逆三角関数
arctan  z = ∑ n = 0 ∞ ( − 1 ) n 2 n + 1 z 2 n + 1 = z − z 3 3 + z 5 5 − z 7 7 + ⋯ ; | z | ≤ 1 , z ≠ ± i {\displaystyle {\begin{aligned}\arctan z&=\textstyle \sum \limits _{n=0}^{\infty }{\dfrac {(-1)^{n}}{2n+1}}z^{2n+1}\\&=z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\dotsb ;\quad |z|\leq 1,z\neq \pm i\end{aligned}}}
arccot  z = π 2 − arctan  z = π 2 − ∑ n = 0 ∞ ( − 1 ) n 2 n + 1 z 2 n + 1 = π 2 − ( z − z 3 3 + z 5 5 − z 7 7 + ⋯ ) ; | z | ≤ 1 , z ≠ ± i {\displaystyle {\begin{aligned}\operatorname {arccot} z&={\dfrac {\pi }{2}}-\arctan z\\&={\frac {\pi }{2}}-\textstyle \sum \limits _{n=0}^{\infty }{\dfrac {(-1)^{n}}{2n+1}}z^{2n+1}\\&={\frac {\pi }{2}}-\left(z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\dotsb \right);\quad |z|\leq 1,z\neq \pm i\end{aligned}}}
逆正接関数の冪級数の2つの代わりはこれらの一般化連分数(英語版)である:
arctan  z = z 1 + ( 1 z ) 2 3 − 1 z 2 + ( 3 z ) 2 5 − 3 z 2 + ( 5 z ) 2 7 − 5 z 2 + ( 7 z ) 2 9 − 7 z 2 + ⋱ = z 1 + ( 1 z ) 2 3 + ( 2 z ) 2 5 + ( 3 z ) 2 7 + ( 4 z ) 2 9 + ⋱ {\displaystyle {\begin{aligned}\arctan z&={\cfrac {z}{1+{\cfrac {(1z)^{2}}{3-1z^{2}+{\cfrac {(3z)^{2}}{5-3z^{2}+{\cfrac {(5z)^{2}}{7-5z^{2}+{\cfrac {(7z)^{2}}{9-7z^{2}+\ddots }}}}}}}}}}\\&={\cfrac {z}{1+{\cfrac {(1z)^{2}}{3+{\cfrac {(2z)^{2}}{5+{\cfrac {(3z)^{2}}{7+{\cfrac {(4z)^{2}}{9+\ddots \,}}}}}}}}}}\end{aligned}}}