Connection between Li(x)-Pi(x) and x-theta(x)

mart_r · 2011-01-25, 20:04

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?")

R.D. Silverman · 2011-01-26, 13:52

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)???

mart_r · 2011-01-26, 16:57

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.

mart_r · 2011-02-27, 21:14

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.

mart_r · 2011-03-18, 16:52

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.)

fivemack · 2011-03-18, 17:25

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

mart_r · 2011-03-18, 18:00

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.

CRGreathouse · 2011-03-18, 19:57

I still can't quite understand what function you're getting at.

mart_r · 2011-03-18, 22:34

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]

fivemack · 2011-03-19, 18:51

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.

mart_r · 2011-03-19, 20:18

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).

2011-01-25, 20:04	#1
mart_r Dec 2008 you know...around... 1236₈ 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?")

2011-03-18, 16:52	#5
mart_r Dec 2008 you know...around... 1236₈ 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, 18:00	#7
mart_r Dec 2008 you know...around... 2×5×67 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

2011-03-19, 18:51	#10
fivemack (loop (#_fork)) Feb 2006 Cambridge, England 3·2,141 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... 2·5·67 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

2011-03-18, 19:57	#8
CRGreathouse Aug 2006 3·1,993 Posts	I still can't quite understand what function you're getting at.

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