円周率を含む数式

出典: フリー百科事典『ウィキペディア(Wikipedia)』

円周率
使用
円の面積 円周 他の数式での使用
特性
無理性 超越性
数値
22/7より小さい 近似 覚え方
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関孝和 建部賢弘 金田康正 岩尾エマはるか
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アルキメデス 劉徽 祖沖之 アーリヤバタ マーダヴァ ルドルフ・ファン・コーレン ウィリアム・ジョーンズ ジョン・マチン ウィリアム・シャンクス シュリニヴァーサ・ラマヌジャン ジョン・レンチ (英語版) チュドノフスキー兄弟 (英語版)
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円積問題 バーゼル問題 ファインマン・ポイント
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円周率を含む数式 (えんしゅうりつをふくむすうしき)を分野別にまとめる。数式自体または円周率、円周率の近似のいずれかの記事において重要性が確立されているものだけを述べる。

円の外周(円周) C と直径 d の関係

${\displaystyle C=\pi d}${\displaystyle C=\pi d}

円の面積 A と半径 r の関係

${\displaystyle A=\pi r^{2}}${\displaystyle A=\pi r^{2}}

球の体積 V と半径 r の関係

${\displaystyle V={\frac {4}{3}}\pi r^{3}}${\displaystyle V={\frac {4}{3}}\pi r^{3}}

球の表面積 S と半径 r の関係

${\displaystyle S=4\pi r^{2}}${\displaystyle S=4\pi r^{2}}

宇宙定数

${\displaystyle \Lambda ={{8\pi G} \over {3c^{2}}}\rho }${\displaystyle \Lambda ={{8\pi G} \over {3c^{2}}}\rho }

ハイゼンベルクの不確定性原理

${\displaystyle \Delta x,円\Delta p\geq {\frac {h}{4\pi }}}${\displaystyle \Delta x,円\Delta p\geq {\frac {h}{4\pi }}}

一般相対性理論のアインシュタイン方程式

${\displaystyle R_{\mu \nu }-{\frac {1}{2}}g_{\mu \nu }R+\Lambda g_{\mu \nu }={8\pi G \over c^{4}}T_{\mu \nu }}${\displaystyle R_{\mu \nu }-{\frac {1}{2}}g_{\mu \nu }R+\Lambda g_{\mu \nu }={8\pi G \over c^{4}}T_{\mu \nu }}

クーロンの法則

${\displaystyle F={\frac {|q_{1}q_{2}|}{4\pi \varepsilon _{0}r^{2}}}}${\displaystyle F={\frac {|q_{1}q_{2}|}{4\pi \varepsilon _{0}r^{2}}}}

振幅が小さい範囲での振り子の周期

${\displaystyle T\approx 2\pi {\sqrt {\frac {L}{g}}}}${\displaystyle T\approx 2\pi {\sqrt {\frac {L}{g}}}}

座屈のオイラーの式

${\displaystyle F={\frac {\pi ^{2}EI}{L^{2}}}}${\displaystyle F={\frac {\pi ^{2}EI}{L^{2}}}}

${\displaystyle \int _{-\infty }^{\infty }\operatorname {sech} (x),円dx=\pi }${\displaystyle \int _{-\infty }^{\infty }\operatorname {sech} (x),円dx=\pi }

${\displaystyle \int _{-\infty }^{\infty }\int _{t}^{\infty }e^{-1/2t^{2}-x^{2}+xt},円dx,円dt=\int _{-\infty }^{\infty }\int _{t}^{\infty }e^{-t^{2}-1/2x^{2}+xt},円dx,円dt=\pi }${\displaystyle \int _{-\infty }^{\infty }\int _{t}^{\infty }e^{-1/2t^{2}-x^{2}+xt},円dx,円dt=\int _{-\infty }^{\infty }\int _{t}^{\infty }e^{-t^{2}-1/2x^{2}+xt},円dx,円dt=\pi }

${\displaystyle \int _{-1}^{1}{\sqrt {1-x^{2}}},円dx={\frac {\pi }{2}}}${\displaystyle \int _{-1}^{1}{\sqrt {1-x^{2}}},円dx={\frac {\pi }{2}}}

