The Large Number on Prime Numbers:
Skewes Number
@Riemann_Zeta_F
\displaystyle \pi (x) = \sum_{m = 1}^{\infty} \frac{\mu(m)}{m} \left(\mathrm{li} (x^{\frac{1}{m}}) - \sum_{\rho} \mathrm{li} (x^{\frac{\rho}{m}}) + \int_{x^{\frac{1}{m}}}^{\infty} \frac{1}{t(t^{2} - 1)} dt - \log 2\right)
π
(
x
)
=
∑
m
=
1
∞
μ
(
m
)
m
(
l
i
(
x
1
m
)
−
∑
ρ
l
i
(
x
ρ
m
)
+
∫
x
1
m
∞
1
t
(
t
2
−
1
)
d
t
−
log
2
)
\displaystyle \pi (x) = \sum_{m = 1}^{\infty} \frac{\mu(m)}{m} \left(\mathrm{li} (x^{\frac{1}{m}}) - \sum_{\rho} \mathrm{li} (x^{\frac{\rho}{m}}) + \int_{x^{\frac{1}{m}}}^{\infty} \frac{1}{t(t^{2} - 1)} dt - \log 2\right)
π
(
x
)
=
m
=
1
∑
∞
m
μ
(
m
)
(
l
i
(
x
m
1
)
−
ρ
∑
l
i
(
x
m
ρ
)
+
∫
x
m
1
∞
t
(
t
2
−
1
)
1
d
t
−
lo
g
2
)
\displaystyle \pi (x) < \mathrm{Li} (x)
π
(
x
)
<
L
i
(
x
)
\displaystyle \pi (x) < \mathrm{Li} (x)
π
(
x
)
<
L
i
(
x
)
\displaystyle \mathrm{Li} (x) - \frac{1}{2} \mathrm{Li} (x^{\frac{1}{2}}) - \frac{1}{3} \mathrm{Li} (x^{\frac{1}{3}}) - \frac{1}{5} \mathrm{Li} (x^{\frac{1}{5}}) + \frac{1}{6} \mathrm{Li} (x^{\frac{1}{6}}) - \cdots
L
i
(
x
)
−
1
2
L
i
(
x
1
2
)
−
1
3
L
i
(
x
1
3
)
−
1
5
L
i
(
x
1
5
)
+
1
6
L
i
(
x
1
6
)
−
⋯
\displaystyle \mathrm{Li} (x) - \frac{1}{2} \mathrm{Li} (x^{\frac{1}{2}}) - \frac{1}{3} \mathrm{Li} (x^{\frac{1}{3}}) - \frac{1}{5} \mathrm{Li} (x^{\frac{1}{5}}) + \frac{1}{6} \mathrm{Li} (x^{\frac{1}{6}}) - \cdots
L
i
(
x
)
−
2
1
L
i
(
x
2
1
)
−
3
1
L
i
(
x
3
1
)
−
5
1
L
i
(
x
5
1
)
+
6
1
L
i
(
x
6
1
)
−
⋯
\displaystyle \zeta (s) = \sum_{k = 1}^{\infty} \frac{1}{k^{s}}
ζ
(
s
)
=
∑
k
=
1
∞
1
k
s
\displaystyle \zeta (s) = \sum_{k = 1}^{\infty} \frac{1}{k^{s}}
ζ
(
s
)
=
k
=
1
∑
∞
k
s
1
\displaystyle \pi (e^{e^{e^{79}}}) > \mathrm{Li} (e^{e^{e^{79}}})
π
(
e
e
e
79
)
>
L
i
(
e
e
e
79
)
\displaystyle \pi (e^{e^{e^{79}}}) > \mathrm{Li} (e^{e^{e^{79}}})
π
(
e
e
e
7
9
)
>
L
i
(
e
e
e
7
9
)
?
?
?
?
\displaystyle \pi (x) \sim \frac{x}{\log x}
π
(
x
)
∼
x
log
x
\displaystyle \pi (x) \sim \frac{x}{\log x}
π
(
x
)
∼
lo
g
x
x
2017. 10. 08.
