Should tetration be a multivalued function?

[marraco]

.... No proof, but numerically:
\( \\[15pt]

{(^{r+1}a)^{^{-(r+1)}a}=(^{-r}a)^{^{r}a}} \)

ok, first lets define y and z as follows:
\( y=\; ^{-(r+1)}a \;\;\; \)
\( z=\; ^{r}a \;\;\; \)

Then substitute these values of y and z into the Op's equation above, noting that
\( ^{(r+1)}a = a^z\;\;\;\; ^{-r}a = a^y \)

\( (a^z)^{y}=(a^y)^z\;\;\; \) This is the Op's equations with the substitutions
\( (a^z)^{y}=(a^y)^z=a^{(y\cdot z)}\;\;\; \) this equation holds for all values of a,y,z

^^ Good. That's far more elegant that my reasoning, which used the commutative x^ln(y) and goofy induction.

Maybe is the key to calculate the derivative at the origin? (taking limit of r → 0)

Here is a graphic illustrating the equality. The exponentiation of the blue arrow is equal to the one in the red arrow.
The expression inside the rectangles are equal.



By induction, if we invert the red arrow, it is equal to a larger arrow; I mean \( \\[20pt]

{(^{r+(n+1)}a)^{^{-(r+(n+1))}a}=(^{-(r+n)}a)^{^{r+n}a}} \). I wrote it thinking that n was entire, but since r is real, and °a is arbitrary, it is obvious that is for any real n.

But is even more general, because this is valid on all branches, and for any definition of °a.
Graphically, you can move the arrows to the left or right, and the equality remains valid.

For example, \( \\[20pt]

{^2a\,^{^3a}\,=\,^4a\,^{^1a}} \)



and also \( \\[20pt]

{^1a\,^{^4a}\,=\,^5a\,^{^0a}} \), \( \\[20pt]

{^0a\,^{^5a}\,=\,^6a\,^{^{-1}a}} \), etc.