${\displaystyle \int _{0}^{1}{1 \over 1+x^{2}},円dx={\frac {\pi }{4}}}${\displaystyle \int _{0}^{1}{1 \over 1+x^{2}},円dx={\frac {\pi }{4}}}

${\displaystyle \int _{-1}^{1}{\frac {dx}{\sqrt {1-x^{2}}}}=\pi }${\displaystyle \int _{-1}^{1}{\frac {dx}{\sqrt {1-x^{2}}}}=\pi }

${\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}=\pi }${\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}=\pi } (逆正接関数)

${\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}},円dx={\sqrt {\pi }}}${\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}},円dx={\sqrt {\pi }}}  (ガウス積分)

${\displaystyle \oint {\frac {dz}{z}}=2\pi i}${\displaystyle \oint {\frac {dz}{z}}=2\pi i} (コーシーの積分定理)

${\displaystyle \int _{-\infty }^{\infty }{\frac {\sin x}{x}},円dx=\pi }${\displaystyle \int _{-\infty }^{\infty }{\frac {\sin x}{x}},円dx=\pi }

${\displaystyle \int _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}},円dx={22 \over 7}-\pi }${\displaystyle \int _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}},円dx={22 \over 7}-\pi } (円周率が22/7より小さいことの証明)

${\displaystyle \sum _{k=0}^{\infty }{\frac {k!}{(2k+1)!!}}=\sum _{k=0}^{\infty }{\frac {2^{k}k!^{2}}{(2k+1)!}}={\frac {\pi }{2}}}${\displaystyle \sum _{k=0}^{\infty }{\frac {k!}{(2k+1)!!}}=\sum _{k=0}^{\infty }{\frac {2^{k}k!^{2}}{(2k+1)!}}={\frac {\pi }{2}}}

${\displaystyle 12\sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k+3/2}}}={\frac {1}{\pi }}}${\displaystyle 12\sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k+3/2}}}={\frac {1}{\pi }}} (Chudnovsky algorithm)

${\displaystyle {\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}={\frac {1}{\pi }}}${\displaystyle {\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}={\frac {1}{\pi }}} (シュリニヴァーサ・ラマヌジャン)

${\displaystyle {\frac {12}{\sqrt {(1249638720+159999840{\sqrt {61}})^{3}}}}\sum _{n=0}^{\infty }{\frac {(-1)^{n}(6n)!(1657145277365+212175710912{\sqrt {61}}+(107578229802750+3773980892672{\sqrt {61}})n)}{(3n)!(n!)^{3}(1249638720+159999840{\sqrt {61}})^{n}}}={\frac {1}{\pi }}}${\displaystyle {\frac {12}{\sqrt {(1249638720+159999840{\sqrt {61}})^{3}}}}\sum _{n=0}^{\infty }{\frac {(-1)^{n}(6n)!(1657145277365+212175710912{\sqrt {61}}+(107578229802750+3773980892672{\sqrt {61}})n)}{(3n)!(n!)^{3}(1249638720+159999840{\sqrt {61}})^{n}}}={\frac {1}{\pi }}}(Borwein)

${\displaystyle {\frac {\sqrt {3}}{6^{5}}}\sum _{k=0}^{\infty }{\frac {((4k)!)^{2}(6k)!}{9^{k+1}(12k)!(2k)!}}\left({\frac {127169}{12k+1}}-{\frac {1070}{12k+5}}-{\frac {131}{12k+7}}+{\frac {2}{12k+11}}\right)=\pi }${\displaystyle {\frac {\sqrt {3}}{6^{5}}}\sum _{k=0}^{\infty }{\frac {((4k)!)^{2}(6k)!}{9^{k+1}(12k)!(2k)!}}\left({\frac {127169}{12k+1}}-{\frac {1070}{12k+5}}-{\frac {131}{12k+7}}+{\frac {2}{12k+11}}\right)=\pi }^[1]

