Prime Density Correction Terms? - Page 2

Zeta-Flux · 2011-09-06, 19:21

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.

Zeta-Flux · 2011-09-06, 20:44

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.

Zeta-Flux · 2011-09-06, 21:02

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 $1/\ln^{2}(x)$ .

CRGreathouse · 2011-09-06, 23:39

Quote:

Originally Posted by Zeta-Flux

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.

Zeta-Flux · 2011-09-07, 00:05

CRGreathouse,

What is the definition of "instantaneous density" in this context?

CRGreathouse · 2011-09-07, 00:19

Quote:

Originally Posted by Zeta-Flux

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 $f(x)=\pi(x).$

Zeta-Flux · 2011-09-07, 00:30

Can you give an example of a suitable function e?

CRGreathouse · 2011-09-07, 01:42

Quote:

Originally Posted by Zeta-Flux

Can you give an example of a suitable function e?

e(x) = 1/sqrt(x).

Zeta-Flux · 2011-09-07, 02:42

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)).

CRGreathouse · 2011-09-07, 21:45

Quote:

Originally Posted by Zeta-Flux

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.

2011-09-06, 19:21	#12
Zeta-Flux May 2003 7·13·17 Posts	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

2011-09-06, 20:44	#13
Zeta-Flux May 2003 7×13×17 Posts	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

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
@ Supermods: Error Correction please	Raman	Forum Feedback	72	2013-06-22 07:24
Asymptotic density of k-almost primes	CRGreathouse	Math	1	2010-08-22 23:47
Prime k-value density rating and factorizations	gd_barnes	No Prime Left Behind	25	2009-07-30 12:06
Density of Mersenne divisors	Random Poster	Math	1	2008-12-15 01:14
prime density vs. sieving vs. prp time	jasong	Math	18	2006-03-31 03:14

2011-09-06, 21:02	#14
Zeta-Flux May 2003 7·13·17 Posts	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 $1/\ln^{2}(x)$ .

2011-09-07, 00:05	#16
Zeta-Flux May 2003 11000001011₂ Posts	CRGreathouse, What is the definition of "instantaneous density" in this context?

2011-09-07, 00:30	#18
Zeta-Flux May 2003 7×13×17 Posts	Can you give an example of a suitable function e?

2011-09-07, 02:42	#20
Zeta-Flux May 2003 7·13·17 Posts	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)).