本文是针对视频
https://www.bilibili.com/video/BV1ZF411f7Xv
整理的笔记
椭圆的参数方程：
{ x = a cos θ y = b sin θ \begin{cases}
x = a \cos \theta \\
y = b \sin \theta
\end{cases}
{ x = a cos θ y = b sin θ
对参数方程求微分：
{ d x = − a sin θ d θ d y = b cos θ d θ \begin{cases}
dx = -a \sin \theta d\theta \\
dy = b \cos \theta d\theta
\end{cases}
{ d x = − a sin θ d θ d y = b cos θ d θ
则周长可以表示为：
C = ∫ 0 2 π d x 2 + d y 2 = ∫ 0 2 π ( − a sin θ ) 2 + ( b cos θ ) 2 d θ = ∫ 0 2 π a 2 ( 1 − c o s 2 θ ) + b 2 c o s 2 θ d θ = ∫ 0 2 π a 2 − ( a 2 − b 2 ) c o s 2 θ d θ = a ∫ 0 2 π 1 − a 2 − b 2 a 2 c o s 2 θ d θ = 4 a ∫ 0 π 2 1 − a 2 − b 2 a 2 c o s 2 θ d θ \begin{aligned}
C &= \int_0^{2\pi} \sqrt{dx^2 + dy^2} \\
&= \int_0^{2\pi} \sqrt{(-a \sin \theta)^2 + (b \cos \theta)^2} d\theta \\
&= \int_0^{2\pi} \sqrt{a^2 (1-cos^2 \theta) + b^2 cos^2 \theta} d\theta \\
&= \int_0^{2\pi} \sqrt{a^2 - (a^2 - b^2) cos^2 \theta} d\theta \\
&= a\int_0^{2\pi} \sqrt{1 - \frac{a^2 - b^2}{a^2} cos^2 \theta} d\theta \\
&= 4a \int_0^{\frac{\pi}{2}} \sqrt{1 - \frac{a^2 - b^2}{a^2} cos^2 \theta} d\theta \\
\end{aligned}
C = ∫ 0 2 π d x 2 + d y 2 = ∫ 0 2 π ( − a sin θ ) 2 + ( b cos θ ) 2 d θ = ∫ 0 2 π a 2 ( 1 − c o s 2 θ ) + b 2 c o s 2 θ d θ = ∫ 0 2 π a 2 − ( a 2 − b 2 ) c o s 2 θ d θ = a ∫ 0 2 π 1 − a 2 a 2 − b 2 c o s 2 θ d θ = 4 a ∫ 0 2 π 1 − a 2 a 2 − b 2 c o s 2 θ d θ
由椭圆的离心率的定义可得：
e = c a = a 2 − b 2 a e = \frac{c}{a} = \frac{\sqrt{a^2 - b^2}}{a}
e = a c = a a 2 − b 2
则有：
C = 4 a ∫ 0 π 2 1 − e 2 c o s 2 θ d θ = 4 a ∫ 0 π 2 1 − e 2 c o s 2 x d x C = 4a \int_0^{\frac{\pi}{2}} \sqrt{1 - e^2 cos^2 \theta} d\theta = 4a \int_0^{\frac{\pi}{2}} \sqrt{1 - e^2 cos^2 x} dx
C = 4 a ∫ 0 2 π 1 − e 2 c o s 2 θ d θ = 4 a ∫ 0 2 π 1 − e 2 c o s 2 x d x
根据二项式定理，可以将一个函数( x + y ) n (x+y)^n ( x + y ) n
( x + y ) n = ∑ k = 0 n C ( n , k ) x n − k y k (x+y)^n = \sum_{k=0}^{n} C(n,k) x^{n-k} y^k
( x + y ) n = k = 0 ∑ n C ( n , k ) x n − k y k
实际上，二项式定理可以将指数解析延拓到任意实数n n n
( x + y ) n = ∑ k = 0 ∞ n ( n − 1 ) ( n − 2 ) . . . ( n − k + 1 ) k ! x n − k y k \begin{aligned}