@Romantic Mathnight
in
\displaystyle \mathrm{MATH~ POWE}\mathbb{R}^{2017}
M
A
T
H
P
O
W
E
R
2017
\displaystyle \mathrm{MATH~ POWE}\mathbb{R}^{2017}
M
A
T
H
P
O
W
E
R
2
0
1
7
\displaystyle \mathrm{def}
d
e
f
\displaystyle \mathrm{def}
d
e
f
\displaystyle L \left(s + \frac{n - 1}{2};~\left(\mathrm{Sym}^{n - 1} \rho_{E, \ell}\right)|_{F'}\right) = L (s; \mathrm{\Pi}_{m, n})
L
(
s
+
n
−
1
2
;
(
S
y
m
n
−
1
ρ
E
,
ℓ
)
∣
F
′
)
=
L
(
s
;
Π
m
,
n
)
\displaystyle L \left(s + \frac{n - 1}{2};~\left(\mathrm{Sym}^{n - 1} \rho_{E, \ell}\right)|_{F'}\right) = L (s; \mathrm{\Pi}_{m, n})
L
(
s
+
2
n
−
1
;
(
S
y
m
n
−
1
ρ
E
,
ℓ
)
∣
F
′
)
=
L
(
s
;
Π
m
,
n
)
自己紹介.
・14歳, 中学2年生.
・去年の7月にある2人の日曜数学者のブログを見て,
数学に興味を持つ.
・去年の12月24日, 今年の7月1日のロマ数でプレゼン.
・最近は佐藤-Tate予想周辺を勉強中.
・グレブナー基底にはポン酢派.
Skewes Number
・めっちゃでかい
・素数定理と関係がある
・Riemann zetaの と関係がある
\displaystyle \rho
ρ
\displaystyle \rho
ρ
・具体的な値は未だ知られていない
でかい数
ある数をでかくするためにいろいろな演算子が導入されてきた:
たしざん
a + b
a
+
b
a + b
a
+
b
かけざん
\displaystyle a \times b = a + a + \cdots + a
a
×
b
=
a
+
a
+
⋯
+
a
\displaystyle a \times b = a + a + \cdots + a
a
×
b
=
a
+
a
+
⋯
+
a
(
\displaystyle b
b
\displaystyle b
b
回
)
\displaystyle +
+
\displaystyle +
+
\displaystyle \times
×
\displaystyle \times
×
べきじょう
\displaystyle \uparrow
↑
\displaystyle \uparrow
↑
\displaystyle a \uparrow b = a \times a \times \cdots \times a
a
↑
b
=
a
×
a
×
⋯
×
a
\displaystyle a \uparrow b = a \times a \times \cdots \times a
a
↑
b
=
a
×
a
×
⋯
×
a
(
\displaystyle b
b
\displaystyle b
b
回
)
\displaystyle a^{b}
a
b
\displaystyle a^{b}
a
b
テトレーション
\displaystyle \uparrow \uparrow
↑
↑
\displaystyle \uparrow \uparrow
↑
↑
\displaystyle a \uparrow \uparrow b = a \uparrow a \uparrow \cdots \uparrow a
a
↑
↑
b
=
a
↑
a
↑
⋯
↑
a
\displaystyle a \uparrow \uparrow b = a \uparrow a \uparrow \cdots \uparrow a
a
↑
↑
b
=
a
↑
a
↑
⋯
↑
a
(
\displaystyle b
b
\displaystyle b
b
回
)
\displaystyle ^{b}a
b
a
\displaystyle ^{b}a
b
a
\displaystyle \cdots
⋯
\displaystyle \cdots
⋯
でかい数
\displaystyle \uparrow
↑
\displaystyle \uparrow
↑
を増やすとつよい
\displaystyle 3 + 3 = 6
3
+
3
=
6
\displaystyle 3 + 3 = 6
3
+
3
=
6
\displaystyle 3 \times 3 = 3 + 3 + 3 = 9
3
×
3
=
3
+
3
+
3
=
9
\displaystyle 3 \times 3 = 3 + 3 + 3 = 9
