20110125, 20:04  #1 
Dec 2008
you know...around...
2^{2}×163 Posts 
Connection between Li(x)Pi(x) and xtheta(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(x) from known values of Li(x)(x). (x), some of you may recognize, is Chebyshev's sum of log(p) for all primes p<=x. I found x(x) = log(x)*[Li(x)(x)]+1 [[log(y+1)log(y)]*[Li(y)(y)]] To simplify the summation term, I split log(y+1)log(y) into , and thus only the most significant terms have to be added, since the remaining terms converge for appropriately large x (to give a ballpark figure, y>10^{8} to get at least five correct decimal digits): x(x) ~ log(x)*[Li(x)(x)]+1.279143 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 (x) surpass x for the first time?") 
20110126, 13:52  #2  
Nov 2003
2^{2}·5·373 Posts 
Quote:
summation)??? 

20110126, 16:57  #3 
Dec 2008
you know...around...
652_{10} Posts 

20110227, 21:14  #4  
Dec 2008
you know...around...
2^{2}·163 Posts 
Quote:
Now I wondered if there's a way to get any kind of error bound. 

20110318, 16:52  #5 
Dec 2008
you know...around...
2^{2}·163 Posts 
May I ask one concise question:
is there a fuction f(x) such that , i.e. ? 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 20110318 at 17:04 
20110318, 17:25  #6  
(loop (#_fork))
Feb 2006
Cambridge, England
2^{4}×3×7×19 Posts 
Quote:
And should the righthand integral also be from 2 to infinity? Plugging things into Wolfram Alpha, so you're asking about the asymptotic behaviour of the exponentialintegral function 

20110318, 18:00  #7 
Dec 2008
you know...around...
2^{2}×163 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 20110318 at 18:01 
20110318, 19:57  #8 
Aug 2006
1011101100001_{2} Posts 
I still can't quite understand what function you're getting at.

20110318, 22:34  #9  
Dec 2008
you know...around...
1214_{8} Posts 
Quote:
~ ? Well... okay, I'll try it again tomorrow. I see I need to get rid of comparing two infinities. Again, I highly recommend the wellknown example: ~ [Quote Jake Long] Aw maan! [\Quote] Last fiddled with by mart_r on 20110318 at 22:40 

20110319, 18:51  #10 
(loop (#_fork))
Feb 2006
Cambridge, England
6384_{10} Posts 
So you want
f(n) = I'm not sure that this is not equal to 1 for all n. Last fiddled with by fivemack on 20110319 at 18:52 
20110319, 20:18  #11 
Dec 2008
you know...around...
652_{10} Posts 
Well, firstly I notice that one of these integrals wasn't necessary at all:
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 20110319 at 20:19 
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