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Let $x \in \mathbb{R}$. Then I have the following \begin{align} e^x (12 - 6x + x^2) &= \left((1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 + \mathcal{O}(x^5)\right)(12 - 6x + x^2) \\ &= 12 - 6x + 6x^2 + \mathcal{O}(x^5). \end{align}

However, I don't get the last equation. Isn't $ x \mathcal{O}(x^n) = \mathcal{O}(x^{n+1})$?

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    $\begingroup$ Because $12\mathcal{O}(x^5)=\mathcal{O}(x^5)$, so you're still stuck with this power (the smallest power). $\endgroup$ Jul 1 '18 at 12:49
  • $\begingroup$ @ThePhenotype But isn't f.e. $x^2 \mathcal{O}(x^5) = \mathcal{O}(x^7)$ and $ \mathcal{O}(x^5) \in \mathcal{O}(x^7)$?. $\endgroup$
    – user397268
    Jul 1 '18 at 12:59
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    $\begingroup$ That's if you consider large $x$. Here the behaviour near $0$ is considered, and there $O(x^{n+1}) \subset O(x^n)$. $\endgroup$ Jul 1 '18 at 13:02
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    $\begingroup$ @Diamir It's the other way around here. $\mathcal{O}(x^5)$ estimates $x^5,x^6,x^7,\ldots$ (as you can see in the expansion of $e^x$), which means that you're considering $x$ close to $0$ here. $\endgroup$ Jul 1 '18 at 13:02
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Note that the coefficient of $x^5$ in $$ e^x (12 - 6x + x^2) $$

$$= \left(1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 +\frac{1}{120}x^5+...\right)(12 - 6x + x^2) \\ $$

$$= 12 - 6x + 6x^2 + \frac {1}{60}x^5+...$$

$$= 12 - 6x + 6x^2 + \mathcal{O}(x^5)$$

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