User:IssaRice/Little o notation

Definition

Definition (little o near a point). Let $f:\mathbf{R}\to \mathbf{R}$ and $g:\mathbf{R}\to \mathbf{R}$ be two functions, and let $a\in \mathbf{R}$. We say that $f$ is little o of $g$ near $a$ iff for every $ϵ>0$ there exists $\delta >0$ such that $|x-a|<\delta$ implies $|f(x)|<ϵ|g(x)|$. Some equivalent ways to say the same thing are:

Notation	Comments
$f$ is little o of $g$ near $a$
$f(x)\in o(g(x))$ as $x\to a$	In this notation, we think of $o(g(x))$ as a set.
$f(x)=o(g(x))$ as $x\to a$
$f\in o(g)$ near $a$
$f=o(g)$ near $a$

Definition (little o at infinity). Let $f:\mathbf{R}\to \mathbf{R}$ and $g:\mathbf{R}\to \mathbf{R}$ be two functions. We say that $f$ is little o of $g$ at infinity iff for every $ϵ>0$ there exists $M$ such that for all $x$, $x>M$ implies $|f(x)|<ϵ|g(x)|$.

Exercise. Can we write just $f\in o(g)$ or $f=o(g)$ or $f(x)\in o(g(x))$ or $f(x)=o(g(x))$?

Expand to see solution:

In general we can't because for this notation to make sense, we also need to know where the argument $x$ is going. In algorithms, we have $x\to \mathrm{\infty }$, but in analysis (e.g. in some definitions of differentiability) we have $x\to 0$.

Exercise. If we are being a little pedantic, what is wrong with saying "$f\in o(g)$ as $x\to a$"?

Expand to see solution:

We are saying $x\to a$, but we haven't clarified what $x$ is. Instead, we are relying on the reader to assume that $x$ is an argument to $f$ and $g$.

Exercise. Interpret the meaning of $x^2\in o(x)$.

Expand to see solution:

It depends on where $x$ is going. We want $\left|x^2\right|<ϵ\left|x\right|$ whenever $|x-a|<\delta$, so this is only true when $a=0$.

Properties

Proposition. Let $f:\mathbf{R}\to \mathbf{R}$ and $g:\mathbf{R}\to \mathbf{R}$ be two functions, and suppose $g(x)\ne 0$ for all $x\in \mathbf{R}$. Then f is little o of g near a if and only if $\underset{x\to a}{lim}\frac{f(x)}{g(x)}=0$.

Proposition. transitivity

Proposition. we can replace the $<$ in the definition with $\le$, right?

Exercise. Let $c,x_0\in \mathbf{R}$ be constants. Interpret the statement "$o(c(x-x_0)+o(x-x_0))\in o(x-x_0)$ as $x\to x_0$".

Expand to see solution:

The statement is saying $f(x)\in o(x-x_0)$ where $f$ is some function such that $\underset{x\to x_0}{lim}\frac{f(x)}{c(x-x_0)+o(x-x_0)}=0$.

Because of the nested little o, we need to expand again and introduce $g$, where $g(x)\in o(x-x_0)$ so $\underset{x\to x_0}{lim}\frac{g(x)}{x-x_0}=0$.

Now we verify:

$\underset{x\to x_0}{lim}\frac{f(x)}{x-x_0}=lim\frac{f(x)}{c(x-x_0)+g(x)}lim\frac{c(x-x_0)+g(x)}{x-x_0}=0\cdot (c+0)=0$

Could $g$ have been arbitrary? In other words, could we have said $o(c(x-x_0)+o(h(x)))\in o(x-x_0)$ for arbitrary $h(x)$? To compute the limit $\underset{x\to x_0}{lim}\frac{f(x)}{x-x_0}$ we actually used the limit laws, which require that the right hand limit exist. This means that we needed $\underset{x\to x_0}{lim}g(x)/h(x)$ to exist.