08/19/2014, 07:56 PM
(This post was last modified: 08/19/2014, 11:23 PM by sheldonison.)
(08/19/2014, 05:31 PM)jaydfox Wrote: ....Too bad you can't find the original paper; \)40 is pretty steep. I'm not in a math program, so I don't think I have access to the local university library without a student ID; never tried.
And there it is: C:≈0.923307±0.000001, where C is 1/a_1 as I defined it.
That abstract looks familiar. I'm pretty sure it's the same paywalled paper I linked to before. It was written the year I was born, so it just seems a shame that there's still such a hefty fee to access it (over \(40).
How large a value of "k" have you calculated A(k) for, and how long does it take? I know you were working on a logarithmic algorithm in terms of memory requirements... but I thought the time was still proportional to "k". I didn't follow where you got to after that. Your chart shows 2^240 which is pretty big ... impressive!
Also, I think \( \alpha_1 \approx 1.0830628782 \pm 2\cdot 10^{-8}\;\;C \approx 0.9233074276 \pm 2\cdot 10^{-8} \), but I'm only using k=800,000, along with some questionable accelerator techniques to estimate the function, also assuming that it converges to a particular periodic function which I posted earlier, fairly quickly.
\( \text{zerocount}(\rho(x)) \approx \frac{2x}{f(x)} \cdot \frac{d}{dx} f(x)
\;\;\;\rho(k) = \frac{\alpha_1 A(k)}{f(k)}-1\;\;\; f(x) = \sum_{k=0}^{\infty}\frac{1}{2^{k(k-1)/2} k!} x^k\ \)
Then I took the inverse of the zerocount function at 16, 15.5, 15, 14.5. 14, 13.5, 13, 12.5 and averaged all of the alpha_1 values with a binomial weighting. I estimated the error term by varying the starting point from 16 to 15.5. Of course, it may be an illusion that this sum converges to alpha_1... I could try more terms though since I'm only using k=800000
- Sheldon