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 2011-01-25, 20:04 #1 mart_r     Dec 2008 you know...around... 63910 Posts Connection between Li(x)-Pi(x) and x-theta(x) I'm still not quite satisfied with what I find when looking for this one... Once upon a time, I think it was about two years ago, I tried to find a way to get the value of x-$\theta$(x) from known values of Li(x)-$\pi$(x). $\theta$(x), some of you may recognize, is Chebyshev's sum of log(p) for all primes p<=x. I found x-$\theta$(x) = log(x)*[Li(x)-$\pi$(x)]+1-$\sum_{y=2}^{x-1}$ [[log(y+1)-log(y)]*[Li(y)-$\pi$(y)]] To simplify the summation term, I split log(y+1)-log(y) into $\frac{1}{y}-\frac{1}{2y^2}+\frac{1}{3y^3}-...$, and thus only the most significant terms $\frac{Li(y)-\pi(y)}{y}$ have to be added, since the remaining terms converge for appropriately large x (to give a ballpark figure, y>108 to get at least five correct decimal digits): x-$\theta$(x) ~ log(x)*[Li(x)-$\pi$(x)]+1.279143-$\sum_{y=2}^{x-1}$ $\frac{Li(y)-\pi(y)}{y}$ Yet I don't exactly know how to proceed with the summation term. Any comments/suggestions? Precisely, the question is "What accuracy is feasible without having to compute every single value of the sum?" (And ultimately "When will $\theta$(x) surpass x for the first time?")
2011-01-26, 13:52   #2
R.D. Silverman

Nov 2003

1D2416 Posts

Quote:
 Originally Posted by mart_r I'm still not quite satisfied with what I find when looking for this one... Once upon a time, I think it was about two years ago, I tried to find a way to get the value of x-$\theta$(x) from known values of Li(x)-$\pi$(x). $\theta$(x), some of you may recognize, is Chebyshev's sum of log(p) for all primes p<=x. I found x-$\theta$(x) = log(x)*[Li(x)-$\pi$(x)]+1-$\sum_{y=2}^{x-1}$ [[log(y+1)-log(y)]*[Li(y)-$\pi$(y)]] To simplify the summation term, I split log(y+1)-log(y) into $\frac{1}{y}-\frac{1}{2y^2}+\frac{1}{3y^3}-...$, and thus only the most significant terms $\frac{Li(y)-\pi(y)}{y}$ have to be added, since the remaining terms converge for appropriately large x (to give a ballpark figure, y>108 to get at least five correct decimal digits): x-$\theta$(x) ~ log(x)*[Li(x)-$\pi$(x)]+1.279143-$\sum_{y=2}^{x-1}$ $\frac{Li(y)-\pi(y)}{y}$ Yet I don't exactly know how to proceed with the summation term. Any comments/suggestions? Precisely, the question is "What accuracy is feasible without having to compute every single value of the sum?" (And ultimately "When will $\theta$(x) surpass x for the first time?")
Have you tried a Stieltje's integral approximation (i.e. Euler-MacLauren
summation)???

2011-01-26, 16:57   #3
mart_r

Dec 2008
you know...around...

32×71 Posts

Quote:
 Originally Posted by R.D. Silverman Have you tried a Stieltje's integral approximation (i.e. Euler-MacLauren summation)???
No, but I'll see what I can do. Will probably take a few days to catch up on those integrals.

2011-02-27, 21:14   #4
mart_r

Dec 2008
you know...around...

10011111112 Posts

Quote:
 Originally Posted by R.D. Silverman Have you tried a Stieltje's integral approximation (i.e. Euler-MacLauren summation)???
I'm afraid these don't help much. Li(x)-Pi(x) is an erratic function, and about the only approximation I don't have too many troubles working with is sqrt(x)/log(x), as extracted from Riemanns approach to Pi(x). I would want to sum the exact terms up to a given number (maybe z = 109 or 1012) and then use a rough approx value $\sum_{y=z}^{x-1}\frac{1}{log(y)*sqrt y}$ ~ $\frac{2*(sqrt x-sqrt z)}{log(x)}$ for the gap in-between (and if x is appropriately large, I could go directly for $\frac{2*sqrt x}{log(x)}$, which I know is an underestimate for the sum, but grows asymptotically).
Now I wondered if there's a way to get any kind of error bound.

