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(li(x1m)−∑ρli(xρm)+∫x1m∞1t(t2−1)dt−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)
\displaystyle \pi (x) < \mathrm{Li} (x)
π(x)&lt;Li(x)\displaystyle \pi (x) &lt; \mathrm{Li} (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
Li(x)−12Li(x12)−13Li(x13)−15Li(x15)+16Li(x16)−⋯\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
\displaystyle \zeta (s) = \sum_{k = 1}^{\infty} \frac{1}{k^{s}}
ζ(s)=∑k=1∞1ks\displaystyle \zeta (s) = \sum_{k = 1}^{\infty} \frac{1}{k^{s}}
\displaystyle \pi (e^{e^{e^{79}}}) > \mathrm{Li} (e^{e^{e^{79}}})
π(eee79)&gt;Li(eee79)\displaystyle \pi (e^{e^{e^{79}}}) &gt; \mathrm{Li} (e^{e^{e^{79}}})
?
??
\displaystyle \pi (x) \sim \frac{x}{\log x}
π(x)∼xlog⁡x\displaystyle \pi (x) \sim \frac{x}{\log x}

2017. 10. 08.

@Romantic Mathnight

in

\displaystyle \mathrm{MATH~ POWE}\mathbb{R}^{2017}
MATH POWER2017\displaystyle \mathrm{MATH~ POWE}\mathbb{R}^{2017}
\displaystyle \mathrm{def}
def\displaystyle \mathrm{def}
\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−12; (Symn−1ρE,ℓ)∣F′)=L(s;Πm,n)\displaystyle L \left(s + \frac{n - 1}{2};~\left(\mathrm{Sym}^{n - 1} \rho_{E, \ell}\right)|_{F&#x27;}\right) = L (s; \mathrm{\Pi}_{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

・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+ba + 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

(

\displaystyle b
b\displaystyle 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

(

\displaystyle b
b\displaystyle b

回

)

\displaystyle a^{b}
ab\displaystyle 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

(

\displaystyle b
b\displaystyle b

回

)

\displaystyle ^{b}a
ba\displaystyle ^{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
\displaystyle 3 \times 3 = 3 + 3 + 3 = 9
3×3=3+3+3=9\displaystyle 3 \times 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
\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
\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

でかい数

\displaystyle \uparrow
↑\displaystyle \uparrow

を

\displaystyle x
x\displaystyle 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
\displaystyle \mathrm{def}
def\displaystyle \mathrm{def}
\displaystyle f (x)
f(x)\displaystyle 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

(

\displaystyle x
x\displaystyle x

回

)

(

\displaystyle 3 \uparrow \uparrow \uparrow 3
3↑↑↑3\displaystyle 3 \uparrow \uparrow \uparrow 3

回

)

\displaystyle =
=\displaystyle =

ウワーッ

でかい数

\displaystyle f (x)
f(x)\displaystyle f (x)

を 64 layers の 入れ子 にしたやつ

x = 4
x=4x = 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

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,
\displaystyle 49 = 7 \times 7,
49=7×7,\displaystyle 49 = 7 \times 7,
\displaystyle 91 = 7 \times 13.
91=7×13.\displaystyle 91 = 7 \times 13.

Prime Number

Conjecture. (Gauß, 1792?)

\displaystyle \pi (x) = \# \{p \leqq x ~ | ~ p : \mathrm{prime} \}
π(x)=#{p≦x ∣ p:prime}\displaystyle \pi (x) = \# \{p \leqq x ~ | ~ p : \mathrm{prime} \}
\displaystyle \mathrm{def}
def\displaystyle \mathrm{def}

に対して,

\displaystyle \pi (x) \sim \frac{x}{\log x} ~~~ (x \to \infty).
π(x)∼xlog⁡x   (x→∞).\displaystyle \pi (x) \sim \frac{x}{\log x} ~~~ (x \to \infty).

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

という意味.

