最优化理论与算法(16)

4/9 2t

§ (t)=f(x(1)+td(1))=(4/9 2t)2+2( 1/9+

1/9+t

t)2§ (t)= 4(4/9 2t)+4( 1/9+t)= 20/9+12t"- (t)=0 (t) ²-:t1=5/27"du (t1)=12>0§Ïdt1=5/27´min (t)=f(x(1)+td(1)) `)"

t≥0 4/9 10/272/27

# S“:µx(2)=x(1)+t1d(1)=="

1/9+5/272/27

14/274√4(2)(2)

≥ε§ d(2)=§ f(x) =1ngS“µ f(x)==

272278/27

1

" 2

2/27 t

‘|¢µx(2)+td(2)=§ (t)=f(x(2)+td(2))=(2/27 t)2+2(2/27

2/27 2t

2t)2§ (t)= 2(2/27 t) 8(2/27 2t)= 20/27+18t"- (t)=0 (t) ²-:t2=10/243"du (t2)=18>0§Ïdt2=10/243´min (t)=f(x(2)+td(2)) `)"

t≥0 2/27 10/2438/243

# S“:µx(3)=x(2)+t2d(2)=="

2/27 20/243 2/243

‘|¢µx(1)+td(1)=1ogS“µ f(x(2))=

8/243

"

2/243

16/243 8/243

=

8243

28√

§ f(x(3)) =<ε§ x =x(3)=

243 1

n! ÛÂñ5

eü{3 ½^ eäk ÛÂñ5"

½n3.3.5 f:Rn→R gëY § 2f(x)k.§f(x)ke.§x(0)∈Rn"^ eü{¦

)(UNP)§Ù¥ε=0§¿… 3~êδ>0§¦(3.3.1)é k¤á§

£1¤e )k S x(1),···,x(k)§K f(x(k))=0¶£2¤e )à S x(1),x(2),···§Klim f(x(k)) =0"

k→∞

y²£1¤ â { ª O^ = "

£2¤ék=0,1,···§ â { ª O^ §d(k)= f(x(k))=0§x(k+1)=x(k)+tkd(k)§Ù¥tk¦(3.3.1)¤á"

Äk§du f(x(k))Td(k)= f(x(k)) 2<0§ âíØ ^ (3.3.1)(¢¤á"d(k)´f3x(k)

? eü §=é¿© t>0§k

f(x(k)+td(k))<f(x(k)),

Ïd

f(x(k+1))=f(x(k)+tkd(k))=minf(x(k)+td(k))<f(x(k)),

t≥0