Difference between revisions of "Euclidean norm"

Latest revision as of 07:23, 27 April 2015

This is an example of a Norm

Definition

The Euclidean norm is denoted [math]\|\cdot\|_2[/math] is a norm on [math]\mathbb{R}^n[/math]

Here for [math]x\in\mathbb{R}^n[/math] we have:

[math]\|x\|_2=\sqrt{\sum^n_{i=1}x_i^2}[/math]

Proof that it is a norm

TODO: proof

Part 4 - Triangle inequality

Let [math]x,y\in\mathbb{R}^n[/math]

[math]\|x+y\|_2^2=\sum^n_{i=1}(x_i+y_i)^2[/math] [math]=\sum^n_{i=1}x_i^2+2\sum^n_{i=1}x_iy_i+\sum^n_{i=1}y_i^2[/math] [math]\le\sum^n_{i=1}x_i^2+2\sqrt{\sum^n_{i=1}x_i^2}\sqrt{\sum^n_{i=1}y_i^2}+\sum^n_{i=1}y_i^2[/math] using the Cauchy-Schwarz inequality

[math]=\left(\sqrt{\sum^n_{i=1}x_i^2}+\sqrt{\sum^n_{i=1}y_i^2}\right)^2[/math] [math]=\left(\|x\|_2+\|y\|_2\right)^2[/math]

Thus we see: [math]\|x+y\|_2^2\le\left(\|x\|_2+\|y\|_2\right)^2[/math], as norms are always [math]\ge 0[/math] we see:

[math]\|x+y\|_2\le\|x\|_2+\|y\|_2[/math] - as required.