 2011-03-18, 16:52 #5 mart_r     Dec 2008 you know...around... 32×71 Posts May I ask one concise question: is there a fuction f(x) such that $\int_2^\infty \frac{n^x}{\log n}dn=\frac{\int n^x}{\log n-f(x)}$, i.e. $f(x)=\log n_{\lim n\rightarrow\infty}-\frac{\int n^x}{\int_2^\infty \frac{n^x}{\log n}dn}$ ? Darn it, that can't be quite right. May I ask another question: does someone at least understand this equation? (Silverman: No. This is gibberish. You compare infinity to some function whose n isn't defined. me: I know, I just have extreme difficulty to get things that I want to say into precise mathematical depictions.) Last fiddled with by mart_r on 2011-03-18 at 17:04
2011-03-18, 17:25   #6
fivemack
(loop (#_fork))

Feb 2006
Cambridge, England

13×491 Posts

Quote:
 Originally Posted by mart_r May I ask one concise question: is there a fuction f(x) such that $\int_2^\infty \frac{n^x}{\log n}dn=\frac{\int n^x}{\log n-f(x)}$, i.e. $f(x)=\log n_{\lim n\rightarrow\infty}-\frac{\int n^x}{\int_2^\infty \frac{n^x}{\log n}dn}$ ? Darn it, that can't be quite right.
Do you really want the infinities there? At the moment the left-hand side is only defined for x<-1.

And should the right-hand integral also be from 2 to infinity?

Plugging things into Wolfram Alpha,

$\int_2^N \frac {n^x} {\log n} dn = Ei( (1+e) \log N ) - Ei(2^{1+e})$

so you're asking about the asymptotic behaviour of the exponential-integral function

 2011-03-18, 18:00 #7 mart_r     Dec 2008 you know...around... 32·71 Posts What I meant is that for a given value of x the equation approaches some constant if n goes to infinity. To give some numerical examples: for x=0, this would be f(0)=1, since the prime counting functions Li(x) and x/(log(x)-1) both are asymptotic to the number of primes up to x. If I'm not mistaken, f(1/2) should be somewhere near 2.24. Last fiddled with by mart_r on 2011-03-18 at 18:01
 2011-03-18, 19:57 #8 CRGreathouse     Aug 2006 32×5×7×19 Posts I still can't quite understand what function you're getting at.
2011-03-18, 22:34   #9
mart_r

Dec 2008
you know...around...

11778 Posts

Quote:
 Originally Posted by CRGreathouse I still can't quite understand what function you're getting at.
Does it make the situation any better if I write
$\int_2^m \frac{n^x}{\log n}dn$ ~ $\frac{\int m^x}{\log m-f(x)}$ ?
$\small \lim m\rightarrow \infty$

Well... okay, I'll try it again tomorrow. I see I need to get rid of comparing two infinities.

Again, I highly recommend the well-known example:
Quote:
 Originally Posted by mart_r For x=0, this would be f(0)=1, since Li(m) ~ m/(log(m)-1)
$\int_2^m \frac{n^0}{\log n}dn$ ~ $\frac{m}{(\log m)-1}$
$\small \lim m\rightarrow \infty$

[Quote Jake Long] Aw maan! [\Quote]

Last fiddled with by mart_r on 2011-03-18 at 22:40

 2011-03-19, 18:51 #10 fivemack (loop (#_fork))     Feb 2006 Cambridge, England 13×491 Posts So you want f(n) = $\lim_{m \rightarrow \infty} \left( m^{n+1} \left(\int_{u=2}^m \frac{x^n}{\log x} {\mathrm d}x \right) - (n+1) \log m \right)$ I'm not sure that this is not equal to 1 for all n. Last fiddled with by fivemack on 2011-03-19 at 18:52
 2011-03-19, 20:18 #11 mart_r     Dec 2008 you know...around... 32·71 Posts Well, firstly I notice that one of these integrals wasn't necessary at all: $f(x)=\log n-\frac{\frac{n^{x+1}}{x+1}}{\int_{m=2}^n \frac{m^x}{\log m}dm}$ $\lim n\rightarrow\infty$ That looks more like it. And now I see that I may have made things too complicated to begin with. f(x) is, judging by the values I get with MathCad, nothing more than 1/(x+1). Last fiddled with by mart_r on 2011-03-19 at 20:19

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