jaydfox Wrote:Essentially, once \( {\Large \log_2}\left(\frac{b^z}{\log_b(\log_b{a})}\right) \) is greater than the number of bits of machine precision we're using, there's no point in further exponentiation, because addition will underflow and the result won't be affected.I should point out that we need to add an absolute value to the equation.
Also, it's more useful from an implementation standpoint to solve for z in \( {\Large \log_2}\left(\frac{b^z}{\left\|\log_b(\log_b{a})\right\|}\right) \). By solving for z, we can find a "cap", a value below which we continue exponentiating. Once we've exceeded the cap, there's no point to exponentiate further. Having a cap allows us to perform a simple comparison (internally, a subtraction), rather than exponentiating and then testing if we've exceeded precision.
\(
\text{To find the cap, }\mathcal{Z}\text{, given }n\text{ bits of precision:}\\
\begin{eqnarray}
{\Large \log_2}\left(\frac{b^{\mathcal{Z}}}{\left\|\log_b(\log_b{a})\right\|}\right) & = & n\\
\log_2(b^{\mathcal{Z}})-\log_2\left(\left\|\log_b(\log_b{a})\right\|\right) & = & n\\
\log_2(b^{\mathcal{Z}}) & = & n+\log_2\left(\left\|\log_b(\log_b{a})\right\|\right) \\
\mathcal{Z}\log_2(b) & = & n+\log_2\left(\left\|\log_b(\log_b{a})\right\|\right) \\
\mathcal{Z} & = & \frac{n+\log_2\left(\left\|\log_b(\log_b{a})\right\|\right)}{\log_2(b)}
\end{eqnarray}
\)
Knowing the precision in bits n, and the bases a and b, we can set a cap for our iterative exponentiations. Once we pass the cap, we begin doing iterated logarithms in the other base (remember, all this is in the context of performing a change of base using my formula).
With these tools in place, we can solve a change of base. The only hole I've identified in the formula is that it is only provably correct for integer iteration counts. In between, it gives us infinitely differentiably results which satisfy the iterative exponential property, but there's no guarantee that it's "the" correct solution for fractional iterations. Given the simplicity and the direct correlation between the superlogarithmic "constant" and the logarithmic constant, I had been lulled into assuming that if it was correct for all integer iterations, it would be correct for fractional iterations as well. To assume otherwise would lead to cyclic (wavy) curves if you graphed \( \text{slog}_b\left({}^{x} a\right) \), when with my formula you would get an asymptotically straight line. It was certainly unexpected to me that such a graph would be wavy, as it would imply that tetration "knows" where it is.
To put this into perspective, think about the iterated multiplication formula (you know, exponentation). Let's say that we know that 2^4 equals 4^2 and 2^6 equals 4^3. In fact, let's say that for all integers k, we know that 2^2k = 4^k.
Wouldn't you expect 2^3 to equal 4^1.5? That's essentially the basis for my initial confidence that my change of base formula was correct. I had found that it was correct for all integer tetrations, so why should I have the slightest concern that it wouldn't be correct for fractional tetrations? It would be as absurd as doubting that 2^3 equals 4^1.5.
And yet, Andrew's solution does indeed show this very problem. It behaves as though 2^3 equaled 4^1.487 or something like that. Sure, the error is small, but it's also unexpected.
Barring a really good reason, I'd be perfectly fine saying that Andrew's solution was in error, not mine. And yet the positive/convex nature of the odd derivatives of his solution is so beautiful as to make all my doubts melt away. I cannot fathom that Andrew's solution is wrong.
Which means that my change of base formula is on the right track, but for whatever reason it requires a cyclic shifting constant, a constant which "knows" the underlying exponent.
It'd be like saying that \( 4^x=2^{x\log_2(4,x)} \), where log_2(4,x) is no longer a constant, but a function of x that is cyclic though very nearly constant. Absurd!
But if I had to choose between this absurdity and denying the beauty of Andrew's solution, I'll take the absurdity. The superlogarithmic constant would appear to be a function of the underlying tetrational exponent, cyclic though very nearly constant.
Of course, further proof is needed. Beauty is fine and dandy, but my solution had a beauty of its own, and I consider it now wrong. Beauty alone cannot prove Andrew's solution either. We must find some underlying reason why having all the odd derivatives be convex is a desirable property, besides the fact that there is (almost certainly) only one such solution per base. Uniqueness alone is insufficient, because my solution is unique in its own way and based on "the" unique solution for base eta.
~ Jay Daniel Fox
