08/17/2014, 05:54 PM
(This post was last modified: 08/18/2014, 01:40 PM by sheldonison.)
(08/09/2014, 10:51 PM)jaydfox Wrote: [quote='jaydfox' pid='7372' dateline='1407607336']
I'm also working on a second-order approximation. This would consist of two parts. The first is to fix oscillations in the error term (1/1.083...)*f(k)-A(k). ....
\( \rho(k) = \frac{\alpha_1 A(k)}{f(k)}-1\;\;\; f(x) = \sum_{k=0}^{\infty}\frac{1}{2^{k(k-1)/2} k!} x^k\;\; \alpha_1 \approx 1.083063 \)
Then \( \rho \) has zeros where f(n) crisscrosses the A(n) partition function. Here is a possible estimate of the zero count of rho:
\( \text{zerocount}(\rho(\ln(x))) \approx 2 \frac{d}{dx}\ln(f(\exp(x))) \)
\( \text{zerocount}(\rho(x)) \approx \frac{2x}{f(x)} \cdot \frac{d}{dx} f(x)
\;\;\; \) Calculus simplifications, Tommy's next post
My conjecture is that an ideal entire asymptotic of a function will have a zero count for ln(x) as x goes to infinity, of twice this derivative. The derivative function is also a guide to where each Taylor series coefficient of f(x) is best approximated by the asymptotic; see http://math.eretrandre.org/tetrationforu...863&page=8, post#76. The zeros of (exp^{0.5}/asymptotic-1 )also appears to have that same pseudo period; which I will post later.
For the partition function A(k), the zerocount approximation zerocount((ln(100000))-zerocount(ln(70))=8.95, which is a reasonable approximation for the number of pseudo periodic cycles for \( \rho(z) \) between 70 and 100000.
I'm doing some numerical calculations, and I have one more conjecture. As x gets arbitrarily large. the number and location of the zeros of f(x) at the negative real axis can also be approximated by the same function, with half the period! Actually, it may be that the approximation converges quicker for the zeros at the negative real axis. The zeros occur where the derivative is ~= n+0.50146
\( \text{zerocount}(f(-\ln(x))) \approx \frac{d}{dx}\ln(f(\exp(x))) \)
\( \text{zerocount}(f(-x)) \approx \frac{x}{f(x)} \cdot \frac{d}{dx} f(x)
\;\;\; \) Calculus simplifications, Tommy's next post
Anyway, this is all very fascinating to me, and interesting enough that I need to formalize these ideas more, and do some similar numerical calculations for the exp^{0.5} asymptotic and the 2sinh^{0.5} asymtotic. But there is no more time today ....
- Sheldon

