算術幾何平均

${\displaystyle a_{0}=a,\quad b_{0}=b}$
${\displaystyle a_{n+1}={\frac {a_{n}+b_{n}}{2}},\quad b_{n+1}={\sqrt {a_{n}b_{n}}}\quad (n\geq 0)}$

${\displaystyle M(a,b)=\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }b_{n}}$

${\displaystyle M(a,b)={\frac {\pi }{2}}{\bigg /}\int _{0}^{\pi /2}{\frac {d\theta }{\sqrt {a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta }}}}$

${\displaystyle {\begin{aligned}M(a,b)&={\frac {\pi }{2}}{\bigg /}\int _{0}^{\pi /2}{\frac {d\theta }{\sqrt {\left({\frac {a+b}{2}}\right)^{2}\cos ^{2}\theta +ab\sin ^{2}\theta }}}\\&={\frac {\pi }{2}}{\bigg /}\int _{0}^{\pi /2}{\frac {d\theta }{\sqrt {\left({\frac {a+b}{2}}\right)^{2}-\left({\frac {a-b}{2}}\right)^{2}\sin ^{2}\theta }}}\\&={\frac {\pi }{2}}{\bigg /}\int _{0}^{\pi /2}{\frac {d\theta }{\left({\frac {a+b}{2}}\right){\sqrt {1-\left({\frac {a-b}{a+b}}\right)^{2}\sin ^{2}\theta }}}}\\\end{aligned}}}$

${\displaystyle a_{n+1}={\frac {a_{n}+b_{n}}{2}}\geq {\sqrt {a_{n}\cdot b_{n}}}=b_{n+1}}$

${\displaystyle a_{n}\geq a_{n+1},}$
${\displaystyle b_{n+1}\geq b_{n}}$

${\displaystyle a_{0}\geq a_{1}\geq a_{2}\geq \cdots \geq b_{2}\geq b_{1}\geq b_{0}}$

${\displaystyle \alpha ={\frac {\alpha +\beta }{2}}}$
${\displaystyle \beta ={\sqrt {\alpha \beta }}}$

${\displaystyle M(ca,cb)=cM(a,b)}$

${\displaystyle |a_{n+1}-b_{n+1}|={\frac {(a_{n}-b_{n})^{2}}{2({\sqrt {a_{n}}}+{\sqrt {b_{n}}})^{2}}}\leq C(a_{n}-b_{n})^{2}}$

${\displaystyle {\frac {\pi }{2}}=M(1,{\sqrt {1-k^{2}}})\int _{0}^{1}{\frac {dz}{\sqrt {(1-z^{2})(1-k^{2}z^{2})}}}.}$

${\displaystyle a_{n}^{\;2}-b_{n}^{\;2}=(a_{n}+b_{n})(a_{n}-b_{n})}$
${\displaystyle a_{n+1}^{\;2}-b_{n+1}^{\;2}=\left({\frac {a_{n}+b_{n}}{2}}\right)^{2}-a_{n}b_{n}={\frac {(a_{n}-b_{n})^{2}}{4}}}$

${\displaystyle {\frac {\left|a_{n+1}^{\;2}-b_{n+1}^{\;2}\right|}{\left|a_{n}^{\;2}-b_{n}^{\;2}\right|}}={\frac {\left|a_{n}-b_{n}\right|}{4\left|a_{n}+b_{n}\right|}}<{\frac {1}{4}}}$

${\displaystyle d_{n}=a_{n+1}-a_{n}=-{\frac {a_{n}-b_{n}}{2}}}$
${\displaystyle {\frac {\left|d_{n+1}\right|}{\left|d_{n}\right|}}={\frac {\sqrt {\left|a_{n+2}^{\;2}-b_{n+2}^{\;2}\right|}}{\sqrt {\left|a_{n+1}^{\;2}-b_{n+1}^{\;2}\right|}}}<{\frac {1}{2}}}$

${\displaystyle {\begin{aligned}I(a,b)&=\int _{0}^{\pi /2}{\frac {d\theta }{\sqrt {a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta }}}\\&=\int _{0}^{\pi /2}{\frac {d\theta }{\sqrt {(a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta )(\cos ^{2}\theta +\sin ^{2}\theta )}}}\\&=\int _{0}^{\pi /2}{\frac {d\theta }{{\cos ^{2}\theta }{\sqrt {(a^{2}+b^{2}\tan ^{2}\theta )(1+\tan ^{2}\theta )}}}}\\\end{aligned}}}$

${\displaystyle {\begin{aligned}I(a,b)&=\int _{0}^{\infty }{\frac {dx}{\sqrt {(a^{2}+b^{2}x^{2})(1+x^{2})}}}\\&=\int _{0}^{\infty }{\frac {dx}{\sqrt {a^{2}x^{2}+b^{2}x^{2}+b^{2}x^{4}+a^{2}}}}\\&=\int _{0}^{\infty }{\frac {dx}{\sqrt {(a+b)^{2}x^{2}+(bx^{2}-a)^{2}}}}\\&={\frac {1}{2}}\int _{0}^{\infty }{\frac {dx}{x{\sqrt {\left(\displaystyle {\frac {a+b}{2}}\right)^{2}+\left(\displaystyle {\frac {bx^{2}-a}{2x}}\right)^{2}}}}}\\\end{aligned}}}$

