平方根の入れ子について

${\begin{aligned}&\arctan x=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}\cdots \\=&x{\sqrt {1-{\frac {2}{3}}x^{2}{\sqrt {1-{\frac {23}{15}}x^{2}{\sqrt {1-{\frac {11623}{4830}}x^{2}{\sqrt {\cdots }}}}}}}}\end{aligned}}$

${\sqrt {1-{\frac {2}{3}}{\sqrt {1-{\frac {23}{15}}{\sqrt {1-{\frac {11623}{4830}}{\sqrt {\cdots }}}}}}}}={\frac {\pi }{4}}$

${\sqrt {1-{\frac {2}{3}}\cdot {\frac {1}{3}}{\sqrt {1-{\frac {23}{15}}\cdot {\frac {1}{3}}{\sqrt {1-{\frac {11623}{4830}}\cdot {\frac {1}{3}}{\sqrt {\cdots }}}}}}}}={\frac {\pi }{2{\sqrt {3}}}}$

${\begin{aligned}&{\sqrt[{3}]{\displaystyle 3\left(\left(\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}\right)\left(\sum _{\begin{smallmatrix}(m,n)\in \mathbb {O} ^{2}&\\m\neq n&\end{smallmatrix}}{\frac {(-1)^{\frac {m+n}{2}}}{mn}}\right)+\sum _{\begin{smallmatrix}(l,m,n)\in \mathbb {O} ^{3}&\\l\neq m\neq n\neq l&\end{smallmatrix}}{\frac {(-1)^{\frac {l+m+n-1}{2}}}{lmn}}\right)}}\\&{\overline {+{\sqrt[{3}]{\displaystyle 3\left(\left(\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)^{3}}}\right)\left(\sum _{\begin{smallmatrix}(m,n)\in \mathbb {O} ^{2}&\\m\neq n&\end{smallmatrix}}{\frac {(-1)^{\frac {m+n}{2}}}{m^{3}n^{3}}}\right)+\sum _{\begin{smallmatrix}(l,m,n)\in \mathbb {O} ^{3}&\\l\neq m\neq n\neq l&\end{smallmatrix}}{\frac {(-1)^{\frac {l+m+n-1}{2}}}{l^{3}m^{3}n^{3}}}\right)}}}}\\&{\overline {\overline {+{\sqrt[{3}]{\displaystyle 3\left(\left(\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)^{9}}}\right)\left(\sum _{\begin{smallmatrix}(m,n)\in \mathbb {O} ^{2}&\\m\neq n&\end{smallmatrix}}{\frac {(-1)^{\frac {m+n}{2}}}{m^{9}n^{9}}}\right)+\sum _{\begin{smallmatrix}(l,m,n)\in \mathbb {O} ^{3}&\\l\neq m\neq n\neq l&\end{smallmatrix}}{\frac {(-1)^{\frac {l+m+n-1}{2}}}{l^{9}m^{9}n^{9}}}\right)+\cdots }}}}}\\=&{\frac {\pi }{4}}\ (\mathbb {O} =\{n=1({\text{mod}}2)|n\in \mathbb {N} \})\end{aligned}}$

$x^{\log(1+y)}=x^{y-{\frac {y^{2}}{2}}+{\frac {y^{3}}{3}}-\cdots }=x^{y(1-{\frac {y}{2}}(1-{\frac {2}{3}}y\cdots ))}=x{\sqrt[{\frac {1}{y}}]{x{\sqrt[{-{\frac {2}{y}}}]{x{\sqrt[{-{\frac {3}{y}}}]{x^{2}{\sqrt[{-{\frac {4}{y}}}]{\cdots }}}}}}}}$

${\frac {\pi ^{2}}{6}}=\left({\sqrt[{4}]{e{\sqrt[{8}]{e{\sqrt[{12}]{e^{2}{\sqrt[{16}]{e^{3}\cdots }}}}}}}}\right)\left({\sqrt[{9}]{e{\sqrt[{18}]{e{\sqrt[{27}]{e^{2}{\sqrt[{36}]{e^{3}\cdots }}}}}}}}\right)\left({\sqrt[{25}]{e{\sqrt[{50}]{e{\sqrt[{75}]{e^{2}{\sqrt[{100}]{e^{3}\cdots }}}}}}}}\right)\cdots$

${\frac {\pi }{2}}=\left({\sqrt[{4}]{e{\sqrt[{8}]{e{\sqrt[{12}]{e^{2}{\sqrt[{16}]{e^{3}\cdots }}}}}}}}\right)\left({\sqrt[{16}]{e{\sqrt[{32}]{e{\sqrt[{48}]{e^{2}{\sqrt[{64}]{e^{3}\cdots }}}}}}}}\right)\left({\sqrt[{36}]{e{\sqrt[{72}]{e{\sqrt[{108}]{e^{2}{\sqrt[{144}]{e^{3}\cdots }}}}}}}}\right)\cdots$