Definition:O Notation/Big-O Notation

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Definition

Big-O notation occurs in a variety of contexts.

Sequences

Let $\left \langle {a_n} \right \rangle$ and $\left \langle {b_n} \right \rangle$ be sequences of real or complex numbers.


$a_n$ is big-O of $b_n$ if and only if

$\exists c \in \R: c \ge 0 : \exists n_0 \in \N : \left({n \ge n_0 \implies \left\vert{a_n}\right\vert \le c \cdot \left\vert{b_n}\right\vert}\right)$

That is:

$\left\vert{a_n}\right\vert \le c \cdot \left\vert{b_n}\right\vert$

for all sufficiently large $n$.


Real Analysis

Let $f$ and $g$ be real-valued or complex-valued functions defined on a neighborhood of $+ \infty$ in $\R$.


The statement:

$\map f x = \map {\mathcal O} {\map g x}$ as $x \to \infty$

is equivalent to:

$\exists c \in \R: c \ge 0: \exists x_0 \in \R: \forall x \in \R: \paren {x \ge x_0 \implies \size {\map f x} \le c \cdot \size {\map g x} }$


That is:

$\size {\map f x} \le c \cdot \size {\map g x}$

for $x$ sufficiently large.


This statement is voiced $f$ is big-O of $g$ or simply $f$ is big-O $g$.


Complex Analysis

Let $f$ and $g$ be complex functions defined for all complex numbers whose modulus is sufficiently large.


The statement:

$f(z) = \mathcal O \left({g(z)}\right)$ as $|z|\to\infty$

is equivalent to:

$\displaystyle \exists c\in \R: c\ge 0 : \exists r_0 \in \R : \forall z \in \C : (|z| \geq r_0 \implies |f(z)| \leq c \cdot |g(z)|)$


That is:

$|f(z)| \leq c \cdot |g(z)|$

for all $z$ in a neighborhood of infinity in $\CC$.


General Defintion

Let $\left({X, \tau}\right)$ be a topological space.

Let $V$ be a normed vector space over $\R$ or $\C$ with norm $\left\Vert{\,\cdot\,}\right\Vert$.

Let $f, g : X \to V$ be functions.


The statement:

$f \left({x}\right) = \mathcal O \left({g \left({x}\right)}\right)$ as $x \to \infty$

is equivalent to:

There exists a neighborhood of infinity $U \subset X$ such that:
$\exists c \in \R: c \ge 0: \forall x \in U: \left\Vert{f \left({x}\right)}\right\Vert \le c \cdot \left\Vert{g \left({x}\right)}\right\Vert$


That is:

$\Vert f \left({x}\right) \Vert \le c \cdot \Vert g \left({x}\right) \Vert$

for all $x$ in a neighborhood of infinity.


Dependence on Parameters

Let $X$ be a topological space.

Let $V$ be a normed vector space over $\R$ or $\C$ with norm $\left\Vert{\,\cdot\,}\right\Vert$

Let $x_0\in X\cup\{\infty\}$.

Let $A$ be a set.

Let $X_\alpha$ be a subset of $X$ for every $\alpha \in A$, which, if $x_0\neq\infty$, contains $x_0$.

Let $f_\alpha : X_\alpha\setminus \{x_0\} \to V$ be a function for every $\alpha \in A$.

Let $g : X \to V$ be a function.

The statement:

$f_\alpha = O_\alpha(g)$ as $x \to x_0$

is equivalent to:

$\forall \alpha \in A : f_\alpha = O(g)$ as $x \to x_0$


The $O$-estimate is said to be independent of $\alpha \in A$ if:

There exists a neighborhood $U$ of $x_0$ in $X$ such that $\displaystyle \exists c\in \R: c\ge 0 : \forall \alpha \in A : \forall x\in (U\setminus\{x_0\}) \cap X_\alpha : \Vert f_\alpha(x)\Vert \leq c\cdot\Vert g(x)\Vert$

That is, if the implied constant and implied neighborhood can be chosen the same for all $\alpha \in A$.


Uniform Estimates

Let $X$ be a set.

Let $V$ be a normed vector space over $\R$ or $\C$ with norm $\left\Vert{\,\cdot\,}\right\Vert$.

Let $f,g : X \to V$ be mappings.


Then $f$ is big O of $g$ uniformly if and only if:

$\exists c>0 : \forall x \in X : \Vert f(x) \Vert \leq c \cdot \Vert g(x) \Vert$

This is denoted: $f=O(g)$.


Also denoted as

In analytic number theory, sometimes Vinogradov's notations $f \ll g$ or $g \gg f$ are used to mean $f = \mathcal O \left({g}\right)$.

This is clearer for estimates leading to typographically complex error terms.


Some sources use an ordinary $O$:

$f = O \left({g}\right)$


Also defined as

Some authors require that the inequality be valid on the entire domain of definition.

On $\mathsf{Pr} \infty \mathsf{fWiki}$, this is known as a uniform big-O estimate.


The statement $f = \mathcal O \left({g}\right)$ is sometimes seen to be defined as:

$\displaystyle \exists \alpha \in \R: \alpha \ge 0 : \lim_{x \to \infty} \frac{f \left({x}\right)}{g \left({x}\right)} = \alpha$

But requiring that the limit exists is too restrictive.


Implied Constant

From the definition of the limit of a function, it can be seen that this is also equivalent to:

$\exists c \in \R: c > 0, k \ge 0: \forall n > k, f \left({n}\right) \le c g \left({n}\right)$

For some fixed $k$ (appropriate to the function under consideration) the infimum of such $c$ is called the implied constant.


Also see


Sources