12/20/2009, 07:50 AM
(12/20/2009, 06:27 AM)Ansus Wrote: Did you try Mueller formula with conventional bases 1<b<e^(1/e) ?
Yes. It converges to a value that looks to agree with the regular iteration at the attracting fixpoint, though the convergence (of the iterating method) is really slow at \( b = e^{1/e} \). But the upshot, however, is that it doesn't seem to have as strongly escalating precision requirements using the regular iteration limit formula does.
For Mueller's formula with a fixed point we have
\( \sum_{n=0}^{z-1} f(n) = zL + f(0) + \sum_{n=1}^{\infty} f(n) - f(n + z - 1) \)
where \( \lim_{z \rightarrow \infty} f(z) = L \) in the right halfplane. This allows us to run the tetration for those bases via a Taylor series expanded at \( z = 0 \). For computation, I compute \( f(n + z - 1) \) recursively by power series exponential.
Using 64 terms of power series, 38 digits of precision, Mueller sum up to 256 terms, and 128 iterations starting with an initial guess on the series of 1, I get \( ^{1/2} \sqrt{2} \approx 1.24362162766852180429 \) (I discard the rest of the digits beyond where the residual (difference between \( \sqrt{2}^{F(-0.5)} \) and \( F(0.5) \)) is of order of the first discarded digit. The residual gives an idea of the accuracy of the approximation.). Bumping the power series term count up to 128 terms yet with no further increase of the numerical precision yields \( ^{1/2} \sqrt{2} \approx 1.2436216276685218042950989836094029317 \), which agrees completely with the regular iteration limit formula when rounded to 38 digits of precision, suggesting that the regular iteration and Mueller sum approaches yield the same function. And the best part about this that it looks more efficient for computation. Namely, it gives us a Taylor expansion we can reuse (so that once we "initialize", we can do any real height without so many expensive computations), and it requires no increase in precision beyond the level we need (I think even the series formula for regular iteration requires ever more "slop" precision like the limit formula does.).
And if you want a graph...
(that is \( y = {}^{x} \sqrt{2} \) obtained from the Mueller-summed continuum sum formula)