3
×
3
=
3
+
3
+
3
=
9
\displaystyle 3 \uparrow 3 = 3 \times 3 \times 3 = 27
3
↑
3
=
3
×
3
×
3
=
27
\displaystyle 3 \uparrow 3 = 3 \times 3 \times 3 = 27
3
↑
3
=
3
×
3
×
3
=
2
7
\displaystyle 3 \uparrow \uparrow 3 = 3 \uparrow 3 \uparrow 3 = 3 \uparrow 27 = 7625597484987
3
↑
↑
3
=
3
↑
3
↑
3
=
3
↑
27
=
7625597484987
\displaystyle 3 \uparrow \uparrow 3 = 3 \uparrow 3 \uparrow 3 = 3 \uparrow 27 = 7625597484987
3
↑
↑
3
=
3
↑
3
↑
3
=
3
↑
2
7
=
7
6
2
5
5
9
7
4
8
4
9
8
7
\displaystyle 3 \uparrow \uparrow \uparrow 3 = 3 \uparrow \uparrow 7625597484987 = \cdots
3
↑
↑
↑
3
=
3
↑
↑
7625597484987
=
⋯
\displaystyle 3 \uparrow \uparrow \uparrow 3 = 3 \uparrow \uparrow 7625597484987 = \cdots
3
↑
↑
↑
3
=
3
↑
↑
7
6
2
5
5
9
7
4
8
4
9
8
7
=
⋯
でかい数
\displaystyle \uparrow
↑
\displaystyle \uparrow
↑
を
\displaystyle x
x
\displaystyle x
x
個並べるともっとつよそう
\displaystyle f (x) = 3 \uparrow \uparrow \cdots \uparrow 3
f
(
x
)
=
3
↑
↑
⋯
↑
3
\displaystyle f (x) = 3 \uparrow \uparrow \cdots \uparrow 3
f
(
x
)
=
3
↑
↑
⋯
↑
3
\displaystyle \mathrm{def}
d
e
f
\displaystyle \mathrm{def}
d
e
f
\displaystyle f (x)
f
(
x
)
\displaystyle f (x)
f
(
x
)
を 入れ子 みたいにするともっとつよい
\displaystyle f (f(3)) = 3 \uparrow \uparrow \cdots \uparrow 3
f
(
f
(
3
)
)
=
3
↑
↑
⋯
↑
3
\displaystyle f (f(3)) = 3 \uparrow \uparrow \cdots \uparrow 3
f
(
f
(
3
)
)
=
3
↑
↑
⋯
↑
3
(
\displaystyle x
x
\displaystyle x
x
回
)
(
\displaystyle 3 \uparrow \uparrow \uparrow 3
3
↑
↑
↑
3
\displaystyle 3 \uparrow \uparrow \uparrow 3
3
↑
↑
↑
3
回
)
\displaystyle =
=
\displaystyle =
=
ウワーッ
でかい数
\displaystyle f (x)
f
(
x
)
\displaystyle f (x)
f
(
x
)
を 64 layers の 入れ子 にしたやつ
x = 4
x
=
4
x = 4
x
=
4
(
)
Graham Number
数学の証明で使われた一番大きい数としてギネス世界記録に認定
(1980)
http://math.ucsd.edu/~fan/ron/images/record.jpg
Skewes Number
"The largest number which has ever served any definite purpose in mathematics."
G. H. Hardy
1877 - 1947
Prime Number
Prime Number
Definition.
\displaystyle 1
1
\displaystyle 1
1
とその数自身の2つ以外に約数をもたない自然数のことを,
素数とよぶ.
Theorem. (Euclid, 3 BC.)
素数は無数に存在する.
Prime Number
Theorem. (Euclid, 3 BC.)
全ての自然数は, 素数の積としてただ一通りに
表すことができる.
Prime Number
Theorem. (Euclid, 3 BC.)
全ての自然数は, 素数の積としてただ一通りに
表すことができる.
Example.
