02/25/2008, 09:57 AM
quickfur Wrote:This argument also works if applied to exponential functions, using the property that \( e^{x+1}=e\cdot e^x \). In other words, this property is not sufficient to uniquely determine a unique (up to scalar multiple of the exponent) exponential function.
Now, the traditional way (or at least, one of the usual ways) of defining the exponential function is by its inverse: \( L(x+y)=L(x)L(y) \). This gives us a logarithm function unique up to change of base. This seems to suggest that perhaps we need to discover a property of the tetralog (tetrational inverse) function that will lead us to a unique (up to change of base) tetration. Unfortunately, the relationship of tetration to lower operations is non-trivial at best, and it's not obvious which property should be used as our basis.
I does not so much matter whether you use the logarithm or the exponential itself, what makes exponentiation and multiplication unique are the "distributive laws":
\( a^{x+y}=a^xa^y \) and \( a(x+y)=ax+ay \) which are an extension of
\( a^{x+1}=aa^x \) and \( a(x+1)=ax+a \), however we know it doesnt work for tetration. Thatswhy I decided to develop a new more general number domain, arborescent numbers, in which we have this distributivity a[[n+1]](x+y)=(a[[n+1]]x)[[n]](a[[n+1]]y) for arbitrary high operations, as we just discussed here.
However it would be good if you have a read through the Tetration FAQ, as uniqueness of exponentiation and multiplication are already explained there.