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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 09/30/2007, 10:29 PM bo198214 Wrote: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 ${}^nx$. Let me try to explain the "obvious" rule behind such a definition. Call the inverse operator of multiplication @t. Then the inverse of multiplication must satisfy: $(x*n)@t=x$ consequently it is easily seen that $@t=*(1/n)$, which is the only operator which fits the bill. Similarly, call the inverse operator of exponentiation @@k. Then the inverse of exponentiation must satisfy: $(x^n)@@k=x$ consequently it is esily seen that $@@k={\^}(1/n)$, which is the only operator which fits the bill. Again similarly, call the inverse operator of tetration @@@m. Then the inverse of tetration must satisfy: $(@@@m)({^n}x)=x$ from which it follows that $@@@m=$tetraroot of order n of x, since the tetraroot is the only operator which satisfies: $(tetraroot-n)(^{n}x)=x$ Now, the very subtle problem which I guess nobody sees (for some strange reason) is that if the operator for tetrating to (1/n) *is NOT* the same operator as that of the tetraroot of order n, then we have an operator discrepancy at a very low level in the hierarchy of operators: ${^{1/n}}({^{n}x)=/= x$ I am not going to argue more about it. Whoever "sees" it, great. Whoever doesn't, great again. bo198214 Wrote:If you not even demand that ${}^xe$ is continuous, what will then remain?Sorry, "demanding" and "constructing" are not the same as "existing". Besides, there's no teling whether tetration as defined using tetraroots is or is not continuous. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/30/2007, 11:10 PM UVIR Wrote:$(x*n)@t=x$ consequently it is easily seen that $@t=*(1/n)$ As I already explained before it is not "consequently easily seen" but a consequence of the law $(xa)b=x(ab)$ and similarly for exponentiation of the law $(x^a)^b=x^{ab}$. This law holds for natural $a$ and $b$ and if we demand it to hold for fractional $a$ and and $b$ too, then necessarily $\frac{1}{n} x$ is the inverse of $nx$ and $x^{1/n}$ is the inverse of $x^n$ by: $(x\frac{1}{n})n=x(\frac{1}{n}n)=x1=x$ and $(x^{1/n})^n=x^{(1/n)n}=x^1=x$ but we have not ${^a(^bx)}={^{ab}x}$ for natural a and b and hence can not generally demand it for fractional a and b. Quote:Now, the very subtle problem which I guess nobody sees (for some strange reason) is that if the operator for tetrating to (1/n) *is NOT* the same operator as that of the tetraroot of order n, then we have an operator discrepancy at a very low level in the hierarchy of operators: ${^{1/n}}({^{n}x)\neq x$ Yes, well we have to live with it. But except that we have to deal with more different operations on the tetra level, I see no problems arising from the inequality. We also have to live with for example: ${^2}({^3 x})\neq {^6x}$. Quote:Besides, there's no teling whether tetration as defined using tetraroots is or is not continuous. There is a telling. It is not continuous at (the exponent) 0 (if we assume that tetration is defined on natural numbered exponents in the usual way, particularly ${^0x}=1$). You showed already that $\lim_{n\to\infty} {^{1/n}x}=x^{1/x}\neq 1={^0x}$ for $x>1$ in your definition. GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 10/01/2007, 12:55 AM Dear Participants, Concerning Towers versus Tetraroots with rational exponents, I should like to attach an additional comment, in support of the last BO comments. Yes, we have to live with that! GFR Attached Files   TVT.pdf (Size: 7.83 KB / Downloads: 344) UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 10/01/2007, 05:56 PM bo198214 Wrote:Quote:Now, the very subtle problem which I guess nobody sees (for some strange reason) is that if the operator for tetrating to (1/n) *is NOT* the same operator as that of the tetraroot of order n, then we have an operator discrepancy at a very low level in the hierarchy of operators: ${^{1/n}}({^{n}x)\neq x$ Yes, well we have to live with it. But except that we have to deal with more different operations on the tetra level, I see no problems arising from the inequality. We also have to live with for example: ${^2}({^3 x})\neq {^6x}$. That's right. There is a BIG difference however: The law ${^m}({^n x})\neq {^{mn}x}$ fails because of the fundamental LAW of tetration for naturals and there's no remedy for the failure. ${^{1/n}}({^n x})\neq x$ doesn't HAVE to fail. It fails because of the way *WE* have defined tetration for the (1/n)-th iterate. That's an additional *introduced* discrepancy, which does NOT depend on the fundamental law of tetration for naturals, and which HAS a remedy. Do you see the difference? Quote:There is a telling. It is not continuous at (the exponent) 0 (if we assume that tetration is defined on natural numbered exponents in the usual way, particularly ${^0x}=1$). You showed already that $\lim_{n\to\infty} {^{1/n}x}=x^{1/x}\neq 1={^0x}$ for $x>1$ in your definition. *IF* we assume that tetration is defined as ${^0x}=1$. If we don't assume that (in particular if we assume that ${^0x}=x^{1/x}$), nothing has been shown. In particular, you haven't shown anything about continuity of the proposed way to do tetration (inside and outside the interval [0,1]). bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 10/01/2007, 06:24 PM Quote:*IF* we assume that tetration is defined as ${^0x}=1$. If we don't assume that (in particular if we assume that ${^0x}=x^{1/x}$) ... ... then we have a more serious problem. If you define ${^0x}$ to be different from 1, say ${^0x}=a$, then ${^1x}=x^a$. And in common use of the word tetration ${^1x}=x$ (hopefully you dont want to change this for the sake of continuity of your construction). So if you accept this then any (real) value of $a$ different from 1 poses the contradiction $x^a=x$. You see that it is necessary to define ${^0x}=1$. I mean there is no need to define it for ${^0x}=1$, it rather shows the pattern. Your solution is also not continuous at exponent 1. Consider the sequence $1+1/n$, by your definition ${^{1+1/n}x}=x^{^{1/n}x}$. Then $\lim_{n\to\infty} {^{1+1/n}x}=x^{x^{1/x}}\neq x={^1x}$. UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 10/01/2007, 07:32 PM bo198214 Wrote:... then we have a more serious problem. If you define ${^0x}$ to be different from 1, say ${^0x}=a$, then ${^1x}=x^a$. And in common use of the word tetration ${^1x}=x$ (hopefully you dont want to change this for the sake of continuity of your construction). So if you accept this then any (real) value of $a$ different from 1 poses the contradiction $x^a=x$. You see that it is necessary to define ${^0x}=1$. I see your point. My definition forces: ${^0}x=x^{1/x}\\ {^1}x=x^{x^{1/x}}\\ {^2}x=x^{x^{x^{1/x}}}$ ... which, I guess goes against standard notation. I am leaving for vacations tomorrow, so I will try to examine the function defined as above and see if there's anything more interesting about it. If there is, I will report it back in a week or so. Thanks to all who participated. GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 10/02/2007, 01:40 PM Yes, Friends! I think that we must try to preserve the systematic structural scheme: x+x = x.2 x*x = x^2 x^x = x#2 (x-tower-2, a must) x#x = x§2 (x-penta-2, a ... future dream) ............. (and ... so on) I have also another crucial example, but I promised (Henryk) to keep quiet for a while! Best wishes to all of you. Keep it going! GFR andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/07/2007, 03:37 PM Agreed. Numerical evidence has shown that ${}^{1/n}x \ne srt_n(x)$. Andrew Robbins andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/20/2007, 07:57 PM However, the new function described above is equivalent to: $f(b, t) = \exp_b^{t+2}(-1)$ which is very interesting, since it's in the orbit from -1 of an exponential, instead of tetration, which is the orbit from 1. Andrew Robbins UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 10/20/2007, 09:05 PM andydude Wrote:However, the new function described above is equivalent to: $f(b, t) = \exp_b^{t+2}(-1)$ which is very interesting, since it's in the orbit from -1 of an exponential, instead of tetration, which is the orbit from 1. Andrew Robbins Andrew, which function are you talking about? Do you mind elaborating a bit more? Thanks, « Next Oldest | Next Newest »

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