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RE: Infinite tetration and superroot of infinitesimal - Ivars - 12/24/2007

jaydfox Wrote:If you want to treat the infinitesimal as always positive, then it's simply the logical end of the sequence dx, dx^2, dx^3, ..., dx^(1/dx). In ordinal notation, it's 1/(omega^omega), and omega^omega is a rather small ordinal in the hierarchy of countable infinities. (Not that I particularly like the hierarchy.)


(1/dx) is bigger than any n. It lies outside sequence of infinite natural numbers. (please look into non-standard analysis). It is a qualitatively different number. And after 1/dx, we still have infinite INTEGERS. What are they?


RE: Infinite tetration and superroot of infinitesimal - jaydfox - 12/24/2007

But omega IS qualitatively different. It is neither odd nor even. It is neither prime nor composite (though a statistical argument can be made that it's almost certainly composite). It IS greater than all natural numbers, by defintion. As such, if you want to relate 1/dx to existing models, the natural fit is omega. You could choose a larger (uncountable) infinity, but I don't see this as necessary, unless there is a strong reason to.


RE: Infinite tetration and superroot of infinitesimal - Ivars - 12/25/2007

jaydfox Wrote:But omega IS qualitatively different. It is neither odd nor even. It is neither prime nor composite (though a statistical argument can be made that it's almost certainly composite). It IS greater than all natural numbers, by defintion. As such, if you want to relate 1/dx to existing models, the natural fit is omega. You could choose a larger (uncountable) infinity, but I don't see this as necessary, unless there is a strong reason to.

I was thinking of dx as a line segment, infinitesimal. A tangent to line x=x. Not a point. And not possible to create from finite number of points. As line segments that has direction but no magnitude. That are infinitely divisible but never You get a point as a result. So they are kind of extremely scalable . Would that qualify for omega?

Please see this- the only place I found so far something written what
I can also accept about infinitesimals. They in totality create a tangent space in all scales which exists, or can be assumed to exist.

http://publish.uwo.ca/~jbell/invitation%20to%20SIA.pdf

Although I do not think stopping at eliminating squares (nilsquared) is good enough-all infinite number of scales has to be involved simultaneously- therefore my interest in infinite tetration.

May be You can suggest other references (no points and sets, please), may be very old.

Ivars


RE: Infinite tetration and superroot of infinitesimal - andydude - 12/25/2007

Before proceeding further, I would like to be precise about what an "infinitesimal" number is.
  • There exists no z such that \( 0 < z < a \) for all real \( 0 < a \)
  • There exists dz such that \( 0 < dz < a \) for all real \( 0 < a \)
An infinitesimal is a number that satisfies the second property, is this correct?
How does this not form a contradiction with the first property?

Andrew Robbins


RE: Infinite tetration and superroot of infinitesimal - Ivars - 12/25/2007

andydude Wrote:Before proceeding further, I would like to be precise about what an "infinitesimal" number is.
  • There exists no z such that \( 0 < z < a \) for all real \( 0 < a \)
  • There exists dz such that \( 0 < dz < a \) for all real \( 0 < a \)
An infinitesimal is a number that satisfies the second property, is this correct?
How does this not form a contradiction with the first property?

Andrew Robbins

Infinitesimal is NOT a Real number at the scale we are looking at things. At the scale where Reals are Reals, Infinitesimals are not Real. Their Real part is 0, but there is more in them. Infinitesimal is definitely closer to zero, as well as closer (surrounding) to any Real than any other Real. It may be that application of relations < ; > is misleading here.

Considering Infinitesimal to be somehow Real just smaller than any Real , or have any Real part ( if it is multidimensional) in the scale we are in leads to contradiction You pointed out.


RE: Infinite tetration and superroot of infinitesimal - jaydfox - 12/26/2007

Andrew, methinks that infinitesimals would necessarily have density measured with 2^C, using a Cantorian system. C is 2^omega or aleph-null or whatever. The reals are infinitely more dense than the rationals, but that's not really enough to describe them. Between any two rationals, there are an infinite number of rationals. Thus, there are an infinite number of rationals between 0 and any rational. There are an infinite number of reals between 0 and any rational. Yet somehow the reals are more dense.