\displaystyle 57 = 3 \times 19,
57
=
3
×
19
,
\displaystyle 57 = 3 \times 19,
5
7
=
3
×
1
9
,
\displaystyle 49 = 7 \times 7,
49
=
7
×
7
,
\displaystyle 49 = 7 \times 7,
4
9
=
7
×
7
,
\displaystyle 91 = 7 \times 13.
91
=
7
×
13.
\displaystyle 91 = 7 \times 13.
9
1
=
7
×
1
3
.
Prime Number
Conjecture. (Gauß, 1792?)
\displaystyle \pi (x) = \# \{p \leqq x ~ | ~ p : \mathrm{prime} \}
π
(
x
)
=
#
{
p
≦
x
∣
p
:
p
r
i
m
e
}
\displaystyle \pi (x) = \# \{p \leqq x ~ | ~ p : \mathrm{prime} \}
π
(
x
)
=
#
{
p
≦
x
∣
p
:
p
r
i
m
e
}
\displaystyle \mathrm{def}
d
e
f
\displaystyle \mathrm{def}
d
e
f
に対して,
\displaystyle \pi (x) \sim \frac{x}{\log x} ~~~ (x \to \infty).
π
(
x
)
∼
x
log
x
(
x
→
∞
)
.
\displaystyle \pi (x) \sim \frac{x}{\log x} ~~~ (x \to \infty).
π
(
x
)
∼
lo
g
x
x
(
x
→
∞
)
.
この ``ニョロン" は
\displaystyle \lim_{x \to \infty} \frac{\pi (x)}{x / \log x} = 1
lim
x
→
∞
π
(
x
)
x
/
log
x
=
1
\displaystyle \lim_{x \to \infty} \frac{\pi (x)}{x / \log x} = 1
x
→
∞
lim
x
/
lo
g
x
π
(
x
)
=
1
という意味.
Prime Number
\displaystyle \pi (x)
π
(
x
)
\displaystyle \pi (x)
π
(
x
)
\displaystyle x
x
\displaystyle x
x
比
\displaystyle \frac{\pi (x)}{x/\log x}
π
(
x
)
x
/
log
x
\displaystyle \frac{\pi (x)}{x/\log x}
x
/
lo
g
x
π
(
x
)
\displaystyle 10
10
\displaystyle 10
1
0
\displaystyle 4
4
\displaystyle 4
4
\displaystyle 0.921
0.921
\displaystyle 0.921
0
.
9
2
1
\displaystyle 10^{2}
1
0
2
\displaystyle 10^{2}
1
0
2
\displaystyle 10^{3}
1
0
3
\displaystyle 10^{3}
1
0
3
\displaystyle 10^{9}
1
0
9
\displaystyle 10^{9}
1
0
9
\displaystyle 10^{24}
1
0
24
\displaystyle 10^{24}
1
0
2
4
\displaystyle 25
25
\displaystyle 25
2
5
\displaystyle 168
168
\displaystyle 168
1
6
8
\displaystyle 50847534
50847534
\displaystyle 50847534
5
0
8
4
7
5
3
4
\displaystyle 18435599767349200867866
18435599767349200867866
\displaystyle 18435599767349200867866
1
8
4
3
5
5
9
9
7
6
7
3
4
9
2
0
0
8
6
7
8
6
6
\displaystyle 1.151
1.151
\displaystyle 1.151
1
.
1
5
1
\displaystyle 1.161
1.161
\displaystyle 1.161
1
.
1
6
1
\displaystyle 1.054
1.054
\displaystyle 1.054
1
.
0
5
4
\displaystyle 1.019
1.019
\displaystyle 1.019
1
.
0
1
9
Prime Number
Theorem. (Hadamard-de la Valèe Poussin, 1896)
予想は正しい.
これを 素数定理 とよぶ.
証明には Riemann zeta が使われた.
Riemann zeta
Bernhard Riemann
1826 - 1866
``Über die Anzahl der Primzahlen unter einer gegebenen Größe"
Riemann zeta
Definition.
