Джумадильдаев А.С. Элементы дискретной математики (часть 1)
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H
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2n +2
C
n
H
2n+2
C
n
H
2n
643189
n
n
n
n
f : A → A, A = {1, 2,...,n},
i<j f(i) ≤ f(j).
{1, 2}
x
1
+···+x
k
= n, x
i
∈{1, 2},i=1, 2,...,k, k =
0, 1, 2,...
A =(a
i,j
) n.
det A =
n
i=1
a
i,j
∆
i,j
=
n
j=1
a
i,j
∆
i,j
,
∆
i,j
=(−1)
i+j
det A
i,j
A =(a
i,j
) ∈ Mat
n
n ≥ 3,
1 ≤ j ≤ n
det A =
n
i=1
λ
i,j
a
i,j
∆
i,j
,
1 ≤ i ≤ n
det A =
n
j=1
¯
λ
i,j
a
i,j
∆
i,j
,
λ
i,j
=(
s=i
a
s,j
)(
n
r=1
a
−1
1,r
···a
−1
i−1,r
a
−1
i+1,r
···a
−1
n,r
),
¯
λ
i,j
=(
s=j
a
i,s
)(
n
r=1
a
−1
r,1
···a
−1
r,j−1
a
−1
r,j+1
···a
−1
r,n
).
A
(−1)
A =(a
i,j
),a
i,j
=0, A
(−1)
=(a
−1
i,j
).
A 3
AA
t
= E,a
i,j
=0. A
(−1)
(x
1
+ ···x
k
)
n
=
i
1
,...,i
k
n
i
1
,...,i
k
x
i
1
···x
i
k
k
,
n
i
1
,...,i
k
=
n!
i
1
! ···i
k
!
x, y
[x, y]=x
yx = x(y − 1).
(x + y)
n
.
(x + y)
n
=
n
i=0
λ
i
n
S
i
(x, y),
S
0
(x, y)=1,
S
i
(x, y)=
i−1
j=0
i
j
x
i
(y +1)···(y + i − j)+x
j
λ
0
0
=1,λ
i
n
= λ
i−1
n−1
−(i +1)λ
i
n−1
,i=0, 1,...,n.
λ
i
n
=(−1)
n+i
¯s
i+1
n+1
,
¯s
i
n
, 1 ≤ i ≤ n,
(x + y)
2
=(x + y)(x + y)=xx + xy + yx + yy =
(
)=x
2
+2xy + y
2
.
(x + y)
2
= x
2
+2xy + y
2
− [x, y],
[x, y]=xy − yx
(x + y)
2
= x
2
+2xy + y
2
−x = S
0
(x, y) −3S
1
(x, y)+S
2
(x, y),
S
0
(x, y)=1,S
1
(x, y)=y +1+x, S
2
(x, y)=(y +1)(y +2)+2x(y +1)+x
2
.
(xx)x = x(xx),
x
3
=(xx)x = x(xx)
(x + y)
n
yx = xy − x.
n
a(a(aa)), (aa)(aa), ((aa)a)a, (a(aa))a, a((aa)a).
1
n
2(n − 1)
n − 1
n
ab = ba, ∀a, b.
c
n
n
c
4
=2 2
a(a(aa)), (aa)(aa).
((aa)a)a = a((aa)a)=a(a(aa)),
(a(aa))a = a(a(aa)),
a((aa)a)=a(a(aa)).
n ≤ 10
n 123456 7 8 9 10
c
n
11123611234698
c
n
c
n
x, y, z,
(x + y)
n
=
n
k=0
n
k
x(x − kz)
k−1
(y + kz)
n−k
,
4×4 1, 2,...,15.
⇒ ⇒
⇒ ?
d
n
dt
n
{t
n−1
f(
1
t
)} =
(−1)
n
t
n+1
f
(n)
(
1
t
),
f
(n)
(1/t) n f 1/t
p (p − 1)! + 1 p
k
k 0 9
a
1729
≡ a(mod 1729) a ∈ Z.
x!+1=y
2
x!+1=y
2
(x, y)=(4, 5), (5, 11), (7, 71).
x!+a = y
2
a x!+a = y
2
, a
x! − a = y
2
a.
n.
A ◦. A
a ◦ (b ◦ c)=(a ◦ b) ◦ c, ∀a, b, c ∈ A.
A [ , ]
[a, b]=a ◦ b − b ◦ a.
[a, b]=−[b, a],
[[a, b],c]+[[b, c],a]+[[c, a],b]=0.
A ◦ A
{a, b},
{a, b} = a ◦ b + b ◦a.
{a, b} = {b, a},
{{a, a}, {b, a}} = {{{a, a},b},a}.
q
A C
◦ q ∈ C. A ◦
q
q
a ◦
q
b = a ◦ b + qb◦ a.
◦
q
(q − 1)
2
(a, c, b)+q [c, [a, b]] = 0,
(a, b, c)=a ◦
q
(b ◦
q
c) − (a ◦
q
b) ◦
q
c,
[a, b]=a ◦
q
b − b ◦
q
a.
A = C[x]
a ◦ b = ∂(ab).
∂ =
∂
∂x
ab
∂(x
5
)=5x
4
x
3
◦ x
5
=5x
7
.
(a ◦ b) ◦ (c ◦ d) − (a ◦ d) ◦ (b ◦ c)=(a, b, c) ◦ d − (a, d, c) ◦ b,
(a, b, c)=a ◦ (b ◦ c) − (a ◦ b) ◦ c.
a = x
2
+ y
2
,b= z
2
+ t
2
⇒ ab =(x
2
+ y
2
)(z
2
+ t
2
)=(xz)
2
+(yz)
2
+(xt)
2
+(yt)
2
=(xz)
2
+2xyzt +(yt)
2
+(xt)
2
− 2xyzt +(yz)
2
=
(xz + yt)
2
+(xt − yz)
2
.
A = {x
2
+ y
2
|x, y ∈ Z}
a, b ∈ A ⇒ ab ∈ A.
k
A = {x
2
1
+ ···+ x
2
k
|x
1
,...,x
k
∈ Z}
•
•
2(a − b)(b − c)(c − d)(d − a)=(b − c)
2
(a − d)
2
−(c − a)
2
(b − d)
2
+(a − b)
2
(c − d)
2
S[a, b, c] S[d, c]+S[b, d, c] S[a, c]+S[d, a, c] S[b, c]=0,
S[a, b, c]=(a − b)(b − c)(c − a),
S[a, b]=(a − b)(b − a).