Формула
Результат
0 \leqslant r < b
0
⩽
r
<
b
{\displaystyle 0\leqslant r<b}
\exists\,r \in \mathbb{Z}
∃
r
∈
Z
{\displaystyle \exists \,r\in \mathbb {Z} }
r \notin \mathbb{Z}
r
∉
Z
{\displaystyle r\notin \mathbb {Z} }
r
∉
Z
{\displaystyle r\not \in \mathbb {Z} }
\nexists\,r \in \mathbb{Z}
∄
r
∈
Z
{\displaystyle \nexists \,r\in \mathbb {Z} }
∄
r
∈
Z
{\displaystyle \not \exists \,r\in \mathbb {Z} }
\forall\,r \in \mathbb{Z}
∀
r
∈
Z
{\displaystyle \forall \,r\in \mathbb {Z} }
\mathbb{Z} \ni r
Z
∋
r
{\displaystyle \mathbb {Z} \ni r}
\mathbb{Z} \not\ni r
Z
∌
r
{\displaystyle \mathbb {Z} \not \ni r}
A \cup B
A
∪
B
{\displaystyle A\cup B}
A \cap B
A
∩
B
{\displaystyle A\cap B}
A \setminus B
A
∖
B
{\displaystyle A\setminus B}
A \bigtriangleup B
A
△
B
{\displaystyle A\bigtriangleup B}
A \subset B
A
⊂
B
{\displaystyle A\subset B}
A \not\subset B
A
⊄
B
{\displaystyle A\not \subset B}
B \supset A
B
⊃
A
{\displaystyle B\supset A}
B \not\supset A
B
⊅
A
{\displaystyle B\not \supset A}
a \land b \equiv b \land a
a
∧
b
≡
b
∧
a
{\displaystyle a\land b\equiv b\land a}
a \lor b \equiv b \lor a
a
∨
b
≡
b
∨
a
{\displaystyle a\lor b\equiv b\lor a}
\left[\left(a \to b\right) \land \left(b \to c\right)\right] \to \left(a \to c\right)
[
(
a
→
b
)
∧
(
b
→
c
)
]
→
(
a
→
c
)
{\displaystyle \left[\left(a\to b\right)\land \left(b\to c\right)\right]\to \left(a\to c\right)}
\lnot\left(a \land b\right) \equiv \left(\lnot a \lor \lnot b\right)
¬
(
a
∧
b
)
≡
(
¬
a
∨
¬
b
)
{\displaystyle \lnot \left(a\land b\right)\equiv \left(\lnot a\lor \lnot b\right)}
α
{\displaystyle \alpha \,}
π
12
=
15
∘
{\displaystyle {\frac {\pi }{12}}=15^{\circ }}
π
10
=
18
∘
{\displaystyle {\frac {\pi }{10}}=18^{\circ }}
π
8
=
22.5
∘
{\displaystyle {\frac {\pi }{8}}=22.5^{\circ }}
π
5
=
36
∘
{\displaystyle {\frac {\pi }{5}}=36^{\circ }}
3
π
10
=
54
∘
{\displaystyle {\frac {3\,\pi }{10}}=54^{\circ }}
3
π
8
=
67.5
∘
{\displaystyle {\frac {3\,\pi }{8}}=67.5^{\circ }}
2
π
5
=
72
∘
{\displaystyle {\frac {2\,\pi }{5}}=72^{\circ }}
sin
α
{\displaystyle \sin \alpha \,}
3
−
1
2
2
{\displaystyle {\frac {{\sqrt {3}}-1}{2\,{\sqrt {2}}}}}
5
−
1
4
{\displaystyle {\frac {{\sqrt {5}}-1}{4}}}
2
−
2
2
{\displaystyle {\frac {\sqrt {2-{\sqrt {2}}}}{2}}}
5
−
5
2
2
{\displaystyle {\frac {\sqrt {5-{\sqrt {5}}}}{2\,{\sqrt {2}}}}}
5
+
1
4
{\displaystyle {\frac {{\sqrt {5}}+1}{4}}}
2
+
2
2
{\displaystyle {\frac {\sqrt {2+{\sqrt {2}}}}{2}}}
5
+
5
2
2
{\displaystyle {\frac {\sqrt {5+{\sqrt {5}}}}{2\,{\sqrt {2}}}}}
