# Function with zeros plus/minus the primes

While playing with Cohen's pari script prodeulerrat found a function.

For $$s \in \mathbb{C}$$ define $$f(s) = \prod_{p \text{ prime}} (1-\frac{s^2}{p^2})$$

The product converges everywhere, no poles and the zeros are $$\pm p$$.

At integers one can tell if $$f(n)=0$$ via primality testing.

Cohen's script computes $$f(s)$$ in $$O(|s|)$$ and it iterates over primes.

Q1 Is there an alternative way to compute $$f(s)$$?

Q2 An explicit series converging to $$f(s)$$?

$$f(1)=1/\zeta(2)$$.

Q3 Is there closed form for $$f$$ at integers?

• Joro, the addtional values I found are: $f(0)=1$ and $f(i)=\sum_{n=1}^{\infty} \frac{|\mu(n)|}{n^{2}} =\frac{\zeta(2)}{\zeta(4)}$. Also note that $f(1-p)$ only has two single zeros at $p=3, p=-2$. Since your function has only zeros and no poles, what would be nice (but unlikely), is that your formula is a 'disguised' Weierstrass product of an entire function of the shape: $f(z) = z^m e^{g(z)} \prod_p^\infty E_{p_n}(z/a_n)$, i.e. an entire function that can be fully expressed by its infinite prime roots. The trick obviously is to find the right parameters... – Agno Sep 6 '13 at 14:31
• @Agno Thanks. Stopple's answer appear interesting, though slowly converging. Looks like the primality of say 11 depends on the factorization of numbers above 10^7. – joro Sep 6 '13 at 14:51
• @Agno: isn't the original definition already an (undisguised) Weierstrass product? – Greg Martin Sep 6 '13 at 22:56
• @Greg: agree. I guess it is an entire function in its current form already. – Agno Sep 7 '13 at 13:50
• @Joro. The difficult task is to find a function that has $s=\pm$ the primes as its roots. In this question, I asked something similar (scroll to the bottom): mathoverflow.net/questions/122582/… Maybe we could tweak Wilson's formula by adding a cosine and using $\Gamma(s)$ instead of the factorial. There are also other prime functions that become 1 or 0 if prime (mathworld.wolfram.com/PrimeFormulas.html). Maybe those using the Floor function could be simplified (eg using $\lfloor x \rfloor = x-\{x\}$). – Agno Sep 7 '13 at 14:12

For Q2 I'm not sure what you expect other than the obvious series which comes from multiplying out the product: $$f(s)=\sum_{n=1}^\infty \mu(n)\frac{s^{2\omega(n)}}{n^2}$$ where $\omega(n)$ is the number of distinct prime divisors of $n$.
• Hm, I can't compute $f(11)$ using your series, it is not near zero for $N=10^7$. Does this mean the primality of 11 is related to the factorization of numbers above $10^7$? – joro Sep 6 '13 at 14:52
• The series in powers of $s$ will automatically be the Taylor expansion at $0$, good for small $s$ only. – Stopple Sep 6 '13 at 20:00
• Since the average order of $\omega(n)$ is $\log(\log(n))$, letting $k=\log(\log(n))$ the series should behave roughly like $\sum_k s^{2k}/\exp(\exp(k))$. So I expect the radius of convergence to be infinite. – Stopple Sep 7 '13 at 18:02
A standard method to improve the speed of convergence is to look for approximations, which can be explicitly evaluated. In this case one would take $f(s)\approx\zeta(2)^{-s^2}$. We have $$f(s)\zeta(2)^{s^2} = \exp\left(\sum_p s^2\log (1-\frac{1}{p^2}) + \log(1-\frac{s^2}{p^2})\right).$$ Each summand is now of magnitude $\frac{s^4}{p^4}$, thus the speed of convergence has improved, since you have to add up primes substantially larger then $s$ anyway, unless you only look for very rough bounds.
If this is not enough, you can develop $\log(1-\frac{1}{p^2})$ and $\log(1-\frac{s^2}{p^2})$ into a Taylor series, and pull out another power of $\zeta$. The next term will probably give $\zeta(4)^{(s^4-s^2)/2}$, and the series should then converge like $\frac{s^6}{p^6}$.