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Old 2021-06-23, 16:29   #1
mart_r
 
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Default 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
\[c = \lim_{m\rightarrow\infty} \frac{1}{m} \sum_{x=2}^m \log(\lvert\frac{Li(x)-\pi(x)}{R(x)-\pi(x)}\rvert)\]
?
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Old 2021-07-19, 18:48   #2
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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?
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Old 2022-03-05, 21:13   #3
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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)
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Old 2022-03-08, 21:06   #4
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Quote:
Originally Posted by dbaugh View Post
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.
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Old 2022-03-08, 23:30   #5
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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.
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Old 2022-03-10, 20:20   #6
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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.
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Old 2022-03-28, 17:39   #7
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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.
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Old 2022-04-05, 19:23   #8
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Quote:
Originally Posted by mart_r View Post
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)
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Old 2022-04-06, 15:39   #9
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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.
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Old 2022-04-07, 15:23   #10
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Quote:
Originally Posted by Till View Post
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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