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 2009-05-28, 17:30 #1 CRGreathouse     Aug 2006 135338 Posts Finite prime zeta? Is there a good estimate for $P_n(s)=\sum_{p=2}^np^{-s}$? This is a generalization of the prime zeta function, as $\lim_{n\to\infty}P_n(s)=\sum_{p=2}^np^{-s}=P(s).$ I'm particularly interested in n around 10^6 and s around 2, if it helps. And yes, I mean high precision. Saying $P_n(s)\approx P(s)$ for large n isn't helpful since I'm calculating $P(s)-P_n(s)$ (the high-order terms of P(s)).
 2009-05-28, 18:43 #2 CRGreathouse     Aug 2006 3×1,993 Posts If nothing else I can do $P_n(s)-P(s)\approx\int_n^\infty\frac{dx}{x^s\log^sx}$ which Mathematica can probably integrate. But this doesn't seem to give good results. Last fiddled with by CRGreathouse on 2009-05-28 at 18:43
2009-05-28, 22:05   #3
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by CRGreathouse If nothing else I can do $P_n(s)-P(s)\approx\int_n^\infty\frac{dx}{x^s\log^sx}$ which Mathematica can probably integrate. But this doesn't seem to give good results.

Try it as a Stieltje's integral. Integrate with respect to
d pi(x) instead of dx. Use repeated integration by parts along with
pi(x) ~ li(x).

I don't know how this will turn out. I've never worked with this
version of the zeta function. You might also try estimating the
function itself using Euler-Maclauren summation with respect to
d pi(x) instead of dx.

 2009-05-29, 20:38 #4 CRGreathouse     Aug 2006 3·1,993 Posts Hmm, not a bad idea. Actually I think it would be best to apply this to the original problem directly; the source of the error may be in the use of the approximation leading to my use of the (truncated) prime zeta function. Perhaps not, but even so reducing error would be good.

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