Элементы дискретной математики. Часть I - 29 стр.

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f
exp f(x)=e
f(x)
=(1)
f(x)
.
(exp f(a))
aE
n
2
exp f(x)
H
2
n
f Z(f)=(Z
a
(f))
aE
n
2
f
f
Z
a
(f)= Σ
xE
n
2
(1)
f(x)(x,a)
xE
n
2
(1)
f(x)(a,x)
.
C(f)=
=(C
a
(f))
aV
n
Z(f)=(Z
a
(f))
aV
n
f a E
n
2
Z
a
(f)=
2 · 2
n
C
a
(f), a =0;
2
n
2 ·2
n
C
a
(f), a =0.
a =0
C
0
(f)=1/2
n
·||f||,Z
0
(f)=(2
n
−||f||) −||f || =2
n
2||f||.
f g α β E
2
N
α,β
(f,g)
{x |f (x)=α & g(x)=β }.
a =0
2
n
· C
a
(f)=N
1,0
(f,(a, x)) N
1,1
(f,(a, x)),
Z
a
(f)=N
0,0
(f,(a, x)) + N
1,1
(f,(a, x)) N
0,1
(f,(a, x)) N
1,0
(f,(a, x)).
N
0,0
(f,(a, x)) + N
1,0
(f,(a, x)) = 2
n1
,
N
1,0
(f,(a, x)) + N
1,1
(f,(a, x)) = ||f||,
N
0,1
(f,(a, x)) + N
1,1
(f,(a, x)) = 2
n1
,
N
0,1
(f,(a, x)) = 2
n1
−||f|| + N
1,0
(f,(a, x)).