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# 34 45 steepest gradient descent convergence let f rn r

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Unformatted text preview: 0 f ( x0 + α d 0 ) = f = 0 + 2(−3α)2 + 0 + 3(−3α) = −3α 18α2 − 9α d 2 d α (18α ￿ α0 = 1 4 ￿ ￿ x1 = ￿ − 9α) = 36α − 9 = 0 0 −3 4 ￿ 33 / 45 Steepest Gradient Descent Example ￿ ￿x0 − x1 ￿2 = 3 4 &gt; ￿ so we continue Second Iteration ￿ ￿ 2x 1 + x 2 ￿ ∇ f ( x) = 4x 2 + x 1 + 3 ￿ ￿ ￿3￿ 3 2(0) − 4 1 1) = − 4 ￿ d = −∇f (x = 0 4(− 3 ) + 0 + 3 ￿3 ￿4 9 9 4α ￿ f ( x1 + α d 1 ) = f = 16 α2 + 9 − 16 α − 9 8 4 −3 4 d 92 d α ( 16 α ￿ α1 = 1 2 ￿ ￿ x2 = ￿ + 3 8 −3 4 ￿x1 − x2 ￿2 = 3 8 9 8 − 9 16 α − 9) = 9α − 4 8 9 16 =0 ￿ &gt; ￿ so we would keep going. 34 / 45 Steepest Gradient Descent Convergence Let f : Rn → R. If for some x0 : ￿ ￿ The sublevel set X0 = {x ∈ R : f (x) ≤ f (x0 } is bounded; and f is continuously diﬀerentiable on the convex hull of X0 then the steepest gradient descent algorithm initiated at x0 will converge to a solution for the KKT conditions for f Recall from last week that a solution for the KKT conditions can be either a local or a global optimum. But if your optimization problem is convex, then it is guaranteed to be...
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