\displaystyle \zeta (s) = \sum_{k = 1}^{\infty} \frac{1}{k^{s}} ~~~~~ (\mathrm{Re} (s) > 1).
ζ
(
s
)
=
∑
k
=
1
∞
1
k
s
(
R
e
(
s
)
>
1
)
.
\displaystyle \zeta (s) = \sum_{k = 1}^{\infty} \frac{1}{k^{s}} ~~~~~ (\mathrm{Re} (s) > 1).
ζ
(
s
)
=
k
=
1
∑
∞
k
s
1
(
R
e
(
s
)
>
1
)
.
Theorem. (Euler, 1734.)
\displaystyle \zeta (2) = \frac{\pi^{2}}{6}.
ζ
(
2
)
=
π
2
6
.
\displaystyle \zeta (2) = \frac{\pi^{2}}{6}.
ζ
(
2
)
=
6
π
2
.
\displaystyle \mathrm{def}
d
e
f
\displaystyle \mathrm{def}
d
e
f
Riemann zeta
Theorem. (Apéry, 1978.)
\displaystyle \zeta (3) \notin \mathbb{Q}.
ζ
(
3
)
∉
Q
.
\displaystyle \zeta (3) \notin \mathbb{Q}.
ζ
(
3
)
∈
/
Q
.
Theorem. (Zudilin, 2001.)
\displaystyle \zeta (5), ~ \zeta (7), ~ \zeta (9), ~ \zeta (11)
ζ
(
5
)
,
ζ
(
7
)
,
ζ
(
9
)
,
ζ
(
11
)
\displaystyle \zeta (5), ~ \zeta (7), ~ \zeta (9), ~ \zeta (11)
ζ
(
5
)
,
ζ
(
7
)
,
ζ
(
9
)
,
ζ
(
1
1
)
のどれか1つは無理数.
Riemann zeta
Theorem. (Euler, 1737.)
\displaystyle \zeta (s) = \prod_{p} \frac{1}{1 - \frac{1}{p^{s}}}.
ζ
(
s
)
=
∏
p
1
1
−
1
p
s
.
\displaystyle \zeta (s) = \prod_{p} \frac{1}{1 - \frac{1}{p^{s}}}.
ζ
(
s
)
=
p
∏
1
−
p
s
1
1
.
Riemann zeta
Theorem. (Riemann, 1859.)
\displaystyle \pi^{-\frac{s}{2}} \Gamma \left(\frac{s}{2}\right) \zeta (s) = \pi^{-\frac{1 - s}{2}} \Gamma \left(\frac{1 - s}{2}\right) \zeta (1 - s).
π
−
s
2
Γ
(
s
2
)
ζ
(
s
)
=
π
−
1
−
s
2
Γ
(
1
−
s
2
)
ζ
(
1
−
s
)
.
\displaystyle \pi^{-\frac{s}{2}} \Gamma \left(\frac{s}{2}\right) \zeta (s) = \pi^{-\frac{1 - s}{2}} \Gamma \left(\frac{1 - s}{2}\right) \zeta (1 - s).
π
−
2
s
Γ
(
2
s
)
ζ
(
s
)
=
π
−
2
1
−
s
Γ
(
2
1
−
s
)
ζ
(
1
−
s
)
.
Riemann zeta
Theorem. (Riemann, 1859.)
\displaystyle \displaystyle \zeta (s) = 2^s\pi^{s-1}\sin \frac{\pi s}{2}\Gamma (1-s)\zeta (1-s).
ζ
(
s
)
=
2
s
π
s
−
1
sin
π
s
2
Γ
(
1
−
s
)
ζ
(
1
−
s
)
.
\displaystyle \displaystyle \zeta (s) = 2^s\pi^{s-1}\sin \frac{\pi s}{2}\Gamma (1-s)\zeta (1-s).
ζ
(
s
)
=
2
s
π
s
−
1
sin
2
π
s
Γ
(
1
−
s
)
ζ
(
1
−
s
)
.
\displaystyle -2n \cdots
−
2
n
⋯
\displaystyle -2n \cdots
−
2
n
⋯
自明な零点.