cos
α
{\displaystyle \cos \alpha \,}
3
+
1
2
2
{\displaystyle {\frac {{\sqrt {3}}+1}{2\,{\sqrt {2}}}}}
5
+
5
2
2
{\displaystyle {\frac {\sqrt {5+{\sqrt {5}}}}{2\,{\sqrt {2}}}}}
2
+
2
2
{\displaystyle {\frac {\sqrt {2+{\sqrt {2}}}}{2}}}
5
+
1
4
{\displaystyle {\frac {{\sqrt {5}}+1}{4}}}
5
−
5
2
2
{\displaystyle {\frac {\sqrt {5-{\sqrt {5}}}}{2\,{\sqrt {2}}}}}
2
−
2
2
{\displaystyle {\frac {\sqrt {2-{\sqrt {2}}}}{2}}}
5
−
1
4
{\displaystyle {\frac {{\sqrt {5}}-1}{4}}}
tg
α
{\displaystyle \operatorname {tg} \,\alpha }
2
−
3
{\displaystyle 2-{\sqrt {3}}}
1
−
2
5
{\displaystyle {\sqrt {1-{\frac {2}{\sqrt {5}}}}}}
2
−
1
2
+
1
{\displaystyle {\sqrt {\frac {{\sqrt {2}}-1}{{\sqrt {2}}+1}}}}
5
−
2
5
{\displaystyle {\sqrt {5-2\,{\sqrt {5}}}}}
1
+
2
5
{\displaystyle {\sqrt {1+{\frac {2}{\sqrt {5}}}}}}
2
+
1
2
−
1
{\displaystyle {\sqrt {\frac {{\sqrt {2}}+1}{{\sqrt {2}}-1}}}}
5
+
2
5
{\displaystyle {\sqrt {5+2\,{\sqrt {5}}}}}
ctg
α
{\displaystyle \operatorname {ctg} \,\alpha }
2
+
3
{\displaystyle 2+{\sqrt {3}}}
5
+
2
5
{\displaystyle {\sqrt {5+2\,{\sqrt {5}}}}}
2
+
1
2
−
1
{\displaystyle {\sqrt {\frac {{\sqrt {2}}+1}{{\sqrt {2}}-1}}}}
1
+
2
5
{\displaystyle {\sqrt {1+{\frac {2}{\sqrt {5}}}}}}
5
−
2
5
{\displaystyle {\sqrt {5-2\,{\sqrt {5}}}}}
2
−
1
2
+
1
{\displaystyle {\sqrt {\frac {{\sqrt {2}}-1}{{\sqrt {2}}+1}}}}
1
−
2
5
{\displaystyle {\sqrt {1-{\frac {2}{\sqrt {5}}}}}}
β
{\displaystyle \beta \,}
π
2
+
α
{\displaystyle {\frac {\pi }{2}}+\alpha }
π
+
α
{\displaystyle \pi +\alpha \,}
3
π
2
+
α
{\displaystyle {\frac {3\,\pi }{2}}+\alpha }
π
2
−
α
{\displaystyle {\frac {\pi }{2}}-\alpha }
π
−
α
{\displaystyle \pi -\alpha \,}
3
π
2
−
α
{\displaystyle {\frac {3\,\pi }{2}}-\alpha }
2
π
−
α
{\displaystyle 2\,\pi -\alpha }
sin
β
{\displaystyle \sin \beta \,}
cos
α
{\displaystyle \cos \alpha \,}
−
sin
α
{\displaystyle -\sin \alpha \,}
−
cos
α
{\displaystyle -\cos \alpha \,}
cos
α
{\displaystyle \cos \alpha \,}
sin
α
{\displaystyle \sin \alpha \,}
−
cos
α
{\displaystyle -\cos \alpha \,}
−
sin
α
{\displaystyle -\sin \alpha \,}
cos
β
{\displaystyle \cos \beta \,}
−
sin
α
{\displaystyle -\sin \alpha \,}
−
cos
α
{\displaystyle -\cos \alpha \,}
sin
α
{\displaystyle \sin \alpha \,}
sin
α
{\displaystyle \sin \alpha \,}
−
cos
α
{\displaystyle -\cos \alpha \,}
−
sin
α
{\displaystyle -\sin \alpha \,}
cos
α
{\displaystyle \cos \alpha \,}
tg
β
{\displaystyle \operatorname {tg} \,\beta }
−
ctg
α
