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By Sven O. Krumke

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45) Φk (0) − Φk (sk ) ≥ σ max {−μ1 , 0} Δ2k . for some 0 < τ < 1 and 0 < σ < 1. 32) satisfies the above inequalities with τ = σ = 1/2. 23 Let f ∈ C2 (Rn ) and lub2 (∇2 f(x)) ≤ M for all x ∈ Rn . 6 on page 38: 1. We have c0 > 0 and the accuracy parameter ε to is set to ε := 0. 2. 44) Φk (0) − Φk (sk ) ≥ τ gk min Δk , gk lub2 (Bk ) . 3. The matrices Bk = BTk have bounded norms: lub2 (Bk ) ≤ M for all k. 4. We have infk f(x(k) ) > −∞. Then limk→ ∞ g(x(k) ) = 0, in particular, every accumulation point of (x (k) )k is a stationary point of f.

2) where g(x) = ∇f(x). 3. 2). If g(x) = 0 and Dg(x)−1 exists, then the Newton step Δx at x is given by Δx := −Dg(x)−1 g(x). 3) Δx = −(∇2 f(x))−1 ∇f(x). The resulting algorithm, Newton’s Method, should be well-known to you. 1 for completeness. 4) 1 Φ(s) := f(x) + ∇f(x)T s + sT ∇2 f(x)s 2 at x. If we search for a stationary point of Φ the equation ∇Φ(s) = 0 leads to s = −(∇2 f(x))−1 ∇f(x). 3) as a solution. 1 the linearization of the gradient implies the locally quadratic convergence of Newton’s method, it is possible to derive some descent properties from the approximation of f by the quadratic model above.

Since f is strictly convex, H := H(x) is positive definite for every x ∈ Rn and z H := (zT H(x)z)1/2 is a norm for every x ∈ Rn . If we linearize f, f(x + s) ≈ l(s) := f(x) + ∇f(x)T s, then on the boundary of the ellipse s H ≤ Δ the difference between f(x + s) and l(s) is constant File: Ò ÛØÓÒºØ Ü Revision: ½º¾¼ Date: ¾¼¼ »¼ »¿¼ ¼ ½¾ ÅÌ 49 50 Newton-Like Methods in first-order approximation (the error is Δ2 /2). Thus, if one optimizes the linearization of f or the quadratic approximation on the ellipse, one obtains the same search direction.

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