582 5 + 9 2 π = ∑ n = 1 ∞ 8 n n ! ( 3 n ) ! ( 4 n ) ! ( 43 n 2 − 624 n + 48 ) {\displaystyle \frac{582}{5}+\frac{9}{2}\pi = \sum_{n=1}^\infty 8^n\frac{n!(3n)!}{(4n)!}(43n^2-624n+48)}
24516 − 360 π 3 = ∑ n = 0 ∞ 9 n n ! ( 3 n ) ! ( 4 n ) ! ( 2743 n 2 − 130971 n − 12724 ) {\displaystyle 24516-360\pi\sqrt{3} = \sum_{n=0}^\infty 9^n\frac{n!(3n)!}{(4n)!}(2743n^2-130971n-12724)}
1872 5 + 8 π 3 = ∑ n = 0 ∞ 3 n n ! ( 3 n ) ! ( 4 n ) ! ( 1435 n 2 − 3403 n + 96 ) {\displaystyle \frac{1872}{5}+8\pi\sqrt{3} = \sum_{n=0}^\infty 3^n\frac{n!(3n)!}{(4n)!}(1435n^2-3403n+96)}
324 + 288 π 3 − 576 log 2 = ∑ n = 0 ∞ ( 9 8 ) n n ! ( 3 n ) ! ( 4 n ) ! ( − 5415 n 2 + 6335 n + 5692 ) {\displaystyle 324+288\pi {\sqrt {3}}-576{\text{log}}2=\sum _{n=0}^{\infty }({\frac {9}{8}})^{n}{\frac {n!(3n)!}{(4n)!}}(-5415n^{2}+6335n+5692)}
7582 + 1008 π 3 − 576 log 2 = ∑ n = 0 ∞ ( 9 8 ) n n ! ( 3 n ) ! ( 4 n ) ! ( 18050 n 2 + 1145 n + 7517 ) {\displaystyle 7582+1008\pi {\sqrt {3}}-576{\text{log}}2=\sum _{n=0}^{\infty }({\frac {9}{8}})^{n}{\frac {n!(3n)!}{(4n)!}}(18050n^{2}+1145n+7517)}
− 264 + 15 2 π + 120 t e x t l o g 2 = ∑ n = 1 ∞ ( − 1 2 ) n ( 2 n ) ! ( 3 n ) ! ( 5 n ) ! ( 44506 n 3 − 38681 n 2 + 241 n + 1514 ) {\displaystyle -264+{\frac {15}{2}}\pi +120text{log}2=\sum _{n=1}^{\infty }(-{\frac {1}{2}})^{n}{\frac {(2n)!(3n)!}{(5n)!}}(44506n^{3}-38681n^{2}+241n+1514)}
20 π 3 + 89 = ∑ n = 1 ∞ ( − 1 3 ) n ( 2 n ! ) 2 ( 3 n ) ! n ! ( 6 n ) ! ( − 22100 n + 4123 ) {\displaystyle 20\pi {\sqrt {3}}+89=\sum _{n=1}^{\infty }(-{\frac {1}{3}})^{n}{\frac {(2n!)^{2}(3n)!}{n!(6n)!}}(-22100n+4123)}
40 27 π 3 + 9 = ∑ n = 1 ∞ ( − 27 ) n ( 2 n ! ) 2 ( 3 n ) ! n ! ( 6 n ) ! ( − 40 n + 3 ) {\displaystyle \frac{40}{27}\pi\sqrt{3}+9 = \sum_{n=1}^\infty (-27)^n\frac{(2n!)^2(3n)!}{n!(6n)!}(-40n+3)}
15 π 2 + 27 = ∑ n = 0 ∞ 8 n ( 2 n ! ) 2 ( 3 n ) ! n ! ( 6 n ) ! ( 350 n − 17 ) {\displaystyle 15\pi\sqrt{2}+27 = \sum_{n=0}^\infty 8^n\frac{(2n!)^2(3n)!}{n!(6n)!}(350n-17)}
