数值最优化(李董辉)课后答案_第2章.pdf

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Y 9

∇f (x∗) = (0, 0)T −400 ∇f (x) = 8 + 2x1 200 ! x∗ = (1, 1)T f (x) kgV > ∇2f (x) = 802 −400 _0℄v 3;gV f (x)jUbW x = (−4, 3)T .kgV gV ∇2f (x) x = (−4, 3)T f (x)va ℄v 4;E x∗ f (x) 4℄v U (x∗)G j if x∗V f (x)_0℄v 12 − 4x2 ! =⇒ ( x1 = −4 ( 8 + 2x1 = 0 12 − 4x2 = 0 ∇2f (x) = 2 0 −4 ! 0 ∀x ∈ U (x∗), f (x) > f (x∗) x2 = 3 2

S = {xn}, {n = 1, · · · , k} f (x1) = f (x2) = · · · = f (xk) = f (x∗) 5;ET! f (x)℄=w bV x∗,i3 2.1.5 x∗V f (x)0yh. ∀x ∈ Dj f (x) ≥ f (x∗) R℄"V 2 j kgV SVT f (αxi + (1 − α)xj) ≤ αf (xi) + (1 − α)f (xj) = f (x∗) ∀α ∈ [0, 1], ∀xi, xj ∈ S αxi + (1 − α)xj ∈ D f (αxi + (1 − α)xj) ≥ f (x∗) f (αxi + (1 − α)xj) = f (x∗) αxi + (1 − α)xj ∈ S 3

f (xk) = |xk| f (xk+1) =   1 2 |xk + 1| xk > 1 1 2 |xk| xk ≤ 1 1 < (xk + 1) < 1 2 (xk + xk) = xk f (xk+1) < xk = f (xk) f (xk+1) = 1 2 |xk| ≤ |xk| = f (xk) f (xk+1) ≤ f (xk) 1 2 6;gV (1) xk > 1F f (xk) = xk (2) xk ≤ 1F 7; (1) gV > f (xk) = 1 2 xT k xk = 1 2 (1 + 1 2k )2, k = 0, 1, · · · f (xk+1) = 1 2 (1 + 1 2k+1 )2, k = 0, 1, · · · 4

1 2k+1 1 2 (1 + 1 2k+1 )2 f (xk) > f (xk+1) > 1 2 (1 + 1 2k 1 2k )2 > kgV (2) {x|kxk2 = 1}DAeb x0,j xT kgV 2 x0 {xk}1 8; (1) kgV xk = (1 + lim k→∞ lim k→∞ 0 x0 = 1 1 2k ) cos k sin k ! xT k xk = (1 + xT k xk = lim k→∞ (1 + 1 2k )2 1 2k )2 = 1 xk = x0 ∇f (x) = 2(x1 + x2 2) 4x2(x1 + x2 2) ! = 2 0 ! ∇f (x)T p = (2, 0)(−1, 1)T = −2 < 0 5

f (x + αp) = ((1 − α) + α2) = α4 − 2α3 + 3α2 − 2α + 1 a0 = 0, b0 = 1, ε = 0.01 [0, 1] 0.382 0.618 f (uk) f (vk) 0.512 0.292 uk vk (2) 0 ... k [ak, bk] p xbZ\ ? 7 9. (1) kgV d0 f (x) x0Z\ (2)gV Armijo[^`:xSV f (x0 + αd0) − f (x0) = 6 ∇f (x) = (x1, 4x2)T , ∇f (x0) = (1, 4)T ∇f (x0)T d0 = (1, 4)(−1, −1)T = −5 < 0 x0 + αd0 = (1 − α, 1 − α)T (1 − α)2 − 5 2 5 2 = 5 2 α2 − 5α

5 2 0 < α ≤ = 0.2 1 5 10 α2 − 5α ≤ 0.9α∇f (x0)T d0 α0 = 0.53 = 0.125 11i#/Bd7 12;i k = 0F k = 1F 6 c1 = max{1, c},Ae kj k∇f (xk)k ≤ c1kdkk k∇f (x0)k · k∇f (x1)k ≤ ckd0kkd1k k∇f (xi)k ≤ ckdik, i = 1, 2, · · · k∇f (x1)k ≤ ckd1k k∇f (x0)k ≤ kd0k k∇f (xi)k ≤ ck kdik k Yi=0 k Yi=0 ... 7

mkdk2 ≤ dT Bkd ≤ M kdk2 mkdk2 ≤ kdT Bkdk ≤ M kdk2 i3 2.4.4,2 8 4 13;gV kgV > αki Armijoi Wolfe-Powelli3 2.4.2,3 2.4.3 kgV kgV k∇f (xk)k = kBkkkdkk ≤ M kdkk ∇f (xk)T dk k∇f (xk)kkdkk k∇f (xk)k2cos2 θk < ∞ Bkdk + ∇f (xk) = 0 ∞ Xk=0 cos θk = − (1) (2) m ≤ kBkk ≤ M ∇f (xk) = −Bkdk Bkd + ∇f (xk) = 0 dT BT k d + ∇f (xk)T d = 0 8

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