Salamin-Brent算法
$\qquad$Arithmetic-Geometric Mean:给定两个正数$a$和$b$$,\ $且有$0<b<a.\;$令$a_0=a,\ b_0=b,\ $并按照递推公式
$$
a_n=\frac{a_{n-1}+b_{n-1}}{2},\;
b_n=\sqrt{a_{n-1}b_{n-1}},\;
n\in \mathbb{N_+}
$$
定义数列$\{a_n\}$和$\{b_n\}.\;$证明这两个数列收敛于同一个极限$,\ $此极限称为$a$与$b$的算术几何平均值$,\ $且其具有解析表达式
$$AGM(a,b)=\frac{\pi}{2I(a,b)}\\
\quad\\
其中\;I(a,b)=\int^{\frac{\pi}{2}}_{0}\frac{\text{d}x}{\sqrt{a^2\cos^2{x}+b^2\sin^2{x}}}$$
$\nabla$证
$$ 使用数学归纳法\\\\ \quad\\\\ a_0>b_0\\\\ \quad\\\\ a_n^2-b_n^2=(\frac{a_{n-1}-b_{n-1}}{2})^2>0\\\\ \quad\\\\ 则有 b = b_0 < b_1 < b_2 < \cdots < b_n < a_n < \cdots < a_2 < a_1 < a_0 = a\\\\ \quad\\\\ \{a_n\}与\{b_n\}单调有界，设其极限分别为A与B\\\\ \quad\\\\ 由递推式a_n=\frac{a_{n-1}+b_{n-1}}{2}有A=\frac{A+B}{2}\\\\ \quad\\\\ 即\{a_n\}与\{b_n\}收敛于同一极限A=B，设其为AGM(a,b)\\\\ \quad\\\\ \text{Gauss}在14岁(1791年)时求出其解析表达式为\\\\ \quad\\\\ AGM(a,b)=\frac{\pi}{2I(a,b)},其中I(a,b)=\int^{\frac{\pi}{2}}_{0}\frac{\text{d}x}{\sqrt{a^2\cos^2{x}+b^2\sin^2{x}}}\\\\ \quad\\\\ 主要思路是证明I(a,b)=I(a_1,b_1),不断迭代,由I函数的连续性得\\\\ \quad\\\\ I=\lim \limits_{n \to \infty}I(a_n,b_n)=I(AGM,AGM)=\int^{\frac{\pi}{2}}_{0}\frac{\text{d}x}{AGM}=\frac{\pi}{2AGM}\\\\ \quad\\\\ 他首先引进变量x',使得\sin{x'}=\frac{2a\sin{x}}{a+b+(a-b)\sin^2{x}}\\\\ \quad\\\\ \text{Gauss then asserts}\\\\ \quad\\\\ ``\text{after the development has been made correctly, it will be seen}"\\\\ \quad\\\\ \text{that} \quad \frac{\text{d}x}{\sqrt{a^2\cos^2{x}+b^2\sin^2{x}}}=\frac{\text{d}x'}{\sqrt{a_1^2\cos^2{x'}+b_1^2\sin^2{x'}}}\\\\ \quad\\\\ 除了\text{Gauss}的方法，不妨试下\text{D.J. Newmann}的代换 \quad\\\\ t=b\tan{x},t'=\frac{1}{2}(t-\frac{b_1^2}{t})\\\\ \quad\\\\ 则\text{d}x=\frac{\cos^2{x}\text{d}t}{b},\text{d}t=\frac{2t^2\text{d}t'}{t^2+b_1^2}\\\\ \quad\\\\ \begin{aligned} I(a,b)&=\int^{+\infty}_{0}\frac{\cos^2{x}\text{d}t}{b\sqrt{a^2\cos^2{x}+b^2\sin^2{x}}}\\\\ &=\int^{+\infty}_{0}\frac{\text{d}t}{\sqrt{\frac{b^2}{\cos^2{x}}(a^2+b^2\tan^2{x})}}\\\\ &=\int^{+\infty}_{0}\frac{\text{d}t}{\sqrt{(b^2+t^2)(a^2+t^2)}}\\\\ &=\int^{+\infty}_{-\infty}\frac{2t^2\text{d}t'}{(t^2+b_1^2)\sqrt{b_1^4+(4a_1^2-2b_1^2)t^2+t^4}}\\\\ &=\int^{+\infty}_{-\infty}\frac{\text{d}t'}{\frac{t^2+b_1^2}{t}\sqrt{[b_1^4+(4a_1^2-2b_1^2)t^2+t^4]/4t^2}}\\\\ &=\int^{+\infty}_{-\infty}\frac{\text{d}t'}{2\sqrt{(a_1^2+t'^2)(b_1^2+t'^2)}}\\\\ &=I(a_1,b_1)\\\\ \end{aligned} $$
$\qquad$Landen变换:先引进如下两个完全椭圆积分
$$ K(k)=\int^{\frac{\pi}{2}}_{0}\frac{\text{d}x}{\sqrt{1-k^2\sin^2{x}}},\; E(k)=\int^{\frac{\pi}{2}}_{0}\sqrt{1-k^2\sin^2{x}}\;\text{d}x\\\\ \quad\\\\ 当k^2+k'^2=1时\\\\ \quad\\\\ K'(k)=K(k'),\;E'(k)=E(k')为另外两个椭圆积分\\\\ \quad\\\\ 这些椭圆积分的对称形式可写为\\\\ \quad\\\\ I(a,b)=\int^{\frac{\pi}{2}}_{0}\frac{\text{d}x}{\sqrt{a^2\cos^2{x}+b^2\sin^2{x}}}\\\\ \quad\\\\ J(a,b)=\int^{\frac{\pi}{2}}_{0}\sqrt{a^2\cos^2{x}+b^2\sin^2{x}}\;\text{d}x\\\\ \quad\\\\ 很明显\;I(a,b)=\frac{K'(\frac{b}{a})}{a},\;J(a,b)=aE'(\frac{b}{a})\\\\ \quad\\\\ 进行\text{Landen}变换可得到如下的式子\\\\ \quad\\\\ I(a_n,b_n)=I(a_{n+1},b_{n+1})\\\\ \quad\\\\ J(a_n,b_n)=2J(a_{n+1},b_{n+1})-a_n b_n I(a_{n+1},b_{n+1})\\\\ $$
$\nabla$证
$$ 第一个式子在算术几何平均值中已经证过\\\\ \quad\\\\ J(a,b)关于a与b的对称性是显然的\\\\ \quad\\\\ 这样第二个式子同样可以运用\text{D.J. Newmann}的代换得出\\\\ \quad\\\\ t=b\tan{x},t'=\frac{1}{2}(t-\frac{b_1^2}{t})\\\\ \quad\\\\ \begin{aligned} J(a,b)&=\int^{\frac{\pi}{2}}_{0}\sqrt{a^2\cos^2{x}+b^2\sin^2{x}}\;\text{d}x\\\\ &=\int^{+\infty}_{0}\frac{\sqrt{(a^2+t^2)(b^2+t^2)}}{(b+\frac{t^2}{b})^2}\;\text{d}t\\\\ &=\int^{+\infty}_{-\infty}\frac{\sqrt{(a^2+t^2)(b^2+t^2)}}{(b+\frac{t^2}{b})^2}\times\frac{2t^2}{b_1^2+t^2}\;\text{d}t'\\\\ &=\int^{+\infty}_{-\infty}\frac{(b_1^2+t^2)\sqrt{(a^2+t^2)(b^2+t^2)}}{2t^2(b_1+\frac{t'^2}{b_1})^2}\times[\frac{2t^2(b_1+\frac{t'^2}{b_1})}{(b_1^2+t^2)(b+\frac{t^2}{b})}]^2\;\text{d}t'\\\\ &[注意到\; b_1^2+t'^2=\frac{(b_1^2+t^2)^2}{4t^2},\;a_1^2+t'^2=\frac{(a^2+t^2)(b^2+t^2)}{4t^2}]\\\\ &=\int^{+\infty}_{-\infty}\frac{2\sqrt{(a_1^2+t'^2)(b_1^2+t'^2)}}{(b_1+\frac{t'^2}{b_1})^2}\times \frac{t^2(b_1+\frac{t'^2}{b_1})^2}{(b_1^2+t'^2)(b+\frac{t^2}{b})^2}\;\text{d}t'\\\\ &=\int^{+\infty}_{-\infty}\frac{2\sqrt{(a_1^2+t'^2)(b_1^2+t'^2)}}{(b_1+\frac{t'^2}{b_1})^2}\times \frac{b^2t^2(b_1+\frac{t'^2}{b_1})^2(a^2+t^2)}{4t^2(b_1^2+t'^2)(a_1^2+t'^2)(b^2+t^2)}\;\text{d}t'\\\\ &=\int^{+\infty}_{-\infty}\frac{2\sqrt{(a_1^2+t'^2)(b_1^2+t'^2)}}{(b_1+\frac{t'^2}{b_1})^2}\times \frac{ab(b_1+\frac{t'^2}{b_1})^2}{4(a_1^2+t'^2)(b_1^2+t'^2)}\times \frac{b(a^2+t^2)}{a(b^2+t^2)}\;\text{d}t'\\\\ &=\int^{+\infty}_{-\infty}\frac{2\sqrt{(a_1^2+t'^2)(b_1^2+t'^2)}}{(b_1+\frac{t'^2}{b_1})^2}\times \frac{ab(b_1+\frac{t'^2}{b_1})^2}{4(a_1^2+t'^2)(b_1^2+t'^2)}\\\\ &\times[-1+\frac{(a+b)ab+(a+b)t^2}{a(b^2+t^2)}]\;\text{d}t'\\\\ &=-abI(a_1,b_1)+\int^{+\infty}_{-\infty}\frac{2\sqrt{(a_1^2+t'^2)(b_1^2+t'^2)}}{(b_1+\frac{t'^2}{b_1})^2}\times \frac{b_1^2+t'^2}{4(a_1^2+t'^2)}\\\\ &\times\frac{(a+b)ab+(a+b)t^2}{a(b^2+t^2)}]\;\text{d}t'\\\\ &=-abI(a_1,b_1)+\int^{+\infty}_{-\infty}\frac{\sqrt{(a_1^2+t'^2)(b_1^2+t'^2)}}{(b_1+\frac{t'^2}{b_1})^2}\times \frac{a_1(b_1^2+t^2)(b_1^2+t'^2)}{a(b^2+t^2)(a_1^2+t'^2)}\;\text{d}t'\\\\ &[利用J(a,b)关于a与b的对称性,由于t是t'与b_1的函数,故无需改变]\\\\ &=-abI(a_1,b_1)+\int^{+\infty}_{-\infty}\frac{\sqrt{(a_1^2+t'^2)(b_1^2+t'^2)}}{(b_1+\frac{t'^2}{b_1})^2}\times \frac{a_1(b_1^2+t^2)(b_1^2+t'^2)}{(a_1^2+t'^2)}\\\\ &\times[\frac{1}{2a{(b^2+t^2)}}+\frac{1}{2b(a^2+t^2)}]\;\text{d}t'\\\\ &=-abI(a_1,b_1)+\int^{+\infty}_{-\infty}\frac{\sqrt{(a_1^2+t'^2)(b_1^2+t'^2)}}{(b_1+\frac{t'^2}{b_1})^2}\times \frac{a_1^2(b_1^2+t'^2)^2}{b_1^2(a_1^2+t'^2)^2}\;\text{d}t'\\\\ &=-abI(a_1,b_1)+\int^{+\infty}_{-\infty}\frac{\sqrt{(a_1^2+t'^2)(b_1^2+t'^2)}}{(a_1+\frac{t'^2}{a_1})^2}\;\text{d}t'\\\\ &[利用J(a_1,b_1)关于a_1与b_1的对称性]\\\\ &=2J(a_1,b_1)-abI(a_1,b_1)\\\\ \end{aligned} $$
$$
于是I(a,b)=\frac{\pi}{2AGM(a,b)}\\
\quad\\
\begin{aligned}
