I was trying to create a formula for the Prime Zeta function and I partially succeeded except for one frustrating error. I was only able to formulate an approximation.

Consider the following sum: $$f_k(s)=\sum_{q\space\nmid\space{p_{n\le k}}}^\infty \frac 1{q^s}$$

This is the infinite sum of the reciprocals of all the numbers, raised to the power s, that are not divisible by any primes less than or equal to the kth prime.

For example, $f_1(s)=1+\frac 1{3^s}+\frac 1{5^s}+\frac 1{7^s}+\frac 1{9^s}+\frac 1{11^s}+\frac 1{13^s}+\frac 1{15^s}+\cdots$

$\qquad\qquad\quad$ $f_2(s)=1+\frac 1{5^s}+\frac 1{7^s}+\frac 1{11^s}+\frac 1{13^s}+\frac 1{17^s}+\frac 1{19^s}+\frac 1{23^s}+\frac 1{25^s}+\cdots$

By taking $f_k(s)-({p_{k+1}}^{-s})f_k(s)$ we end up with $f_{k+1}(s)$. In other words $$f_k(s)\left(\frac {{p_{k+1}}^s-1}{{p_{k+1}}^s}\right)=f_{k+1}(s)$$

Therefore, since we define $f_0(s)=\zeta(s)$, it's clear that $$f_k(s)=\zeta(s)\left(\prod_{n=1}^k\frac {{p_n}^s-1}{{p_n}^s}\right)$$

It can also be shown that $$f_k(s)=\zeta(s)\left(\frac {P(k,s)^2-P(k,2s)}2-P(k,s)+1\right)$$ where $P(k,s)$ is the partial Prime Zeta Function: $$P(k,s)=\sum_{p\in primes}^k \frac 1{p^s}$$

This is where I hit a snag. I want to equate the above two equations for $f_k(s)$ like this: $$f_k(s)=\zeta(s)\left(\prod_{n=1}^k\frac {{p_n}^s-1}{{p_n}^s}\right)=\zeta(s)\left(\frac {P(k,s)^2-P(k,2s)}2-P(k,s)+1\right)$$ In theory they should be equal, but when I test them, they show to be slightly different, especially for larger k. Why is this? I can't figure out a single reason why they should be unequal.

If they were perfectly equal, then $$P(s)=\lim\limits_{x \to \infty} \left(1-\sqrt{\frac 2{\zeta(2^0s)}-\sqrt{\frac 2{\zeta(2^1s)}-\sqrt{\frac 2{\zeta(2^2s)}-\cdots-\sqrt{\frac 2{\zeta(2^xs)}-1}}}}\right)$$ However, they are only almost equal and thus this is simply an approximation.


The "it can also be shown" expression is wrong. It only contains terms with up to two different prime factors. Write it out for $k=3$ to see that it differs from the correct expression by a term $\zeta(s)(2\cdot3\cdot5)^{-s}$.

  • $\begingroup$ I'm sorry @joriki. I still don't quite see how that expression is wrong. I can see how it differs there for k=3, but I don't understand the significance in how you said "it only contains terms with up to two different prime factors". I am aware of this. I derived it that way. $\endgroup$ – tyobrien Jul 2 '15 at 13:53
  • $\begingroup$ Consider you start at $\zeta(s)$ and subtract $(2^{-s})\zeta(s)$ then add $((3*2)^{-s})\zeta(s)$ and subtract $((3)^{-s})\zeta(s)$ then add $((5*2)^{-s})\zeta(s)$, add $((5*3)^{-s})\zeta(s)$, and subtract $((5)^{-s})\zeta(s)$ then add $((7*2)^{-s})\zeta(s)$, add $((7*3)^{-s})\zeta(s)$, add $((7*5)^{-s})\zeta(s)$, and subtract $((7)^{-s})\zeta(s)$ and so on... This process is defined by the ""it can also be shown"" expression. It makes perfect sense to me... @joriki $\endgroup$ – tyobrien Jul 2 '15 at 13:54
  • $\begingroup$ @TyO'Brien: You're only considering pairs of primes, but the correct expression rightly contains products for any number of primes, not just pairs. If you subtract for $2$, $3$ and $5$, then add for $2\cdot3$, $2\cdot5$ and $3\cdot5$, you've subtracted the term for $2\cdot3\cdot5$ a net total of $0$ times, where in fact you should be subtracting it once, so you need to subtract it once more. Perhaps looking at the Wikipedia article on inclusion/exclusion will help to make this clearer. $\endgroup$ – joriki Jul 2 '15 at 13:58
  • $\begingroup$ I think I see what you're saying. But you're not saying that if I were to subtract for $2$, $3$, and $5$ once more, the value will be correct, right? Because I tried that and I still don't get the right answer. What, then, would be the right action to take to make the formula correct? @joriki $\endgroup$ – tyobrien Jul 2 '15 at 16:00
  • $\begingroup$ @TyO'Brien: I do think that you should get $f_3$ right if you subtract for $2$, $3$ and $5$ once more -- what did you get instead? (BTW, you don't need to ping me under my own answer; that notification is generated automatically.) $\endgroup$ – joriki Jul 2 '15 at 16:03

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.