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 Infinite tetration and superroot of infinitesimal Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/16/2007, 06:02 PM In non-standard analysis,hyperreal numbers are introduced, such that around 0, hyperreal is smaller than any real. ( Let us not involve infinite hyperreals now, if not needed). This is called infinitesimal which is smaller than any real and is not 0. Non-standard analysis is quite legitimite, albeit different way of doing and understanding calculus (at least). So there migth be potential for insights if non-standard is used instead of limit based math in tetration. Question is: If we can not find h(0) as its value oscillates, may be we can find: h( infinitesimal) =? and Superroot of infinitesimal = ? andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 12/17/2007, 02:27 AM Ivars Wrote:h( infinitesimal) =? I thought we already answered this question here. The problem is not that we can't find an answer for this. We can find h(0) (if we define it from the Lambert W function). We can also find $h_O(0) = 0$ and we can find $h_E(0) = 1$. So the problem is not that we can't find its value, the problem is that there are 3 extensions of h(x) beyond the interval $[e^{-e}, e^{1/e}]$. Each extension has some very nice properties to it, and a reason for its existence. The "nice" extension is the one that is differentiable across $e^{-e}$, which is the definition of h(x) in terms of the Lambert W function, or equivalently, as the inverse function of $x^{1/x}$. Using this definition, h(0) = 0, and thats final, there is no oscillation. Asking about super-real numbers near zero is, at best, going to give a fourth extension of h(x), which is not going to bring us any closer to having single definition for h(x) near zero, as it is going to increase the number of definitions. Ivars Wrote:Superroot of infinitesimal = ? Did you mean the 2nd super-root? or the infinite super-root? The infinite super-root is just $\text{srt}_{\infty}(x) = x^{1/x}$, and $\lim_{x \rightarrow 0} x^{1/x} = 0$ which may answer your question. The 2nd super-root (sometimes called the super-sqrt) is the inverse function of $x^x$ and since that function has a minimum at 1/e which gives the value $(1/e)^{(1/e)} = 0.6922$. This means that the 2nd super-root $\text{srt}_{2}(x) = \frac{\log(x)}{W(\log(x))}$ is real-valued for $x \ge 0.6922$, but would be complex-valued from that value down to zero. Using the 2nd super-root defined by the Lambert W function, $\lim_{x \rightarrow 0} \text{srt}_{2}(x) = -\infty$ which can be verified independently. Andrew Robbins Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/18/2007, 02:59 PM (This post was last modified: 12/19/2007, 08:03 AM by Ivars.) hej Andy Thanks for careful examination and corrections. It seems so that h(infinitesimal) does not contain much new information, at least I do not feel so . But superroot is more interesting andydude Wrote:Did you mean the 2nd super-root? or the infinite super-root? The infinite super-root is just $\text{srt}_{\infty}(x) = x^{1/x}$, and $\lim_{x \rightarrow 0} x^{1/x} = 0$ which may answer your question. I am not sure You can apply limit to hyperreals. Infinitesimal I mean in this case is by definition smaller than any real, but NOT 0. So infinitesimal^(1/infinitesimal) = (infinitesimal^(infinity of the same grade) ) IS NOT 0. I would rather say it is e, or e^pi/2. andydude Wrote:The 2nd super-root (sometimes called the super-sqrt) is the inverse function of $x^x$ and since that function has a minimum at 1/e which gives the value $(1/e)^{(1/e)} = 0.6922$. This means that the 2nd super-root $\text{srt}_{2}(x) = \frac{\log(x)}{W(\log(x))}$ is real-valued for $x \ge 0.6922$, but would be complex-valued from that value down to zero. Using the 2nd super-root defined by the Lambert W function, $\lim_{x \rightarrow 0} \text{srt}_{2}(x) = -\infty$ which can be verified independently. Andrew Robbins This is very interesting as well is it returns complex values from real numbers. I have to learn more about superroots. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/19/2007, 08:42 AM May be this makes it easier to move forward with the problem at hand: I have come to one conclusion based on my tetration thinking about what is infinitesimals and i. Let me put it like this: 1) Imagine infinitesimal of some grade, structure. Then (1/this infinitesimal) is infinity of the same grade, so that Infinitesimal* ( 1/infinitesimal) is = 1. This is simple stuff known from Euler's times, and accepted in non-standard analysis. Now ask a Question: what is Infinitesimal^(1/infinitesimal) = infinitesimal^ (infinity of the same grade)? It is not 0 as infinitesimal is NOT 0, as it is bigger than infinitesimal squared, which is also not zero. Or, alternatively, what is (1/infinitesimal) ^ (infinitesimal) = (infinity of the same grade)^ infinitesimal? It is not 1 since infinitesimal is NOT 0. 