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As we all (should) know, the Maclaurin series is a special case of the Taylor series when the Taylor series is centered around 0. This is the canonical definition of the Maclaurin series:

$$ f(x) = \sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)}{n!}x^n $$

My question is this. Should it be that:

$$ \dfrac{f(x)}{g(x)} = {\dfrac{\sum^{\infty}_{n=0}{\dfrac{f^{(n)}(0)}{n!}}x^n}{\sum^{\infty}_{n=0}{\dfrac{g^{(n)}(0)}{n!}}x^n}} $$

or that:

$$ \sum^\infty_{n=0}{\dfrac{\dfrac{d^{(n)}}{d^{(n)}x}{\dfrac{f(x)}{g(x)}}}{n!}}x^n $$

Intuitively speaking, I find the former equation much more appealing, not only because it seems logical, but also because the latter equation would be much much harder to solve for quotients.

The same question applies to products.

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  • $\begingroup$ Both can be hard. Division of power series (assuming say $g(0)\ne 0$) is feasible for obtaining the first few terms, but general expression will often not be accessible. Already $\tan x=\frac{\sin x}{\cos x}$ poses difficulties. $\endgroup$ Commented Jun 8, 2013 at 15:21
  • $\begingroup$ I should point out that you've only stated the coefficients of the Maclaurin series. For it to make sense you need to multiply each coefficient by $x^n$. $\endgroup$
    – Zen
    Commented Jun 8, 2013 at 15:23
  • $\begingroup$ Yes, thanks for that! I missed it out. $\endgroup$
    – Yuruk
    Commented Jun 9, 2013 at 0:41

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