${\displaystyle x={\sqrt {ab}}\;t+{\sqrt {ab+abt^{2}}},\quad dx=\left({\sqrt {ab}}+{\frac {\sqrt {ab}}{\sqrt {ab+abt^{2}}}}\right)dt={\frac {x}{\sqrt {1+t^{2}}}}dt}$
${\displaystyle {\begin{aligned}I(a,b)&={\frac {1}{2}}\int _{-\infty }^{\infty }{\frac {dt}{\sqrt {\left(\left({\frac {a+b}{2}}\right)^{2}+abt^{2}\right)\left(1+t^{2}\right)}}}\\&=\int _{0}^{\infty }{\frac {dt}{\sqrt {\left(\left({\frac {a+b}{2}}\right)^{2}+abt^{2}\right)\left(1+t^{2}\right)}}}\\&=I\left({\frac {a+b}{2}},{\sqrt {ab}}\right)\end{aligned}}}$

${\displaystyle {\begin{aligned}I(a,b)&=I(a_{1},b_{1})=\lim _{n\to \infty }I(a_{n},b_{n})=\lim _{n\to \infty }I\left(M(a,b),M(a,b)\right)\\&=\int _{0}^{\pi /2}{\frac {d\theta }{\sqrt {M(a,b)^{2}(\cos ^{2}\theta +\sin ^{2}\theta )}}}={\frac {\pi }{2M(a,b)}}\\\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {t}{\frac {a+b}{2{\sqrt {ab}}}}}&={\frac {(b/a)x^{2}-1}{(1+b/a)x}}={\frac {(u+iv)x^{2}-1}{(1+u+iv)x}}\\&={\frac {(ux^{2}-1+ivx^{2})(1+u-iv)}{\left((1+u)^{2}+v^{2}\right)x}}\\&={\frac {(u+u^{2}+v^{2})x^{2}-(1+u)+ivx^{2}+iv}{(1+2u+u^{2}+v^{2})x}}\\\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {t}{\frac {a+b}{2{\sqrt {ab}}}}}&={\frac {iv{\frac {1+2u+^{2}+v^{2}}{u^{2}+v^{2}+u}}}{(1+2u+u^{2}+v^{2}){\sqrt {\frac {1+u}{u+u^{2}+v^{2}}}}}}\\&={\frac {iv}{\sqrt {(1+u)(u+u^{2}+v^{2})}}}\\\end{aligned}}}$

${\displaystyle {\begin{aligned}t&={\frac {{\sqrt {b/a}}\;x^{2}-{\sqrt {a/b}}}{2x}}={\frac {(u'+iv')^{2}x^{2}-1}{(u'+iv')x}}\\&={\frac {(u'^{2}+v'^{2})(u'+iv')x^{2}-(u'-iv')}{2(u'^{2}+v'^{2})x}}\\\end{aligned}}}$

${\displaystyle {\begin{aligned}t&={\frac {iv'}{\sqrt {u'^{2}+v'^{2}}}}\\\end{aligned}}}$

${\displaystyle a_{0}=a,\quad b_{0}=b}$
${\displaystyle a_{n+1}={\frac {a_{n}+b_{n}}{2}},\quad b_{n+1}={\frac {2a_{n}b_{n}}{a_{n}+b_{n}}}\quad (n\geq 0)}$
の極限について ${\displaystyle \operatorname {AHM} (a,b)=\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }b_{n}}$

${\displaystyle a_{n+1}b_{n+1}={\frac {a_{n}+b_{n}}{2}}\cdot {\frac {2a_{n}b_{n}}{a_{n}+b_{n}}}=a_{n}b_{n}}$
${\displaystyle \operatorname {AHM} (a,b)=\lim _{n\to \infty }{\sqrt {a_{n}b_{n}}}={\sqrt {ab}}}$

${\displaystyle a_{0}=a,\quad b_{0}=b}$
${\displaystyle a_{n+1}={\frac {2a_{n}b_{n}}{a_{n}+b_{n}}},\quad b_{n+1}={\sqrt {a_{n}b_{n}}}\quad (n\geq 0)}$
の極限について ${\displaystyle \operatorname {HGM} (a,b)=\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }b_{n}}$

${\displaystyle (1/a_{n+1})=(1/a_{n})+(1/b_{n}),\quad (1/b_{n+1})={\sqrt {(1/a_{n})(1/b_{n})}}}$
${\displaystyle \operatorname {HGM} (a,b)={\frac {1}{\operatorname {AGM} \left({\frac {1}{a}},{\frac {1}{b}}\right)}}={\frac {ab}{\operatorname {AGM} \left(b,a\right)}}}$