A more consistent definition of tetration of tetration for rational exponents
#15
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 \( \lim_{n\to\infty} {^{1+1/n}x}=x^{x^{1/x}}\neq x={^1x} \).
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RE: A more consistent definition of tetration of tetration for rational exponents - by bo198214 - 10/01/2007, 06:24 PM

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