{\displaystyle -\operatorname {ctg} \,\alpha }
tg
α
{\displaystyle \operatorname {tg} \,\alpha }
−
ctg
α
{\displaystyle -\operatorname {ctg} \,\alpha }
ctg
α
{\displaystyle \operatorname {ctg} \,\alpha }
−
tg
α
{\displaystyle -\operatorname {tg} \,\alpha }
ctg
α
{\displaystyle \operatorname {ctg} \,\alpha }
tg
α
{\displaystyle \operatorname {tg} \,\alpha }
ctg
β
{\displaystyle \operatorname {ctg} \,\beta }
−
tg
α
{\displaystyle -\operatorname {tg} \,\alpha }
ctg
α
{\displaystyle \operatorname {ctg} \,\alpha }
−
tg
α
{\displaystyle -\operatorname {tg} \,\alpha }
tg
α
{\displaystyle \operatorname {tg} \,\alpha }
−
ctg
α
{\displaystyle -\operatorname {ctg} \,\alpha }
tg
α
{\displaystyle \operatorname {tg} \,\alpha }
ctg
α
{\displaystyle \operatorname {ctg} \,\alpha }
sin
2
A
=
2
sin
A
cos
A
=
2
tg
A
1
+
tg
2
A
{\displaystyle \sin {2A}=2\sin A\cos A={\frac {2\operatorname {tg} A}{1+\operatorname {tg} ^{2}A}}\,}
sin
3
A
=
3
sin
A
−
4
sin
3
A
{\displaystyle \sin {3A}=3\sin A-4\sin ^{3}A\,}
sin
4
A
=
cos
A
(
4
sin
A
−
8
sin
3
A
)
{\displaystyle \sin {4A}=\cos A(4\sin A-8\sin ^{3}A)\,}
sin
5
A
=
5
sin
A
−
20
sin
3
A
+
16
sin
5
A
{\displaystyle \sin {5A}=5\sin A-20\sin ^{3}A+16\sin ^{5}A\,}
sin
6
A
=
cos
A
(
6
sin
A
−
32
sin
3
A
+
32
sin
5
A
)
{\displaystyle \sin {6A}=\cos A(6\sin A-32\sin ^{3}A+32\sin ^{5}A)\,}
sin
7
A
=
7
sin
A
−
56
sin
3
A
+
112
sin
5
A
−
64
sin
7
A
{\displaystyle \sin {7A}=7\sin A-56\sin ^{3}A+112\sin ^{5}A-64\sin ^{7}A\,}
Для целого положительного чётного n
sin
n
A
=
(
−
1
)
n
2
+
1
cos
A
[
2
n
−
1
sin
A
−
(
n
−
2
)
1
!
2
n
−
3
sin
n
−
3
A
+
(
n
−
3
)
(
n
−
4
)
2
!
2
n
−
5
sin
n
−
5
A
−
(
n
−
4
)
(
n
−
5
)
(
n
−
6
)
3
!
2
n
−
7
sin
n
−
7
A
+
…
]
,
{\displaystyle \sin {nA}=(-1)^{{\frac {n}{2}}+1}\cos A\left[2^{n-1}\sin A-{\frac {(n-2)}{1!}}2^{n-3}\sin ^{n-3}A+{\frac {(n-3)(n-4)}{2!}}2^{n-5}\sin ^{n-5}A-{\frac {(n-4)(n-5)(n-6)}{3!}}2^{n-7}\sin ^{n-7}A+\ldots \right],}
ряд обрывается, когда коэффициент обращается в нуль.
Другой ряд:
sin
n
A
=
n
cos
A
[
sin
A
−
(
n
2
−
2
2
)
3
!
sin
3
A
+
(
n
2
−
2
2
)
(
n
2
−
4
2
)
5
!
sin
5
A
−
(
n
2
−
2
2
)
(
n
2
−
4
2
)
(
n
2
−
6
2
)
7
!
sin
7
A
+
…
]
{\displaystyle \sin {nA}=n\cos A\left[\sin A-{\frac {(n^{2}-2^{2})}{3!}}\sin ^{3}A+{\frac {(n^{2}-2^{2})(n^{2}-4^{2})}{5!}}\sin ^{5}A-{\frac {(n^{2}-2^{2})(n^{2}-4^{2})(n^{2}-6^{2})}{7!}}\sin ^{7}A+\ldots \right]}
n чётное и >0
Для нечётного целого
n
>
1
{\displaystyle n>1\,}
sin
n
A
=
(
−
1
)
n
−
1
2
[
2
n
−
1
sin
n
A
−
n
1
!