\displaystyle \rho \cdots
ρ
⋯
\displaystyle \rho \cdots
ρ
⋯
非自明な零点.
\displaystyle 0 \leqq \mathrm{Re}(\rho) \leqq 1.
0
≦
R
e
(
ρ
)
≦
1.
\displaystyle 0 \leqq \mathrm{Re}(\rho) \leqq 1.
0
≦
R
e
(
ρ
)
≦
1
.
Theorem.
Riemann zeta
Conjecture. (Riemann, 1859.)
\displaystyle \mathrm{Re} ( \rho) = \frac{1}{2}.
R
e
(
ρ
)
=
1
2
.
\displaystyle \mathrm{Re} ( \rho) = \frac{1}{2}.
R
e
(
ρ
)
=
2
1
.
Riemann zeta
Theorem. (Riemann, 1859.)
\displaystyle \pi (x) = \sum_{m \leqq \log_{2} x} \frac{\mu(m)}{m} \left(\mathrm{li} (x^{\frac{1}{m}}) - \sum_{\rho} \mathrm{li} (x^{\frac{\rho}{m}}) + \int_{x^{\frac{1}{m}}}^{\infty} \frac{1}{t(t^{2} - 1)} dt - \log 2\right).
π
(
x
)
=
∑
m
≦
log
2
x
μ
(
m
)
m
(
l
i
(
x
1
m
)
−
∑
ρ
l
i
(
x
ρ
m
)
+
∫
x
1
m
∞
1
t
(
t
2
−
1
)
d
t
−
log
2
)
.
\displaystyle \pi (x) = \sum_{m \leqq \log_{2} x} \frac{\mu(m)}{m} \left(\mathrm{li} (x^{\frac{1}{m}}) - \sum_{\rho} \mathrm{li} (x^{\frac{\rho}{m}}) + \int_{x^{\frac{1}{m}}}^{\infty} \frac{1}{t(t^{2} - 1)} dt - \log 2\right).
π
(
x
)
=
m
≦
lo
g
2
x
∑
m
μ
(
m
)
(
l
i
(
x
m
1
)
−
ρ
∑
l
i
(
x
m
ρ
)
+
∫
x
m
1
∞
t
(
t
2
−
1
)
1
d
t
−
lo
g
2
)
.
Prime Number Theorem
Theorem. (Hadamard-de la Valeè Poussin, 1896)
に対して
\displaystyle \mathrm{Re} (s) = 1
R
e
(
s
)
=
1
\displaystyle \mathrm{Re} (s) = 1
R
e
(
s
)
=
1
\displaystyle \zeta (s) \neq 0.
ζ
(
s
)
≠
0.
\displaystyle \zeta (s) \neq 0.
ζ
(
s
)
̸
=
0
.
この事実から素数定理が証明される.
Prime Number Theorem
その後の影響.
Deligne (1974) - Weil Conjectures
Taylor et al. (2011) - Sato-Tate Conjecture
絶対収束域ギリギリの線上での非零性から,
数論的な帰結を得る.
Prime Number Theorem
素数はとてもおもしろい. それは,
・とても不規則に並んでいて, 次に何が出てくる
というところと,
かわからない,
・とても規則的に並んでいる
というところにある.
Don Zagier:
Prime Number Theorem
Dirichlet達は次の関数を考えた:
\displaystyle \mathrm{Li} (x) = \int_{2}^{x} \frac{1}{\log t} dt.
L
i
(
x
)
=
∫
2
x
1
log
t
d
t
.
\displaystyle \mathrm{Li} (x) = \int_{2}^{x} \frac{1}{\log t} dt.
L
i
(
x
)
=
∫
2
x
lo
g
t
1
d
t
.