以下は、円周率の任意の桁を2進数で求められる効率的な数式である。

${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{16^{k}}}\left({\frac {4}{8k+1}}-{\frac {2}{8k+4}}-{\frac {1}{8k+5}}-{\frac {1}{8k+6}}\right)=\pi }${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{16^{k}}}\left({\frac {4}{8k+1}}-{\frac {2}{8k+4}}-{\frac {1}{8k+5}}-{\frac {1}{8k+6}}\right)=\pi } (ベイリー=ボールウェイン=プラウフの公式)

${\displaystyle {\frac {1}{2^{6}}}\sum _{n=0}^{\infty }{\frac {{(-1)}^{n}}{2^{10n}}}\left(-{\frac {2^{5}}{4n+1}}-{\frac {1}{4n+3}}+{\frac {2^{8}}{10n+1}}-{\frac {2^{6}}{10n+3}}-{\frac {2^{2}}{10n+5}}-{\frac {2^{2}}{10n+7}}+{\frac {1}{10n+9}}\right)=\pi }${\displaystyle {\frac {1}{2^{6}}}\sum _{n=0}^{\infty }{\frac {{(-1)}^{n}}{2^{10n}}}\left(-{\frac {2^{5}}{4n+1}}-{\frac {1}{4n+3}}+{\frac {2^{8}}{10n+1}}-{\frac {2^{6}}{10n+3}}-{\frac {2^{2}}{10n+5}}-{\frac {2^{2}}{10n+7}}+{\frac {1}{10n+9}}\right)=\pi }

${\displaystyle \zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}}${\displaystyle \zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}} (バーゼルの問題、リーマンゼータ関数)

${\displaystyle \zeta (4)={\frac {1}{1^{4}}}+{\frac {1}{2^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{4^{4}}}+\cdots ={\frac {\pi ^{4}}{90}}}${\displaystyle \zeta (4)={\frac {1}{1^{4}}}+{\frac {1}{2^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{4^{4}}}+\cdots ={\frac {\pi ^{4}}{90}}}

${\displaystyle \zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}},円={\frac {1}{1^{2n}}}+{\frac {1}{2^{2n}}}+{\frac {1}{3^{2n}}}+{\frac {1}{4^{2n}}}+\cdots =(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}}${\displaystyle \zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}},円={\frac {1}{1^{2n}}}+{\frac {1}{2^{2n}}}+{\frac {1}{3^{2n}}}+{\frac {1}{4^{2n}}}+\cdots =(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}} , ただし B_2n は ベルヌーイ数)

${\displaystyle \sum _{n=1}^{\infty }{\frac {3^{n}-1}{4^{n}}},円\zeta (n+1)=\pi }${\displaystyle \sum _{n=1}^{\infty }{\frac {3^{n}-1}{4^{n}}},円\zeta (n+1)=\pi }^[2]

${\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{1}={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\arctan {1}={\frac {\pi }{4}}}${\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{1}={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\arctan {1}={\frac {\pi }{4}}} (ライプニッツ公式)

${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n^{2}}}={\frac {1}{1^{2}}}-{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}-{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{12}}}${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n^{2}}}={\frac {1}{1^{2}}}-{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}-{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{12}}}

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{(2n)^{2}}}={\frac {1}{2^{2}}}+{\frac {1}{4^{2}}}+{\frac {1}{6^{2}}}+{\frac {1}{8^{2}}}+\cdots ={\frac {\pi ^{2}}{24}}}${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{(2n)^{2}}}={\frac {1}{2^{2}}}+{\frac {1}{4^{2}}}+{\frac {1}{6^{2}}}+{\frac {1}{8^{2}}}+\cdots ={\frac {\pi ^{2}}{24}}}

${\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{2}={\frac {1}{1^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{5^{2}}}+{\frac {1}{7^{2}}}+\cdots ={\frac {\pi ^{2}}{8}}}${\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{2}={\frac {1}{1^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{5^{2}}}+{\frac {1}{7^{2}}}+\cdots ={\frac {\pi ^{2}}{8}}}

