Ларин Р.М., Пяткин А.В., Плясунов А.В. Методы оптимизации примеры и задачи
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°
Z
E
n
n
x =
x
1
x
2
. . .
x
n
E
n
x
>
= (x
1
, x
2
, . . . , x
n
) >
{x ∈ X| Q} X Q X = E
n
{x | Q}
[x, y] = {z | z = αx + (1 − α)y, 0 ≤ α ≤ 1} x y
(x, y) = [x, y] \ {x, y} = {z | z = αx + (1 − α)y, 0 < α < 1}
(x, y] [x, y)
hx, yi = x
>
y =
P
n
i=1
x
i
y
i
x y
kxk =
p
hx, xi x
ρ(x, y) = kx − yk x y
U
ε
(x) = {y | ρ(x, y) ≤ ε} ε x ε
x
X ⊂ E
n
x ∈ E
n
x X
ε > 0 X ∩ U
ε
(x) X
X X x
X ε
X x X ε > 0,
U
ε
(x) ⊂ X X
X X x ∈ X
X
X X = X \ X
X
x X
x
1
, x
2
∈ X, x
1
6= x
2
x = (x
1
+ x
2
)/2
x y E
n
x ≥ y E
n
x
i
≥ y
i
, (i = 1, n)
”i = 1, n” i ∈ {1, . . . , n}
x > y x
i
> y
i
, (i = 1, n)
x 6= y x
i
6= y
i
i
f
0
(x) f(x) x
(f
0
(x))
>
= (
∂f (x)
∂x
1
,
∂f (x)
∂x
2
, . . . ,
∂f (x)
∂x
n
).
X
X ⊂ X
f : X −→ E
1
f(x) −→ min
x∈X⊂X
(f(x) −→ max
x∈X⊂X
).
X
f(x) −→ min
x∈X
(f(x) −→ max
x∈X
). (1)
x
∗
∈ X f(x
∗
) ≤ f(x) x ∈ X
f(x
∗
) = min
x∈X
f(x);
x
∗
f
∗
= inf
x∈X
f(x);
f(x) X, f
∗
= −∞;
X = ∅
max
x∈X
f(x) = −min
x∈X
(−f(x)),
f(x) −→ extr
X
f(x)
X
X = E
n
X
X = {x | ϕ
j
(x) ≤ 0, j = 1, m}, (2)
ϕ
j
(x), j = 1, m
ϕ
j
(x) ≤ 0,
f(x)
x
∗
f(x) X
x
∗
= {f(x) | x ∈ X}
X
∗
= {f(x) | x ∈ X}
X
∗
= {x
∗
∈ X | f(x
∗
) = min
x∈X
f(x)}.
x ∈ X f
ε > 0, f(x) ≤ f(y) y ∈ X ∩U
ε
(x)
X = E
n
X
X E
n
X = {x | − x
2
≤ 0}
f(x)
X
x
0
x
00
X f(x
0
) ≤ f(x) ≤ f(x
00
) x ∈ X.
a b f(x, y) = x + ay
X = {(x, y) | bx
2
− 2xy + y
2
≤ 1}
f a X = {(x, y) | (x −
y)
2
+ (b − 1)x
2
≤ 1} b > 1 X
f a = −1 b = 1 f(x, y) = x − y
X
=
{
(
x, y
)
| −
1
≤
(
x
−
y
)
≤
1
} −
1
a b
b ≤ 1 a 6= −1 y = x
x f(x, x) = (a + 1)x a < −1,
f(x, x) −→ −∞ x −→ +∞ a > −1, f(x, x) −→ −∞ x −→ −∞
b < 1, a = −1 y = (1 −
√
1 − b)x
(x −y)
2
+ (b −1)x
2
= (1 −b)x
2
+ (b −1)x
2
= 0
x f(x, y) = x − y =
√
1 − bx −→ −∞ x −→ −∞
f X b > 1
a b = 1 a = −1
f(x) g(x) X
h(x) = f(x) + g(x)
a f(x) = (a + 1)x + a ln x −2 sin x
a f(x, y) = (a −2)(a −3)e
x
+ |y + 10|/(y
2
+ 1)
X = {(x, y) | (a − 4)x
2
+ y
2
≤ 1}
f(x) −→
x∈X
,
X ⊂ E
1
f(x)
X x
∗
f
0
(x
∗
) = 0
f(x) X ⊂ E
1
X
1
= {x ∈ X | f
0
(x) = 0}
X
2
= {x ∈ X | f (x) }
X
3
= {x ∈ X | f
0
(x) }
±∞
X
4
= FrX = X \ IntX
X X
f(x) = x
2/3
− (x
2
− 1)
1/3
f
0
(x) =
2
3
(x
2
− 1)
2/3
− x
4/3
x
1/3
(x
2
− 1)
2/3
x = 0 x = ±1
x = ±
√
2/2
X
1
= {−
√
2/2,
√
2/2} X
3
= {−1, 0, 1}
X
2
= ∅ X = E
1
lim
x→+∞
f(x) = lim
x→−∞
f(x) = 0.
