1
$\begingroup$

Question follow the one answered already, zeros about Riemann Zeta function and some L-function

Let's me try my best to make it clear on what I am asking.

In his 1859 paper "On the Number of Primes Less Than a Given Magnitude" Riemann found an explicit formula for the number of primes $\pi(x)$ less than a given number x. His formula was given in terms of the related function,

$\Pi(x)=\pi(x)+1/2\pi(x^{1/2})$ +1/3$\pi(x^{1/3})$ +1/4$\pi(x^{1/4})$+1/5$\pi(x^{1/5})$+...

which counts primes where a prime power $p^n$ counts as $\frac{1}{n}$ of a prime. With help of Möbius function, Riemann gave an explicit formula for prime counting function $\pi(x)$, which relating with non-trivial zeros of $\zeta$ function.

$\Pi(x)=\mathrm{Li}(x)-\sum_{\rho}(\mathrm{Li}(x^\rho)) - log(2) +\int_x^{\infty} \frac{1}{t(t^2-1)log(t)}dt$

One can find details info from wiki,

http://en.wikipedia.org/wiki/Riemann_hypothesis

Following Riemann's way, we can define $\Pi_q$, which counts primes where a prime power $p^n$ counts as $\frac{1}{n}$ of a prime. Here $(p,q)=1$.

$\Pi_q$ goes with following series,

$\zeta_q=\sum _{n =1}^{\infty}\frac{1}{n^{s}}, (q,n)=1$

We can see $\zeta_q$ has functional equation, too. Further more, we should be able to get similar $\Pi_q$ explicit formula for prime counting function $\pi(x)$, which exclude prime(s) (q's prime factor).

Here $\Pi_q$ should be much similar with $\Pi$, that it has also $Li(x)-\sum_\rho(Li(x^\rho)$, if $\zeta$ has exact the same non-trivial zeros with $\zeta_q$. The differences are only on other part, which relating with $q$.

If $q<x^\frac{1}{2}$, as $x$ is big enough but $x<\infty$, we can see,

<1> $\Pi(x)-\Pi_q(x) = 1+1/2+1/3+1/4+1/5+...1/n+...$

In <1>, $Li(x)-\sum_\rho(Li(x^\rho)$ is gone but remains function relation with $q$ only, here, $q$ has only 1 prime factor $p$.

so, we find another way for Euler–Mascheroni constant,

<2> $\gamma = \lim_{x\to\infty}(\Pi(x)-\Pi_q(x) - log(log_p(x))$, as $n->\infty$, $q<x^\frac{1}{2}$

If it is correct, $\gamma$ in <2> is relating with $q$ only. If so, can we go further for $\gamma$'s character?

Further more, we can see $\Pi(x)>\Pi_2(x)>\Pi_{2*3}(x)>\Pi_{2*3*5}(x)>\Pi_{2*3*5*7}(x)>...>\Pi_q(x)$, when $q$ includes prime factor ->$\infty$, $0<\Pi_q(x)\le{1}$.

Hense, we have relationship for $Li(x)-\sum_\rho(Li(x^\rho)$ and function with $q$ only, we should be able to find boundary of $\sum_\rho(Li(x^\rho)$.

My question,

(1) What the RH equivalent for Riemann prime formula $\Pi(x)$?

(2) What the formula should be for $\Pi_q(x)$?

$\Pi_q(x)=\mathrm{Li}(x)-\sum_{\rho}(\mathrm{Li}(x^\rho))+???$

$\endgroup$
3
$\begingroup$

The usual zeta function is $$ \prod_p \big( 1 - p^{-s} \big)^{-1} = \zeta(s). $$

The zeta function where you omit the single prime $q$ is $$ \prod_{p\ne q} \big( 1 - p^{-s} \big)^{-1} = \zeta(s) \big( 1-q^{-s} \big). $$

The zeta function where you omit all the primes up to and including $q$ is $$ \prod_{p>q} \big( 1 - p^{-s} \big)^{-1} = \zeta(s) \prod_{p\le q} \big( 1-p^{-s} \big). $$

Any function of the form $1-q^{-s}$ has zeros only when the real part of $s$ equals $0$. Therefore the zeros of all three zeta functions strictly inside the critical strip are identical, and so the Riemann hypothesis is exactly the same for all three functions.

$\endgroup$
  • $\begingroup$ Thanks Greg! Can you please answer my 2 questions? Note - the function $\Pi(x)$ is counting primes and prime power ($p^n$ is counting as $\frac{1}{n}$), which has explicit formula as per Riemann. $\endgroup$ – Ocean Yu Jan 13 '14 at 8:15
  • $\begingroup$ The easiest way is to just adjust the corresponding $\Pi$ function by hand. For example, if you omit simply one prime $q$: use the classical formula for $\Pi(x)$, and then subtract $\sum_{r\le (\log x)/\log q} \frac1r$ to account for the powers of $q$ that are no longer present. $\endgroup$ – Greg Martin Jan 13 '14 at 9:00
  • $\begingroup$ It is correct if doing this by the easiest way. Is this the only way? If recall Riemann's paper, he got following explicit formula with $\zeta$ function, $\Pi(x)=\mathrm{Li}(x)-\sum_{\rho}(\mathrm{Li}(x^\rho)) - log(2) +\int_x^{\infty} \frac{1}{t(t^2-1)log(t)}dt$, I am curious if we can get similar formula for $\zeta_q$ by following Riemann's way, instead of just subtracting $\sum_{r\le (\log x)/\log q} \frac1r$. If we can, we will break through RH with further discussion I believe. $\endgroup$ – Ocean Yu Jan 16 '14 at 1:45
  • $\begingroup$ It is painful that I realized that I was in a round circle, as seems we can not find a good control-able formula, even by following Riemann's way. The questions seem have no a single value, that can be closed. $\endgroup$ – Ocean Yu Jan 16 '14 at 2:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.