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?

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.

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

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

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

\( a\,\oplus_\gamma\,q \neq a + q \)

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:

\( (a\,\,\otimes_\gamma\,\, b)\,\oplus_\gamma\,(a\,\,\otimes_\gamma\,\, c) = a\,\,\otimes_\gamma\,\, (b\,\oplus_\gamma\,c) \)

I wonder, are these suggested different operators incapable of being analytic? Or are the normal addition and normal multiplication and normal exponentiation the only ones that work? Thanks for any help, James

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?

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.

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

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

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

\( a\,\oplus_\gamma\,q \neq a + q \)

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:

\( (a\,\,\otimes_\gamma\,\, b)\,\oplus_\gamma\,(a\,\,\otimes_\gamma\,\, c) = a\,\,\otimes_\gamma\,\, (b\,\oplus_\gamma\,c) \)

I wonder, are these suggested different operators incapable of being analytic? Or are the normal addition and normal multiplication and normal exponentiation the only ones that work? Thanks for any help, James