2) By some logic, we would expect that ( I can not prove it yet): infinitesimal^(1/infinitesimal) = 1/infinitesimal^infinitesimal ( all infinitesimals infinities are of the same grade, in the SAME SCALE of infinitesimals). Now look at this- these are simple identities once You use i=e^(ipi/2), -i = e^-i(pi/2): i^-i = i^(1/i) = -i^i = (1/i)^i = e^( pi/2) We also know that (1+ 1/infinitesimal)^infinitesimal = e, so (1-i)^i =e as well. Is it? What is the conclusion/conjecture? At any given scale (grade) of infinitesimals , what ever it is, i is a notation for infinitesimal, and it has a property in this scale that i^1/i = infinitesimal^ 1/infinitesimal= e^pi/2. So complex numbers, or imaginary unit appear as a result of MANDATORY need to involve infinitesimals in any GIVEN scale we operate our math. Since we use to jump over scales of infinitesimals in mathematics ( everything was just small, in case of limit approach, we do not even notice them at all), we could not decipher it earlier. 3) At any given infinitesimal scaleSCALE is THE word, and SCALES are relative): i^2=-1 i^3=-i i^4=1 etc telling that by definition, infinitesimals of the same grade has very limiting constrictions of how they can be organized. And that scales 1 and -1 are created with some meaning at ANY scale - may be in the one we happen to be in, may be lower, may be higher. Even if infinitesimal scale is something like infinity^sgrt(7) smaller than some other, once we enter that scale, and define the size of infinitesimal by dividing previous scale with infinity^sgrt(7) it becomes i on that scale, and the cyclical properties of i holds. 4) The difference between scales of infinitesimals is the level of complexity. In our case it is Monster. In lower scales- less. In higher? Are there scales more complex than Human? I do not know. 5) Huh, finally I see the light once imaginary unit is somehow placed where it should be. I ( imaginary unit) is a universal notation of infinitesimal in ONE infinitesimal scale which is more or less INFINITY away from next scale. From here we can notice that i^3= -i = infinity in that scale. 1 is clear, more or less, - 1 still a little bit of a mystery. 6) As we go further with powers of i knowing that i is infinitesimal, we may ask whether or not as we rotate something via e^i(phi) = e^ h( e^pi/2) *phi what we are really doing? Are we not adding infinitesimal 3RD dimension on top of complex plane by each rotation, so that it is UNNOTICEABLE in the scale we operate, but in infinitesimal space, there is a spiral structure growing which we do not notice directly, but it INFLUENCES what happens in our scale. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/19/2007, 04:03 PM (1-i)^i=e ln(1-i)=1/i=-i is this true? if we expand ln(1-i) = -i -i^2/2+i^3/3-i^4/4........since i is infinitisemal, i^2, i^3, i^4 etc vanishes in summation compared to -i, if we stay in the same scale with our calculations so : ln(1-i)=-i=1/i and (1-i)^i=e Main idea was that +i is in fact, infinitesimal of any scale provided we do calculations while staying in that scale. Then -i is infinity of the same grade. jaydfox Long Time Fellow Posts: 423 Threads: 30 Joined: Aug 2007 12/19/2007, 06:29 PM (This post was last modified: 12/19/2007, 06:31 PM by jaydfox.) Ivars Wrote:(1-i)^i=e ln(1-i)=1/i=-i is this true? The absolute value of (1-i) is sqrt(2), and the angle from the origin is -pi/4, so the natural logarithm is ln(sqrt(2))-pi*I/4, or 0.5*ln(2)-(pi/4)*I if you prefer it that way. ~ Jay Daniel Fox Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/19/2007, 08:55 PM jaydfox Wrote:The absolute value of (1-i) is sqrt(2), and the angle from the origin is -pi/4, so the natural logarithm is ln(sqrt(2))-pi*I/4, or 0.5*ln(2)-(pi/4)*I if you prefer it that way. Yes I know I am outstepping convetions, probably this forum is not the right place to do it, but: Argand diagram is not applicable to my idea. The fact that mod(i) = 1 is not obvious from definition of i as sqrt(-1). The fact that (sgrt^(-1))^2 = 1 does not mean module of a negative number squared is 1. At least when applied to physics since negative numbers as such are artificial construct:If someone ows me money, he has negative money in balance sheet. But in fact he does not have negative money, just more or less positive. So negative numebr as such involves delay, some time period over which it can be represented back into positives, and this period may be such that the scale of negative numbers is totally different from positives- e.g. interest or inflation may reduce the scale of negative money because of time. I think that the diagram would be right, but i being infintiesimal and -i being infinity of the same grade, and -1 also having different modulus than 1. mod(1-i) = infinity so ln (1-i) = infinity as well (1/i) mod(1+i)= 1 so ln (1+i) = i ( not 0). mod(1)=1 mod(-1) = mod(i^2) = second order infinitesimal - smaller than i mod(1)=mod(i^4) = 4th order infinitesimal even smaller than i In this case the diagram would have totally different scales on i, -1, -i and 1 axis. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 12/20/2007, 03:16 AM Yes, super-roots are very interesting. In fact, I would really like to know if there is a recurrence equation for the coefficients of super-roots in general. I have done some initial research into series-expansions of super-roots, but I have not found any nice results yet. The only fruit of my research into this subject is the Taylor-Puiseux conversion that I found and talked about in Abel Functional Equation since it was useful to use it there. I use this term since a Puiseux series is any series involving logarithms, and that theorem allows one to convert between Taylor series and series involving logarithms in a very interesting way. Although it was much easier to prove than I thought at first, it is not trivial, since it confused me for the longest time. As an example of the research that I did on super-roots, it is best to start with the research of Ioannis Galidakis, as he found (with the help of Leroy Quet) the first recurrence equation for the coefficients of integer tetration. Although he does not use the same terminology, Ioannis Galidakis uses a Puiseux series for tetration, for example, the Puiseux series for x^x^x is: $x^{x^x} = 1 + {\ln(x)} + \frac{3}{2}{\ln(x)}^2 + \frac{8}{3}{\ln(x)}^3 + \frac{101}{24}{\ln(x)}^4 + \cdots$ now, for comparison, the Taylor series for x^x^x is: $x^{x^x} = 1 + {(x-1)} + {(x-1)}^2 +\frac{3}{2}{(x-1)}^3 + \frac{4}{3}{(x-1)}^4 + \cdots$ Now if we use Langrange series inversion to invert the first series, we will get "ln(x) = something", which is not what we want, so it is actually easier to use Lagrange series inversion to find the inverse of the second series to obtain: $\text{srt}_3(z) = 1 + {(z - 1)} - {(z - 1)}^2 + \frac{1}{2}{(z - 1)}^3 + \frac{7}{6}{(z - 1)}^4 - \frac{17}{4}{(z - 1)}^5 + \frac{821}{120}{(z - 1)}^6 - \frac{25}{12}{(z - 1)}^7 + \cdots$ and since the first series had a constant term of 1, this inversion is a Taylor series about 1! This is the first time I've ever seen anyone ever define super-roots (other than 2 and infinity) so I might have been the first to do this. But now that we can find the coefficients of super-roots through Lagrange inversion, what I am interested in is in a recurrence equation that allows us to calculate these coefficients much faster. Who knows what we'll find... Andrew Robbins Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/20/2007, 08:23 AM (This post was last modified: 12/20/2007, 08:31 AM by Ivars.) Excuse me for continuing along the same lines- redefining complex numbers in terms of hyperreals: In non-standard analysis, there are new "numbers" added to real line, which is subset of hyperreal set. Let us denote any infinitesimal in hyperreal set such that 0< infinitesimal than any INTEGER ( that follows from defintion that for any real there exists integer bigger than any real, and in non-standard analysis, all statements about reals are applied to hypperreals) . Let us define first of these integers as 1=i^4, second as 1=i^8 etc. Now we have : 1) hyperreal infinitesimals = I ( but the I includes too much information, however, giving correctly via complex analysis the general properties of infinitesimals related to reals) 2) hyperreal infinities 1/I ( the same comment applies here, they are all different, but 1/I captures their properties as long as You stay within the same order of infinitesimals) 3) Hyperreal infinite integers = I^4, I^8, I^12 ...........etc. 4) Then statement like x+I*Y is true and finite as long as I is infinitesimal, Y= Y*1, where 1- hyperreal infinite integer, Y- just a number , coefficient. So far I do not see any problems with proceeding with this construct. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 12/20/2007, 06:31 PM First, if I were you, I would not use i for this infinitesimal construction, as it is not related to the imaginary unit at all. I would use z instead, to be clear, and since the numbers are close to zero. Second, it sounds like your definition of infinitesimals are equivalent to the reciprocals of ordinal numbers, since you talk about $0 < z < \mathbb{R}^+$ (mirroring $\omega > \mathbb{N}$) and "infinities of the same order", so you might be better off, and able to communicate your ideas better if you simply define your transfinitesimals in terms of Cantor's ordinal numbers instead. So you could define your infinitesimals as: $z := \frac{1}{\omega}$ in terms of the first transfinite ordinal number $\omega$. My only problem with this is that I have a deep-seated hatred of Cantor's numbers. I'll save that talk for later. Andrew Robbins « Next Oldest | Next Newest »

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