把求π的公式翻个底朝天
2026-01-10 21:04:52
发布于：浙江
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$\frac{1}{\pi}=\frac{2\sqrt{2}}{9801}\sum\limits_{k=0}^{\infty}\frac{(4k)!(1103+26390k)}{(k!)^4396^{4k}}\\ \frac{\pi}{4}=\sum\limits_{k=0}^{\infty}\frac{(-1)^k}{2k+1}\\ a_0=1,b_0=\frac{1}{\sqrt{2}},t_0=\frac{1}{4},p_0=1\\ a_{n+1}=\frac{a_n+b_n}{2},b_{n+1}=\sqrt{a_nb_n},t_{n+1}=t_n-p_n(a_n-a_{n+1})^2,p_{n+1}=2p_n\\ \pi\approx\frac{(a_{n+1}+b_{n+1})^2}{4t_{n+1}}\\ \Large\pi=\frac{426880\sqrt{10005}}{\sum\limits_{k=0}^{\infty}\frac{(6k)!(545140134k+13591409)}{(3k)!(k!)^3(-640320)^{3k}}}\\ \normalsize \frac{\pi}{2}=\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdots\\ \arctan{x}=\sum\limits_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{2k+1}\\ \pi=16\arctan{\frac{1}{5}}-4\arctan{\frac{1}{239}}\\ \pi=24\arctan{\frac{1}{8}}+8\arctan{\frac{1}{57}}+4\arctan{\frac{1}{239}}\\ \pi=48\arctan{\frac{1}{18}}+32\arctan{\frac{1}{57}}-20\arctan{\frac{1}{239}}\\ \pi=32\arctan{\frac{1}{10}}-4\arctan{\frac{1}{239}}-16\arctan{\frac{1}{515}}\\ \pi=\frac{\ln{-1}}{\sqrt{-1}}\\ \frac{\pi^2}{6}=\sum\limits_{k=1}^{\infty}\frac{1}{k^2}\\ \huge\pi=3+\frac{1}{7+\frac{1}{15+\frac{1}{1+\frac{1}{292+\frac{1}{1+\cdots}}}}} $1周前 来自 浙江
0
π
1
9801
2
2
k=0
∑
∞
(k!)
4
396
4k(4k)!(1103+26390k)
4
π
k=0
∑
∞
2k+1
(−1)
k
a
0
=1,b
0
2
1
,t
0
4
1
,p
0
=1
a
n+1
2
a
n
+b
n
,b
n+1
a
n
b
n
,t
n+1
=t
n
−p
n
(a
n
−a
n+1
)
2
,p
n+1
=2p
n
π≈
4t
n+1
(a
n+1
+b
n+1
)
2
π=
k=0
∑
∞
(3k)!(k!)
3
(−640320)
3k(6k)!(545140134k+13591409)
426880
10005
2
π
1
2
⋅
3
2
⋅
3
4
⋅
5
4
⋅
5
6
⋅
7
6
⋯
arctanx=
k=0
∑
∞
(−1)
k2k+1
x
2k+1
π=16arctan
5
1
−4arctan
239
1
π=24arctan
8
1
+8arctan
57
1
+4arctan
239
1
π=48arctan
18
1
+32arctan
57
1
−20arctan
239
1
π=32arctan
10
1
−4arctan
239
1
−16arctan
515
1
π=
−1
ln−1
6
π
2
k=1
∑
∞
k
21
π=3+
7+
15+
1+
292+
1+⋯
1
1
1
1
1
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