f(−1) = f(0) = f(1) = 1, f(−
√
2/2) =
f(
√
2/2) = 2/
3
√
2 f(x)
x = ±
√
2/2 inf f(x) = 0
f(x) f(x) = f(−x)
[0, ∞)
f(x) =
x
2
− 5x + 6
x
2
+ 1
.
f(x)
f
0
(x) =
5x
2
− 10x − 5
(x
2
+ 1)
2
.
X
2
= X
3
= ∅ X
1
= {1 −
√
2, 1 +
√
2} f(1 −
√
2) ≈ 7.04
f(1 +
√
2) ≈ −0.03
lim
x→+∞
f(x) = lim
x→−∞
f(x) = 1.
f(x) x
1
= 1 −
√
2
x
2
= 1 +
√
2
f(x) = cos
3
x + sin
3
x x ∈ E
1
f(x) = (1 + x
2
)e
−x
2
x ∈ E
1
f(x) = (x
2
− 3x + 2)/(x + 1)
2
[−2, 2]
f(x) = x − 2 sin x x ∈ [0, +∞)
f(x) = x
2/3
e
−x
x ∈ [−1, +∞)
f(x) = |x|e
−|x|
x ∈ [−2, 2]
f(x) = x
3
√
x − 1 x ∈ [−7, 2]
f(x) = x
3
p
|x| − 1 x ∈ [−7, 2]
f(x) = x(x − 2)
2/3
x ∈ [−1, +∞)
|3x − x
3
| ≤ 2 |x| ≤ 2.
f(x)
x ∈ E
n
f(x)
x
∗
x
∗
f
0
(x
∗
) = 0
n
X
i=1
n
X
j=1
a
ij
y
i
y
j
(1)
y
1
, . . . , y
n
, a
ij
= a
ji
y
1
= . . . = y
n
= 0.
f(x)
x
∗
a
ij
= f
00
x
i
x
j
(x
∗
)
x
∗
A = (a
ij
)
i,j=1,n
A
f(x) x
∗
f(x) = x
1
x
2
2
+x
2
1
x
2
−3x
2
1
−3x
2
2
½
f
0
x
1
(x) = x
2
2
+ 2x
1
x
2
− 6x
1
= 0,
f
0
x
2
(x) = x
2
1
+ 2x
1
x
2
− 6x
2
= 0.
(x
2
− x
1
)(x
1
+ x
2
+ 6) = 0.
x
1
= x
2
3x
2
1
− 6x
1
= 0
x
1
= x
2
= 0 x
1
= x
2
= 2
µ
2x
2
− 6 2x
1
+ 2x
2
2x
1
+ 2x
2
2x
1
− 6
¶
.
x
1
= x
2
= 0
(0, 0)
>
x = (2, 2)
>
x
1
+ x
2
+ 6 = 0 x
2
x
1
x
2
1
+ 6x
1
−36 = 0 ±3
√
5 −3
(3
√
5 −3, −3
√
5 −3) (−3
√
5 −3, 3
√
5 −3)
f(x) x
1
x
2
µ
−6
√
5 − 12 −12
−12 6
√
5 − 12
¶
.
−6
√
5 − 12 < 0 6
√
5 − 12 > 0
(0, 0)
>
f(x) = (1 + e
x
2
) cos x
1
− x
2
e
x
2
½
f
0
x
1
(x) = −(1 + e
x
2
) sin x
1
= 0,
f
0
x
2
(x) = e
x
2
(cos x
1
− x
2
− 1) = 0.
x
1
∈ {kπ | k ∈ Z}
x
2
= 0 k x
2
= −2
(2kπ, 0) ((2k + 1)π, −2) k ∈ Z