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Old 2011-09-06, 19:21   #12
Zeta-Flux
 
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William,

By prime density I am assuming you simply mean \pi(x)/x. If you mean something else, please let me know. If that is the case, then yes there are some improvements available. My favorite source currently for clean inequalities is in the work of Pierre Dusart. From his paper "Sharper bounds for \psi,\theta,\pi, p_{k}" we have

\pi(x)\geq \frac{x}{\ln(x)}\left(1+\frac{1}{\ln(x)}+\frac{1.8}{\ln^{2}(x) \right) for x\geq 32299 and

\pi(x)\leq \frac{x}{\ln(x)}\left(1+\frac{1}{\ln(x)}+\frac{2.51}{\ln^{2}(x)} \right) for x\geq 355991.

Better bounds than these are also available (which are asymptotically equal to Li(x)), using zero-free regions for the zeta function, but they are not nearly as clean. In other words, you can get \pi(x)-Li(x)=O(f(x)) for x\geq x_0 where f(x) is an explicit function which is better than x/ln^{k}(x) for any k, and x_0 is also explicit.

Last fiddled with by Zeta-Flux on 2011-09-06 at 20:07
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Old 2011-09-06, 20:44   #13
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On a second reading, I'm not entirely sure what you are asking. Let me ask some follow-up questions.

It is known that \pi(x)-\int_{2}^{x}\frac{1}{\ln(t)}dt has an error term which is o(x/\ln^{k}(x)) for all k. Under the Riemann hypothesis, the error term is O(\sqrt{x}\ln(x)).

Is your question whether or not we can replace 1/\ln(x) inside the integrand with another function, and get an even better error term? If so, the answer is yes. You just replace it with the step function which when integrated matches the prime counting function. On the other hand, if you are asking whether or not there is a simple smooth function which improves the integrand, I think the answer is unknown. At any rate, the difference between that function and 1/ln(x) would be small, and up to O(\sqrt{x}\ln(x) (after integration) would depend only on the zero-free region of the zeta-function.

Last fiddled with by Zeta-Flux on 2011-09-06 at 20:45
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Old 2011-09-06, 21:02   #14
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In particular, note that if you replace the integrand 1/\ln(t) by a better approximation to the prime density, like Li(t)/t, then you do not necessarily get a "better" approximation to the prime counting function.

In this particular case, if I plugged things into Mathematica correctly, I think that the error term got worse. The error term is now asymptotically 1/\ln^{2}(x).
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Old 2011-09-06, 23:39   #15
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Quote:
Originally Posted by Zeta-Flux View Post
In particular, note that if you replace the integrand 1/\ln(t) by a better approximation to the prime density, like Li(t)/t, then you do not necessarily get a "better" approximation to the prime counting function.

In this particular case, if I plugged things into Mathematica correctly, I think that the error term got worse. The error term is now asymptotically 1/\ln^{2}(x).
I would expect so. 1/log x is a good approximation to the instantaneous density at x, while Li(x) is a good approximation to the average density over 1..x. If you use one in place of the other you should get a worse result.
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Old 2011-09-07, 00:05   #16
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CRGreathouse,

What is the definition of "instantaneous density" in this context?
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Old 2011-09-07, 00:19   #17
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Quote:
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CRGreathouse,

What is the definition of "instantaneous density" in this context?
I didn't have a particular definition in mind, but let's say
\lim_{x\to\infty}\frac{f(x\cdot(1+e(x)))-f(x\cdot(1-e(x)))}{2e(x)}
for some suitable e:\mathbb{R}^+\to\mathbb{R}^+ such that \lim_{x\to\infty}e(x)=0, in this case taking f(x)=\pi(x).
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Old 2011-09-07, 00:30   #18
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Can you give an example of a suitable function e?
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Old 2011-09-07, 01:42   #19
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Quote:
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Can you give an example of a suitable function e?
e(x) = 1/sqrt(x).
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Old 2011-09-07, 02:42   #20
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I'm having a hard time understanding your definition. First, if I take e(x)=1/sqrt{x}, and plot the function inside your limit, it heads off to infinity. So you don't get a function with growth 1/ln(x).

Second, if you actually do the limit then you get a constant (which in our case is infinity) and not a function (which in our case we want 1/ln(x)).
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Old 2011-09-07, 21:45   #21
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Quote:
Originally Posted by Zeta-Flux View Post
I'm having a hard time understanding your definition. First, if I take e(x)=1/sqrt{x}, and plot the function inside your limit, it heads off to infinity. So you don't get a function with growth 1/ln(x).

Second, if you actually do the limit then you get a constant (which in our case is infinity) and not a function (which in our case we want 1/ln(x)).
Oh yeah, you probably need to split it into two variables and take only the y limit.
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