Suggest edit — asymtotic analysis

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---
title: "asymtotic analysis"
type: concept
source: https://www.jemoka.com/posts/kbhasymtotic_analysis/
confidence: high
status: active
---

Intuition:
\(O\): \(\leq\) \(\theta\): \(=\) \(\Omega\): \(\geq\) Definitions:
\(f(n) = O(g(n)) \implies \exists n_{0}: \forall n &gt; n_0, f(n) \leq c (g(n))\) \(f(n) = \Omega(g(n)) \implies \exists n_{0}: \forall n &gt; n_0, f(n) \geq c (g(n))\) \(f(n) = \theta(g(n)) \implies \exists n_{0}: \forall n &gt; n_0, f(n) \geq 1 (g(n)), f(n) \leq c (g(n))\) ~ Given functions \(f(n)\) and \(g(n)\), if:
\begin{equation} \lim_{n\to \infty} \left(\frac{f(n)}{g(n)}\right) = 1 \end{equation}
we say that \(f \sim g\).
That &ndash; the relationship between \(f\) and \(g\) grows in a similar fashion as \(n\) increases. For instance:
\(f(n) = n+1\) \(g(n) = n+2\) Therefore:
\begin{equation} f\sim g = \lim_{n\to \infty} \frac{f(n)}{g(n)} = \lim_{n\to \infty} \frac{n+1}{n+2} = 1 \end{equation}
The \(\sim\) operator is commutative (\(f \sim g \Rightarrow g\sim f\)) and transitive (\(f\sim g, g\sim h \Rightarrow f \sim h\)).
o(n) Given two functions \(f(n)\), \(g(n)\), if their relationship shows:
\begin{equation} \lim_{n \to \infty} \frac{f(n)}{g(n)} = 0 \end{equation}
we can write it as
\begin{equation} f = o(g) \end{equation}
This tells us that if \(n\) becomes very large, \(g\) becomes much larger than \(f\). \(f\) does not grow nearly as fast as \(g\).
The operation is not commutative, but is transitive (\(f = o(g), g = o(h) \Rightarrow f = o(h)\))
O(n) Given two functions \(f(n)\), \(g(n)\).
\begin{equation} \lim_{n \to \infty} \frac{f(n)}{g(n)} &lt; \infty \end{equation}
that the relationship between \(f(n)\) and \(g(n)\) is countable as \(n\) trends to infinity.
We can also say that, given \(n\), \(n_0\), and some \(c\) which \(\forall n, n &gt; n_0\), there is:
\begin{equation} |f(n)| &lt; |cg(n)| \end{equation}
This tells us that \(f(n)\) does not grow much much faster than \(g(n)\).
Therefore:
If \(f \sim g\), \(f = O(g)\) (as they grow together, \(f\) is not much faster) If \(f = o(g)\), \(f=O(g)\) (as \(f\) does not grow at all, \(f\) is not faster) \(\theta\)(n) \(f=\theta(g)\) IFF \(f=O(g)\) and \(g=O(f)\), its essentially \(\sim\) but without the strict requirement of a 1:1 ratio.
\(\omega\)(n) and \(\Omega\)(n) The inverses of \(O\) and \(o\):
\(f(n) = O(g(n)) \Rightarrow g(n) = \omega(f(n))\) \(f(n) = o(g(n)) \Rightarrow g(n) = \Omega(f(n))\)