2
n
−
3
sin
n
−
2
A
+
n
(
n
−
3
)
2
!
2
n
−
5
sin
n
−
4
A
−
n
(
n
−
4
)
(
n
−
5
)
3
!
2
n
−
7
sin
n
−
6
A
+
…
]
,
{\displaystyle \sin {nA}=(-1)^{\frac {n-1}{2}}\left[2^{n-1}\sin ^{n}A-{\frac {n}{1!}}2^{n-3}\sin ^{n-2}A+{\frac {n(n-3)}{2!}}2^{n-5}\sin ^{n-4}A-{\frac {n(n-4)(n-5)}{3!}}2^{n-7}\sin ^{n-6}A+\ldots \right],}
ряд обрывается, когда коэффициент обращается в нуль.
Другой ряд:
sin
n
A
=
n
sin
A
−
(
n
2
−
1
2
)
3
!
sin
3
A
+
(
n
2
−
1
2
)
(
n
2
−
3
2
)
5
!
sin
5
A
−
…
{\displaystyle \sin {nA}=n\sin A-{\frac {(n^{2}-1^{2})}{3!}}\sin ^{3}A+{\frac {(n^{2}-1^{2})(n^{2}-3^{2})}{5!}}\sin ^{5}A-\ldots }
n нечётное и >0
cos
2
A
=
cos
2
A
−
sin
2
A
=
2
cos
2
A
−
1
=
1
−
2
sin
2
A
=
1
−
tg
2
A
1
+
tg
2
A
=
ctg
A
−
tg
A
ctg
A
+
tg
A
{\displaystyle \cos {2A}=\cos ^{2}A-\sin ^{2}A=2\,\cos ^{2}A-1=1-2\,\sin ^{2}A={\frac {1-\operatorname {tg} ^{2}\,A}{1+\operatorname {tg} ^{2}\,A}}={\frac {\operatorname {ctg} \,A-\operatorname {tg} \,A}{\operatorname {ctg} \,A+\operatorname {tg} \,A}}}
cos
3
A
=
4
cos
3
A
−
3
cos
A
{\displaystyle \cos {3A}=4\,\cos ^{3}A-3\,\cos A}
cos
4
A
=
8
cos
4
A
−
8
cos
2
A
+
1
{\displaystyle \cos {4A}=8\,\cos ^{4}A-8\,\cos ^{2}A+1}
cos
5
A
=
16
cos
5
A
−
20
cos
3
A
+
5
cos
A
{\displaystyle \cos {5A}=16\,\cos ^{5}A-20\,\cos ^{3}A+5\,\cos A}
cos
6
A
=
32
cos
6
A
−
48
cos
4
A
+
18
cos
2
A
−
1
{\displaystyle \cos {6A}=32\,\cos ^{6}A-48\,\cos ^{4}A+18\,\cos ^{2}A-1}
cos
7
A
=
64
cos
7
A
−
112
cos
5
A
+
56
cos
3
A
−
7
cos
A
{\displaystyle \cos {7A}=64\,\cos ^{7}A-112\,\cos ^{5}A+56\,\cos ^{3}A-7\,\cos A}
cos
n
A
=
2
n
−
1
cos
n
A
−
n
1
!
2
n
−
3
cos
n
−
2
A
+
n
(
n
−
3
)
2
!
2
n
−
5
cos
n
−
4
A
−
n
(
n
−
4
)
(
n
−
5
)
3
!
2
n
−
7
cos
n
−
6
A
+
n
(
n
−
5
)
(
n
−
6
)
(
n
−
7
)
4
!
2
n
−
9
cos
n
−
8
A
+
⋯
{\displaystyle \cos {nA}=2^{n-1}\,\cos ^{n}A-{\frac {n}{1!}}\,2^{n-3}\,\cos ^{n-2}A+{\frac {n(n-3)}{2!}}\,2^{n-5}\,\cos ^{n-4}A-{\frac {n(n-4)(n-5)}{3!}}\,2^{n-7}\,\cos ^{n-6}A+{\frac {n(n-5)(n-6)(n-7)}{4!}}\,2^{n-9}\,\cos ^{n-8}A+\cdots }
Формула обрывается, когда коэффициент обращается в нуль (n целое и >2).
