# Rules for Landau Symbols

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 $x \mathcal{O}(x^n) = \mathcal{O}(x^{n+1})$?

• Because $12\mathcal{O}(x^5)=\mathcal{O}(x^5)$, so you're still stuck with this power (the smallest power). Jul 1 '18 at 12:49
• @ThePhenotype But isn't f.e. $x^2 \mathcal{O}(x^5) = \mathcal{O}(x^7)$ and $\mathcal{O}(x^5) \in \mathcal{O}(x^7)$?.
– user397268
Jul 1 '18 at 12:59
• That's if you consider large $x$. Here the behaviour near $0$ is considered, and there $O(x^{n+1}) \subset O(x^n)$. Jul 1 '18 at 13:02
• @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. Jul 1 '18 at 13:02

Note that the coefficient of $x^5$ in $$e^x (12 - 6x + x^2)$$
$$= \left(1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 +\frac{1}{120}x^5+...\right)(12 - 6x + x^2) \\$$
$$= 12 - 6x + 6x^2 + \frac {1}{60}x^5+...$$
$$= 12 - 6x + 6x^2 + \mathcal{O}(x^5)$$