# Thread: Growth rate of Prime zeta function

1. ## Growth rate of Prime zeta function

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

2. ## Re: Growth rate of Prime zeta function

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

By definition is...

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

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

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

... so that is...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ## Re: Growth rate of Prime zeta function

Originally Posted by chisigma
By definition is...

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

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

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

... so that is...

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

Kind regards

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

Thanks

4. ## Re: Growth rate of Prime zeta function

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

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

... then the condition is approximately...

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

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

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

... and the condition is approximately...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. ## Re: Growth rate of Prime zeta function

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 $\displaystyle < 10^{-D}$. If You write...

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

... then the condition is approximately...

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

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

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

... and the condition is approximately...

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

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

Kind regards

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

6. ## Re: Growth rate of Prime zeta function

Originally Posted by mathematicaphoenix
Isn't there a way to derive a simple formula such as that for the $\displaystyle \zeta (s)$
Of course the approximate relation $\displaystyle n \approx 10^{\frac{D}{s}}$ valid for $\displaystyle \zeta(s)$ is valid for $\displaystyle P(s)$ using p instead of n...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

7. ## Re: Growth rate of Prime zeta function

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

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

Kind regards

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

Thanks