15 π + 42 = ∑ n = 1 ∞ ( − 4 ) n ( 2 n ! ) 2 ( 3 n ) ! n ! ( 6 n ) ! ( − 952 n + 201 ) {\displaystyle 15\pi +42 = \sum_{n=1}^\infty (-4)^n\frac{(2n!)^2(3n)!}{n!(6n)!}(-952n+201)}
sin 2 x 2 = 1 − cos x 2 = 1 − 1 − sin 2 x 2 {\displaystyle \sin^2\frac{x}{2}=\frac{1-\cos x}{2}=\frac{1-\sqrt{1-\sin^2 x}}{2}}
4 sin 2 x 2 = 2 − 4 − 4 sin 2 x {\displaystyle 4\sin ^{2}{\frac {x}{2}}=2-{\sqrt {4-4\sin ^{2}x}}}
2 sin x 2 = 2 − 4 − ( 2 sin x ) 2 {\displaystyle 2\sin {\frac {x}{2}}={\sqrt {2-{\sqrt {4-(2\sin x)^{2}}}}}}
これより、帰納的に C n = 2 sin 1 2 n arcsin C 0 2 = 2 sin 1 2 n arcsin 1 2 = 2 sin π 6 ⋅ 2 n {\displaystyle C_n=2\sin \frac{1}{2^n}\arcsin \frac{C_0}{2}=2\sin \frac{1}{2^n}\arcsin \frac{1}{2}=2\sin \frac{\pi}{6\cdot 2^n}}
π 6 ⋅ 2 n = x {\displaystyle \frac{\pi}{6\cdot 2^n}=x} とすると、
lim n → ∞ 3 ⋅ 2 n C n = lim n → ∞ 6 ⋅ 2 n sin π 6 ⋅ 2 n = lim x → 0 π sin x x = π {\displaystyle \lim_{n\to\infty}3\cdot 2^nC_n=\lim_{n\to\infty}6\cdot 2^n \sin\frac{\pi}{6\cdot 2^n}=\lim_{x\to 0}\pi\frac{\sin x}{x}=\pi}
なお、 C n {\displaystyle C_n} は半径2の円に内接する正 6 ⋅ 2 n {\displaystyle 6\cdot 2^n} の一辺の長さを表している。
sin x 2 = 1 − 1 − ( sin x ) 2 2 {\displaystyle \sin\frac{x}{2}=\sqrt{\frac{1-\sqrt{1-(\sin x)^2}}{2}}}
これより、帰納的に U n = sin 1 2 n arcsin U 0 = sin 1 2 n arcsin 1 2 = sin π 6 ⋅ 2 n {\displaystyle U_n=\sin \frac{1}{2^n}\arcsin U_0=\sin \frac{1}{2^n}\arcsin \frac{1}{2}=\sin \frac{\pi}{6\cdot 2^n}}
lim n → ∞ 6 ⋅ 2 n U n = lim n → ∞ 6 ⋅ 2 n sin π 6 ⋅ 2 n = lim x → 0 π sin x x = π {\displaystyle \lim_{n\to\infty}6\cdot 2^nU_n=\lim_{n\to\infty}6\cdot 2^n \sin\frac{\pi}{6\cdot 2^n}=\lim_{x\to 0}\pi\frac{\sin x}{x}=\pi}
また、 tan x = 2 tan x 2 1 − tan 2 x 2 {\displaystyle \tan x=\frac{2\tan \frac{x}{2}}{1-\tan^2\frac{x}{2}}}
tan x 2 = − 1 + 1 + tan 2 x tan x {\displaystyle \tan \frac{x}{2}=\frac{-1+\sqrt{1+\tan^2 x}}{\tan x}}
これより、帰納的に V n = tan 1 2 n arctan U 0 = sin 1 2 n arctan 1 2 = sin π 6 ⋅ 2 n {\displaystyle V_n=\tan \frac{1}{2^n}\arctan U_0=\sin \frac{1}{2^n}\arctan \frac{1}{2}=\sin \frac{\pi}{6\cdot 2^n}}
