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I'm interested in application of the Weierstrass factorization theorem to the primality function.

Let $np(x)\colon \mathbb N\to \mathbb N$ is a "not-prime" function: $$ np(x) = \begin{cases}1, & \text{$x$ is not prime}\\0, & \text{$x$ is prime}\end{cases} $$ Obviously, prime numbers are zeros of $np$. Therefore it seems reasonable to use Weierstrass factorization theorem to express $np$. But this factorization is not unique, and I don't know how to do it. I tried to use this product: $$ \prod_{k=1}^\infty \left(1-\frac{z}{p_k}\right), $$ but it's obviously diverges. Since we can assume that if $p$ is a prime then $-p$ is also is prime, and series $\sum \frac{1}{p_k^2}$ converges, I think that we can use $$ \prod_{k=1}^\infty \left(1-\frac{z^2}{p_k^2}\right)\tag1\label{1} $$ for $np$ (for its continuation to $\mathbb R$ or $\mathbb C$). Function $np$ may be represented as $$ np(z)=e^{g(z)}\prod_{k=1}^\infty \left(1-\frac{z^2}{p_k^2}\right), $$ but that little use even if we use that $np(z)=1$ for $z=0,1,\ldots$.

My question: is there Weierstrass factorization of $np$? Can it finded by using \eqref{1}? Is it this factorization unique? (I think it's not unique in $\mathbb R$, but in $\mathbb C$ we can use analytic continuation.) What about multiplicity of roots (I mean can be there smth like $\left(1-\frac{z^6}{p_1^6}\right)^2 \left(1-\frac{z^2}{p_2^2}\right)^4$ in product)?

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Since the sum of $(r/p_n)^2$ converges for any $r,$ we can use $E_1(z)=(1-z)e^z$ as our function so the product $$ f(x):=\prod_p(1-\frac zp)e^{z/p} $$ is an entire function which has zeros exactly at the prime numbers.

Note that the Weierstrass factorization theorem does not let you control the values outside the zeros, so this function is not your function np. Since that function is not entire, the Weierstrass factorization theorem has nothing to say about it.

Edit: The OP has asked if it is possible to continue the function np to an entire function. By the strong form of the Weierstrass factorization theorem, any such function must have the form $$ f(z)e^{g(z)} $$ where $f$ is defined above and $g$ is entire. So since $f(1)=0.72926\ldots, f(4)=2.3482402\ldots,f(6)=-31.225098\ldots,$ etc., you need an entire function $g$ with $g(6)=-\log(31.225098\ldots) + \pi(2n+1)i$ for some $n$, $g(4)=-\log(2.3482402\ldots) + 2n\pi i$ for some (possibly different) $n$, etc. I don't know of a way of building such an entire function.

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  • $\begingroup$ Ok, well, but my question was about "not-prime function". We have infinitely functions which has zeros exactly at the prime numbers. $\endgroup$ – Michael Galuza Jul 13 '15 at 15:15
  • $\begingroup$ @MichaelGaluza: I edited the answer. Since your function is not entire, the Weierstrass factorization theorem has nothing to say about it. $\endgroup$ – Charles Jul 13 '15 at 15:17
  • $\begingroup$ why $np$ is not entire function? Ok, $np$ is not entire; but I assummed that we can continue it to $\mathbb C$ like $n!$ continues to $\Gamma (n)$ $\endgroup$ – Michael Galuza Jul 13 '15 at 15:19
  • $\begingroup$ @MichaelGaluza: The Weierstrass factorization theorem doesn't let you specify values at the non-zeros, so if you're going to continue it you'll need other tools. $\endgroup$ – Charles Jul 13 '15 at 15:22
  • $\begingroup$ I understand that Weierstrass can't help with continuation. May be I should reformulate question? Can we continue $np$ to an entire function, for example? $\endgroup$ – Michael Galuza Jul 13 '15 at 15:24

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