J(a,b)&=2J(a_1,b_1)-abI(a,b)\\
\quad\\
&=2J(a_1,b_1)+(a^2-2a_1^2+2a_1^2-a^2-b_1^2)I(a,b)\\
\quad\\
\end{aligned}
\quad\\
\Downarrow\\
\begin{aligned}
J(a,b)-a^2I(a,b)&=2(J(a_1,b_1)-a_1^2I(a_1,b_1))-\frac{a^2-b^2}{2}I(a,b)\\
\quad\\
&=\lim\limits_{n \to \infty}[2^n(J(a_n,b_n)-a_n^2I(a_n,b_n))]\\
\quad\\
&-\lim\limits_{n \to \infty}[\sum\limits_{i=0}^{n-1}2^i\frac{a_i^2-b_i^2}{2}I(a_i,b_i)]\\
\quad\\
&=\lim \limits_{n \to \infty}\int_{0}^{2^{n-1}\pi}\sqrt{a_n^2\cos^2{x}+b_n^2\sin^2{x}}\;\text{d}x\\
\quad\\
&-\lim \limits_{n \to \infty}\int_{0}^{2^{n-1}\pi}\frac{a_n^2}{\sqrt{a_n^2\cos^2{x}+b_n^2\sin^2{x}}}\;\text{d}x\\
\quad\\
&-\lim\limits_{n \to \infty}[\sum\limits_{i=0}^{n-1}2^i\frac{a_i^2-b_i^2}{2}I(a_i,b_i)]\\
\quad\\
&=\int_{0}^{+\infty}(AGM-AGM)\;\text{d}x\\
\quad\\
&-\sum\limits_{i=0}^{\infty}2^i\frac{a_i^2-b_i^2}{2}I(a_i,b_i)\\
\quad\\
&=-\sum\limits_{i=0}^{\infty}2^i\frac{a_i^2-b_i^2}{2}I(a_i,b_i)\\
\quad\\
\end{aligned}
\quad\\
故J(a,b)=[a^2-\frac{1}{2}\sum\limits_{i=0}^{\infty}2^i(a_i^2-b_i^2)]I(a,b)
$$
$\qquad$Legendre关系:对于上述两种完全椭圆积分成立等式
$$
K(k)E’(k)+K’(k)E(k)-K(k)K’(k)=\frac{\pi}{2}
$$
$\nabla$证
$$ \begin{aligned} 令P(k)&=K(k)E'(k)+K'(k)E(k)-K(k)K'(k)\\\\ &=\int_{0}^{\frac{\pi}{2}}\frac{\text{d}x}{\sqrt{1-k^2\sin^2{x}}}\int_{0}^{\frac{\pi}{2}}\sqrt{1-(1-k^2)\sin^2{x}}\;\text{d}x\\\\ &+\int_{0}^{\frac{\pi}{2}}\frac{\text{d}x}{\sqrt{1-(1-k^2)\sin^2{x}}}\int_{0}^{\frac{\pi}{2}}\sqrt{1-k^2\sin^2{x}}\;\text{d}x\\\\ &-\int_{0}^{\frac{\pi}{2}}\frac{\text{d}x}{\sqrt{1-k^2\sin^2{x}}}\int_{0}^{\frac{\pi}{2}}\frac{\text{d}x}{\sqrt{1-(1-k^2)\sin^2{x}}}\\\\ \end{aligned} \quad\\\\ 易得P(0)=P(1)=\frac{\pi}{2}\\\ \quad\\\\ 只需\frac{\text{d}P}{\text{d}k}=0即可证明等式成立\\\\ \quad\\\\ \begin{aligned} 又\frac{\text{d}K}{\text{d}k}&=\int_{0}^{\frac{\pi}{2}}\frac{k\sin^2{x}\text{d}x}{\sqrt{1-k^2\sin^2{x}}^3}\\\\ &=\int_{0}^{\frac{\pi}{2}}\frac{(k\sin^2{x}-\frac{1}{k}+\frac{1}{k})\text{d}x}{\sqrt{1-k^2\sin^2{x}}^3}\\\\ &=-\frac{1}{k}K+\frac{1}{k}\int_{0}^{\frac{\pi}{2}}\frac{\text{d}x}{\sqrt{1-k^2\sin^2{x}}^3}\\\\ &[泰勒展开，逐项积分]\\\\ &=-\frac{1}{k}K+\frac{\pi}{2k}[1+3(\frac{1}{2})^2k^2+5(\frac{1\cdot3}{2\cdot4})^2k^4+\cdots]\\\\ &=-\frac{1}{k}K+\frac{\pi}{2k}[1+\frac{1\cdot3}{1}(\frac{1}{2})^2k^2+\frac{3\cdot5}{3}(\frac{1\cdot3}{2\cdot4})^2k^4+\cdots]\\\\ &=-\frac{1}{k}K+\frac{\pi}{2kk'^2}[1+\frac{1\cdot3-2\cdot2}{1}(\frac{1}{2})^2k^2+\frac{3\cdot5-4\cdot4}{3}(\frac{1\cdot3}{2\cdot4})^2k^4+\cdots]\\\\ &=\frac{1}{kk'^2}E-\frac{1}{k}K\\\\ \end{aligned} \quad\\\\ \begin{aligned} \frac{\text{d}E}{\text{d}k}&=-\int_{0}^{\frac{\pi}{2}}\frac{k\sin^2{x}\text{d}x}{\sqrt{1-k^2\sin^2{x}}}\\\\ &=-\int_{0}^{\frac{\pi}{2}}\frac{(k\sin^2{x}-\frac{1}{k}+\frac{1}{k})\text{d}x}{\sqrt{1-k^2\sin^2{x}}}\\\\ &=\frac{1}{k}E-\frac{1}{k}K\\\\ \end{aligned} \quad\\\\ 由对称性\\\\ \quad\\\\ \begin{aligned} \frac{\text{d}K'}{\text{d}k}&=\frac{\text{d}K'}{\text{d}k'}\frac{\text{d}k'}{\text{d}k}\\\\ &=(\frac{1}{k'k^2}E'-\frac{1}{k'}K')(-\frac{k}{k'})\\\\ &=-\frac{1}{kk'^2}E'+\frac{k}{k'^2}K'\\\\ \end{aligned} \quad\\\\ \begin{aligned} \frac{\text{d}E'}{\text{d}k}&=\frac{\text{d}E'}{\text{d}k'}\frac{\text{d}k'}{\text{d}k}\\\\ &=(\frac{1}{k'}E'-\frac{1}{k'}K')(-\frac{k}{k'})\\\\ &=-\frac{k}{k'^2}E'+\frac{k}{k'^2}K'\\\\ \end{aligned} \quad\\\\ \begin{aligned} 于是\frac{\text{d}P}{\text{d}k}&=\frac{\text{d}K}{\text{d}k}E'+K\frac{\text{d}E'}{\text{d}k}+ \frac{\text{d}K'}{\text{d}k}E+K'\frac{\text{d}E}{\text{d}k}-\frac{\text{d}K'}{\text{d}k}K-K'\frac{\text{d}K}{\text{d}k}\\\\ &=(\frac{1}{kk'^2}E-\frac{1}{k}K)E'+K(-\frac{k}{k'^2}E'+\frac{k}{k'^2}K')+(-\frac{1}{kk'^2}E'+\frac{k}{k'^2}K')E\\\\ &+K'(\frac{1}{k}E-\frac{1}{k}K)-(-\frac{1}{kk'^2}E'+\frac{k}{k'^2}K')K-K'(\frac{1}{kk'^2}E-\frac{1}{k}K)\\\\ &=(\frac{1}{kk'^2}-\frac{1}{kk'^2})EE'+(\frac{k}{k'^2}-\frac{1}{k}-\frac{k}{k'^2}+\frac{1}{k})KK'\\\\ &+(\frac{k}{k'^2}+\frac{1}{k}-\frac{1}{kk'^2})EK'+(-\frac{1}{k}-\frac{k}{k'^2}+\frac{1}{kk'^2})KE'\\\\ &=0 \end{aligned} $$