(x+y)^n &= \sum_{k=0}^{\infty} \frac{n(n-1)(n-2)...(n-k+1)}{k!} x^{n-k} y^k \\
\end{aligned}
( x + y ) n = k = 0 ∑ ∞ k ! n ( n − 1 ) ( n − 2 ) . . . ( n − k + 1 ) x n − k y k
当 x=1, y=z, n=p 时，得到：
( 1 + z ) p = 1 + ∑ n = 1 ∞ z n n ! ∏ k = 0 n − 1 ( p − k ) (1+z)^p = 1 + \sum_{n=1}^{\infty} \frac{z^n}{n!} \prod_{k=0}^{n-1} (p-k)
( 1 + z ) p = 1 + n = 1 ∑ ∞ n ! z n k = 0 ∏ n − 1 ( p − k )
只有当∣ z ∣ < 1 |z|<1 ∣ z ∣ < 1
令z = − e 2 c o s 2 x z = -e^2 cos^2 x z = − e 2 c o s 2 x ∣ z ∣ < 1 |z|<1 ∣ z ∣ < 1
C = 4 a ∫ 0 π 2 1 − e 2 c o s 2 x d x \begin{aligned}
C &= 4a \int_0^{\frac{\pi}{2}} \sqrt{1 - e^2 cos^2 x} dx \\
\end{aligned}
C = 4 a ∫ 0 2 π 1 − e 2 c o s 2 x d x
1 − e 2 c o s 2 x = ( 1 − e 2 c o s 2 x ) 1 2 = 1 + ∑ n = 1 ∞ ( − e 2 c o s 2 x ) n n ! ∏ k = 0 n − 1 ( 1 2 − k ) = 1 + ∑ n = 1 ∞ ( − 1 ) n e 2 n c o s 2 n x n ! ∏ k = 0 n − 1 1 − 2 k 2 \begin{aligned}
\sqrt{1 - e^2 cos^2 x} &= (1 - e^2 cos^2 x)^{\frac{1}{2}} \\
&=1 + \sum_{n=1}^{\infty} \frac{(-e^2 cos^2 x)^n}{n!} \prod_{k=0}^{n-1} (\frac{1}{2}-k) \\
&=1 + \sum_{n=1}^{\infty} (-1)^n\frac{e^{2n} cos^{2n} x}{n!} \prod_{k=0}^{n-1} \frac{1-2k}{2} \\
\end{aligned}
1 − e 2 c o s 2 x = ( 1 − e 2 c o s 2 x ) 2 1 = 1 + n = 1 ∑ ∞ n ! ( − e 2 c o s 2 x ) n k = 0 ∏ n − 1 ( 2 1 − k ) = 1 + n = 1 ∑ ∞ ( − 1 ) n n ! e 2 n c o s 2 n x k = 0 ∏ n − 1 2 1 − 2 k
∏ k = 0 n − 1 1 − 2 k 2 = ( − 1 ) n ∏ k = 0 n − 1 2 k − 1 2 = ( − 1 ) n ⋅ 0 − 1 2 ⋅ 2 − 1 2 ⋅ 4 − 1 2 ⋯ 2 n − 3 2 = ( − 1 ) n + 1 2 n ⋅ 1 ⋅ 3 ⋯ ( 2 n − 3 ) = ( − 1 ) n + 1 2 n ⋅ ( 2 n − 3 ) ! ! \begin{aligned}
\prod_{k=0}^{n-1} \frac{1-2k}{2}
&= (-1)^n \prod_{k=0}^{n-1} \frac{2k-1}{2} \\
&= (-1)^n \cdot \frac{0-1}{2} \cdot \frac{2-1}{2} \cdot \frac{4-1}{2} \cdots \frac{2n-3}{2} \\
&= \frac{(-1)^{n+1}}{2^n} \cdot 1 \cdot 3 \cdots (2n-3) \\
&= \frac{(-1)^{n+1}}{2^n} \cdot (2n-3)!! \\
\end{aligned}
k = 0 ∏ n − 1 2 1 − 2 k = ( − 1 ) n k = 0 ∏ n − 1 2 2 k − 1 = ( − 1 ) n ⋅ 2 0 − 1 ⋅ 2 2 − 1 ⋅ 2 4 − 1 ⋯ 2 2 n − 3 = 2 n ( − 1 ) n + 1 ⋅ 1 ⋅ 3 ⋯ ( 2 n − 3 ) = 2 n ( − 1 ) n + 1 ⋅ ( 2 n − 3 ) ! !
