[split] Understanding Kneser Riemann method

[marraco]

@marraco, @tommy
That certainly is a cool equation, even if it is easily provable.

@everyone
Also, I think I can express my earlier comment in different words now. Tetration is defined as the 1-initialized superfunction of exponentials. The previous functions discussed earilier are 3-initialized and 5-initialized, which makes them, not tetration, by definition. However, if there is an analytic continuation of the 1-initialized superfunction that overlaps with the 3-initialized superfunction, AND if on the overlap f(0) = 3, then they can be considered branches of the same function. But until that is proven, I don't think it's accurate to say that they're all "tetration". They are, however, iterated exponentials in the sense that they extend \( \exp_b^n(3) \) to non-integer n. And so I would probably write these functions as \( \exp_b^n(1),\, \exp_b^n(3),\, \exp_b^n(5) \) instead of saying that \( {}^{n}b \) is a multivalued function that returns all three.

That makes sense, but on other side, we need to solve equations like \( \\[22pt]

{^x \left(\sqrt[2]{2} \right)=3} \) (3>2, the asymptotic limit).
Similarly, when we solve \( \\[14pt]

{e^{x}=-1} \) (-1<0, the asymptotic limit), we do not say that is is a function different than exponentiation. We just extend the domain to complex numbers.

but this equation \( \\[22pt]

{^x \left(\sqrt[2]{2} \right)=3} \) has real solutions, unless we consider the pair \( \\[16pt]

{(x,\,^0a)} \) a new kind of number.

If we use a new kind of number, then, for the main branch, \( \\[16pt]

{^0a=(1,1)} \); no more a real number.