lim n → ∞ 6 ⋅ 2 n U n = lim n → ∞ 6 ⋅ 2 n tan π 6 ⋅ 2 n = lim x → 0 π tan x x = π {\displaystyle \lim_{n\to\infty}6\cdot 2^nU_n=\lim_{n\to\infty}6\cdot 2^n \tan\frac{\pi}{6\cdot 2^n}=\lim_{x\to 0}\pi\frac{\tan x}{x}=\pi}
なお、 U n , V n {\displaystyle U_n,V_n} は半径1の円にそれぞれ内接、外接する正 6 ⋅ 2 n {\displaystyle 6\cdot 2^n} の一辺の長さを表している。
cos 2 x 2 = 1 2 ( 1 + cos x ) {\displaystyle \cos^2\frac{x}{2}=\frac{1}{2}(1+\cos x)}
および
sin x 2 sin x 2 = 2 sin x 2 cos x 2 2 sin x 2 = cos x 2 {\displaystyle \frac{\sin x}{2\sin\frac{x}{2}}=\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{2\sin \frac{x}{2}}=\cos\frac{x}{2}}
より、
cos x 2 = 1 2 ( 1 + cos x ) {\displaystyle \cos\frac{x}{2}=\sqrt{\frac{1}{2}(1+\cos x)}}
sin x 2 sin x 2 = 1 2 ( 1 + sin 2 x 2 sin x ) {\displaystyle \frac{\sin x}{2\sin\frac{x}{2}}=\sqrt{\frac{1}{2}(1+\frac{\sin 2x}{2\sin x})}}
2 sin x 2 sin x = 2 ( 2 sin x 2 sin x + sin 2 x ) {\displaystyle \frac{2\sin\frac{x}{2}}{\sin x}=\sqrt{2(\frac{2\sin x}{2\sin x+\sin 2x})}}
2 sin x 2 = sin x 2 sin x sin x + 1 2 sin 2 x {\displaystyle 2\sin\frac{x}{2}=\sin x\sqrt{\frac{2\sin x}{\sin x+\frac{1}{2}\sin 2x}}}
これより、帰納的に x n = 2 n sin z 2 n {\displaystyle x_n=2^n\sin\frac{z}{2^n}} と表せる。
x 1 = 2 sin z 2 = 1 {\displaystyle x_1=2\sin\frac{z}{2}=1} より、 z = π {\displaystyle z=\pi}
よって、 x n = 2 n sin π 2 n {\displaystyle x_n=2^n\sin\frac{\pi}{2^n}}
π ⋅ 2 n = p {\displaystyle {\frac {\pi }{\cdot 2^{n}}}=p} とすると、
lim n → ∞ x n = lim n → ∞ 2 n sin π 2 n = lim x → 0 π sin x x = π {\displaystyle \lim_{n\to\infty} x_n=\lim_{n\to\infty}2^n \sin\frac{\pi}{2^n}=\lim_{x\to 0}\pi\frac{\sin x}{x}=\pi}
3 n ∑ k = 1 n 1 + ( 4 n 2 − k 2 − 4 n 2 − ( k − 1 ) 2 ) 2 = 3 n ∑ k = 1 n 1 + ( 4 n 2 − k 2 2 − 4 n 2 − ( k − 1 ) 2 2 4 n 2 − k 2 + 4 n 2 − ( k − 1 ) 2 ) 2 = 3 n ∑ k = 1 n 1 + ( − 2 k + 1 4 n 2 − k 2 + 4 n 2 − ( k − 1 ) 2 ) 2 = 3 n ∑ k = 1 n 1 + ( − 2 k n + 1 n 4 − ( k n ) 2 + 4 − ( k n − 1 n ) 2 ) 2 = 3 ∫ 0 1 1 + ( − 2 x 2 4 − x 2 ) 2 d x = 3 ∫ 0 1 1 + x 2 4 − x 2 d x = 3 ∫ 0 1 4 4 − x 2 d x = 3 ∫ 0 1 2 2 4 4 − 4 t 2 d t ( t = 1 2 x ) = 6 ∫ 0 1 2 1 1 − t 2 