1 − e 2 c o s 2 x = 1 + ∑ n = 1 ∞ ( − 1 ) 2 n + 1 e 2 n c o s 2 n x 2 n ⋅ n ! ⋅ ( 2 n − 3 ) ! ! = 1 − ∑ n = 1 ∞ e 2 n c o s 2 n x 2 n ⋅ n ! ⋅ ( 2 n − 3 ) ! ! \begin{aligned}
\sqrt{1 - e^2 cos^2 x}
&= 1 + \sum_{n=1}^{\infty} \frac{(-1)^{2n+1} e^{2n} cos^{2n} x}{2^n \cdot n!} \cdot (2n-3)!! \\
&= 1-\sum_{n=1}^{\infty} \frac{e^{2n} cos^{2n} x}{2^n \cdot n!} \cdot (2n-3)!! \\
\end{aligned}
1 − e 2 c o s 2 x = 1 + n = 1 ∑ ∞ 2 n ⋅ n ! ( − 1 ) 2 n + 1 e 2 n c o s 2 n x ⋅ ( 2 n − 3 ) ! ! = 1 − n = 1 ∑ ∞ 2 n ⋅ n ! e 2 n c o s 2 n x ⋅ ( 2 n − 3 ) ! !
2 n ⋅ n ! = 2 n ⋅ 1 ⋅ 2 ⋯ n = 2 ⋅ 4 ⋯ 2 n = ( 2 n ) ! ! \begin{aligned}
2^n \cdot n!
&= 2^n \cdot 1 \cdot 2 \cdots n \\
&= 2 \cdot 4 \cdots 2n \\
&= (2n)!! \\
\end{aligned}
2 n ⋅ n ! = 2 n ⋅ 1 ⋅ 2 ⋯ n = 2 ⋅ 4 ⋯ 2 n = ( 2 n ) ! !
双阶乘说明：
k = { k ⋅ ( k − 2 ) ⋯ 3 ⋅ 1 , k 为奇数 k ⋅ ( k − 2 ) ⋯ 4 ⋅ 2 , k 为偶数 k = \begin{cases}
k \cdot (k-2) \cdots 3 \cdot 1, & k \text{ 为奇数} \\
k \cdot (k-2) \cdots 4 \cdot 2, & k \text{ 为偶数}
\end{cases}
k = { k ⋅ ( k − 2 ) ⋯ 3 ⋅ 1 , k ⋅ ( k − 2 ) ⋯ 4 ⋅ 2 , k 为奇数 k 为偶数
规定0 ! ! = 1 0!! = 1 0 ! ! = 1
由于k ! ! = k ⋅ ( k − 2 ) ! ! k!!=k \cdot (k-2)!! k ! ! = k ⋅ ( k − 2 ) ! !
1 ! ! = 1 ⋅ ( − 1 ) ! ! = 1 ⇒ ( − 1 ) ! ! = 1 ( − 1 ) ! ! = 1 ⋅ ( − 3 ) ! ! ⇒ ( − 3 ) ! ! = − 1 ( − 3 ) ! ! = ( − 1 ) ⋅ ( − 5 ) ! ! ⇒ ( − 5 ) ! ! = 1 \begin{aligned}
1!!=1 \cdot (-1)!! = 1 &\Rightarrow (-1)!! = 1 \\
(-1)!! = 1 \cdot (-3)!! &\Rightarrow (-3)!! = -1 \\
(-3)!! = (-1) \cdot (-5)!! &\Rightarrow (-5)!! = 1 \\
\end{aligned}
1 ! ! = 1 ⋅ ( − 1 ) ! ! = 1 ( − 1 ) ! ! = 1 ⋅ ( − 3 ) ! ! ( − 3 ) ! ! = ( − 1 ) ⋅ ( − 5 ) ! ! ⇒ ( − 1 ) ! ! = 1 ⇒ ( − 3 ) ! ! = − 1 ⇒ ( − 5 ) ! ! = 1
当 n=0 时，有
e 2 n c o s 2 n x ( 2 n ) ! ! ⋅ ( 2 n − 3 ) ! ! = 1 0 ! ! ⋅ ( − 3 ) ! ! = − 1 \frac{e^{2n} cos^{2n} x}{(2n)!!} \cdot (2n-3)!! = \frac{1}{0!!} \cdot (-3)!! = -1
( 2 n ) ! ! e 2 n c o s 2 n x ⋅ ( 2 n − 3 ) ! ! = 0 ! ! 1 ⋅ ( − 3 ) ! ! = − 1
故可以继续化简：
1 − e 2 c o s 2 x = 1 − ∑ n = 1 ∞ e 2 n c o s 2 n x ( 2 n ) ! ! ⋅ ( 2 n − 3 ) ! ! = − ∑ n = 0 ∞ e 2 n c o s 2 n x ( 2 n ) ! ! ⋅ ( 2 n − 3 ) ! ! = − ∑ n = 0 ∞ ( 2 n − 3 ) ! ! ( 2 n ) ! ! ⋅ e 2 n c o s 2 n x \begin{aligned}
