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[IML] early sections update
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\textbf{Problem}: Finding $\hat{w}$ for $l$ with no closed form solution.\\
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\subtext{or if the closed form solution is too expensive to compute.}
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\textbf{Solution}: Iterative optimization methods.
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\begin{algorithm}
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\caption{Iterative Optimization}
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$t \gets 0$ \;
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$w^{(0)} \gets w_\text{initial}$ \;
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\SetKwRepeat{Do}{repeat}{until}
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\Do{\text{Stopping Criterion}}{
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$w^{(t+1)} \gets w^{(t)} + \tilde{\eta}_t v^{(t)}$ \;
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$t \gets t+1$
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}
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\textbf{return} $w^{(t)}$
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\end{algorithm}
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{\footnotesize
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\notation The update takes the form $\tilde{\eta}_t v^{(t)}$. $v^{(t)}$ is the update direction, $\tilde{\eta}_t$ is the step size.
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}
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\subsection{Gradient Descent}
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Intuitively: go in the direction $v^{(t)}$ where $L$ decreases most.
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\lemma $-\nabla L(w^{(t)})$ is the direction of steepest descent.\\
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\subtext{Assuming diff.-able $L$. Provable via Taylor expansion \& Cauchy-Schwarz.}
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\definition \textbf{Gradient Descent Update Step}
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\begin{align*}
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w^{(t+1)} &= w^{(t)} - \tilde{\eta}_t\cdot \frac{\nabla L(w^{(t)})}{\Vert \nabla L(w^{(t)}) \Vert} & (\text{Normalized}) \\
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w^{(t+1)} &= w^{(t)} - \eta\cdot \nabla L(w^{(t)}) & (\text{Unnormalized})
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\end{align*}
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Unnormalized gradient descent takes advantage of $\Vert \nabla L(w^{(t)}) \Vert$:
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{\small
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\begin{itemize}
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\item $\Vert \nabla L(w^{(t)}) \Vert$ small $\mapsto$ close to stat. point $\mapsto$ small steps.
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\item $\Vert \nabla L(w^{(t)}) \Vert$ large $\mapsto$ far from stat. point $\mapsto$ large steps.
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\end{itemize}
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}
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Stopping criterion uses the same idea: $\Vert w^{t} - w^{t+1} \Vert < \epsilon$ or equivalently $\Vert \nabla L(w^{(t)}) \Vert < \epsilon$.
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\begin{algorithm}
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\caption{Gradient Descent}
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$t \gets 0$ \;
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$w^{(0)} \gets w_\text{initial}$ \;
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\SetKwRepeat{Do}{repeat}{until}
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\Do{\text{$\Vert w^{t} - w^{t+1} \Vert < \epsilon$}}{
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$w^{(t+1)} \gets w^{(t)} - \eta \nabla L(w^{(t)})$ \;
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$t \gets t+1$
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}
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\textbf{return} $w^{(t)}$
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\end{algorithm}
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