${\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{3}={\frac {1}{1^{3}}}-{\frac {1}{3^{3}}}+{\frac {1}{5^{3}}}-{\frac {1}{7^{3}}}+\cdots ={\frac {\pi ^{3}}{32}}}${\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{3}={\frac {1}{1^{3}}}-{\frac {1}{3^{3}}}+{\frac {1}{5^{3}}}-{\frac {1}{7^{3}}}+\cdots ={\frac {\pi ^{3}}{32}}}

${\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{4}={\frac {1}{1^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{5^{4}}}+{\frac {1}{7^{4}}}+\cdots ={\frac {\pi ^{4}}{96}}}${\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{4}={\frac {1}{1^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{5^{4}}}+{\frac {1}{7^{4}}}+\cdots ={\frac {\pi ^{4}}{96}}}

${\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{5}={\frac {1}{1^{5}}}-{\frac {1}{3^{5}}}+{\frac {1}{5^{5}}}-{\frac {1}{7^{5}}}+\cdots ={\frac {5\pi ^{5}}{1536}}}${\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{5}={\frac {1}{1^{5}}}-{\frac {1}{3^{5}}}+{\frac {1}{5^{5}}}-{\frac {1}{7^{5}}}+\cdots ={\frac {5\pi ^{5}}{1536}}}

${\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{6}={\frac {1}{1^{6}}}+{\frac {1}{3^{6}}}+{\frac {1}{5^{6}}}+{\frac {1}{7^{6}}}+\cdots ={\frac {\pi ^{6}}{960}}}${\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{6}={\frac {1}{1^{6}}}+{\frac {1}{3^{6}}}+{\frac {1}{5^{6}}}+{\frac {1}{7^{6}}}+\cdots ={\frac {\pi ^{6}}{960}}}

${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(4n+1)(4n+3)}}={\frac {1}{1\cdot 3}}+{\frac {1}{5\cdot 7}}+{\frac {1}{9\cdot 11}}+\cdots ={\frac {\pi }{8}}}${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(4n+1)(4n+3)}}={\frac {1}{1\cdot 3}}+{\frac {1}{5\cdot 7}}+{\frac {1}{9\cdot 11}}+\cdots ={\frac {\pi }{8}}}

${\displaystyle \pi ={1}+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}-{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}-{\frac {1}{10}}+{\frac {1}{11}}+{\frac {1}{12}}-{\frac {1}{13}}+\cdots }${\displaystyle \pi ={1}+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}-{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}-{\frac {1}{10}}+{\frac {1}{11}}+{\frac {1}{12}}-{\frac {1}{13}}+\cdots }   (オイラー、1748年)

この式では、最初の2つの項の後、符号は次のように決定される。分母が4m - 1で表される素数である場合、符号は正であり。分母が4m + 1で表される素数である場合、符号は負である。 合成数の場合、符号はその素因数分解した素数の符号の積に等しい^[3] 。

また

${\displaystyle \sum _{n=1}^{\infty }{\frac {F_{2n}}{n^{2}{\binom {2n}{n}}}}={\frac {4\pi ^{2}}{25{\sqrt {5}}}}}${\displaystyle \sum _{n=1}^{\infty }{\frac {F_{2n}}{n^{2}{\binom {2n}{n}}}}={\frac {4\pi ^{2}}{25{\sqrt {5}}}}}

ただし ${\displaystyle F_{n}}${\displaystyle F_{n}} はn番目のフィボナッチ数。

${\displaystyle {\frac {\pi }{4}}=\arctan 1}${\displaystyle {\frac {\pi }{4}}=\arctan 1}

${\displaystyle {\frac {\pi }{4}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{3}}}${\displaystyle {\frac {\pi }{4}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{3}}}

${\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{2}}-\arctan {\frac {1}{7}}}${\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{2}}-\arctan {\frac {1}{7}}}

${\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}}${\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}}

${\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}}${\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}} (マチンの公式)

${\displaystyle {\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}}${\displaystyle {\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}}