\sqrt{1 - e^2 cos^2 x}
&= 1-\sum_{n=1}^{\infty} \frac{e^{2n} cos^{2n} x}{(2n)!!} \cdot (2n-3)!! \\
&= -\sum_{n=0}^{\infty} \frac{e^{2n} cos^{2n} x}{(2n)!!} \cdot (2n-3)!! \\
&= -\sum_{n=0}^{\infty} \frac{(2n-3)!!}{(2n)!!} \cdot e^{2n} cos^{2n} x \\
\end{aligned}
1 − e 2 c o s 2 x = 1 − n = 1 ∑ ∞ ( 2 n ) ! ! e 2 n c o s 2 n x ⋅ ( 2 n − 3 ) ! ! = − n = 0 ∑ ∞ ( 2 n ) ! ! e 2 n c o s 2 n x ⋅ ( 2 n − 3 ) ! ! = − n = 0 ∑ ∞ ( 2 n ) ! ! ( 2 n − 3 ) ! ! ⋅ e 2 n c o s 2 n x
现在椭圆周长公式变为了：
C = 4 a ∫ 0 π 2 1 − e 2 c o s 2 x d x = − 4 a ∫ 0 π 2 ∑ n = 0 ∞ ( 2 n − 3 ) ! ! ( 2 n ) ! ! ⋅ e 2 n c o s 2 n x d x = − 4 a ∑ n = 0 ∞ ( 2 n − 3 ) ! ! ( 2 n ) ! ! ⋅ e 2 n ∫ 0 π 2 c o s 2 n x d x \begin{aligned}
C
&= 4a \int_0^{\frac{\pi}{2}} \sqrt{1 - e^2 cos^2 x} dx \\
&= -4a \int_0^{\frac{\pi}{2}} \sum_{n=0}^{\infty} \frac{(2n-3)!!}{(2n)!!} \cdot e^{2n} cos^{2n} x dx \\
&= -4a \sum_{n=0}^{\infty} \frac{(2n-3)!!}{(2n)!!} \cdot e^{2n} \int_0^{\frac{\pi}{2}} cos^{2n} x dx \\
\end{aligned}
C = 4 a ∫ 0 2 π 1 − e 2 c o s 2 x d x = − 4 a ∫ 0 2 π n = 0 ∑ ∞ ( 2 n ) ! ! ( 2 n − 3 ) ! ! ⋅ e 2 n c o s 2 n x d x = − 4 a n = 0 ∑ ∞ ( 2 n ) ! ! ( 2 n − 3 ) ! ! ⋅ e 2 n ∫ 0 2 π c o s 2 n x d x
研究积分∫ 0 π 2 c o s n x d x \int_0^{\frac{\pi}{2}} cos^{n} x dx ∫ 0 2 π c o s n x d x
I k = ∫ 0 π 2 c o s k x d x = ∫ 0 π 2 c o s k − 1 x ⋅ c o s x d x = ∫ 0 π 2 c o s k − 1 x ⋅ d ( s i n x ) = [ c o s k − 1 x ⋅ s i n x ] 0 π 2 − ∫ 0 π 2 s i n x ⋅ d ( c o s k − 1 x ) = 0 − ∫ 0 π 2 s i n x ⋅ ( − s i n x ) ⋅ ( k − 1 ) c o s k − 2 x d x = ∫ 0 π 2 ( k − 1 ) s i n 2 x ⋅ c o s k − 2 x d x = ( k − 1 ) ∫ 0 π 2 ( 1 − c o s 2 x ) c o s k − 2 x d x = ( k − 1 ) [ ∫ 0 π 2 c o s k − 2 x d x − ∫ 0 π 2 c o s k x d x ] = ( k − 1 ) [ I k − 2 − I k ] \begin{aligned}
I_k
&= \int_0^{\frac{\pi}{2}} cos^{k} x dx \\
&= \int_0^{\frac{\pi}{2}} cos^{k-1} x \cdot cos x dx \\
&= \int_0^{\frac{\pi}{2}} cos^{k-1} x \cdot d (sinx) \\
&= \left[ cos^{k-1} x \cdot sin x \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} sin x \cdot d (cos^{k-1} x) \\