\displaystyle \mathrm{def}
d
e
f
\displaystyle \mathrm{def}
d
e
f
†
\displaystyle \pi (x) \sim \frac{x}{\log x} ~~~ (x \to \infty) \Longleftrightarrow \pi (x) \sim \mathrm{Li} (x) ~~~ (x \to \infty)
π
(
x
)
∼
x
log
x
(
x
→
∞
)
⟺
π
(
x
)
∼
L
i
(
x
)
(
x
→
∞
)
\displaystyle \pi (x) \sim \frac{x}{\log x} ~~~ (x \to \infty) \Longleftrightarrow \pi (x) \sim \mathrm{Li} (x) ~~~ (x \to \infty)
π
(
x
)
∼
lo
g
x
x
(
x
→
∞
)
⟺
π
(
x
)
∼
L
i
(
x
)
(
x
→
∞
)
Prime Number Theorem
\displaystyle \mathrm{Li} (x)
L
i
(
x
)
\displaystyle \mathrm{Li} (x)
L
i
(
x
)
は
を非常に良く近似する:
\displaystyle \pi (x)
π
(
x
)
\displaystyle \pi (x)
π
(
x
)
\displaystyle \pi (10) = 4
π
(
10
)
=
4
\displaystyle \pi (10) = 4
π
(
1
0
)
=
4
\displaystyle \mathrm{Li} (10) = 5
L
i
(
10
)
=
5
\displaystyle \mathrm{Li} (10) = 5
L
i
(
1
0
)
=
5
\displaystyle \pi (100) = 25
π
(
100
)
=
25
\displaystyle \pi (100) = 25
π
(
1
0
0
)
=
2
5
\displaystyle \mathrm{Li} (100) = 29
L
i
(
100
)
=
29
\displaystyle \mathrm{Li} (100) = 29
L
i
(
1
0
0
)
=
2
9
\displaystyle \pi (1000000000) = 50847534
π
(
1000000000
)
=
50847534
\displaystyle \pi (1000000000) = 50847534
π
(
1
0
0
0
0
0
0
0
0
0
)
=
5
0
8
4
7
5
3
4
\displaystyle \mathrm{Li} (1000000000) = 50849234
L
i
(
1000000000
)
=
50849234
\displaystyle \mathrm{Li} (1000000000) = 50849234
L
i
(
1
0
0
0
0
0
0
0
0
0
)
=
5
0
8
4
9
2
3
4
\displaystyle \pi (x) < \mathrm{Li} (x)
π
(
x
)
<
L
i
(
x
)
\displaystyle \pi (x) < \mathrm{Li} (x)
π
(
x
)
<
L
i
(
x
)
にみえる.
Gaußはこの関係が常に成り立つと予想した.
まで正しい.
\displaystyle x = 3000000
x
=
3000000
\displaystyle x = 3000000
x
=
3
0
0
0
0
0
0
Prime Number Theorem
Riemann:
Prime Number Theorem
Riemann:
Prime Number Theorem
Theorem. (Littlewood, 1914.)
は無限回符号を変える.
\displaystyle \pi (x) - \mathrm{Li} (x)
π
(
x
)
−
L
i
(
x
)
\displaystyle \pi (x) - \mathrm{Li} (x)
π
(
x
)
−
L
i
(
x
)
John Edensor Littlewood
(1875 - 1977)
Skewesの指導教官.
Prime Number Theorem
Theorem. (Skewes, 1933.)
は
\displaystyle \pi (x) > \mathrm{Li} (x)
π
(
x
)
>
L
i
(
x
)
\displaystyle \pi (x) > \mathrm{Li} (x)
π
(
x
)
>
L
i
(
x
)
なる
\displaystyle x
x
\displaystyle x
x
\displaystyle e^{e^{e^{79}}} = 10^{10^{10^{34}}}
e
e
e
79
=
1
0
1
0
1
0
34
\displaystyle e^{e^{e^{79}}} = 10^{10^{10^{34}}}
e
e
e
7
9
=
1
0
1
0
1
0
3
4
以下に存在する.
これがSkewes number.
Prime Number Theorem
評価の改良:
Skewes (1955)
\displaystyle x \leqq e^{e^{e^{e^{7.705}}}} = 10^{10^{10^{963}}}
x
≦
e
e
e
e
7.705
=
1
0
1
0
1
0
963
\displaystyle x \leqq e^{e^{e^{e^{7.705}}}} = 10^{10^{10^{963}}}
x
≦
e
e
e
e
7
.
