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Show that $(1 + \mathcal{O}(\epsilon))(1 + \mathcal{O}(\epsilon)) = (1 + \mathcal{O}(\epsilon)) . $ The precise meaning of this statement is that if $f$ is a function satisfying $f(\epsilon) = (1 + \mathcal{O}(\epsilon))(1 + \mathcal{O}(\epsilon)$ as $\epsilon \to 0$, then $f$ also satisfies $f(\epsilon) = (1 + \mathcal{O}(\epsilon))$ as $\epsilon \to 0$.

However, what does it mean for $f$ to satisfy $f(\epsilon) = (1 + \mathcal{O}(\epsilon))(1 + \mathcal{O}(\epsilon)$ ?

All I know is that $f$ satisfies $f(\epsilon) =\mathcal{O}(\epsilon)$, whenever there exists a $C$ such that $|f(\epsilon)| \leq C \epsilon$.

Jeroen
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  • Why $\epsilon \to 0$? It must tends to infinity. – Leox Oct 05 '14 at 17:53
  • Hey @Leox, whay you are commenting if you have no concept of asymptotics? You can do asymptotics for $\epsilon\to 0$ or $\infty$. Different problems, different solutions, but both reasonable. – Karl Oct 05 '14 at 18:00

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I means that $f(\epsilon)$ is $(1+g(\epsilon))(1+h(\epsilon))$ where $g$ and $h$ are functions such that, for $\epsilon\to 0$, $g(\epsilon)\leq C_1\epsilon$ and $h(\epsilon)\leq C_2\epsilon$, so, for $\epsilon\to 0$, $f(\epsilon)=1+g(\epsilon)+h(\epsilon)+g(\epsilon)h(\epsilon)\leq1+(C_1+C_2)\epsilon+C_1C_2\epsilon^2\leq 1+(C_1+C_2+C_1C_2)\epsilon$, therefore $f(\epsilon)=1+\mathcal{O}(\epsilon)$, where $f(\epsilon)=1+\mathcal{O}(\epsilon)$ means that there is a function $k$ such that $f(\epsilon)=1+k(\epsilon)$ and $k(\epsilon)\leq K\epsilon$ for $\epsilon\to 0$. In this case $k(\epsilon)=g(\epsilon)+h(\epsilon)+g(\epsilon) h(\epsilon)$ satisfies the $\mathcal{O}$ property.

Self-teaching worker
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