&= 0 - \int_0^{\frac{\pi}{2}} sin x \cdot (-sin x) \cdot (k-1) cos^{k-2} x dx \\
&= \int_0^{\frac{\pi}{2}} (k-1) sin^2 x \cdot cos^{k-2} x dx \\
&= (k-1) \int_0^{\frac{\pi}{2}} (1-cos^2 x) cos^{k-2} x dx \\
&= (k-1) \left[ \int_0^{\frac{\pi}{2}} cos^{k-2} x dx - \int_0^{\frac{\pi}{2}} cos^{k} x dx \right] \\
&= (k-1) \left[ I_{k-2} - I_k \right] \\
\end{aligned}
I k = ∫ 0 2 π c o s k x d x = ∫ 0 2 π c o s k − 1 x ⋅ c o s x d x = ∫ 0 2 π c o s k − 1 x ⋅ d ( s i n x ) = [ c o s k − 1 x ⋅ s i n x ] 0 2 π − ∫ 0 2 π s i n x ⋅ d ( c o s k − 1 x ) = 0 − ∫ 0 2 π s i n x ⋅ ( − s i n x ) ⋅ ( k − 1 ) c o s k − 2 x d x = ∫ 0 2 π ( k − 1 ) s i n 2 x ⋅ c o s k − 2 x d x = ( k − 1 ) ∫ 0 2 π ( 1 − c o s 2 x ) c o s k − 2 x d x = ( k − 1 ) [ ∫ 0 2 π c o s k − 2 x d x − ∫ 0 2 π c o s k x d x ] = ( k − 1 ) [ I k − 2 − I k ]
继续计算递推公式：
I k + ( k − 1 ) I k = ( k − 1 ) I k − 2 k I k = ( k − 1 ) I k − 2 I k = k − 1 k I k − 2 \begin{aligned}
I_k + (k-1) I_k &= (k-1) I_{k-2} \\
k I_k &= (k-1) I_{k-2} \\
I_k &= \frac{k-1}{k} I_{k-2} \\
\end{aligned}
I k + ( k − 1 ) I k k I k I k = ( k − 1 ) I k − 2 = ( k − 1 ) I k − 2 = k k − 1 I k − 2
当k = 0 k=0 k = 0 I 0 = ∫ 0 π 2 d x = π 2 I_0 = \int_0^{\frac{\pi}{2}} dx = \frac{\pi}{2} I 0 = ∫ 0 2 π d x = 2 π
当k = 1 k=1 k = 1 I 1 = ∫ 0 π 2 c o s x d x = 1 I_1 = \int_0^{\frac{\pi}{2}} cos x dx = 1 I 1 = ∫ 0 2 π c o s x d x = 1
即得到Wallis公式(俗称点火公式，偶数时可以数到 1，点火成功再补充一个π 2 \frac{\pi}{2} 2 π
I k = { k − 1 k ⋅ k − 3 k − 2 ⋯ 1 2 ⋅ I 0 , k 为偶数 k − 1 k ⋅ k − 3 k − 2 ⋯ 2 3 ⋅ I 1 , k 为奇数 I_k = \begin{cases}
\frac{k-1}{k} \cdot \frac{k-3}{k-2} \cdots \frac{1}{2} \cdot I_0, & k \text{ 为偶数} \\
\frac{k-1}{k} \cdot \frac{k-3}{k-2} \cdots \frac{2}{3} \cdot I_1, & k \text{ 为奇数}
\end{cases}
I k = { k k − 1 ⋅ k − 2 k − 3 ⋯ 2 1 ⋅ I 0 , k k − 1 ⋅ k − 2 k − 3 ⋯ 3 2 ⋅ I 1 , k 为偶数 k 为奇数
当 k=2n 时，可得：
I 2 n = π 2 ⋅ ( 2 n − 1 ) ! ! ( 2 n ) ! ! \begin{aligned}
I_{2n} &= \frac{\pi}{2} \cdot \frac{(2n-1)!!}{(2n)!!} \\
\end{aligned}
I 2 n = 2 π ⋅ ( 2 n ) ! ! ( 2 n − 1 ) ! !