${\displaystyle {\frac {\pi }{4}}=6\arctan {\frac {1}{8}}+2\arctan {\frac {1}{57}}+\arctan {\frac {1}{239}}}${\displaystyle {\frac {\pi }{4}}=6\arctan {\frac {1}{8}}+2\arctan {\frac {1}{57}}+\arctan {\frac {1}{239}}}

${\displaystyle {\frac {\pi }{4}}=12\arctan {\frac {1}{49}}+32\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}+12\arctan {\frac {1}{110443}}}${\displaystyle {\frac {\pi }{4}}=12\arctan {\frac {1}{49}}+32\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}+12\arctan {\frac {1}{110443}}}

${\displaystyle {\frac {\pi }{4}}=44\arctan {\frac {1}{57}}+7\arctan {\frac {1}{239}}-12\arctan {\frac {1}{682}}+24\arctan {\frac {1}{12943}}}${\displaystyle {\frac {\pi }{4}}=44\arctan {\frac {1}{57}}+7\arctan {\frac {1}{239}}-12\arctan {\frac {1}{682}}+24\arctan {\frac {1}{12943}}}

${\displaystyle {\frac {\pi }{2}}=\sum _{n=0}^{\infty }\arctan {\frac {1}{F_{2n+1}}}=\arctan {\frac {1}{1}}+\arctan {\frac {1}{2}}+\arctan {\frac {1}{5}}+\arctan {\frac {1}{13}}+\cdots }${\displaystyle {\frac {\pi }{2}}=\sum _{n=0}^{\infty }\arctan {\frac {1}{F_{2n+1}}}=\arctan {\frac {1}{1}}+\arctan {\frac {1}{2}}+\arctan {\frac {1}{5}}+\arctan {\frac {1}{13}}+\cdots }

ただし ${\displaystyle F_{n}}${\displaystyle F_{n}} はn番目のフィボナッチ数。

円周率を含む級数^[4]

${\displaystyle (x)_{n}}${\displaystyle (x)_{n}} は階乗冪。

${\displaystyle {\frac {\pi }{4}}={\frac {3}{4}}\cdot {\frac {5}{4}}\cdot {\frac {7}{8}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdot {\frac {17}{16}}\cdot {\frac {19}{20}}\cdot {\frac {23}{24}}\cdot {\frac {29}{28}}\cdot {\frac {31}{32}}\cdot ,円\cdots }${\displaystyle {\frac {\pi }{4}}={\frac {3}{4}}\cdot {\frac {5}{4}}\cdot {\frac {7}{8}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdot {\frac {17}{16}}\cdot {\frac {19}{20}}\cdot {\frac {23}{24}}\cdot {\frac {29}{28}}\cdot {\frac {31}{32}}\cdot ,円\cdots } (オイラー)

全ての奇素数を分子とし、それに最も近い4の倍数を分母とした分数の総乗。

${\displaystyle \prod _{n=1}^{\infty }{\frac {4n^{2}}{4n^{2}-1}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots ={\frac {4}{3}}\cdot {\frac {16}{15}}\cdot {\frac {36}{35}}\cdot {\frac {64}{63}}\cdot ,円\cdots ={\frac {\pi }{2}}}${\displaystyle \prod _{n=1}^{\infty }{\frac {4n^{2}}{4n^{2}-1}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots ={\frac {4}{3}}\cdot {\frac {16}{15}}\cdot {\frac {36}{35}}\cdot {\frac {64}{63}}\cdot ,円\cdots ={\frac {\pi }{2}}} (ウォリス積を参照)

${\displaystyle \prod _{n=1}^{\infty }\cos {\frac {90^{\circ }}{2^{n}}}={\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}\cdot ,円\cdots ={\frac {2}{\pi }}}${\displaystyle \prod _{n=1}^{\infty }\cos {\frac {90^{\circ }}{2^{n}}}={\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}\cdot ,円\cdots ={\frac {2}{\pi }}} (ビエトの公式)

${\displaystyle \pi ={3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+{\cfrac {7^{2}}{6+\ddots ,円}}}}}}}}}}${\displaystyle \pi ={3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+{\cfrac {7^{2}}{6+\ddots ,円}}}}}}}}}}

