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 Infinite tetration and superroot of infinitesimal andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 01/06/2008, 09:26 PM $(p \wedge \neg p)$ is a short-hand version of the law of the excluded middle (P or not P), which does not apply in SIA according to the reference you gave. Andrew Robbins Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 01/07/2008, 10:04 PM Thanks, This I have to digest slowly. Where can I read about existing geometrical interpretations of tetration, both finite and infinite, and its opposite log(log(log(log(log(.....) both finite and infinite? If You involve all infinite values of log, this construction expands pretty fast into infinity^infinity trees of values. You mentioned rotation via new dimensions- how does it happen? andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 01/08/2008, 08:28 AM Ivars Wrote:Where can I read about existing geometrical interpretations of tetration, You can't, because I don't think anyone is seriously researching this aspect. The only two places that come close are a link that Gottfried posted, and Tony Smith's "work" (author of Banned by Cornell) which gives an interesting discussion of dimensions. Ivars Wrote:both finite and infinite, and its opposite log(log(log(log(log(.....) both finite and infinite?What? Ivars Wrote:If You involve all infinite values of log, this construction expands pretty fast into infinity^infinity trees of values.What? Ivars Wrote:You mentioned rotation via new dimensions- how does it happen?Well, mathematically we are not limited to 3 dimensions. There are about as many dimensions as you have "commas", meaning, if you can write a "comma", then you can make a new dimension... So for example, (1, 2) is a point 1 unit along the first dimension, then two units along the second dimension. There are many ways of looking at lists of numbers, most are part of what most mathematicians call "vector spaces" all vector spaces have a dimensionality to them, so for example, although (1, 2) and (1, 2, 0) can represent the same polynomial/series (infinite dimensional space), they are two completely different vectors because they come from different vector spaces. When you are interpreting a vector as an hypercube as opposed to a point, then you can consider the hyper-volume (length, area, volume, etc...) of a vector (also called a norm or a measure) as the volume enclosed by the vector's scaled basis vectors in each dimension. This is the sense in which I was talking about "rotating through dimensions" a rotation in this sense is a map from (..., 1, 0, ...) to (..., 0, 1, ...) all else being equal. The hypervolume of two different lengths in two different dimensions (a, b) is a*b, or in other words: Multiplication. The hypervolume of one length rotated through n dimensions (a, a, a, ..., a) with n a's is a^n, or in other words: Exponentiation. So if this is the pattern for exponentiation, and tetration's definition in terms of exponentiation is: ${}^{n}{x} = x^{\left({}^{(n-1)}x\right)}$ then all we have to do is substitute this formula into our interpretation of exponentiation. This lexical substitution gives: ${}^{n}x$ is the hypervolume of a length x rotated through ${}^{(n-1)}x$ dimensions. Although this is not directly interpretable, it is recursive. This is as far as I've gotten in a semi-physical interpretation of tetration. I think trying to understand that statement intuitively is doomed, as recursive interpretation may lead to madness... Andrew Robbins andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 01/08/2008, 08:52 AM andydude Wrote:Ivars Wrote:both finite and infinite, and its opposite log(log(log(log(log(.....) both finite and infinite?What? Oh, I get it now. You are mistaken, $\log^n(x) = {}^{\text{slog}(x)-n}e$ which is not the inverse of tetration... it is tetration. Tetration has two inverses, super-roots and super-logarithms. You might be able to consider super-logarithms as iterated logarithms, but instead of giving the nth iterate of a logarithm, the super-logarithm gives you the n required to produce the given iterate. Please see Wikipedia's super-logarithm and iterated logarithm for more. Andrew Robbins Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 01/08/2008, 03:41 PM But is not applying of infinite log(log(log( .........h(z)) from the right leading back to z? Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 01/08/2008, 03:44 PM andydude Wrote:I think trying to understand that statement intuitively is doomed, as recursive interpretation may lead to madness... Andrew Robbins May be not if You try to understand at least just one: h(e^(pi/2))=-+i. This means e^(-i*(pi/2)) = e^h(e^(pi/2))*(pi/2). So infinitely working beginning with end ( from right) in infinite number of dimensions on e^(pi/2) is equal to -imaginary unit. If we look at e^(-i*pi/2) as turn 90 degrees clockwise in complex plane, this turn involves going through infinite tetration in other dimensions. What could that be? So log(log(log(...... ) is tetration? if we take e^pi/2=i^-i, and log with base i, we get (from right) (only main value) : log i (.................log i ((i^-i)) = log i( ....( log i ( -i) = log i (....(log i (-i))= log i ( ... log i (logi ( -1) = log i ( ...log i ( log i^2) = log i (......log i (2) = log i(...........log i(log i (i) + log i(i)) = log i(..............logi(log i i^2 = log i (......log i (2)=log i(...........log i(log i (i) + log i(i)) = log i(..............logi(log i i^2 = log i (......log i (2)=............ So,if we ever get to the end, we might have either log i(i^2) = 2 or log i (2) = 2ln2/i*pi. Is this the same as tetration of (e^pi/2)? How do we calculate it in this case? andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 01/08/2008, 11:44 PM Ivars Wrote:But is not applying of infinite log(log(log( .........h(z)) from the right leading back to z? No, no, no! What h represents is a fixed point of an exponential function. So if $a = h(z)$ then $a = z^a$ and if you are taking the base-z logarithm of both sides then you can see that $\log_z(a) = a$ which means that a is not only a fixed point of the base-z exponential function, but it is also a fixed point of the base-z logarithm function as well. And if it is a fixed point of the logarithm function, then there is no amount of iteration that will give any other output, so $\log_z^{\infty}(a) = a$, so infinitely iterated logarithms are not the inverse of infinitely iterated exponentials, only $z = a^{1/a}$ is the inverse of infinitely iterated exponentials. Andrew Robbins andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 01/09/2008, 12:50 AM Ivars Wrote:andydude Wrote:I think trying to understand that statement intuitively is doomed, as recursive interpretation may lead to madness... Andrew RobbinsSo infinitely working beginning with end ( from right) in infinite number of dimensions on e^(pi/2) is equal to -imaginary unit. I think you mean: rotating the length $e^{\pi/2}$ through infinite number of dimensions gives a hyper-volume equal to -imaginary unit. I only knew what you were talking about from context. If I were not in this forum, and I saw your words, they would not describe what you are trying to describe. Try and be more careful about how to say things. Ivars Wrote:So log(log(log(...... ) is tetration? if we take e^pi/2=i^-i, and log with base i, we get (from right) (only main value) : $\begin{tabular}{rl}\log_i^{\infty} (e^{\pi/2}) & = \log_i^{\infty} (i^{-i}) \\ & = \log_i^{\infty} (-i) \\ & = \log_i^{\infty} (i^{-1}) \\ & = \log_i^{\infty} (\log_i (-1)) \\ & = \log_i^{\infty} (\log_i i^2) \\ & = \log_i^{\infty} (2) \\ & = \log_i^{\infty} (2 \log_i (i)) \\ & = \log_i^{\infty} (\log_i i^2) \\ & = \log_i^{\infty} (2) \\ & = \log_i^{\infty} (2 \log_i (i)) \\ & = \log_i^{\infty} (\log_i i^2) \\ & = \log_i^{\infty} (2) \\ & = ...\end{tabular}$ So, if we ever get to the end, we might have either $\log_i(i^2) = 2$ or $\log_i(2) = \frac{2 \ln2}{i \pi}$.(sorry, I had to reorganize your math to understand it) True and true. I think your issue here is that you are calculating with base i, and as such, the only way to calculate the infinitely iterated exponential with base i is to use exponentials, not logarithms. The base-b logarithms can be infinitely iterated for all b such that $|\log(a)| > 1$ where a is a fixed point $a = b^a = {}^{\infty}b$. Exponentials on the other hand converge everywhere logs don't. So the base-b exponentials can be infinitely iterated for all b such that $|\log(a)| < 1$. Now the fixed point of base i is ${}^{\infty}i = 0.438283+ 0.360592 i$ which means $|\log({}^{\infty}i)| = 0.891514 < 1$ Since this value (sometimes called the Lyapunov number), means that exponentials will converge, but logarithms will not, because it is less than 1. So what you have found is that when you try and use logarithms for base i, then it doesn't converge, it oscillates between values, as expected. Ivars Wrote:Is this the same as tetration of (e^pi/2)? How do we calculate it in this case?You would do it just as you would with base i, only we are looking at base $e^{\pi/2}$ this time. According to the infinitely iterated exponential (with the usual branch structure) we get ${}^{\infty}(e^{\pi/2}) = -i$. This means that -i is the fixed point of the base-$e^{\pi/2}$ exponential function. To determine whether we should be using exponentials or logarithms to find this fixed point, we do the test: $|\log(-i)| = 1.5708 > 1$ so we should be using logarithms instead of exponentials, because this number (the Lyapunov number) is greater than 1. So with this, we can say that: $\log_{e^{\pi/2}}^n(-i) = -i$ for all n. Since this is true for all n, the version when $n=\infty$ is true. Andrew Robbins Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 01/09/2008, 01:53 PM Thank You Andrew, This is all very clearly explained and very interesting. I will study it for a while, but thanks again. I am happy to have not made mistakes in log base i calculation of (e^(pi/2)), and the conclusion of oscillation between 2 interesting values. I am somewhat attached to this value i^-i more than to general conclusions- may be it is not an exception, but rather first glimpse of some uncovered relations. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 01/10/2008, 09:19 PM (This post was last modified: 01/12/2008, 12:24 PM by Ivars.) Steven in Physorg forum came up with an interesting limit, whihc I extrapolated a little, and will write here just to check if its right, as i used some web complex number calculator to check limits. I wonder if i/x as x-> infinity can be considered as imaginary infinitesimal in the scale where x is. lim ( 1+- (i/x))^ (pi*(x^2 ))= e^(pi/2) as x-> infinity if You substitute x=sgrt(y) than it is valid as well, true for both roots of y. So lim ( 1+-(i/x))^(n*pi*(x^2)) when x-> infinity = e^(n*(pi/2)) for n>=1 Now what happens if n is real? For n=1/2 lim (1+-(i/x))^(pi/2)*(x^2) as x-> infinity e^(pi/4), formula works also for all 1/n lim (1+-(i/x))^ a*(pi/2)*(x^2) = e^(a*(pi/2)) , a= real, including 0. But we know that e^(pi/2) = i^ (1/i) or i-th root of i. So lim (1+-(i/x))^(pi*(x^2)) x-> infinity= i^(1/i) and generally, lim (1+-i/x)^(a*(pi*(x^2)) x-> infinity = (i^(1/i))^a= (i^-i)^a. This is probably well known, but still looks nice. I do not know if it really holds for all a I have indicated. Now if we drop the x-> infinity in a way that we include x in numbers itself, so that i gets "smaller" by infinity and correspondingly pi "longer" by infinity^2, the result would still hold, but in each scale of x ( since we can substitute x with x^2, x^n, x^infinity) the identity will hold, but i and pi will be different, as will e. Ivars « Next Oldest | Next Newest »

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