继续化简椭圆周长公式：
C = − 4 a ∑ n = 0 ∞ ( 2 n − 3 ) ! ! ( 2 n ) ! ! ⋅ e 2 n ⋅ I 2 n = − 4 a ∑ n = 0 ∞ ( 2 n − 3 ) ! ! ( 2 n ) ! ! ⋅ e 2 n ⋅ π 2 ⋅ ( 2 n − 1 ) ! ! ( 2 n ) ! ! = − 4 a ⋅ π 2 ∑ n = 0 ∞ ( 2 n − 3 ) ! ! ( 2 n ) ! ! ⋅ e 2 n ⋅ ( 2 n − 1 ) ! ! ( 2 n ) ! ! = − 4 a ⋅ π 2 ∑ n = 0 ∞ ( 2 n − 3 ) ! ! ⋅ ( 2 n − 1 ) ! ! ( 2 n ) ! ! 2 ⋅ e 2 n = − 2 π a ∑ n = 0 ∞ e 2 n 2 n − 1 [ ( 2 n − 1 ) ! ! ( 2 n ) ! ! ] 2 = 2 π a { 1 − ∑ n = 1 ∞ e 2 n 2 n − 1 [ ( 2 n − 1 ) ! ! ( 2 n ) ! ! ] 2 } = 2 π a { 1 − ∑ n = 1 ∞ e 2 n 2 n − 1 ( ∏ k = 1 n 2 k − 1 2 k ) 2 } \begin{aligned}
C
&= -4a \sum_{n=0}^{\infty} \frac{(2n-3)!!}{(2n)!!} \cdot e^{2n} \cdot I_{2n} \\
&= -4a \sum_{n=0}^{\infty} \frac{(2n-3)!!}{(2n)!!} \cdot e^{2n} \cdot \frac{\pi}{2} \cdot \frac{(2n-1)!!}{(2n)!!} \\
&= -4a \cdot \frac{\pi}{2} \sum_{n=0}^{\infty} \frac{(2n-3)!!}{(2n)!!} \cdot e^{2n} \cdot \frac{(2n-1)!!}{(2n)!!} \\
&= -4a \cdot \frac{\pi}{2} \sum_{n=0}^{\infty} \frac{(2n-3)!! \cdot (2n-1)!!}{(2n)!!^2} \cdot e^{2n} \\
&= -2\pi a \sum_{n=0}^{\infty} \frac{e^{2n}}{2n-1} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \\
&= 2\pi a \left\{1- \sum_{n=1}^{\infty} \frac{e^{2n}}{2n-1} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \right\} \\
&= 2\pi a \left\{1- \sum_{n=1}^{\infty} \frac{e^{2n}}{2n-1} \left(\prod_{k=1}^{n} \frac{2k-1}{2k}\right)^2 \right\}\\
\end{aligned}
C = − 4 a n = 0 ∑ ∞ ( 2 n ) ! ! ( 2 n − 3 ) ! ! ⋅ e 2 n ⋅ I 2 n = − 4 a n = 0 ∑ ∞ ( 2 n ) ! ! ( 2 n − 3 ) ! ! ⋅ e 2 n ⋅ 2 π ⋅ ( 2 n ) ! ! ( 2 n − 1 ) ! ! = − 4 a ⋅ 2 π n = 0 ∑ ∞ ( 2 n ) ! ! ( 2 n − 3 ) ! ! ⋅ e 2 n ⋅ ( 2 n ) ! ! ( 2 n − 1 ) ! ! = − 4 a ⋅ 2 π n = 0 ∑ ∞ ( 2 n ) ! ! 2 ( 2 n − 3 ) ! ! ⋅ ( 2 n − 1 ) ! ! ⋅ e 2 n = − 2 π a n = 0 ∑ ∞ 2 n − 1 e 2 n [ ( 2 n ) ! ! ( 2 n − 1 ) ! ! ] 2 = 2 π a { 1 − n = 1 ∑ ∞ 2 n − 1 e 2 n [ ( 2 n ) ! ! ( 2 n − 1 ) ! ! ] 2 } = 2 π a ⎩ ⎨ ⎧ 1 − n = 1 ∑ ∞ 2 n − 1 e 2 n ( k = 1 ∏ n 2 k 2 k − 1 ) 2 ⎭ ⎬ ⎫
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 from math import pi, sqrtfrom functools import reducedef C (a: float , b: float , m: int )-> float : a, b = (a, b) if a > b else (b, a) c = sqrt(a**2 - b**2 ) e = c / a return 2 * pi * a * (1 -sum ( map ( lambda n: ((e**(2 *n)) / (2 *n - 1 )) * (reduce( lambda x, y: x * y, map ( lambda k: (2 *k - 1 ) / (2 *k), range (1 , n+1 ), ), )) ** 2 , range (1 , m+1 ), ), )) for i in range (1 , 100 ): print (C(3 , 2 , i))
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