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A taylor series is always in the form of a summation. But, is there something that is a product equivalent to a taylor series where you arbitrarily estimate a function with a product of infinite factors? And if so, why aren't those products more commonly used?

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  • $\begingroup$ If $f(z)$ is analytic and doesn't vanish on $\{z, \in \mathbb{C}, |z-a| < r\}$ then $\frac{f'(z)}{f(z)}$ is analytic on $\{z, \in \mathbb{C}, |z-a| < r\}$ and hence $\displaystyle \frac{f'(z)}{f(z)} = \sum_{n=0}^\infty c_n (z-a)^n ,$ $ \log f(z) = \log f(a)+\sum_{n=1}^\infty \frac{c_n}{n+1} (z-a)^{n+1},$ $ f(z) = f(a)\prod_{n=1}^\infty \exp(\frac{c_n}{n+1} (z-a)^{n+1})$ for $|z-a| < r$ $\endgroup$
    – reuns
    Aug 7, 2017 at 3:39
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    $\begingroup$ Incidentally, two examples of infinite products that are sometimes used for approximation are, for small $x$, $$ \frac{1}{1 - x} = (1+x) (1+x^2) (1+x^4) (1+x^8) \ldots $$ and the Euler product for Riemann zeta: $$\zeta(s) = \prod_{\substack{p \\ p \mathrm{\ is\ prime}}} \frac{1}{1 - p^{-s}} $$ $\endgroup$
    – user14972
    Aug 7, 2017 at 3:41
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    $\begingroup$ You are probably interested in Weierstrass factorization theorem. $\endgroup$ Aug 7, 2017 at 4:51
  • $\begingroup$ Another classic result (Euler) is $\sin \pi x=$ $\pi x \prod_{n\in \mathbb N}(1-\frac {x^2}{n^2})$. $\endgroup$ Aug 7, 2017 at 5:52
  • $\begingroup$ @JyrkiLahtonen. That is the "right" answer, I think. $\endgroup$ Aug 7, 2017 at 5:54

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