${\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+{\cfrac {4^{2}}{9+\ddots }}}}}}}}}}}${\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+{\cfrac {4^{2}}{9+\ddots }}}}}}}}}}}

${\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+\ddots }}}}}}}}}}}${\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+\ddots }}}}}}}}}}}

${\displaystyle 2\pi ={6+{\cfrac {2^{2}}{12+{\cfrac {6^{2}}{12+{\cfrac {10^{2}}{12+{\cfrac {14^{2}}{12+{\cfrac {18^{2}}{12+\ddots }}}}}}}}}}}}${\displaystyle 2\pi ={6+{\cfrac {2^{2}}{12+{\cfrac {6^{2}}{12+{\cfrac {10^{2}}{12+{\cfrac {14^{2}}{12+{\cfrac {18^{2}}{12+\ddots }}}}}}}}}}}}

${\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}}${\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}} (スターリングの近似)

${\displaystyle e^{i\pi }+1=0}${\displaystyle e^{i\pi }+1=0} (オイラーの等式)

${\displaystyle \sum _{k=1}^{n}\varphi (k)\sim {\frac {3n^{2}}{\pi ^{2}}}}${\displaystyle \sum _{k=1}^{n}\varphi (k)\sim {\frac {3n^{2}}{\pi ^{2}}}} (オイラーのトーティエント関数)

${\displaystyle \sum _{k=1}^{n}{\frac {\varphi (k)}{k}}\sim {\frac {6n}{\pi ^{2}}}}${\displaystyle \sum _{k=1}^{n}{\frac {\varphi (k)}{k}}\sim {\frac {6n}{\pi ^{2}}}} (オイラーのトーティエント関数)

${\displaystyle \Gamma \left({1 \over 2}\right)={\sqrt {\pi }}}${\displaystyle \Gamma \left({1 \over 2}\right)={\sqrt {\pi }}} (ガンマ関数)

${\displaystyle \pi ={\frac {\Gamma \left({1/4}\right)^{4/3}\operatorname {agm} (1,{\sqrt {2}})^{2/3}}{2}}}${\displaystyle \pi ={\frac {\Gamma \left({1/4}\right)^{4/3}\operatorname {agm} (1,{\sqrt {2}})^{2/3}}{2}}} (agmは算術幾何平均)

${\displaystyle \lim _{n\rightarrow \infty }{\frac {1}{n^{2}}}\sum _{k=1}^{n}(n\;{\bmod {\;}}k)=1-{\frac {\pi ^{2}}{12}}}${\displaystyle \lim _{n\rightarrow \infty }{\frac {1}{n^{2}}}\sum _{k=1}^{n}(n\;{\bmod {\;}}k)=1-{\frac {\pi ^{2}}{12}}} (mod は剰余演算)

${\displaystyle \pi =\lim _{n\rightarrow \infty }{\frac {4}{n^{2}}}\sum _{k=1}^{n}{\sqrt {n^{2}-k^{2}}}}${\displaystyle \pi =\lim _{n\rightarrow \infty }{\frac {4}{n^{2}}}\sum _{k=1}^{n}{\sqrt {n^{2}-k^{2}}}} (単位円の面積とリーマン和を参照)

${\displaystyle \pi =\lim _{n\rightarrow \infty }{\frac {2^{4n}}{n{2n \choose n}^{2}}}}${\displaystyle \pi =\lim _{n\rightarrow \infty }{\frac {2^{4n}}{n{2n \choose n}^{2}}}} (スターリングの近似)

• Peter Borwein, The Amazing Number Pi
• Kazuya Kato, Nobushige Kurokawa, Saito Takeshi: Number Theory 1: Fermat's Dream. American Mathematical Society, Providence 1993, ISBN 0-8218-0863-X.

「https://ja.wikipedia.org/w/index.php?title=円周率を含む数式&oldid=106162774」から取得

カテゴリ:

• 数学の一覧
• 円周率
• 数学に関する記事