1 - Growth of Functions

1 - Growth of Functions#

Big-O Notation#

Big-O Notation describes the tight upper bound:

\[ O(g(n)) = \begin{cases} f(n): \exists \text{ positive constants } c, n_0 \text{ s.t. } 0 \leq f(n) \leq c g(n), \forall n \geq n_0 \end{cases} \]

\(f(n)\) is \(O(g(n))\) iff \(\exists\) positive constants \(c\), \(n_0\) such that \(f(n) \leq cg(n) \ \forall n \geq n_0\).

  • This just means that \(27n^2 + 10n - 5\) is \(O(n^2)\)

Note that Little-o Notation describes the loose upper bound.

Big-Omega Notation#

Big-Omega Notation describes the tight lower bound:

\[ \Omega(g(n)) = \begin{cases} f(n) : \exists \text{ positive constants } c, n_0 \text{ s.t. } 0 \leq cg(n) \leq f(n), \forall n \geq n_0 \end{cases} \]
\[\begin{split} \begin{align*} 27n^2 + 10n - 5 &\leq cn^2 \\ 27 + \frac{10}{n} - \frac{5}{n^2} &\leq c, \text{ for } n \geq 10, \frac{10}{n} \leq 1 \\ 28 &\leq c \end{align*} \end{split}\]

Note that Little-omega Notation describes the loose lower bound.

\(27n^2 + 10n - 5\) is \(\Omega(n^2)\)

\[\begin{split} \begin{align*} 27n^2 + 10n - 5 &\geq cn^2 \\ 27 + \frac{10}{n} - \frac{5}{n^2} &\geq c, \text{ for } n \geq 1, \frac{5}{n^2} \leq 5 \\ 22 &\geq c \end{align*} \end{split}\]

\(13n + 8\) is \(O(n^2)\)

\[\begin{split} \begin{align*} 13n + 8 &\leq cn^2 \\ \frac{13}{n} + \frac{8}{n^2} &\leq c \end{align*} \end{split}\]

\(13n + 8\) is \(\Omega(n^2)\)

\[\begin{split} \begin{align*} 13n + 8 &\geq cn^2 \\ 13 + \frac{8}{n} &\geq cn \end{align*} \end{split}\]

This is not true, hence why it is important to care about both the upper and lower bounds.

\[\begin{split} \begin{align*} an^2 + bn + c \text{ is }& \Theta(n^2) \\ c_1n^2 \leq an^2 + bn + c &\leq c_2n^2 \\ c_1 \leq a + \frac{b}{n} + \frac{c}{n^2} &\leq c_2, \text{ let } c_1 = \frac{a}{4}, c_2 = \frac{7a}{4}, n_0 = 2 \max\left(\frac{|b|}{a}, \sqrt{\frac{|c|}{a}}\right) \\ c_1 \leq a + \frac{b}{2\frac{|b|}{a}} + \frac{c}{4 \frac{|c|}{a}} &\leq c_2 \\ c_1 \leq a \pm \frac{a}{2} \pm \frac{a}{4} &\leq c_2 \\ \end{align*} \end{split}\]