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+--- Thread: A more consistent definition of tetration of tetration for rational exponents (/showthread.php?tid=64)
RE: A more consistent definition of tetration of tetration for rational exponents - UVIR - 09/30/2007
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.
RE: A more consistent definition of tetration of tetration for rational exponents - bo198214 - 09/30/2007
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.
RE: A more consistent definition of tetration of tetration for rational exponents - GFR - 10/01/2007
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
RE: A more consistent definition of tetration of tetration for rational exponents - UVIR - 10/01/2007
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]).
RE: A more consistent definition of tetration of tetration for rational exponents - bo198214 - 10/01/2007
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 \( \lim_{n\to\infty} {^{1+1/n}x}=x^{x^{1/x}}\neq x={^1x} \).
RE: A more consistent definition of tetration of tetration for rational exponents - UVIR - 10/01/2007
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.
RE: A more consistent definition of tetration of tetration for rational exponents - GFR - 10/02/2007
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
RE: A more consistent definition of tetration of tetration for rational exponents - andydude - 10/07/2007
Agreed. Numerical evidence has shown that \( {}^{1/n}x \ne srt_n(x) \).
Andrew Robbins
RE: A more consistent definition of tetration of tetration for rational exponents - andydude - 10/20/2007
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
RE: A more consistent definition of tetration of tetration for rational exponents - UVIR - 10/20/2007
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,