In real analysis, reals are as dense as the number line gets. I'm assuming that with this non-standard analysis, the infinitesimals would create a field that is as dense relative to the reals as the reals are to the rationals. But I don't particularly see how a coherent system could be derived from this. Of course, I'm not a very good topologist, and at any rate, I never bought the distinction in density (perhaps cardinality is the right word?) between rationals and reals, so I'm not really in a position to be convinced of yet another layer of density.


RE: Infinite tetration and superroot of infinitesimal - Ivars - 12/28/2007

I have been looking at relation for infinitesimal exponentation of base a:

by definition, if we accept I as infinitesimal which is not 0, I^I = (1+J) where J is infinitesimal as well (since I^0=1). Then log base I ( 1+J) = I log base I (I) = I.
Euler made a proposition which he proved that J=kI, where k = 1 for base e so that ln(1+I)=I where I is just infinitesimal.

If we take imaginary Unit I as being the infinitesimal in base I , what is k?

From Eulers derivations, k as function of base a is :
k= 2( (a-1)/(a+1) + (a-1)^3/3*(a+1)^3+ (a-1)^5/5*(a+1)^5 ....

As we work with base I, where I = imaginary Unit, we have to evaluate term

(I-1)/(I+1) = (sqrt(2)* e^I*((3pi/4)+2pik)/ sqrt(2)* e^I* ((pi/4)+2pin) = e^I* (3pi/4-pi/4+2pik-2pin) = e^I*pi/2= i as a single value, not periodic, since all 2pik and 2pin cancel out.

So k= 2 ( I+I^3 /3+ I^5/5+I^7/7 +...) = 2*I ( 1-1/3+1/5-1/7.......) = 2*I* pi where I is JUST e^I*pi/2, single value.

Then J as infinitesimal result J = (I*pi/2 ) * I so J is also infinitesimal but turned by 90 degrees via multiplication of I by I plus it has module pi/2 and of I^I =(1+J) = (1+ (I*pi/2)*I) and

But we know that I^I = e^-pi/2 so that ln(I^I) = -pi/2= (I*pi/2)*I:

But we also know that I^I = (1+I*pi/2*I) so that ln(I^I) = ln(1+I^2*pi/2) = k*I= J= I*pi/2*I= -pi/2 if I is imaginary infinitesimal such that I^2 = -1.

So assumption that I may be imaginary infinitesimal works with logarithms, but the unique note here is that we are not dealing with ALL periodic values of I, but just I, such that is linked to 1 in this scale so that I=(1-I)/(1+I); In a similar vein, unique value of -I = (1+I)/(1-I) can be obtained.

Interestingly, first logarithms of Napier ( 1-10-7) and Burgi ( 1+10-4) were also based on small deviations from 1.

The conclusion is ( or proposition was) : Mathematics works in SCALES which we do not notice for a reason that they are imaginary ( meaning in our minds),so we assume the same scale exists for all operations, which is not true, and in each scale the relation ship between Unit lentgh 1 and imaginary infinitesimal is defined by I=(1-I)/(1+I). This imaginary infinitesimal is Unique for that scale and by rotating it by +2pi we move into next scale, perhaps via multiplying by J= I*pi/2 where 1 is also different, but we do not notice it, as in math, we mix all scales in our heads - which is not possible in physics, and therefore the discrepancy.


RE: Infinite tetration and superroot of infinitesimal - Ivars - 01/02/2008

A few more ideas involving tetration as missing operation in the end. It has nothing to do with Cantor, sorry. I am not a fan of neither limits, nor real number line. I hope some one will point to some obvious mistake I am making in logic, not in rigorSmile

What would You say about Pi being infinitely infinite? And i being scalable imaginary infinitesimal with internal linear wave structure (basically, organized frequencies).

