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Consider the function $\sin(x)$ and notice that it can be expressed as an infinite product as shown here: $\sin(x)$ infinite product formula: how did Euler prove it?.

I would like to know when a function can be expressed as an infinite product and when a function can't. To be more specific, for some $f(x)$ with 0 as a root of $f$, does there exist a sequence of complex numbers $r_n$ and complex constant $a$ such that $f(x)=a\prod_{n=1}^{\infty}(1-x/r_n)$. If $f$ has a root at $0$, then we should instead ask if there exists a sequence $r_n$ and constant $a$ such that $f(x)=ax\prod_{n=1}^{\infty}(1-x/r_n)$.

In general, I would like to know for which functions such a representation exists and for which functions no such representation exists.

A-Level Student
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Mathew
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    The Weierstrass factorization theorem gives you that, whenever a complex function is entire, it can be factored as a (possibly infinite) product. It can be generalized to meromorphic functions. – PrincessEev Apr 05 '21 at 17:46
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    Also it often happens that there is an additional $\exp(g(x))$ term with $g$ entire (try with $f(x)=\exp(x)$), and/or some $\exp(x/r_n)$-looking regularization terms (try with $f(x)=\frac{\sin(x^{1/2})}{x^{1/2}}$ the product over its zeros doesn't converge). – reuns Apr 05 '21 at 17:49
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    @EeveeTrainer, Does this apply to the exponential function? The exponential function is entire but has no roots. – Mathew Apr 05 '21 at 17:52
  • The factorization theorem and the form of the result admits for an exponential factor

    $$f(z)=z^m e^{g(z)} \prod_{n=1}^\infty E_{p_n}!!\left(\frac{z}{a_n}\right)$$

    where the $E$'s represent elementary factors. See the Wikipedia article for more.

     – PrincessEev Apr 05 '21 at 18:40

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