properties of $O$ and $o$

The following properties of Landau notation hold:

1. $o(f)$ and $O(f)$ are vector spaces, i.e. if $g,h\in o(f)$ (resp. in $O(f)$) then $\lambda g+\mu h\in o(f)$ (resp. in $O(f)$) whenever $\lambda ,\mu \in ℝ$;In particular $o(f)+o(f)=o(f)$ and $\lambda o(f)=o(f)$;

2. if $\lambda \ne 0$ then $\lambda o(f)=o(f)$ and $\lambda O(f)=O(f)$;

3. $fo(g)=o(fg)$, $fO(g)=O(fg)$;

4. $o{(g)}^{\alpha }=o(g^{\alpha })$, $O{(g)}^{\alpha }=O(g^{\alpha })$;

5. $o(f)\subseteq O(f)$; on the other hand if $f\in o(g)$ then $O(f)\subseteq o(g)$;

6. $o(f)\subseteq o(g)$ if $f\in O(g)$; analogously $O(f)\subseteq O(g)$ if $f\in O(g)$;

7. $o(o(f))=o(f)$, $O(O(f))=O(f)$, $o(O(f))=o(f)$, $O(o(f))=o(f)$.

Here are some examples.First of all we consider Taylor formula.If $x_0\in (a,b)\subset ℝ$ and $f:(a,b)\to ℝ$ has $n$ derivatives, then

$$f(x)\in \sum _{k=0}^n\frac{f^{(k)}(x_0)}{k!}{(x-x_0)}^k+o({(x-x_0)}^n).$$

As a consequence, if $f$ has $n+1$ derivatives, we can replace $o({(x-x_0)}^n)$ with $O({(x-x_0)}^{n+1})$ in the previous formula.

For example:

$$e^x\in 1+x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^4+O(x^5)\subset 1+x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^4+o(x^4).$$

Using the properties stated above we can compose and iterate Taylor expansions.For example from the expansions

$$\sin x\in x+\frac{x^3}{3!}+o(x^4),e^x\in 1+x+\frac{x^2}{2}+O(x^3),$$

$$\cos x\in 1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^5)\subseteq 1-\frac{x^2}{2}+O(x^4),\log (1+x)\in x-\frac{x^2}{2}+o(x^2)$$

we get

$(x\sin x-e^{(x^2)})\log (\cos x)$$\in (x(x-{\displaystyle \frac{x^3}{3!}}+o(x^4))-(1+x^2+{\displaystyle \frac{x^4}{2}}+O({(x^2)}^3))\log (1-{\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{x^4}{4!}}+o(x^5))$

$=\left(x^2-{\displaystyle \frac{x^4}{3!}}+o(x^4)-1-x^2-{\displaystyle \frac{x^4}{2}}+O(x^6)\right)\left(-{\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{x^4}{4!}}+o(x^5)-{\displaystyle \frac{{(-\frac{x^2}{2}+o(x^3))}^2}{2}}+o({(-{\displaystyle \frac{x^2}{2}}+o(x^3))}^2)\right)$

$=(-1-{\displaystyle \frac{2}{3}}{x}^4+o(x^4)+O(x^6))\left(-{\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{x^4}{4!}}+o(x^5)-{\displaystyle \frac{\frac{x^4}{4}-2\frac{x^2}{2}o(x^3)+{(o(x^3))}^2}{2}}+o({\displaystyle \frac{x^4}{4}}+o(x^4))\right)$

$=(-1-{\displaystyle \frac{2}{3}}{x}^4+o(x^4))(-{\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{x^4}{4!}}+o(x^5)+{\displaystyle \frac{x^4}{8}}+o(x^5)+o(x^6)+o(x^4))$

$=(-1-{\displaystyle \frac{2}{3}}{x}^4+o(x^4))(-{\displaystyle \frac{x^2}{2}}+6x^4+o(x^4))$

$=-{\displaystyle \frac{x^2}{2}}-6x^4+o(x^4)+x^{4}O(x^2)+o(x^4)O(x^2)$

$=-{\displaystyle \frac{x^2}{2}}-6x^4+o(x^4)+O(x^6)+o(x^6)$

$=-{\displaystyle \frac{x^2}{2}}-6x^4+o(x^4)$