円周率を含む数式
∑ 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}}}
F = | q 1 q 2 | 4 π ε 0 r 2 {\displaystyle F={\frac {|q_{1}q_{2}|}{4\pi \varepsilon _{0}r^{2}}}}