n
{\displaystyle n\,}
10
n
−
1
{\displaystyle 10^{n}-1\,}
10
n
+
1
{\displaystyle 10^{n}+1\,}
1
9
=
3
2
{\displaystyle 9=3^{2}\,}
11
{\displaystyle 11\,}
#
2
99
=
3
2
⋅
11
{\displaystyle 99=3^{2}\cdot 11}
101
{\displaystyle 101\,}
#
3
999
=
3
3
⋅
37
{\displaystyle 999=3^{3}\cdot 37}
1001
=
7
⋅
11
⋅
13
{\displaystyle 1001=7\cdot 11\cdot 13}
4
9999
=
3
2
⋅
11
⋅
101
{\displaystyle 9999=3^{2}\cdot 11\cdot 101}
10001
=
73
⋅
137
{\displaystyle 10001=73\cdot 137}
5
99999
=
3
2
⋅
41
⋅
271
{\displaystyle 99999=3^{2}\cdot 41\cdot 271}
100001
=
11
⋅
9091
{\displaystyle 100001=11\cdot 9091}
6
999999
=
3
3
⋅
7
⋅
11
⋅
13
⋅
37
{\displaystyle 999999=3^{3}\cdot 7\cdot 11\cdot 13\cdot 37}
1000001
=
101
⋅
9901
{\displaystyle 1000001=101\cdot 9901}
7
9999999
=
3
2
⋅
239
⋅
4649
{\displaystyle 9999999=3^{2}\cdot 239\cdot 4649}
10000001
=
11
⋅
909091
{\displaystyle 10000001=11\cdot 909091}
(
10
n
−
1
)
⋅
(
10
n
+
1
)
=
10
2
n
−
1
{\displaystyle \left(10^{n}-1\right)\cdot \left(10^{n}+1\right)=10^{2n}-1}
n
{\displaystyle n\,}
2
n
−
1
{\displaystyle 2^{n}-1\,}
2
n
+
1
{\displaystyle 2^{n}+1\,}
1
1
{\displaystyle 1\,}
#
3
{\displaystyle 3\,}
#
2
3
{\displaystyle 3\,}
#
5
{\displaystyle 5\,}
#
3
7
{\displaystyle 7\,}
#
9
=
3
2
{\displaystyle 9=3^{2}\,}
4
15
=
3
⋅
5
{\displaystyle 15=3\cdot 5}
17
{\displaystyle 17\,}
#
5
31
{\displaystyle 31\,}
#
33
=
3
⋅
11
{\displaystyle 33=3\cdot 11}
6
63
=
3
2
⋅
7
{\displaystyle 63=3^{2}\cdot 7}
65
=
5
⋅
13
{\displaystyle 65=5\cdot 13}
7
127
{\displaystyle 127\,}
#
129
=
3
⋅
43
{\displaystyle 129=3\cdot 43}
8
255
=
3
⋅
5
⋅
17
{\displaystyle 255=3\cdot 5\cdot 17}
257
{\displaystyle 257\,}
#
9
511
=
7
⋅
73
{\displaystyle 511=7\cdot 73}
513
=
3
3
⋅
19
{\displaystyle 513=3^{3}\cdot 19}
10
1023
=
3
⋅
11
⋅
31
{\displaystyle 1023=3\cdot 11\cdot 31}
1025
=
5
2
⋅
41
{\displaystyle 1025=5^{2}\cdot 41}
11
2047
=
23
⋅
89
{\displaystyle 2047=23\cdot 89}
2049
=
3
⋅
683
{\displaystyle 2049=3\cdot 683}
12
4095
=
3
2
⋅
5
⋅
7
⋅
13
{\displaystyle 4095=3^{2}\cdot 5\cdot 7\cdot 13}
4097
=
17
⋅
241
{\displaystyle 4097=17\cdot 241}
13
8191
{\displaystyle 8191\,}
#
8193
=
3
⋅
2731
{\displaystyle 8193=3\cdot 2731}
14
16383
=
3
⋅
43
⋅
127
{\displaystyle 16383=3\cdot 43\cdot 127}
16385
=
5
⋅
29
⋅
113
{\displaystyle 16385=5\cdot 29\cdot 113}
15
32767
=
7
⋅
31
⋅
151
{\displaystyle 32767=7\cdot 31\cdot 151}
32769
=
3
2
⋅
11
⋅
331
{\displaystyle 32769=3^{2}\cdot 11\cdot 331}
16
65535
=
3
⋅
5
⋅
17
⋅
257
{\displaystyle 65535=3\cdot 5\cdot 17\cdot 257}
65537
{\displaystyle 65537\,}
#
17
131071
{\displaystyle 131071\,}
#
131073
=
3
⋅
43691
{\displaystyle 131073=3\cdot 43691}
18
262143
=
3
3
⋅
7
⋅
19
⋅
73
{\displaystyle 262143=3^{3}\cdot 7\cdot 19\cdot 73}
262145
=
5
⋅
13
⋅
37
⋅
109
{\displaystyle 262145=5\cdot 13\cdot 37\cdot 109}
19
524287
{\displaystyle 524287\,}
#
524289
=
3
⋅
174763
{\displaystyle 524289=3\cdot 174763}
20
1048575
=
3
⋅
5
2
⋅
11
⋅
31
⋅
41
{\displaystyle 1048575=3\cdot 5^{2}\cdot 11\cdot 31\cdot 41}
1048577
=
17
⋅
61681