d t = 6 [ arcsin x ] 0 1 2 = 6 ( π 6 − 0 ) = π {\displaystyle \begin{align} &\frac{3}{n}\sum_{k=1}^n\sqrt{1+(\sqrt{4n^2-k^2}-\sqrt{4n^2-(k-1)^2})^2} \\ &=\frac{3}{n}\sum_{k=1}^n\sqrt{1+(\frac{\sqrt{4n^2-k^2}^2-\sqrt{4n^2-(k-1)^2}^2}{\sqrt{4n^2-k^2}+\sqrt{4n^2-(k-1)^2}})^2} \\ &=\frac{3}{n}\sum_{k=1}^n\sqrt{1+(\frac{-2k+1}{\sqrt{4n^2-k^2}+\sqrt{4n^2-(k-1)^2}})^2} \\ &=\frac{3}{n}\sum_{k=1}^n\sqrt{1+(\frac{-2\frac{k}{n}+\frac{1}{n}}{\sqrt{4-(\frac{k}{n})^2}+\sqrt{4-(\frac{k}{n}-\frac{1}{n})^2}})^2} \\ &=3\int_0^1 \sqrt{1+(\frac{-2x}{2\sqrt{4-x^2}})^2}dx \\ &=3\int_0^1 \sqrt{1+\frac{x^2}{4-x^2}}dx \\ &=3\int_0^1 \sqrt{\frac{4}{4-x^2}}dx \\ &=3\int_0^\frac{1}{2} 2\sqrt{\frac{4}{4-4t^2}}dt(t=\frac{1}{2}x) \\ &=6\int_0^\frac{1}{2} \sqrt{\frac{1}{1-t^2}}dt \\ &=6[\arcsin x]_0^\frac{1}{2} \\ &=6(\frac{\pi}{6}-0) \\ &=\pi \\ \end{align}}
とすると、
は半径2の円に内接する正
の一辺の長さを表している。
とすると、
とすると、
は半径1の円にそれぞれ内接、外接する正の一辺の長さを表している。
と表せる。
より、
とすると、
![{\displaystyle {\begin{aligned}&{\frac {3}{n}}\sum _{k=1}^{n}{\sqrt {1+({\sqrt {4n^{2}-k^{2}}}-{\sqrt {4n^{2}-(k-1)^{2}}})^{2}}}\\&={\frac {3}{n}}\sum _{k=1}^{n}{\sqrt {1+({\frac {{\sqrt {4n^{2}-k^{2}}}^{2}-{\sqrt {4n^{2}-(k-1)^{2}}}^{2}}{{\sqrt {4n^{2}-k^{2}}}+{\sqrt {4n^{2}-(k-1)^{2}}}}})^{2}}}\\&={\frac {3}{n}}\sum _{k=1}^{n}{\sqrt {1+({\frac {-2k+1}{{\sqrt {4n^{2}-k^{2}}}+{\sqrt {4n^{2}-(k-1)^{2}}}}})^{2}}}\\&={\frac {3}{n}}\sum _{k=1}^{n}{\sqrt {1+({\frac {-2{\frac {k}{n}}+{\frac {1}{n}}}{{\sqrt {4-({\frac {k}{n}})^{2}}}+{\sqrt {4-({\frac {k}{n}}-{\frac {1}{n}})^{2}}}}})^{2}}}\\&=3\int _{0}^{1}{\sqrt {1+({\frac {-2x}{2{\sqrt {4-x^{2}}}}})^{2}}}dx\\&=3\int _{0}^{1}{\sqrt {1+{\frac {x^{2}}{4-x^{2}}}}}dx\\&=3\int _{0}^{1}{\sqrt {\frac {4}{4-x^{2}}}}dx\\&=3\int _{0}^{\frac {1}{2}}2{\sqrt {\frac {4}{4-4t^{2}}}}dt(t={\frac {1}{2}}x)\\&=6\int _{0}^{\frac {1}{2}}{\sqrt {\frac {1}{1-t^{2}}}}dt\\&=6[\arcsin x]_{0}^{\frac {1}{2}}\\&=6({\frac {\pi }{6}}-0)\\&=\pi \\\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/0856f4619db6d6b42a947d5ba182d53abe1f7ce0)