円周率を含む数式

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 C = π d {\displaystyle C=\pi d} 円の面積 A と半径 r の関係 A = π r 2 {\displaystyle A=\pi r^{2}} 球の体積 V と半径 r の関係 V = 4 3 π r 3 {\displaystyle V={\frac {4}{3}}\pi r^{3}} 球の表面積 S と半径 r の関係 S = 4 π r 2 {\displaystyle S=4\pi r^{2}} 宇宙定数 Λ = 8 π G 3 c 2 ρ {\displaystyle \Lambda ={{8\pi G} \over {3c^{2}}}\rho } ハイゼンベルクの不確定性原理 Δ x Δ p ≥ h 4 π {\displaystyle \Delta x\,\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 R μ ν − 1 2 g μ ν R + Λ g μ ν = 8 π G c 4 T μ ν {\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 F = | q 1 q 2 | 4 π ε 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 T ≈ 2 π L g {\displaystyle T\approx 2\pi {\sqrt {\frac {L}{g}}}} 座屈のオイラーの式 F = π 2 E I L 2 {\displaystyle F={\frac {\pi ^{2}EI}{L^{2}}}} ∫ − ∞ ∞ sech ⁡ ( x ) d x = π {\displaystyle \int _{-\infty }^{\infty }\operatorname {sech} (x)\,dx=\pi } ∫ − ∞ ∞ ∫ t ∞ e − 1 / 2 t 2 − x 2 + x t d x d t = ∫ − ∞ ∞ ∫ t ∞ e − t 2 − 1 / 2 x 2 + x t d x d t = π {\displaystyle \int _{-\infty }^{\infty }\int _{t}^{\infty }e^{-1/2t^{2}-x^{2}+xt}\,dx\,dt=\int _{-\infty }^{\infty }\int _{t}^{\infty }e^{-t^{2}-1/2x^{2}+xt}\,dx\,dt=\pi } ∫ − 1 1 1 − x 2 d x = π 2 {\displaystyle \int _{-1}^{1}{\sqrt {1-x^{2}}}\,dx={\frac {\pi }{2}}} ∫ − 1 1 d x 1 − x 2 = π {\displaystyle \int _{-1}^{1}{\frac {dx}{\sqrt {1-x^{2}}}}=\pi } ∫ − ∞ ∞ d x 1 + x 2 = π {\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}=\pi }  （逆正接関数） ∫ − ∞ ∞ e − x 2 d x = π {\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}}   （ガウス積分） ∮ d z z = 2 π i {\displaystyle \oint {\frac {dz}{z}}=2\pi i}  （コーシーの積分定理） ∫ − ∞ ∞ sin ⁡ x x d x = π {\displaystyle \int _{-\infty }^{\infty }{\frac {\sin x}{x}}\,dx=\pi } ∫ 0 1 x 4 ( 1 − x ) 4 1 + x 2 d x = 22 7 − π {\displaystyle \int _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}}\,dx={22 \over 7}-\pi }  （円周率が22/7より小さいことの証明） ∑ k = 0 ∞ k ! ( 2 k + 1 ) ! ! = ∑ k = 0 ∞ 2 k k ! 2 ( 2 k + 1 ) ! = π 2 {\displaystyle \sum _{k=0}^{\infty }{\frac {k!}{(2k+1)!!}}=\sum _{k=0}^{\infty }{\frac {2^{k}k!^{2}}{(2k+1)!}}={\frac {\pi }{2}}} 12 ∑ k = 0 ∞ ( − 1 ) k ( 6 k ) ! ( 13591409 + 545140134 k ) ( 3 k ) ! ( k ! ) 3 640320 3 k + 3 / 2 = 1 π {\displaystyle 12\sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k+3/2}}}={\frac {1}{\pi }}}  （Chudnovsky algorithm） 2 2 9801 ∑ k = 0 ∞ ( 4 k ) ! ( 1103 + 26390 k ) ( k ! ) 4 396 4 k = 1 π {\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） 12 ( 1249638720 + 159999840 61 ) 3 ∑ n = 0 ∞ ( − 1 ) n ( 6 n ) ! ( 1657145277365 + 212175710912 61 + ( 107578229802750 + 3773980892672 61 ) n ) ( 3 n ) ! ( n ! ) 3 ( 1249638720 + 159999840 61 ) n = 1 π {\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) 3 6 5 ∑ k = 0 ∞ ( ( 4 k ) ! ) 2 ( 6 k ) ! 9 k + 1 ( 12 k ) ! ( 2 k ) ! ( 127169 12 k + 1 − 1070 12 k + 5 − 131 12 k + 7 + 2 12 k + 11 ) = π {\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 } 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。 ∑ k = 0 ∞ 1 16 k ( 4 8 k + 1 − 2 8 k + 4 − 1 8 k + 5 − 1 8 k + 6 ) = π {\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） 1 2 6 ∑ n = 0 ∞ ( − 1 ) n 2 10 n ( − 2 5 4 n + 1 − 1 4 n + 3 + 2 8 10 n + 1 − 2 6 10 n + 3 − 2 2 10 n + 5 − 2 2 10 n + 7 + 1 10 n + 9 ) = π {\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 } ζ ( 2 ) = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ = π 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） ζ ( 4 ) = 1 1 4 + 1 2 4 + 1 3 4 + 1 4 4 + ⋯ = π 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}}} ζ ( 2 n ) = ∑ k = 1 ∞ 1 k 2 n = 1 1 2 n + 1 2 2 n + 1 3 2 n + 1 4 2 n + ⋯ = ( − 1 ) n + 1 B 2 n ( 2 π ) 2 n 2 ( 2 n ) ! {\displaystyle \zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}\,={\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 B2n は ベルヌーイ数） ∑ n = 1 ∞ 3 n − 1 4 n ζ ( n + 1 ) = π {\displaystyle \sum _{n=1}^{\infty }{\frac {3^{n}-1}{4^{n}}}\,\zeta (n+1)=\pi } ∑ n = 0 ∞ ( ( − 1 ) n 2 n + 1 ) 1 = 1 1 − 1 3 + 1 5 − 1 7 + 1 9 − ⋯ = arctan ⁡ 1 = π 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） ∑ n = 1 ∞ ( − 1 ) n + 1 n 2 = 1 1 2 − 1 2 2 + 1 3 2 − 1 4 2 + ⋯ = π 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}}} ∑ n = 1 ∞ 1 ( 2 n ) 2 = 1 2 2 + 1 4 2 + 1 6 2 + 1 8 2 + ⋯ = π 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}}} ∑ n = 0 ∞ ( ( − 1 ) n 2 n + 1 ) 2 = 1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + ⋯ = π 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}}} ∑ n = 0 ∞ ( ( − 1 ) n 2 n + 1 ) 3 = 1 1 3 − 1 3 3 + 1 5 3 − 1 7 3 + ⋯ = π 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}}} ∑ n = 0 ∞ ( ( − 1 ) n 2 n + 1 ) 4 = 1 1 4 + 1 3 4 + 1 5 4 + 1 7 4 + ⋯ = π 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}}} ∑ n = 0 ∞ ( ( − 1 ) n 2 n + 1 ) 5 = 1 1 5 − 1 3 5 + 1 5 5 − 1 7 5 + ⋯ = 5 π 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}}} ∑ n = 0 ∞ ( ( − 1 ) n 2 n + 1 ) 6 = 1 1 6 + 1 3 6 + 1 5 6 + 1 7 6 + ⋯ = π 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}}} ∑ n = 0 ∞ 1 ( 4 n + 1 ) ( 4 n + 3 ) = 1 1 ⋅ 3 + 1 5 ⋅ 7 + 1 9 ⋅ 11 + ⋯ = π 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}}} π = 1 + 1 2 + 1 3 + 1 4 − 1 5 + 1 6 + 1 7 + 1 8 + 1 9 − 1 10 + 1 11 + 1 12 − 1 13 + ⋯ {\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。 また ∑ n = 1 ∞ F 2 n n 2 ( 2 n n ) = 4 π 2 25 5 {\displaystyle \sum _{n=1}^{\infty }{\frac {F_{2n}}{n^{2}{\binom {2n}{n}}}}={\frac {4\pi ^{2}}{25{\sqrt {5}}}}} ただし  F n {\displaystyle F_{n}} はn番目のフィボナッチ数。 