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For this equation :

$$\parallel x - x^\prime||^2_2$$

From What does double vertical-line means in linear algebra? I understand what $\parallel$ signify.

What does 2 represent in this problem? The upper 2 represents an exponent? What does 2 at lower position represent?

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    $\begingroup$ The upper 2 is an exponent. $||\cdot||_2$ is the Euclidean norm, so altogether it just means $\sum_i{(x_i-x'_i)^2}$ $\endgroup$ – saulspatz Apr 3 '18 at 8:25
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The $2$ at the upper position is just an exponent. The $2$ at the lower position represents the Euclidean norm (frequently called $2$-norm). It is defined as $$\Vert x \Vert_2 = \left( \sum_{i = 1}^n \vert x_i \vert^2\right)^{1/2} \qquad \text{for all } x \in \mathbb R^n.$$ One can generalize this construction for $1 \leq p < \infty$ to get the $p$-norm defined by $$\Vert x \Vert_p = \left( \sum_{i = 1}^n \vert x_i \vert^p\right)^{1/p} \qquad \text{for all } x \in \mathbb R^n.$$ Due to this generalization one calls the Euclidean norm frequently the $2$-norm. I hope things got clearer for you :)

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