A question concerning uniqueness

[BenStandeven]

My question is simple, and I hope somebody has an answer because I am a little confused. How come Tetration has multiple possible extensions to the complex domain that are analytic, but exponentiation only has one? Is it possible to have an alternative extension for exponentiation that is still analytic?

It would have a piecewise definition, the gamma variable is to distinguish it from regular exponentiation:
\( a^{\otimes_\gamma\,\,k} = a\,\cdot\,a\,\cdot... a \) k amount of times if \( k \in N \)

and then, for \( 0<q<1 \)
\( a^{\otimes_\gamma\,\,q} \neq a^q \)

and
\( a^{\otimes_\gamma\,\,k + q} = a\,\cdot\,a\,\cdot\,a\,... \cdot\,a^{\otimes_\gamma\,\, q} \)

Is it possible for \( a^{\otimes_\gamma \,\,x} \) to be analytic? I'm sure it's possible to make alternative models, where \( a^{\otimes_\gamma\,\,q} \) is defined arbitrarily, but is normal exponentiation the only analytic model?

So we would need \( a^{\otimes_\gamma\,\,k + q} = a^k f(q) \) for \( q \) from 0 to 1, and \( k \) an integer, or in other words \( a^{\otimes_\gamma\,\,x} = a^x f(frac(x)) \); then using the same argument as for tetration, we get that \( \log_a a^{\otimes_\gamma\,k+q} = k + \log f(q) = k + q + \log f(q)/a^q \). So \( f(q)/a^q \) is analytic and takes value 1 at both 0 and 1. Then any function \( f \) satisfying these conditions can be used to define a new "exponential" function; for example, we could have \( a^{\otimes_\gamma\,\,x} = a^x cos(2 \pi x) \). Of course, f can also depend on \( a \).

thanks for reading this and I hope someone can clarify

As I think about it, I think this gets a little messy when we consider the laws of exponentiation:

\( a^{\otimes_\gamma\,\, b} \cdot a^{\otimes_\gamma\,\, c} \neq a^{\otimes_\gamma\,\, b+ c} \) because this would imply there is a number seperate from the square root that when squared returns a.

Yeah, the standard exponential functions are the only continuous functions on the reals which satisfy the full laws of exponentiation.

This type of idea could extend to multiplication as well, insofar as we could define a new multiplication:
\( a\,\otimes_\gamma\,k = a_1 + a_2 + a_3 ... + a_k \)

and
\( a\,\otimes_\gamma q \neq a \cdot q \)

For this, the solution would be \( a\,\otimes_\gamma x - a\,\times x = g(frac(x)) \), where \( g \) is zero at 0 and 1.

We could also go to addition, and define a different addition:

\( a \oplus_\gamma\, k = a + 1_1 + 1_2 + 1_3.... + 1_k \)
and
\( a\,\oplus_\gamma\,q \neq a + q \)

For this, the solution would be \( a\,\oplus_\gamma x = a + floor(x)\, \oplus_\gamma frac(x) = a + floor(x) + h(x) \), with \( h(x) \) going from 0 to 1 on the unit interval. Here I am assuming that \( h \) does not depend on \( a \); if it did, it could also depend on \( floor(x) \).

This may let us keep our exponentiation laws, though altered:
\( a^{\otimes_\gamma \,\,b}\,\,\otimes_\gamma\,\,a^{\otimes_\gamma\,\, c} = a^{\otimes_\gamma\,\,b\,\oplus_\gamma\,c} \)


we'd also have a similar law for multiplication maybe:
\( (a\,\,\otimes_\gamma\,\, b)\,\oplus_\gamma\,(a\,\,\otimes_\gamma\,\, c) = a\,\,\otimes_\gamma\,\, (b\,\oplus_\gamma\,c) \)

Sure, just set \( t(x) = x + h(x) \) for some 1-periodic function \( h \), and define \( a \oplus_\gamma b = h(h^{-1}(a) + h^{-1}(b)) \), \( a \otimes_\gamma b = h(h^{-1}(a) \times h^{-1}(b)) \), and so on. I don't know if there is any other way to turn R into a field than this one, let alone a field with an exponential operator.