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Old 2011-01-25, 20:04   #1
mart_r
 
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Default 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?")
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Old 2011-01-26, 13:52   #2
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Quote:
Originally Posted by mart_r View Post
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)???
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Old 2011-01-26, 16:57   #3
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Quote:
Originally Posted by R.D. Silverman View Post
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.
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Old 2011-02-27, 21:14   #4
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Quote:
Originally Posted by R.D. Silverman View Post
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.
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Old 2011-03-18, 16:52   #5
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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
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Old 2011-03-18, 17:25   #6
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Quote:
Originally Posted by mart_r View Post
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
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Old 2011-03-18, 18:00   #7
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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
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Old 2011-03-18, 19:57   #8
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I still can't quite understand what function you're getting at.
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Old 2011-03-18, 22:34   #9
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Quote:
Originally Posted by CRGreathouse View Post
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 View Post
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
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Old 2011-03-19, 18:51   #10
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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
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Old 2011-03-19, 20:18   #11
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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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