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A more consistent definition of tetration of tetration for rational exponents
#11
bo198214 Wrote:It is in the same sense arbitrary as say that I define
. No rule justifies that should be the inverse of .

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:



consequently it is easily seen that , which is the only operator which fits the bill. Similarly, call the inverse operator of exponentiation @@k. Then the inverse of exponentiation must satisfy:



consequently it is esily seen that , 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:



from which it follows that tetraroot of order n of x, since the tetraroot is the only operator which satisfies:



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:



I am not going to argue more about it. Whoever "sees" it, great. Whoever doesn't, great again.Smile

bo198214 Wrote:If you not even demand that 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.
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Messages In This Thread
RE: A more consistent definition of tetration of tetration for rational exponents - by UVIR - 09/30/2007, 10:29 PM

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