(02/23/2008, 12:29 AM)quickfur Wrote: That's interesting, because analysing f(x+y) = f(x)f(y) leads to the exponential function \( e^x \). Unfortunately, we don't seem to have such a handy property involving tetration that we can use, except for \( f(x+1)=b^{f(x)} \), which isn't very helpful for non-integer x.
Now that I think of it, I think the root of the problem is that exponentiation is non-associative, so there's no easy way to algebraically "access the top of the exponential tower", so to speak. In order to derive any useful relations, we need to have some way of "reaching the top of the tower". For example, given b[4]n, if we can somehow reach the top of the tower and add another tower of height m, then we can state the property that J(b[4]n, b[4]m) = b[4](n+m), where the hypothetical J operator attaches the second tower to the top of the first tower.
But I doubt that J is expressible as an algebraic operation (I'm not even sure if it can be consistently defined if b is not fixed!). And so even if we could make such a statement, it wouldn't be of the same utility as \( f(x+1)=b^{f(x)} \) w.r.t. exponentiation.
I'd like to respectfully disagree.
consider:
\( \mu (x) =\, ^x \mu \)
\( a \,\otimes_\mu\,b = \mu(\mu^{-1}(a) + \mu^{-1}(b)) \)
it's very easy to see that:
\( ^c \mu \, \otimes_\mu\,^d \mu = \,^{c+d} \mu \)
I think even, this operator may give some short cuts for algebra involving tetration.
I did a little research into this under the following thread:
http://math.eretrandre.org/tetrationforu...hp?tid=699
The only rule is if we create:
\( \phi (x) = \,^x \phi \)
\( a \,\otimes_\phi \,b = \phi(\phi^{-1}(a) + \phi^{-1}(b)) \)
\( \otimes_\mu \neq \otimes_\phi \)
whereas when we create the same operator with exponentiation instead of tetration, the operators are equivalent and both are multiplication.