{\displaystyle 1048577=17\cdot 61681}
21
2097151
=
7
2
⋅
127
⋅
337
{\displaystyle 2097151=7^{2}\cdot 127\cdot 337}
2097153
=
3
2
⋅
43
⋅
5419
{\displaystyle 2097153=3^{2}\cdot 43\cdot 5419}
22
4194303
=
3
⋅
23
⋅
89
⋅
683
{\displaystyle 4194303=3\cdot 23\cdot 89\cdot 683}
4194305
=
5
⋅
397
⋅
2113
{\displaystyle 4194305=5\cdot 397\cdot 2113}
23
8388607
=
47
⋅
178481
{\displaystyle 8388607=47\cdot 178481}
8388609
=
3
⋅
2796203
{\displaystyle 8388609=3\cdot 2796203}
24
16777215
=
3
2
⋅
5
⋅
7
⋅
13
⋅
17
⋅
241
{\displaystyle 16777215=3^{2}\cdot 5\cdot 7\cdot 13\cdot 17\cdot 241}
16777217
=
97
⋅
257
⋅
673
{\displaystyle 16777217=97\cdot 257\cdot 673}
32
4294967295
=
3
⋅
5
⋅
17
⋅
257
⋅
65537
{\displaystyle 4294967295=3\cdot 5\cdot 17\cdot 257\cdot 65537}
4294967297
=
641
⋅
6700417
{\displaystyle 4294967297=641\cdot 6700417}
(
2
n
−
1
)
⋅
(
2
n
+
1
)
=
2
2
n
−
1
{\displaystyle \left(2^{n}-1\right)\cdot \left(2^{n}+1\right)=2^{2n}-1}
Источник: [1] , [2]
Ueber unendliche, lineare Punktmannichfaltigkeiten.
Von Georg Cantor in Halle a. d. Saale.
1.
In einer, im Borchardt'schen Journale, Bd. 84, pag. 242 herausgegebenen Abhandlung habe ich für ein sehr weitreichendes Gebiet von geometrischen und arithmetischen, sowohl continuirlichen, wie discontinuirlichen Mannichfaltigkeiten den Nachweis geführt, dass sie eindeutig und vollständig einer geraden Strecke oder einem discontinuirliehen Bestandteile von ihr sich zuordnen lassen.
Hierdurch gewinnen die letzteren Mannichfaltigkeiten, wir nennen sie lineare Punktmannichfaltigkeiten oder kürzer lineare Punktmengen , welche also entweder eine continuirliche, endliche oder unendliche, gerade Strecke bilden oder doch mit allen ihren Punkten in einer solchen, als Theile enthalten sind, ein besonderes Interesse, und es dürfte daher nicht unwerth sein, wenn wir denselben eine Reihe von Betrachtungen widmen und zunächst im Folgenden ihre Classiäcation untersuchen wollen. Verschiedene Gesichtspunkte und damit verbundene Classificationspriucipiea führen uns dazu, die linearen Punktmengen in gewisse Gruppen zu fassen. Um mit einem dieser Gesichtspunkte zu beginnen, erinnern wir an den Begriff der Ableitung einer gegebenen Punktmenge P , welcher in einer Arbeit über trigonometrische Reihen (Math. Annaien, Bd. V, pag. 129) dargelegt worden ist; in dem jüngst erschienenen Werke Dini's (Fondamenti per la teorica d. funzioui d. variabili reali, Pisa, 1878) sehen wir diesen Begriff noch weiter entwickelt, indem er als Ausgangspunkt für eine Reihe bemerkenswerther Verallgemeinerungen von bekannten analytischen Sätzen genommen wird.*) Der Begriif der Ableitung einer gegebenen Mannichfaltigkeit ist übrigens nicht auf die linearen Mannichfaltigkeiten beschränkt, sondern gilt in gleicher Weise auch für die ebenen, räumlichen und n-fachen stetigen und unstetigen Mannichfaltigkeiten. Auf ihn wird, wie wir später zeigen wollen, die einfachste und zugleich vollständigste Erklärung resp. Bestimmung eines Continuums gegründet.
