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E.Dickson mentioned the following result by E.Zondadari in his book "History of theory of numbers vol.1"(Chap XVIII):

Consider the following 'function':

$$ \frac{\sin²(πx)}{(πx)²(1-x²)^2} \prod_{n=2}^\infty \frac{(πx)} {n\sin(πx/n)} $$

It's zero if |x|= prime and else otherwise .

How to prove the above identity ?

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  • $\begingroup$ Is $x$ real or complex? $\endgroup$
    – saulspatz
    Commented Jan 31, 2020 at 19:52
  • $\begingroup$ @saulspatz real $\endgroup$
    – bambi
    Commented Jan 31, 2020 at 19:54
  • $\begingroup$ I don't understand. If $x=4$ say, then the factor outside the product is $0$ so if the product converges, the value must be $0$. What am I missing? $\endgroup$
    – saulspatz
    Commented Jan 31, 2020 at 20:00
  • $\begingroup$ @saulspatz the doubt is mutual ; that's why ,earlier ,I added the image of the page of the book in the first place . But got downvotes saying don't add images! $\endgroup$
    – bambi
    Commented Jan 31, 2020 at 20:03
  • $\begingroup$ if n divides x we get a division by 0 under my interprwtation. but it still fails for primes then unless it's an empty product. $\endgroup$
    – user645636
    Commented Jan 31, 2020 at 22:30

1 Answer 1

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A copy of Zondadari's paper can be found here.

The key to understanding the stated result is an earlier one that the infinite product $$ P(x)=\prod_{n=1}^\infty \frac{\sin\frac{\pi x}{n}}{\frac{\pi x}{n}} $$ converges absolutely and uniformly to an analytic function whose zeros are precisely the non-zero integers, with multiplicities equal to the number of their positive divisors. By $\ \frac{\sin\frac{\pi x}{n}}{\frac{\pi x}{n}}\ $ Zondadari obviously here means the analytic function $\ \sum_\limits{i=0}^\infty \frac{(-1)^i}{(2i+1)!}\left(\frac{\pi x}{n}\right)^{2i}\ $, which assumes the value $1$ at $\ x=0\ $, and adopts the convention that an infinite product with only a finite number of zeroes and a tail that converges to a non-zero quantity is convergent.

Enumerating the zeros of this product, we have $\ \pm 1\ $, each of multiplicity $1$, primes (positive or negative), each of multiplicity $2$, and and composites, each of multiplicity $\ d\ $, where $\ d\ $ is the number of its positive divisors (equal to $\ \left(n_1+1\right) \left(n_2+1\right)\dots \left(n_r+1\right)\ $, where $\ p_1^{n_1}p_2^{n_2}\dots p_r^{n_r}\ $ is its prime decomposition).

Now the function $$ f(x) =\frac{\sin^3\pi x}{(\pi x)^3\left(1-x^2\right)^2} $$ has zeros of multiplicty $3$ at every integer, except $0$, where it has value $1$, and $\ \pm1\ $, where it has zeros of multiplicity $1$. Therefore, the function \begin{align} g(x)&= \frac{\sin^3\pi x}{(\pi x)^3\left(1-x^2\right)^2}\cdot\frac{1}{P(x)}\\ &= \frac{\sin^2\pi x}{(\pi x)^2\left(1-x^2\right)^2} \prod_{n=2}^\infty \frac{\pi x}{n\sin\frac{\pi x}{n}} \end{align} has zeros of multiplictity $1$ for prime $\ x\ $, a pole of multiplicity $\ d-3\ $ at any composite $\ x\ $ with $\ d\ $ positive divisors, and has a finite non-zero value at $\ x=0\ $ and $\ x=\pm1\ $.

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  • $\begingroup$ thank you for the answer . Also , I'm asking if the product could be twisted in such a way that it gives zero for primes and finite values for composite ? $\endgroup$
    – bambi
    Commented Feb 2, 2020 at 8:55
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    $\begingroup$ The function $\ \frac{g(x)}{g(x)+x}\ $ will vanish for prime $\ x\ $ and have value $1$ for composite $\ x\ $. $\endgroup$ Commented Feb 2, 2020 at 10:20
  • $\begingroup$ Is this a "good" test ?( Efficiency) $\endgroup$
    – bambi
    Commented Feb 2, 2020 at 11:16
  • $\begingroup$ @lonzaleggiera: I see, sorry, I hadn't reloaded; I've deleted the comment. $\endgroup$
    – joriki
    Commented Feb 2, 2020 at 11:40
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    $\begingroup$ I can't claim any expertise in tests for primality, but it looks to me that to distinguish whether an integer $\ x\ne\pm1, 0\ $ is a zero or a pole of $\ g(x)\ $ you would need to evaluate $\ \sin\frac{\pi x}{n}\ $ at least for $\ n= 2, 3,\dots,\left\lfloor\sqrt{|x|}\right\rfloor\ $, which would be even less efficient than using trial division as a test for primality. $\endgroup$ Commented Feb 2, 2020 at 23:13

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