7
0
5
=
1
0
1
0
1
0
9
6
3
Lehman (1960)
\displaystyle x \leqq 1.65 \times 10^{1065}
x
≦
1.65
×
1
0
1065
\displaystyle x \leqq 1.65 \times 10^{1065}
x
≦
1
.
6
5
×
1
0
1
0
6
5
te Riele (1987)
\displaystyle x \leqq 6.69 \times 10^{370}
x
≦
6.69
×
1
0
370
\displaystyle x \leqq 6.69 \times 10^{370}
x
≦
6
.
6
9
×
1
0
3
7
0
Prime Number Races
Prime Number Races
\displaystyle \mod 3
m
o
d
  
3
\displaystyle \mod 3
m
o
d
3
Race.
Team
\displaystyle 1
1
\displaystyle 1
1
\displaystyle 2, 5, 11, 17, 23, 29, 41, 47\cdots
2
,
5
,
11
,
17
,
23
,
29
,
41
,
47
⋯
\displaystyle 2, 5, 11, 17, 23, 29, 41, 47\cdots
2
,
5
,
1
1
,
1
7
,
2
3
,
2
9
,
4
1
,
4
7
⋯
Team
\displaystyle 2
2
\displaystyle 2
2
\displaystyle 7, 13, 19, 31, 37, 43, 61, \cdots
7
,
13
,
19
,
31
,
37
,
43
,
61
,
⋯
\displaystyle 7, 13, 19, 31, 37, 43, 61, \cdots
7
,
1
3
,
1
9
,
3
1
,
3
7
,
4
3
,
6
1
,
⋯
Prime Number Races
\displaystyle \mod 3
m
o
d
  
3
\displaystyle \mod 3
m
o
d
3
の素数を数える:
\displaystyle \pi_{3, ~ 2}^{\mathrm{mod}} (x)
π
3
,
2
m
o
d
(
x
)
\displaystyle \pi_{3, ~ 2}^{\mathrm{mod}} (x)
π
3
,
2
m
o
d
(
x
)
\displaystyle \pi_{3, ~1}^{\mathrm{mod}} (x)
π
3
,
1
m
o
d
(
x
)
\displaystyle \pi_{3, ~1}^{\mathrm{mod}} (x)
π
3
,
1
m
o
d
(
x
)
\displaystyle x
x
\displaystyle x
x
\displaystyle 100
100
\displaystyle 100
1
0
0
\displaystyle 11
11
\displaystyle 11
1
1
\displaystyle 13
13
\displaystyle 13
1
3
\displaystyle 80
80
\displaystyle 80
8
0
\displaystyle 87
87
\displaystyle 87
8
7
\displaystyle 1000
1000
\displaystyle 1000
1
0
0
0
\displaystyle 611
611
\displaystyle 611
6
1
1
\displaystyle 617
617
\displaystyle 617
6
1
7
\displaystyle 10000
10000
\displaystyle 10000
1
0
0
0
0
\displaystyle 100000
100000
\displaystyle 100000
1
0
0
0
0
0
\displaystyle4784
4784
\displaystyle4784
4
7
8
4
\displaystyle 4807
4807
\displaystyle 4807
4
8
0
7
\displaystyle 1000000
1000000
\displaystyle 1000000
1
0
0
0
0
0
0
\displaystyle 39266
39266
\displaystyle 39266
3
9
2
6
6
\displaystyle 39231
39231
\displaystyle 39231
3
9
2
3
1
Prime Number Races
\displaystyle \mod 3
m
o
d
  
3
\displaystyle \mod 3
m
o
d
3
の素数レースは, 無限回符号が
入れかわることが知られている.
最初に入れかわる数:
\displaystyle x = 608981813029.
x
=
608981813029.
\displaystyle x = 608981813029.
x
=
6
0
8
9
8
1
8
1
3
0
2
9
.
1976年のクリスマスに, Bays, Hudsonによって発見された.
Thank you for your attention!
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