) Man vergleiche auch: Ascoli. Nuove richerche sulla serie di Fourier, Reale Academia dei Lincei (1877—78)
Если множества
A
{\displaystyle A}
B
{\displaystyle B}
⋀
x
[
x
∈
A
≡
x
∈
B
]
→
(
A
=
B
)
{\displaystyle \bigwedge _{x}\left[x\in A\equiv x\in B\right]\to \left(A=B\right)}
Существует такое множество
∅
,
{\displaystyle \emptyset ,}
x
{\displaystyle x}
⋁
P
⋀
x
(
x
∉
P
)
{\displaystyle \bigvee _{P}\bigwedge _{x}\left(x\notin P\right)}
Для произвольных
a
{\displaystyle a}
b
{\displaystyle b}
a
{\displaystyle a}
b
.
{\displaystyle b.}
⋁
P
⋀
x
{
(
x
∈
P
)
≡
[
(
x
=
a
)
∨
(
x
=
b
)
]
}
{\displaystyle \bigvee _{P}\bigwedge _{x}\left\{\left(x\in P\right)\equiv \left[\left(x=a\right)\lor \left(x=b\right)\right]\right\}}
Эта аксиома зависима от остальных.
Для каждого семейства множеств
A
{\displaystyle \mathbf {A} }
S
,
{\displaystyle S,}
X
,
{\displaystyle X,}
A
.
{\displaystyle \mathbf {A} .}
x
∈
S
≡
⋁
X
[
(
x
∈
X
)
∧
(
X
∈
A
)
]
{\displaystyle x\in S\equiv \bigvee _{X}\left[\left(x\in X\right)\land \left(X\in \mathbf {A} \right)\right]}
Множество
S
{\displaystyle S}
суммой (объединением) множеств , принадлежащих семейству
A
,
{\displaystyle \mathbf {A} ,}
S
(
A
)
{\displaystyle S\left(\mathbf {A} \right)}
⋃
X
∈
A
X
.
{\displaystyle \bigcup _{X\in \mathbf {A} }X.}
Для каждого множества
A
{\displaystyle A}
P
,
{\displaystyle \mathbf {P} ,}
A
{\displaystyle A}
X
∈
P
=
(
X
⊂
A
)
{\displaystyle X\in \mathbf {P} =\left(X\subset A\right)}
Семейство
P
{\displaystyle \mathbf {P} }
степенью множества
A
{\displaystyle A}
2
A
.
{\displaystyle 2^{A}.\,}
Существует такое семейство множеств
A
,
{\displaystyle \mathbf {A} ,}
∅
{\displaystyle \emptyset }
X
∈
A
,
{\displaystyle X\in \mathbf {A} ,}
A
{\displaystyle \mathbf {A} }
Y
,
{\displaystyle Y,}
X
{\displaystyle X}
X
.
{\displaystyle X.}
⋁
A
(
(
∅
∈
A
)
∧
⋀
X
∈
A
⋁
Y
∈
A
⋀
x
{
(
x
∈
Y
)
≡
[
(
x
∈
X
)
∨
(
x
=
X
)
]
}
)
{\displaystyle \bigvee _{\mathbf {A} }\left(\left(\emptyset \in \mathbf {A} \right)\land \bigwedge _{X\in \mathbf {A} }\bigvee _{Y\in \mathbf {A} }\bigwedge _{x}\left\{\left(x\in Y\right)\equiv \left[\left(x\in X\right)\lor \left(x=X\right)\right]\right\}\right)}
Таким образом семейству
A
{\displaystyle \mathbf {A} }
∅
,
{\displaystyle \emptyset ,}
N
1
=
{
∅
}
,
{\displaystyle N_{1}=\left\{\emptyset \right\},}
N
2
=
{
∅
,
N
1
}
,
{\displaystyle N_{2}=\left\{\emptyset ,N_{1}\right\},}
N
3
=
{
∅
,
N
1
,
N
2
}
{\displaystyle N_{3}=\left\{\emptyset ,N_{1},N_{2}\right\}}
Для каждого семейства
A
{\displaystyle \mathbf {A} }
B
,
{\displaystyle B,}
X
,
{\displaystyle X,}
A
.