π 4 = arctan ⁡ 1 {\displaystyle {\frac {\pi }{4}}=\arctan 1} π 4 = arctan ⁡ 1 2 + arctan ⁡ 1 3 {\displaystyle {\frac {\pi }{4}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{3}}} π 4 = 2 arctan ⁡ 1 2 − arctan ⁡ 1 7 {\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{2}}-\arctan {\frac {1}{7}}} π 4 = 2 arctan ⁡ 1 3 + arctan ⁡ 1 7 {\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}} π 4 = 4 arctan ⁡ 1 5 − arctan ⁡ 1 239 {\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}}  （マチンの公式） π 4 = 5 arctan ⁡ 1 7 + 2 arctan ⁡ 3 79 {\displaystyle {\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}} π 4 = 6 arctan ⁡ 1 8 + 2 arctan ⁡ 1 57 + arctan ⁡ 1 239 {\displaystyle {\frac {\pi }{4}}=6\arctan {\frac {1}{8}}+2\arctan {\frac {1}{57}}+\arctan {\frac {1}{239}}} π 4 = 12 arctan ⁡ 1 49 + 32 arctan ⁡ 1 57 − 5 arctan ⁡ 1 239 + 12 arctan ⁡ 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}}} π 4 = 44 arctan ⁡ 1 57 + 7 arctan ⁡ 1 239 − 12 arctan ⁡ 1 682 + 24 arctan ⁡ 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}}} π 2 = ∑ n = 0 ∞ arctan ⁡ 1 F 2 n + 1 = arctan ⁡ 1 1 + arctan ⁡ 1 2 + arctan ⁡ 1 5 + arctan ⁡ 1 13 + ⋯ {\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  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 ( x ) n {\displaystyle (x)_{n}}  は階乗冪。Ramanujan–Sato seriesを参照 π 4 = 3 4 ⋅ 5 4 ⋅ 7 8 ⋅ 11 12 ⋅ 13 12 ⋅ 17 16 ⋅ 19 20 ⋅ 23 24 ⋅ 29 28 ⋅ 31 32 ⋅ ⋯ {\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 \,\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。 ∏ n = 1 ∞ 4 n 2 4 n 2 − 1 = 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋅ 8 7 ⋅ 8 9 ⋯ = 4 3 ⋅ 16 15 ⋅ 36 35 ⋅ 64 63 ⋅ ⋯ = π 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 \,\cdots ={\frac {\pi }{2}}} （ウォリス積を参照） Viète's formula: 2 2 ⋅ 2 + 2 2 ⋅ 2 + 2 + 2 2 ⋅ ⋯ = 2 π {\displaystyle {\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}\cdot \,\cdots ={\frac {2}{\pi }}} π = 3 + 1 2 6 + 3 2 6 + 5 2 6 + 7 2 6 + ⋱ {\displaystyle \pi ={3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+{\cfrac {7^{2}}{6+\ddots \,}}}}}}}}}} π = 4 1 + 1 2 3 + 2 2 5 + 3 2 7 + 4 2 9 + ⋱ {\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+{\cfrac {4^{2}}{9+\ddots }}}}}}}}}}} π = 4 1 + 1 2 2 + 3 2 2 + 5 2 2 + 7 2 2 + ⋱ {\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+\ddots }}}}}}}}}}} 2 π = 6 + 2 2 12 + 6 2 12 + 10 2 12 + 14 2 12 + 18 2 12 + ⋱ {\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 }}}}}}}}}}}} （連分数を参照） n ! ∼ 2 π n ( n e ) n {\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}} （スターリングの近似） e i π + 1 = 0 {\displaystyle e^{i\pi }+1=0}  （オイラーの等式） ∑ k = 1 n φ ( k ) ∼ 3 n 2 π 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） ∑ k = 1 n φ ( k ) k ∼ 6 n π 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） Γ ( 1 2 ) = π {\displaystyle \Gamma \left({1 \over 2}\right)={\sqrt {\pi }}} （ガンマ関数） π = Γ ( 1 / 4 ) 4 / 3 agm ⁡ ( 1 , 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） lim n → ∞ 1 n 2 ∑ k = 1 n ( n mod k ) = 1 − π 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） π = lim n → ∞ 4 n 2 ∑ k = 1 n 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） π = lim n → ∞ 2 4 n n ( 2 n 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）