$\qquad$$\pi$的表达式:令$k=\frac{a}{b}$、$k’=\frac{a’}{b’},$将勒让德关系中的$E$、$K$用$I$、$J$代换可得
$$
\frac{a’}{a}I(a’,b’)J(a,b)+\frac{a}{a’}I(a,b)J(a’,b’)-aa’I(a,b)I(a’,b’)=\frac{\pi}{2}\\
\quad\\
再将\text{Landen}变换得出的以下两个式子代入\\
\quad\\
I(a,b)=\frac{\pi}{2AGM(a,b)}\\
\quad\\
J(a,b)=[a^2-\frac{1}{2}\sum\limits_{i=0}^{\infty}2^i(a_i^2-b_i^2)]I(a,b)\\
\quad\\
可得
\quad\\
\begin{aligned}
&\frac{a’}{a}[a^2-\frac{1}{2}\sum\limits_{i=0}^{\infty}2^i(a_i^2-b_i^2)]\frac{1}{2AGM(a,b)AGM(a’,b’)}+\\
\quad\\
&\frac{a}{a’}[a’^2-\frac{1}{2}\sum\limits_{i=0}^{\infty}2^i(a_{i}’^2-b_{i}’^2)]\frac{1}{2AGM(a,b)AGM(a’,b’)}-\\
\quad\\
&aa’\frac{1}{2AGM(a,b)AGM(a’,b’)}\\
\quad\\
&=\frac{1}{\pi}\\
\end{aligned}
\quad\\
若令a=a’=1,\\
\quad\\
b=k,\;b’=k’,\\
\quad\\
a_i^2-b_i^2=c_i^2,\;a_{i}’^2-b_{i}’=c_{i}’^2则有\\
\quad\\
\pi=\frac{4AGM(1,k)AGM(1,k’)}{1-\sum\limits_{i=1}^{\infty}2^i(c_{i}^2+c_{i}’^2)}\\
\quad\\
进一步令k=k’=\frac{1}{\sqrt{2}}则有\\
\quad\\
\pi=\frac{4[AGM(1,\frac{1}{\sqrt{2}})]^2}{1-\sum\limits_{i=1}^{\infty}2^{i+1}c_{i}^2}
$$
$\qquad$Salamin-Brent算法:令$a_0=1,\ b_0=\frac{1}{\sqrt{2}},\ s_0=\frac{1}{2},\ $并用递推公式
$$ \begin{aligned} &a_n=\frac{a_{n-1}+b_{n-1}}{2},\\\\ &b_n=\sqrt{a_{n-1}b_{n-1}},\\\\ &s_n=s_{n-1}-2^n(a_n^2-b_n^2),\\\\ &p_n=\frac{2a_n^2}{s_n} \end{aligned} $$
作迭代$,\ $则$\{p_n\}$收敛于$\pi,\ $这是上文$\pi$表达式的直接结果
$\qquad$误差分析:至此得出了Salamin-Brent算法$,\ $但还需确定该算法中$\pi$的收敛速度$,\ $虽然上文中$\pi$的表达式对任意复数$k$成立$,\ $以下仅就$k$为实数的情况讨论(ToDo)
$$
记a,b第n次与a’,b’第n’迭代的结果为\pi_{nn’}=\frac{4a_n a_{n’}’}{1-\sum\limits_{i=1}^{n}2^{i}c_{i}^2-\sum\limits_{i=1}^{n’}2^{i}c_{i}’^2}\\
\quad\\
引入辅助量\overline\pi_{nn’}=\frac{4AGM(1,k)AGM(1,k’)}{1-\sum\limits_{i=1}^{n}2^{i}c_{i}^2-\sum\limits_{i=1}^{n’}2^{i}c_{i}’^2}\\
\quad\\
我们需要找出误差上界e_{nn’}与\overline e_{nn’}使得\\
\quad\\
0<\pi-\overline\pi_{nn’}<e_{nn’}\\