{\displaystyle \mathbf {A} .}
⋀
X
,
Y
∈
A
{
[
X
≠
∅
]
∧
[
(
X
≠
Y
)
→
(
X
∩
Y
=
∅
)
]
}
→
⋁
B
⋀
X
∈
A
⋁
x
⋀
y
[
(
y
∈
X
∩
Y
)
≡
(
y
=
x
)
]
{\displaystyle \bigwedge _{X,\,Y\in \mathbf {A} }\left\{\left[X\neq \emptyset \right]\land \left[\left(X\neq Y\right)\to \left(X\cap Y=\emptyset \right)\right]\right\}\to \bigvee _{B}\bigwedge _{X\in \mathbf {A} }\bigvee _{x}\bigwedge _{y}\left[\left(y\in X\cap Y\right)\equiv \left(y=x\right)\right]}
![{\displaystyle \left[\left(a\to b\right)\land \left(b\to c\right)\right]\to \left(a\to c\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20eedabbf89f4e48a05e466fc26062c113064dbf)
![{\displaystyle \sin {nA}=(-1)^{{\frac {n}{2}}+1}\cos A\left[2^{n-1}\sin A-{\frac {(n-2)}{1!}}2^{n-3}\sin ^{n-3}A+{\frac {(n-3)(n-4)}{2!}}2^{n-5}\sin ^{n-5}A-{\frac {(n-4)(n-5)(n-6)}{3!}}2^{n-7}\sin ^{n-7}A+\ldots \right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00a2934bfe1ed506420c84020c621925ef09d222)
![{\displaystyle \sin {nA}=n\cos A\left[\sin A-{\frac {(n^{2}-2^{2})}{3!}}\sin ^{3}A+{\frac {(n^{2}-2^{2})(n^{2}-4^{2})}{5!}}\sin ^{5}A-{\frac {(n^{2}-2^{2})(n^{2}-4^{2})(n^{2}-6^{2})}{7!}}\sin ^{7}A+\ldots \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/70ffc07e1a711eb4d767df1ef21d040b93bde00a)
![{\displaystyle \sin {nA}=(-1)^{\frac {n-1}{2}}\left[2^{n-1}\sin ^{n}A-{\frac {n}{1!}}2^{n-3}\sin ^{n-2}A+{\frac {n(n-3)}{2!}}2^{n-5}\sin ^{n-4}A-{\frac {n(n-4)(n-5)}{3!}}2^{n-7}\sin ^{n-6}A+\ldots \right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e39e0b517f17ab8815b81d8c04c922b7fceae489)
![{\displaystyle \bigwedge _{x}\left[x\in A\equiv x\in B\right]\to \left(A=B\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8dc2daf3e8666c0c37c699f8d4c2af0345a0d5d7)
![{\displaystyle \bigvee _{P}\bigwedge _{x}\left\{\left(x\in P\right)\equiv \left[\left(x=a\right)\lor \left(x=b\right)\right]\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/100dba144fe6b5f80d5559b1e559343a122451e9)
![{\displaystyle x\in S\equiv \bigvee _{X}\left[\left(x\in X\right)\land \left(X\in \mathbf {A} \right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/46dd207717b478cd8d5742305cb6a677b6753bb5)
![{\displaystyle \bigvee _{\mathbf {A} }\left(\left(\emptyset \in \mathbf {A} \right)\land \bigwedge _{X\in \mathbf {A} }\bigvee _{Y\in \mathbf {A} }\bigwedge _{x}\left\{\left(x\in Y\right)\equiv \left[\left(x\in X\right)\lor \left(x=X\right)\right]\right\}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38fdd6692a0a680b44d38f36d1444a9676208358)
![{\displaystyle \bigwedge _{X,\,Y\in \mathbf {A} }\left\{\left[X\neq \emptyset \right]\land \left[\left(X\neq Y\right)\to \left(X\cap Y=\emptyset \right)\right]\right\}\to \bigvee _{B}\bigwedge _{X\in \mathbf {A} }\bigvee _{x}\bigwedge _{y}\left[\left(y\in X\cap Y\right)\equiv \left(y=x\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3614d9c77e829cfb4eb0de2930c9a621a591c75c)