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 2021-06-23, 16:29 #1 mart_r     Dec 2008 you know...around... 7×109 Posts A probabilistic measure arising in approximating pi(x) / Question Given the logarithmic integral $Li(x) = \int_2^x \frac{dt}{\log t}$ and the smooth part of Riemann's prime counting formula as the equivalent of the Gram series $R(x) = 1+\sum_{n=1}^\infty \frac{\log^n x}{n\cdot n!\cdot\zeta(n+1)},$ is there a constant c such that $c = \lim_{m\rightarrow\infty} \frac{1}{m} \sum_{x=2}^m \log(\lvert\frac{Li(x)-\pi(x)}{R(x)-\pi(x)}\rvert)$ ?
 2021-07-19, 18:48 #2 mart_r     Dec 2008 you know...around... 7·109 Posts Let me admit at this point that I still don't seem to understand how to approximate pi(x) by employing the nontrivial zeta zeroes, specifically how I achieve sufficient convergence of x^(1/2 ± t i) as t gets larger. I'd like to see how well the above mentioned value c fares when x is large, but for this I need to have a pi(x) approximation that makes use of at least a couple of those zeroes. Are there any freely available programs for this?
 2022-03-05, 21:13 #3 dbaugh     Aug 2005 1718 Posts The book Prime Obsession has a wonderful exposition of the method with worked examples. For a list of zeros I suggest: https://www.lmfdb.org/zeros/zeta/ To compare your results I suggest: http://sweet.ua.pt/tos/primes.html (Estimates of pi(x) for "large" values of x)
2022-03-08, 21:06   #4
mart_r

Dec 2008
you know...around...

7·109 Posts

Quote:
 Originally Posted by dbaugh The book Prime Obsession has a wonderful exposition of the method with worked examples. For a list of zeros I suggest: https://www.lmfdb.org/zeros/zeta/ To compare your results I suggest: http://sweet.ua.pt/tos/primes.html (Estimates of pi(x) for "large" values of x)
I think I'm going to have to order some books anyway. (I may confess that I hardly ever order anything online, only about once or twice a year.)
Do you happen to know the author of "Prime obsession"? I'm imagining to get some NSFW results by simply looking for the title...
In the past I already worked with some of the data from the links you gave.

 2022-03-08, 23:30 #5 dbaugh     Aug 2005 11110012 Posts The author's name is John Derbyshire. The book still sells well and has been translated into over a dozen languages. I'll be very interested in your results. What is your favorite programming language? You will probably want to use at least double precision for your calculations.
 2022-03-10, 20:20 #6 mart_r     Dec 2008 you know...around... 10111110112 Posts OK thanks! This calculation doesn't have top priority at the moment, so it will probably take a while until I get results worth mentioning. My favorite programming language would be the one I'm most familiar with, which is BASIC - I still remember the countless hours on a C128 back in the mid-90s... but I'm going to use Pari for this one. First I have to figure out how to get the numbers right, only then I can investigate things like precision and efficiency and trade-offs.
 2022-03-28, 17:39 #7 mart_r     Dec 2008 you know...around... 7×109 Posts I think I got it now. The paper arXiv: 1407.1914 "On the first sign change of $$\theta$$(x)-x" by Platt and Trudgian was in the scope of my knowledge, but it was not until recently I found the focus to put the pieces together to tackle the $$\pi$$(x) approximation. I mean, I've seen the equation on top of page 6 before, but I can't remember what went wrong last time I tried to use it, maybe I was stupid and mistook log($$\omega$$) for $$\omega$$, I don't know... So from what I've figured out, it appears the constant in my OP exists and it's about 2.32. This finally gives an answer to something I have asked myself for years. Riemann's approximation is, on a logarithmic average scale, about ten times or one significant digit closer to $$\pi$$(x) than Gauss's. Of course now would be the time for more statistics. Like, naïvely speaking, how probable is it that R(x) is 2 or 10 or 100 times closer to $$\pi$$(x) than Li(x), but that's another story for another day. At the very least I have to consult the work of Sarnak and Rubinstein first. I guess it can be interpreted as a good sign that I eventually find the answers to many of my questions on my own.
2022-04-05, 19:23   #8
mart_r

Dec 2008
you know...around...

7×109 Posts

Quote:
 Originally Posted by mart_r So from what I've figured out, it appears the constant in my OP exists and it's about 2.32. This finally gives an answer to something I have asked myself for years. Riemann's approximation is, on a logarithmic average scale, about ten times or one significant digit closer to $$\pi$$(x) than Gauss's.
Correction: c ~ 2.14, and R(x) - also Li(x)-$$\sqrt{x}$$/log(x) - should be ~ 8.5 times closer to $$\pi$$(x) by logarithmic average.
That is, until I find the next mistake in my calculations...

These numbers may also be dangerously premature (and recklessly worded), but I don't know when I'll be able to verify them:
Code:
Li(x) better than R(x)  0.9% of the time
R(x) better than Li(x) 99.1% of the time
R(x) 2x better than Li(x) 94.0% of the time
R(x) 3x better than Li(x) 86.7% of the time
R(x) 10x better than Li(x) 35.9% of the time
R(x) 100x better than Li(x) 3.7% of the time
R(x) 1000x better than Li(x) 0.38% of the time
50% of the time, R(x) is >6.9x better than Li(x)

 2022-04-06, 15:39 #9 Till     "Tilman Neumann" Jan 2016 Germany 1F316 Posts Are you sure about 2.14? Then I would try to get it two more digits accurate, it smells like pi-1... But it seems that your constant c-expression is prone to approximation errors. You are approximating all of Li(x), R(x) and π(x). If either Li(x)−π(x) or R(x)−π(x) gets very close to 0 by some accidental approximation error, then the log-term of this series element might dominate your series.
2022-04-07, 15:23   #10
mart_r

Dec 2008
you know...around...

7×109 Posts

Quote:
 Originally Posted by Till But it seems that your constant c-expression is prone to approximation errors. You are approximating all of Li(x), R(x) and π(x). If either Li(x)−π(x) or R(x)−π(x) gets very close to 0 by some accidental approximation error, then the log-term of this series element might dominate your series.
That's correct, I was considering this as well. And the more I think about it, the less it makes sense to find an average value of this kind.
So I rather retract my earlier statements about said constant.

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