It goes like this:

Look at this link:Continuous variable transmission with flexible shaft

1) From it we can see, that , if the circle is measured from inside with a disc with radius= 1/2 we get ratio circumference/radius of the big gear = pi/2. If it is measured by rotating the same disc in a plane perpendicular to big gear, ratio circumference/radius of the big gear = pi; if measured from outside, ratio is 3pi/2. When it is turned back downside, You get pi again, and if turned back inside ( 4th turn by 90 degrees= totally 360 degrees) , You get Pi/2 again.

2) So in fact, PI is multidimensional, or, infinitely dimensional number as any infinitesimal change of measuring angle will lead to change in ratio. That is why circle is not squarable in the plane, You can only do it in infinite dimensions, each of which is represented by infinitesimal angle.

3) Other infinity involved in PI is the size of measuring disc. If we use 1/2 of radius for measuring disc, we get Pi values I described above. If we use smaller disc, the ratios will be different- if we use infinitesimal disc, the ratios would be the same from all sides=pi , while if we use infinite disc, ratio will always be 1/2.

4) So there is 2 infinite spectra's of values pi can have , based on angle and measuring disc side ( there is no other exact way to measure circumference of a circle with infinitesimally thin boundary as by rotating a tangent disc with a same thickness boundary along it).

5) So we have Pi as function of 2 infinities. They may sometimes cancel out, giving pi, but in most cases, they would not. This leaves Pi as a number of dimension say infinity^infinity.

6) When we differentiate anything involving pi in a continuous way, we reduce the scale (that is the length part of uncertainty in PI) of PI , the dimensionality by 1 infinity, so it becomes infinity^infinity-1. For all practical purposes, differentiation or integration would leave PI infinite compared to any other number, but in fact, its size will be reduced/increased with each such operation. The angle part remains infinitely complicated ( if differentiation is done in Cartesian coordinates, not e.g. in polar and bu angle).

7) Now how to get a finite rotation out of infinitely dimensional number? We know that rotations in each scale are performed by i, so that

e^i* pi/2 always means 90 degrees to the left, anticlockwise. That means that while Pi, i, and perhaps also 2 change in each scale with each differentiation/integration by coordinate, e^I*pi/2 is always the same, finite.So that d(e^i*pi/2)/dxdangled something) = e^ipi/2

Cool As Pi has 2 uncertainties involved ( see above), we can make a conjecture that in each scale ( after each differentiation/integration) e takes care of change in scale (since de^x/dx = e^x) so that e is simpler infinitely dimensional number than Pi, i takes care of angle uncertainty ( so I is also simpler than Pi, but have different dimensionality in each scale as well ) and I am not sure what, but 1/2 must also take care of some thing.

9) So the final conjecture is that Pi is perhaps dimensional in a way (infinity^infinity) infinite times - a tetration of (infinity^infinity). That is not so far fetched, since h( e^pi/2) = i-1/i+1 ; -i +1/-i-1 as can be easily verified.

10) There is no way PI can stay on real axis, since reals are 1 dimensional infinite numbers, working in 1 scale only. The same applies to 1/2 and e. i becomes a hyperreal infinitesimal with structure that is based on symmetries of infinitesimal linear angles in each scale.


RE: Infinite tetration and superroot of infinitesimal - andydude - 01/02/2008

Ivars Wrote:What would You say about Pi being infinitely infinite? And i being scalable imaginary infinitesimal with internal linear wave structure (basically, organized frequencies).

In order for this to be true, one of the following would have to be true:
  • Your definition of \( \pi \) is different than mine.
  • Your definition of infinite is different than mine.
In either case, I have no idea what you're talking about.

Andrew Robbins


RE: Infinite tetration and superroot of infinitesimal - Ivars - 01/02/2008

andydude Wrote:In order for this to be true, one of the following would have to be true:
  • Your definition of \( \pi \) is different than mine.
  • Your definition of infinite is different than mine.
In either case, I have no idea what you're talking about.

Andrew Robbins

I have to prepare better what I want to say, and express shorter. My definitions are still in making...But the most significant difference will appear in imaginary unit, most likely.

Ivars