Prime Number

\displaystyle \pi (x)
π(x)\displaystyle \pi (x)
\displaystyle x
x\displaystyle x

比

\displaystyle \frac{\pi (x)}{x/\log x}
π(x)x/log⁡x\displaystyle \frac{\pi (x)}{x/\log x}
\displaystyle 10
10\displaystyle 10
\displaystyle 4
4\displaystyle 4
\displaystyle 0.921
0.921\displaystyle 0.921
\displaystyle 10^{2}
102\displaystyle 10^{2}
\displaystyle 10^{3}
103\displaystyle 10^{3}
\displaystyle 10^{9}
109\displaystyle 10^{9}
\displaystyle 10^{24}
1024\displaystyle 10^{24}
\displaystyle 25
25\displaystyle 25
\displaystyle 168
168\displaystyle 168
\displaystyle 50847534
50847534\displaystyle 50847534
\displaystyle 18435599767349200867866
18435599767349200867866\displaystyle 18435599767349200867866
\displaystyle 1.151
1.151\displaystyle 1.151
\displaystyle 1.161
1.161\displaystyle 1.161
\displaystyle 1.054
1.054\displaystyle 1.054
\displaystyle 1.019
1.019\displaystyle 1.019

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∞1ks     (Re(s)&gt;1).\displaystyle \zeta (s) = \sum_{k = 1}^{\infty} \frac{1}{k^{s}} ~~~~~ (\mathrm{Re} (s) &gt; 1).

Theorem. (Euler, 1734.)

\displaystyle \zeta (2) = \frac{\pi^{2}}{6}.
ζ(2)=π26.\displaystyle \zeta (2) = \frac{\pi^{2}}{6}.
\displaystyle \mathrm{def}
def\displaystyle \mathrm{def}

Riemann zeta

Theorem. (Apéry, 1978.)

\displaystyle \zeta (3) \notin \mathbb{Q}.
ζ(3)∉Q.\displaystyle \zeta (3) \notin \mathbb{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)

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)=∏p11−1ps.\displaystyle \zeta (s) = \prod_{p} \frac{1}{1 - \frac{1}{p^{s}}}.

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−s2Γ(s2)ζ(s)=π−1−s2Γ(1−s2)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).

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)=2sπs−1sin⁡πs2Γ(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).
\displaystyle -2n \cdots
−2n⋯\displaystyle -2n \cdots

自明な零点.

\displaystyle \rho \cdots
ρ⋯\displaystyle \rho \cdots

非自明な零点.

\displaystyle 0 \leqq \mathrm{Re}(\rho) \leqq 1.
0≦Re(ρ)≦1.\displaystyle 0 \leqq \mathrm{Re}(\rho) \leqq 1.

Theorem.

Riemann zeta

Conjecture. (Riemann, 1859.)

\displaystyle \mathrm{Re} ( \rho) = \frac{1}{2}.
Re(ρ)=12.\displaystyle \mathrm{Re} ( \rho) = \frac{1}{2}.

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⁡2xμ(m)m(li(x1m)−∑ρli(xρm)+∫x1m∞1t(t2−1)dt−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).

Prime Number Theorem

Theorem. (Hadamard-de la Valeè Poussin, 1896)

に対して

\displaystyle \mathrm{Re} (s) = 1
Re(s)=1\displaystyle \mathrm{Re} (s) = 1
\displaystyle \zeta (s) \neq 0.
ζ(s)≠0.\displaystyle \zeta (s) \neq 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.
Li(x)=∫2x1log⁡tdt.\displaystyle \mathrm{Li} (x) = \int_{2}^{x} \frac{1}{\log t} dt.
\displaystyle \mathrm{def}
def\displaystyle \mathrm{def}

†

\displaystyle \pi (x) \sim \frac{x}{\log x} ~~~ (x \to \infty) \Longleftrightarrow \pi (x) \sim \mathrm{Li} (x) ~~~ (x \to \infty)
π(x)∼xlog⁡x   (x→∞)⟺π(x)∼Li(x)   (x→∞)\displaystyle \pi (x) \sim \frac{x}{\log x} ~~~ (x \to \infty) \Longleftrightarrow \pi (x) \sim \mathrm{Li} (x) ~~~ (x \to \infty)

