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 A more consistent definition of tetration of tetration for rational exponents UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 09/29/2007, 11:56 PM I believe I have found a definition for rational hyperexponents which is totally consistent with the notion of the tetraroot. Details here. The computation of this function is very demanding (often exceeding the capabilities of Maple) and I am not sure whether it is even continuous, but some crude tests suggest that it is. I will be leaving for vacations in a few days (but will have net access), and may try to calculate the values of the function ${^{m/100}}e$ for $m\in \{1,2,...99\}$ to get a better feel of how this function behaves. Cheerio. Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 09/30/2007, 10:39 AM (This post was last modified: 09/30/2007, 11:10 AM by Gottfried.) Hi - I'd like to try your computation with Pari/GP. Unfortunately I don't understand your maple-code completely. There are two calls of #-preceded expressions, #flip and #reduce, without parameters. [update] ... I see. These are surely comments.(my god; it's early sunday noon, I seem to not to have my mind up to math already...) [/update] [update 2] This seems to be the continued fraction-process. For irrational parameters one needs infinite expansion. Well; for quadratic characteristics they are at least periodic - but what about higher algebraicities? I once had considered this problem and seem to have found, that - for numbers of quadratic algebraicity the solution can be expressed by a 2x2 matrix and its eigenvalues. This followed from the observation of the usual 2x2-matrix approach for the iterative solution of convergents of continued fractions. - for higher algebraicity of the number in question we are possibly able to express the problem using a matrix of adequate dimension, establish a periodicity this way and solve then for its eigenvalues. However, I could not complete this idea and was not able to find a way to determine the appropriate matrices. May be one should evaluate this idea again and push it up to an actual solution. This idea had, for instance, the problem, that its intermediate steps were *not* the "best approximations" in the sense as this is meant for the usual method of evaluation of continued fractions. It were then interesting, whether this solution would be connected in one way or another to the diagonalization-approaches to tetration... [/update 2] Gottfried Gottfried Helms, Kassel UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 09/30/2007, 11:26 AM Gottfried Wrote:Hi - I'd like to try your computation with Pari/GP. Unfortunately I don't understand your maple-code completely. There are two calls of #-preceded expressions, #flip and #reduce, without parameters. Gottfried Hi Gottfried, Trying the code is almost out of the question (because there is no code). I did all those by hand, by solving the corresponding equation with Maple. The Maple code which is displayed in the web page has nothing to do with it. This code just proves that any rational exponent m/n<1 is always a function of tetraroot exponents. One has to do the reduction to tetraroots by hand first and then solve the corresponding equation numerically. Please look at the example. For the example, the equation to be solved with Maple is: $((a^{(^{1/3}a)})^a)^{a^{(^{1/3}a)}}=e$ Unfortunately, for every rational, the equation to be solved is different. For example, to calculate $^{9/10}e$, you would do: $^{9/10}e=a \Leftrightarrow\\ e={^{10/9}}a \Leftrightarrow\\ e=a^{(^{1/9}a)}$ so the equation to be solved numerically is the last one. GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 09/30/2007, 11:38 AM Dear UVIR and Gottfried, I have examined with interest the "details" shown in the UVIR's posting. Unfortunately, I cannot agree with putting: b # n = y => b = y # (1/n). Moreover, I think that these two "tetration" formulas don't have any point in common. Their sets are disjoint. There is no similarity with the "exponentiation" case, where the power and the root functions are described in the same domain and we can find a root function representable as a power. I know but, unfortunately, ... this is the real problem. GFR UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 09/30/2007, 04:41 PM GFR Wrote:Dear UVIR and Gottfried, I have examined with interest the "details" shown in the UVIR's posting. Unfortunately, I cannot agree with putting: b # n = y => b = y # (1/n). Dear GFR, There has already been at least one peer reviewed paper on this definition of tetraroots, so whether you "agree" or not is irrelevant. GFR Wrote:Moreover, I think that these two "tetration" formulas don't have any point in common. Their sets are disjoint. There is no similarity with the "exponentiation" case, where the power and the root functions are described in the same domain and we can find a root function representable as a power. I know but, unfortunately, ... this is the real problem. GFR Nonsense. The tetraroot is the exact reverse operation of tetration, since: $ {^{n}}({^{1/n}}x)={^{1/n}}({^{n}}x)=x$ Specifically, one can see the connection with infinite tetration, when one writes: $ lim_{n\rightarrow \infty}{^{1/n}}e=y\Leftrightarrow\\ e=lim_{n\rightarrow \infty}{^n} y\Leftrightarrow\\ y=e^{1/e}$ which is absolutely correct. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/30/2007, 06:18 PM (This post was last modified: 09/30/2007, 06:26 PM by bo198214.) GFR Wrote:I have examined with interest the "details" shown in the UVIR's posting. Unfortunately, I cannot agree with putting: b # n = y => b = y # (1/n). Quote:There has already been at least one peer reviewed paper on this definition of tetraroots, so whether you "agree" or not is irrelevant. The tetra root is defined as the inverse of the x#n but it has nothing to do with x#(1/n). GFR Wrote:Moreover, I think that these two "tetration" formulas don't have any point in common. Their sets are disjoint. There is no similarity with the "exponentiation" case, where the power and the root functions are described in the same domain and we can find a root function representable as a power. I know but, unfortunately, ... this is the real problem. I absolutely agree with Gianfranco. The reason why $x^{1/n}$ and $\sqrt[n]{x}$ coincide is that $(x^a)^b=x^{ab}$. If we demand this law to be valid also for rational exponents then $(x^{1/n})^n = x^{n/n}=x^1=x$ and thatswhy $x^{1/n}$ must be the inverse of $x^n$. Unfortunately ${}^a({}^b x)={}^{ab}x$ is no more valid. Quote:The tetraroot is the exact reverse operation of tetration, since Nobody denies that the tetraroot is the inverse of ${}^nx$ (by definition) however to define ${}^{1/n}x$ as being the tetraroot is quite arbitrary and additionally does not coincide with our other methods. As far as I have seen the 3 definitions of ${}^y x$ (via Daniel's continuous iteration at the first fixed point, via Gottfried's matrixoperator method and via Andrew's natural slog) coincide for $1. As an example I graphed ${}^{1/2}x$ (via the method of continuous iteration at the first fixed point) (red) in comparison with the inverse of ${}^{2}x$ (blue):     Curve 1 is ${}^{1/2}x$ Curve 2 is the identity function. Curve 3 is ${}^2x$ and Curve 4 is the inverse of ${}^2x$ (2nd superroot) Quote:$ lim_{n\rightarrow \infty}{^{1/n}}e=y\Leftrightarrow\\ e=lim_{n\rightarrow \infty}{^n} y\Leftrightarrow\\ y=e^{1/e}$ Sorry Ioannis, but this rather proves that it is the wrong definition. As ${}^xe$ should be a function continuous in $x$ it must $\lim_{n\to\infty}{}^{1/n}e={}^0e=1\neq e^{1/e}$ UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 09/30/2007, 07:33 PM bo198214 Wrote:Nobody denies that the tetraroot is the inverse of ${}^nx$ (by definition) however to define ${}^{1/n}x$ as being the tetraroot is quite arbitrary Why? What is "arbitrary" about it? And if it's "arbitrary", in what "sense" is it "arbitrary"? Again, if the definition of the n-th order tetraroot as ${^{1/n}}x$ is "arbitrary", surely the tetraroot can be defined better as ${^{m/n}}x$ for some m,n. Do you care to define m and n in some other consistent way for the tetraroot? bo198214 Wrote:and additionally does not coincide with our other methods. So? Has a divine judge decided on the validity of any of the proposed methods so far? bo198214 Wrote:Sorry Ioannis, but this rather proves that it is the wrong definition. As ${}^xe$ should be a function continuous in $x$ it must $\lim_{n\to\infty}{}^{1/n}e={}^0e=1\neq e^{1/e}$ Sorry, I am not convinced that ${^{x}}e$ should even *be* continuous (even whether it exists), despite the agreement between all the current methods. All the methods so far (including mine), exhibit a certain "artificiality" if you wish, which is apparent from the complexity which reveals itself when one asks a very simple question: HOW do you define the tetration function for RATIONAL values. If you cannot tell me how the tetration function is defined at the rationals, then you *cannot* tell me how it's defined at the reals. If you want to debate the above, then I will ask you the following: how do you define for example ${^{7/11}}e$? or ${^{2/3}}e$? Sorry, definitions via decimal expansions won't cut it, because decimal expansions suffer from non-uniqueness. So, if you tell me for example, take Andrew's or Gottfried's or your method and "input" 0.666... or 0.6363..., and then see what the function outputs, this is already suffering badly as a definition. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/30/2007, 08:06 PM (This post was last modified: 09/30/2007, 08:07 PM by bo198214.) UVIR Wrote:bo198214 Wrote:Nobody denies that the tetraroot is the inverse of ${}^nx$ (by definition) however to define ${}^{1/n}x$ as being the tetraroot is quite arbitrary Why? What is "arbitrary" about it? And if it's "arbitrary", in what "sense" is it "arbitrary"? It is in the same sense arbitrary as say that I define ${}^{1/n}x=1$. No rule justifies that ${}^{1/n}x$ should be the inverse of ${}^nx$. If you not even demand that ${}^xe$ is continuous, what will then remain? Quote:Again, if the definition of the n-th order tetraroot as ${^{1/n}}x$ is "arbitrary", surely the tetraroot can be defined better as ${^{m/n}}x$ for some m,n. Do you care to define m and n in some other consistent way for the tetraroot? The n-th tetraroot is already clearly defined as the inversion of ${^nx}$, no need to redefine it. What we all are seeking for is a sound definition for the tetration for non-integer exponents. Natural demands are ${^xb}$ being analytic, strict increase and ${^{x+1}b}=b^{^xb}$. Quote:So? Has a divine judge decided on the validity of any of the proposed methods so far? The interesting thing is not about validity of each single method, but rather that 3 completely different methods yield identical values. Quote:So, if you tell me for example, take Andrew's or Gottfried's or your method and "input" 0.666... or 0.6363..., and then see what the function outputs, this is already suffering badly as a definition. Can not see how this suffers. Take for example the method of continuous iteration at a fixed point. There you move the fixed point of $b^x$ to 0, say $f(x)=b^{x-a}+a$ where a is the fixed point of $b^x$. Then you compute $g=f^{\circ 1/n}$ by unique analytic solution of the equation $g^{\circ n}=f$ and then you have the value ${^{1/n}b}=g(1+a)-a$. Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 09/30/2007, 09:46 PM Well, in my opinion we are still in a phase of definition. Note, that also a discussion of "the best" way to define the gamma-function can still be led (although I think, that the known alternatives are not as power- and /or useful). It is a matter of which properties seem to be the most useful for number theory, I think. Also I think, that the methods (at least implicitely) based on matrix-diagonalization and/or -logarithm will come out to be the most useful ones - but that's still just an assumtion, with good arguments, though. Another point is, that the three mentioned methods are not different, but essentially the same, as I understand things now - so there is no majority in this competition of methods (just a remark). Kind regards - Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/30/2007, 10:23 PM Gottfried Wrote:Another point is, that the three mentioned methods are not different, but essentially the same Nah, I wouldnt go that far. Remember that we currently can not even prove the equality of any two methods. It is all based on numerical considerations. « Next Oldest | Next Newest »

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