\quad\\
0<\pi_{nn’}-\overline\pi_{nn’}<\overline e_{nn’}\\
\quad\\
\overline e_{nn’}<e_{nn’}\\
\quad\\
由以上三式即可得出|\pi-\pi_{nn’}|<e_{nn’}\\
\quad\\
\begin{aligned}
&\qquad\qquad\qquad\qquad首先计算\pi-\overline\pi_{nn’}\\
\quad\\
&=\frac{4AGM(1,k)AGM(1,k’)}{1-\sum\limits_{i=1}^{\infty}2^i(c_{i}^2+c_{i}’^2)}-\frac{4AGM(1,k)AGM(1,k’)}{1-\sum\limits_{i=1}^{n}2^{i}c_{i}^2-\sum\limits_{i=1}^{n’}2^{i}c_{i}’^2}\\
\quad\\
&[由不等式\frac{1}{x}-\frac{1}{x+y}<\frac{y}{x^2}有]\\
\quad\\
&<4AGM(1,k)AGM(1,k’)\frac{\sum\limits_{i=n+1}^{\infty}2^{i}c_{i}^2+\sum\limits_{i=n’+1}^{\infty}2^{i}c_{i}’^2}
{[1-\sum\limits_{i=1}^{\infty}2^i(c_{i}^2+c_{i}’^2)]^2}\\
\quad\\
&=\frac{\pi^2}{4AGM(1,k)AGM(1,k’)}(\sum\limits_{i=n+1}^{\infty}2^{i}c_{i}^2+\sum\limits_{i=n’+1}^{\infty}2^{i}c_{i}’^2)\\
\quad\\
&[由c_i=\frac{a_{i-1}-b_{i-1}}{2},c_i^2=4a_{i+1}c_{i+1}有]\\
\quad\\
&=\frac{\pi^2}{2AGM(1,k)AGM(1,k’)}(\sum\limits_{i=n+2}^{\infty}2^{i}a_{i}c_{i}+\sum\limits_{i=n’+2}^{\infty}2^{i}a_{i}’c_{i}’)\\
\quad\\
\end{aligned}
\quad\\
于是可记e_{nn’}=\frac{\pi^2}{2AGM(1,k)AGM(1,k’)}(\sum\limits_{i=n+2}^{\infty}2^{i}a_{i}c_{i}+\sum\limits_{i=n’+2}^{\infty}2^{i}a_{i}’c_{i}’)\\
\quad\\
接下来计算\pi_{nn’}-\overline\pi_{nn’}\\
\quad\\
\begin{aligned}
&=\frac{4a_n a_{n’}’}{1-\sum\limits_{i=1}^{n}2^{i}c_{i}^2-\sum\limits_{i=1}^{n’}2^{i}c_{i}’^2}-
\frac{4AGM(1,k)AGM(1,k’)}{1-\sum\limits_{i=1}^{n}2^{i}c_{i}^2-\sum\limits_{i=1}^{n’}2^{i}c_{i}’^2}\\
\quad\\
&<\frac{4}{1-\sum\limits_{i=1}^{\infty}2^i(c_{i}^2+c_{i}’^2)}(a_{n}’a_{n’}’-AGM(1,k)AGM(1,k’))\\
\quad\\
&=\frac{\pi}{AGM(1,k)AGM(1,k’)}(a_{n}’a_{n’}’-AGM(1,k)AGM(1,k’))\\
\quad\\
&=
\end{aligned}
$$
$\qquad$参考与引用来源:
Computation of $\pi$ Using Arithmetic-Geometric Mean
Methods of Mathematical Physics最后一节
From Lintearia to Lemniscate II:Gauss and Landen’s Work
Elliptic Integrals78~80面
The Legendre Relation for Elliptic Integrals
The Quest for Pi
谢惠民第二版上册29与267面