would be ln(e^a+e^-∞), not a.
(03/20/2015, 09:44 AM)marraco Wrote: This definition for zeration has similar properties to the common zeration:
\( a \circ b \,=\, ln(e^a+e^b) \)
\( a \circ -\infty \,=\, a \,+\, ln(1),=\, a \)
\( a \circ a \,=\, a \,+\, ln(2) \)
\( (a_n \circ \,...\,(a_3 \circ (a_2 \circ a_1))) \,=\, a \,+\, ln(n) \)
if b>>a
\( a \circ b \,=\, b \,+\ ln(1+e^{a-b}) \,\approx \, b \)
if b>a
\( a \circ b \,=\, b \,+\ ln(1+e^{a-b}) \,<\, b \,+\, ln(2) \)
\( a \circ -\infty \,=\, -\infty \circ a \,=\, a \)
\( -\infty \circ -\infty \,=\,-\infty \)
\( x \,\circ\, -x \,=\, ln(cosh(x)) \,+\, ln(2) \)
Under that definition of zeration, a°-∞ would be ln(e^a+e^-∞). Not a.
Hey, Catullus. Those are one and the same on the extended real number system.
You are confusing your systems of logic. Yes, in hyperreals \(e^{-\infty}\) is infinitesimal. But in the extended reals it is 0. It depends on how we interpret these equations. And as we are operating under traditional limits, and not hyperreal limits--the result stands.
(06/14/2022, 01:16 AM)JmsNxn Wrote: You are confusing your systems of logic. Yes, in hyperreals \(e^{-\infty}\) is infinitesimal. But in the extended reals it is 0. It depends on how we interpret these equations. And as we are operating under traditional limits, and not hyperreal limits--the result stands.
If e^-∞ = 0, then e^-(-∞) = 1/e^-∞ = 1/0. But you can not divide by zero.
(06/14/2022, 01:16 AM)JmsNxn Wrote: You are confusing your systems of logic. Yes, in hyperreals \(e^{-\infty}\) is infinitesimal. But in the extended reals it is 0. It depends on how we interpret these equations. And as we are operating under traditional limits, and not hyperreal limits--the result stands.
If e^-∞ = 0, then e^-(-∞) = 1/e^-∞ = 1/0. But you can not divide by zero.
Nahhhh, none of that is how things work. The function \(\exp(x) : (-\infty,0] \to (0,1]\), which the limit on the boundary as \(x \to -\infty\), tends to the limit on the boundary of codomain. Which is zero. Same principle here.
(02/14/2008, 06:38 PM)GFR Wrote: a ° b = a + 1 , if a > b
a ° b = b + 1 , if a < b
a ° b = a + 2 = b + 2 , if a = b
Under that definition of zeration, a°-∞ would be equal to max(a,-∞)+1. Not a.
Nope, the definition says nothing about its evaluation at \(-\infty\). That evaluation is derived by another rule.
The main purpose of the Rubtsov-Romerio's approach is to propose a solution that satisfies some constraint, and where it fails to compute to look for more fundamental and natural principles from where we can derive new constraint on the solution.
Their solution to the problem is the original Zeration. I mean that nowadays we use the term zeration for all the 0th ranks operations of a given Hyperoperation family. But the therm Zeration was created original by Rubtsov and Romerio to denote that particular solution.
Note also that their proposed solution, that they extended to a new field \(\mathbb R\subseteq \Delta\), has presents a discontinuity at \(-\infty\), that's why the limit doesn't agree.
The info on this is not well presented in this thread nor in many other threads (except a few where I hope I had given a reasonable review of it). On top of this add that this research around "pre-additions" was always conducted mainly by non professional mathematicians and rejected as trivial/stupid by professional mathematicians. I know it may seem an annoying state of the art but there is lot of hidden folklore around this and the original papers are not easily accessible anymore.
I have much of the material though, I might share some, one day.
To add to the correct JmsNxn remarks you can see it as follows:
evaluating something defined on the reals at infinity means evaluating a function at something that is outside it's domain. It is impossible by definition.
In order to do that you can extend the concept of evaluation or you can extend the domain.
The first solution is the definition by limit or the analytical route. You define the value at infinity as a limit, extending the meaning of function evaluation.
The second solution is by extending the domain, i.e. the algebraic route. You adjoin new elements to the domain, but then all the previous algebraic rules could fail. You can do this in many non-equivalent ways. For example by forming the hyperreals you lose the archimedean property, a fundamental property for analysis but also you lose the uniqueness of infinity because you have different kind of infinities and infinitesimals.
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
06/28/2022, 06:44 AM (This post was last modified: 06/29/2022, 10:50 PM by Catullus.)
(02/19/2008, 09:41 PM)andydude Wrote: In my view, it is not a matter of proof, but a matter of choice. If we choose to hold \( a [N-1] b = a [N] (\text{hy}N\text{log}_a(b) + 1) \) to be true for all N, then the natural consequence is that \( a [0] b = b + 1 \) for all a (because \( a [1] (\text{sub}_a(b) + 1) = a + ((b - a) + 1) = b + 1 \)). If we choose to hold \( a [N-1] a = a [N] 2 \) true for all N, then we *** \( a [0] b = b + 2 \) for a == b. Both cannot be true, so we must make a choice.
Andrew Robbins
Then, why not make a zerated to the a equal a plus 3 halves as a compromise?
