最优化理论与算法(19)

Äk§ f(x)=

2x14x2

§ 2f(x)=

1

201/20

= ½§ 2f(x)"

0401/4

1 gS“µ

√21

f(x(0))==2§ f(x(0)) =2≥ε§d(0)= [ 2f(x(0))] 1 f(x(0))=

42

1/202 10 =§x(1)=x(0)+d(0)="

01/44 10

1 gS“µ

00

f(x1)=§ f(x(1)) =0<ε§ `)x =x(1)="

00

~3.4.2´ à åà g5y¯K§^Newton{¦) §²L gS“= °(

`)"Ù¢§éu à åà g5y¯K(UQP)§^Newton{¦) k Ó (ا=Newton{äk gª 5"

½Â3.4.1e {^u¦)î à g¼ê4 ¯K §l?¿Ð© Ñu§ {²Lk gS“ ¼ê 4 :§K¡T {äk gÂñ5"

3 ½^ e§eЩ:3 `)NC§KNewton{äk Âñ Ý"

½n3.4.3 f3x ∈Rn , S gëY §…x ÷v f(x )=0, 2f(x ) ½§K 3~êδ>0§¦

x(0)∈Uδ(x ) {x| x x <δ} §Úî{

x

(k+1)

(k)

=x f(x

2(k)

1) f(x(k)),

k=0,1,···

) : {x(k)} 5Âñux "e 2f3x ?LipschitzëY§= 3~êL>0§¦

2f(x) 2f(x ) ≤L x x ,

x∈Uδ(x ),

K{x(k)} gÂñux §= 3~êC>0§¦ k¿© §

x(k+1) x ≤C x(k) x 2.

y²df gëY ! 2f(x ) ½

δ1>0,¦ 2f(x) — ½, x∈Uδ1(x ),

M>0,¦ 2f(x) 1 ≤M, x∈Uδ1(x ),

1

, x∈Uδ(x ). δ≤δ1,¦ 2f(x) 2f(x ) ≤4M

x(0)∈Uδ(x ) §k

x(1) x = x(0) x 2f(x(0)) 1 f(x(0))