8.1: Big O
- Page ID
- 28955
Big O
The idea of Big O is to characterize functions according to their growth rates. The O refers to the order of a function. In computer science, Big O is used to classify algorithms for their running time or space requirements.
Notice in the figure below that \(f(x)>g(x)\) right before \(x=1\). However, for \(x>1\), we see that \(g(x)>f(x).\) In the long run (namely after \(x>1\)) \(g(x)\) overtakes \(f(x).\)
We say "\(f\) is of order at most \(g\)" or "\(f(x)\) is Big O of \(g(x)\)" .
We write: \(f(x)= O\big(g(x)\big).\)
In our definition of Big O notation, there are certain parameters.
- We use \(x\) greater than a certain initial value, \(n\); in the diagram above, \(n=1\).
- We use absolute value for both functions.
- We use \(k\) as a constant multiplied by the function inside the O.
Definition: Big O Notation
\[f(x)= O\big(g(x)\big).\]
if and only if there exist real numbers \(k,n \) with \(k>0 , n\geq 0 \) such that
\[|f(x)| \leq k|g(x)| \qquad \forall x>n.\]
Example \(\PageIndex{1}\)
Take this statement and express it in Big O notation: \(|7x^5+4x^3+x| \leq 14|x^5|\) for \(x>1.\)
- Solution
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\((7x^5+4x^3+x) \mbox{ is } O(x^5)\)
Comparing orders of common functions
A constant function, such as \(f(x)=6\) does not grow at all. Logarithmic functions grow very slowly. Here is a list of some common functions in increasing order of growth rates.
constant function, logarithmic function, polynomial function, exponential function
Example \(\PageIndex{2}\)
Put these functions in order of increasing growth rates:
\[\log_6x,\qquad x^5,\qquad 2^x,\qquad x^2,\qquad \log_{15}x,\qquad 100x^4,\qquad 64x+1000,\qquad x^5 \log_6x,\qquad 5^x, \qquad 6\]
- Solution
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\[6, \qquad \log_6x,\qquad \log_{15}x,\qquad 64x+1000,\qquad x^2,\qquad 100x^4,\qquad x^5,\qquad x^5 \log_6x, \qquad 2^x,\qquad 5^x\]
Proofs
We will be using the Triangle Inequality Theorem which is \[|x+y|\ \leq |x|+|y|.\]
Example \(\PageIndex{3}\)
Prove: \(4x^3-11x^2+3x-2=O(x^3)\)
- Proof
-
Choose \(n=1\), i.e. \(x \geq 1.\)
\(|4x^3-11x^2+3x-2 |\leq |4x^3|+|-11x^2|+|3x|+|-2| \qquad \mbox{ by the Triangle Inequality Theorem}\)
\(\qquad \qquad \qquad \qquad \qquad=4x^3+11x^2+3x+2 \qquad \mbox{ applying absolute value; note: }x \mbox{ is positive}\)
\(\qquad \qquad \qquad \qquad \qquad\leq 4x^3+11x^3+3x^3+2x^3 \qquad \mbox{ since }x \mbox{ is positive and greater than 1}\)
\(\qquad \qquad \qquad \qquad \qquad =20x^3\)
\(\qquad \qquad \qquad \qquad \qquad=20|x^3|\qquad \mbox{ since }x \mbox{ is positive and greater than 1}\)
Thus for all \(x \geq 1,\) \(|4x^3-11x^2+3x-2 | \leq 20|x^3|\)
Therefore, using \(n=1\) and \(k=20\), \(4x^3-11x^2+3x-2=O(x^3)\) by the definition of Big O.
Summary and Review
- Big O is used to compare the growth rates of functions.
- Be sure to understand the examples here.
Exercises
exercise \(\PageIndex{1}\)
True or False?
(a) \(11x^3=O(87x^2)\)
(b) \(x^{13}=O(3^x)\)
(c) \(-2x=O(58\log_{35}x)\)
- Answer
-
(a) false
(b) true
(c) false
Exercise \(\PageIndex{2}\)
True or False?
(a) \(4x^3+12x^2+36=O(x^3)\)
(b) \(.01x^5=O(48x^4)\)
(c) \(4^x=O(x^7)\)
(d) \(3x\log_2x=O(25x)\)
Exercise \(\PageIndex{3}\)
True or False?
(a) \(23\ln x=O(3x)\)
(b) \(7x^5=O(x^5)\)
(c) \(x^5=O(7x^5)\)
- Answer
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all true
Exercise \(\PageIndex{4}\)
Prove: \(2x^5+3x^4-x^3+5x=O(x^5)\)
Example \(\PageIndex{5}\)
Put these functions in order of increasing growth rates:
\[x^7,\qquad 6^x,\qquad 78x^2,\qquad x^2\log x,\qquad 1000x,\qquad 7, \qquad \log_{11}x\]
- Answer
-
\[7, \qquad \log_{11}x, \qquad 1000x, \qquad 78x^2,\qquad x^2\log x, \qquad x^7,\qquad 6^x\]
Exercise \(\PageIndex{6}\)
Take this statement and express it in Big O notation: \(|2x^4-5x^3+x^2-5| \leq 13|x^4|\) for \(x>1.\)