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# System prepended metadata

title: Untitled

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$\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$
$\newcommand{\deriv}[1]{\frac{\partial f}{\partial \phi_{#1}}}$
$\newcommand{\result}[0]{(E^TE)^{-1}E^Tx}$

On note $f= \norm{x - E\phi}^2_2$
On constate que $f = \sum_{j=1}^{N}{(x_j - \sum_{i=1}^{M}{\phi_{i}E_{ij}})}$

Calculons $\deriv{1}$:
$$\deriv{1} = -2\sum_{j=1}^N{E_{1j}(x_j - \sum_{i=1}^m{\phi_{i}E_{ij}})}\\
\deriv{1} = -2(\underbrace{\sum_{j=1}^N{E_{ij}x_j}}_{<E_1, x>} -
\underbrace{\sum_{j=1}^{N}E_{1j}\sum_{i=1}^m{\phi_{i}E_{ij}}}_{<E_1, E\phi>})\\
\deriv{1} = -2(E_1^{T}x - E_1^{T}E\phi)\\
\boxed{\deriv{1} = 2E_1^{T}(E\phi - x)}
$$

D'où on obtient:
$$\deriv{} = 2E^{T}(E\phi - x)$$

On cherche les extrêmum de $f$, autrement dit:
$$\deriv{}=0\\
2E^T(E\phi - x) = 0\\
E^TE\phi = E^Tx\\
\boxed{\phi = \result}$$

$E$ étant défini positive, on obtient, de même:
$$\deriv{} < 0 \implies \phi < \result$$
$$\deriv{} > 0 \implies \phi > \result$$

D'où $\result$ le minimum global de la fonction convexe $f$.