Prime Number Theorem

\displaystyle \mathrm{Li} (x)
Li(x)\displaystyle \mathrm{Li} (x)

は

を非常に良く近似する：

\displaystyle \pi (x)
π(x)\displaystyle \pi (x)
\displaystyle \pi (10) = 4
π(10)=4\displaystyle \pi (10) = 4
\displaystyle \mathrm{Li} (10) = 5
Li(10)=5\displaystyle \mathrm{Li} (10) = 5
\displaystyle \pi (100) = 25
π(100)=25\displaystyle \pi (100) = 25
\displaystyle \mathrm{Li} (100) = 29
Li(100)=29\displaystyle \mathrm{Li} (100) = 29
\displaystyle \pi (1000000000) = 50847534
π(1000000000)=50847534\displaystyle \pi (1000000000) = 50847534
\displaystyle \mathrm{Li} (1000000000) = 50849234
Li(1000000000)=50849234\displaystyle \mathrm{Li} (1000000000) = 50849234
\displaystyle \pi (x) < \mathrm{Li} (x)
π(x)&lt;Li(x)\displaystyle \pi (x) &lt; \mathrm{Li} (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

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)−Li(x)\displaystyle \pi (x) - \mathrm{Li} (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)&gt;Li(x)\displaystyle \pi (x) &gt; \mathrm{Li} (x)

なる

\displaystyle x
x\displaystyle x
\displaystyle e^{e^{e^{79}}} = 10^{10^{10^{34}}}
eee79=10101034\displaystyle e^{e^{e^{79}}} = 10^{10^{10^{34}}}

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≦eeee7.705=101010963\displaystyle x \leqq e^{e^{e^{e^{7.705}}}} = 10^{10^{10^{963}}}

Lehman (1960)

\displaystyle x \leqq 1.65 \times 10^{1065}
x≦1.65×101065\displaystyle x \leqq 1.65 \times 10^{1065}

te Riele (1987)

\displaystyle x \leqq 6.69 \times 10^{370}
x≦6.69×10370\displaystyle x \leqq 6.69 \times 10^{370}

Prime Number Races

Prime Number Races

\displaystyle \mod 3
mod&ThinSpace;&ThinSpace;3\displaystyle \mod 3

Race.

Team 

\displaystyle 1
1\displaystyle 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

Team 

\displaystyle 2
2\displaystyle 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

Prime Number Races

\displaystyle \mod 3
mod&ThinSpace;&ThinSpace;3\displaystyle \mod 3

の素数を数える: 

\displaystyle \pi_{3, ~ 2}^{\mathrm{mod}} (x)
π3, 2mod(x)\displaystyle \pi_{3, ~ 2}^{\mathrm{mod}} (x)
\displaystyle \pi_{3, ~1}^{\mathrm{mod}} (x)
π3, 1mod(x)\displaystyle \pi_{3, ~1}^{\mathrm{mod}} (x)
\displaystyle x
x\displaystyle x
\displaystyle 100
100\displaystyle 100
\displaystyle 11
11\displaystyle 11
\displaystyle 13
13\displaystyle 13
\displaystyle 80
80\displaystyle 80
\displaystyle 87
87\displaystyle 87
\displaystyle 1000
1000\displaystyle 1000
\displaystyle 611
611\displaystyle 611
\displaystyle 617
617\displaystyle 617
\displaystyle 10000
10000\displaystyle 10000
\displaystyle 100000
100000\displaystyle 100000
\displaystyle4784
4784\displaystyle4784
\displaystyle 4807
4807\displaystyle 4807
\displaystyle 1000000
1000000\displaystyle 1000000
\displaystyle 39266
39266\displaystyle 39266
\displaystyle 39231
39231\displaystyle 39231

Prime Number Races

\displaystyle \mod 3
mod&ThinSpace;&ThinSpace;3\displaystyle \mod 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.

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!

Skewes Number @.Romantic Mathnight in Math Powe\mathbb{R}^{2017}

By zeta_aniki

Made with Slides.com

Skewes Number @.Romantic Mathnight in Math Powe\mathbb{R}^{2017}

酢kewe酢酢

  • 2,864

More from zeta_aniki