Growth rate of Prime zeta function

**mathematicaphoenix** · January 2nd 2012, 07:22 AM

What would be the asymptotic growth rate of $\zeta _{p}{(s)}$ or for the special cases $\zeta _{p}{(2)}$ or $\zeta _{p}{(3)}$
I tried using the relation ${p}_{n}\sim n\ln (n)$ but couldn't get it to work. How many terms would you require to compute $\zeta _{p}{(2)}$ and $\zeta _{p}{(3)}$ to 20 digits using direct computation??

**chisigma** · January 2nd 2012, 10:13 PM

Originally Posted by mathematicaphoenix

What would be the asymptotic growth rate of $\zeta _{p}{(s)}$ or for the special cases $\zeta _{p}{(2)}$ or $\zeta _{p}{(3)}$
I tried using the relation ${p}_{n}\sim n\ln (n)$ but couldn't get it to work. How many terms would you require to compute $\zeta _{p}{(2)}$ and $\zeta _{p}{(3)}$ to 20 digits using direct computation??

By definition is...

$P(s)=\sum_{p} \frac{1}{p^{s}}$ (1)

... and it is immediate to verify that...

$\lim_{s \rightarrow \infty} 2^{s}\ P(s)=1$ (2)

... so that is...

$P(s) \sim \frac{1}{2^{s}}$ (3)

Kind regards

$\chi$ $\sigma$

**mathematicaphoenix** · January 3rd 2012, 04:04 AM

Originally Posted by chisigma

By definition is...

$P(s)=\sum_{p} \frac{1}{p^{s}}$ (1)

... and it is immediate to verify that...

$\lim_{s \rightarrow \infty} 2^{s}\ P(s)=1$ (2)

... so that is...

$P(s) \sim \frac{1}{2^{s}}$ (3)

Kind regards

$\chi$ $\sigma$

I got that result. But how would you use (3) to calculate the the number of terms required to get $P(s)$ to a given accuracy using direct calculation.
For example, for $\zeta (s)$ I have already found that, to get $D$ digits of accuracy you would require
$n\approx 10^{\frac{D}{s}}$
terms. I want to do a similar thing with $P(s)$ . Can you help?

Thanks

**chisigma** · January 3rd 2012, 04:37 AM

Originally Posted by mathematicaphoenix

I got that result. But how would you use (3) to calculate the the number of terms required to get $P(s)$ to a given accuracy using direct calculation.
For example, for $\zeta (s)$ I have already found that, to get $D$ digits of accuracy you would require
$n\approx 10^{\frac{D}{s}}$
terms. I want to do a similar thing with $P(s)$ . Can you help?

Thanks

If You intend to use that formula, then what You can do is to extimate the minimum value of s that gives an error $< 10^{-D}$ . If You write...

$P(s)= \frac{1}{2^{s}} + \varepsilon (s)$ (1)

... then the condition is approximately...

$\frac{2}{3^{s}}< 10^{-D} \implies s> D\ \frac{\ln 10 + \ln 2}{\ln 3}$ (2)

If You uses two terms of the series writing $P(s) \sim \frac{1}{2^{s}} + \frac{1}{3^{s}}$ then You write...

$P(s)= \frac{1}{2^{s}} + \frac{1}{3^{s}}+ \varepsilon (s)$ (3)

... and the condition is approximately...

$\frac{2}{5^{s}}< 10^{-D} \implies s> D\ \frac{\ln 10 + \ln 2}{\ln 5}$ (4)

Now You can proceed increasing the terms till to arrive to the required accuracy...

Kind regards

$\chi$ $\sigma$

**mathematicaphoenix** · January 6th 2012, 01:05 AM

Originally Posted by chisigma

If You intend to use that formula, then what You can do is to extimate the minimum value of s that gives an error $< 10^{-D}$ . If You write...

$P(s)= \frac{1}{2^{s}} + \varepsilon (s)$ (1)

... then the condition is approximately...

$\frac{2}{3^{s}}< 10^{-D} \implies s> D\ \frac{\ln 10 + \ln 2}{\ln 3}$ (2)

If You uses two terms of the series writing $P(s) \sim \frac{1}{2^{s}} + \frac{1}{3^{s}}$ then You write...

$P(s)= \frac{1}{2^{s}} + \frac{1}{3^{s}}+ \varepsilon (s)$ (3)

... and the condition is approximately...

$\frac{2}{5^{s}}< 10^{-D} \implies s> D\ \frac{\ln 10 + \ln 2}{\ln 5}$ (4)

Now You can proceed increasing the terms till to arrive to the required accuracy...

Kind regards

$\chi$ $\sigma$

Isn't there a way to derive a simple formula such as that for the $\zeta (s)$

**chisigma** · January 6th 2012, 01:33 AM

Originally Posted by mathematicaphoenix

Isn't there a way to derive a simple formula such as that for the $\zeta (s)$

Of course the approximate relation $n \approx 10^{\frac{D}{s}}$ valid for $\zeta(s)$ is valid for $P(s)$ using p instead of n...

$p \approx 10^{\frac{D}{s}}$ (1)

Kind regards

$\chi$ $\sigma$

**mathematicaphoenix** · January 6th 2012, 08:33 PM

Originally Posted by chisigma

Of course the approximate relation $n \approx 10^{\frac{D}{s}}$ valid for $\zeta(s)$ is valid for $P(s)$ using p instead of n...

$p \approx 10^{\frac{D}{s}}$ (1)

Kind regards

$\chi$ $\sigma$

If I apply (1) to get an approximation for 20 digits of $P(2)$ I get a value that's 10 orders of magnitude less. I should get $10^{20}$ (P. Sebah et.al.) What should I do?

Thanks

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