01/13/2016, 08:24 AM
@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 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.