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Old 2009-05-28, 17:30   #1
CRGreathouse
 
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Default 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)).
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Old 2009-05-28, 18:43   #2
CRGreathouse
 
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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
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Old 2009-05-28, 22:05   #3
R.D. Silverman
 
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Quote:
Originally Posted by CRGreathouse View Post
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.
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Old 2009-05-29, 20:38   #4
CRGreathouse
 
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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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