逆双曲線関数

演算子は複素数平面で次のように定義される。

${\displaystyle {\begin{aligned}\operatorname {arsinh} \,z&=\ln(z+{\sqrt {z^{2}+1}}\,)\\[2.5ex]\operatorname {arcosh} \,z&=\ln(z+{\sqrt {z+1}}{\sqrt {z-1}}\,)\\[1.5ex]\operatorname {artanh} \,z&={\tfrac {1}{2}}\ln \left({\frac {1+z}{1-z}}\right)\\\operatorname {arcoth} \,z&={\tfrac {1}{2}}\ln \left({\frac {z+1}{z-1}}\right)\\\operatorname {arcsch} \,z&=\ln \left({\frac {1}{z}}+{\sqrt {{\frac {1}{z^{2}}}+1}}\,\right)\\\operatorname {arsech} \,z&=\ln \left({\frac {1}{z}}+{\sqrt {{\frac {1}{z}}+1}}\,{\sqrt {{\frac {1}{z}}-1}}\,\right)\end{aligned}}}$

上記の平方根は正の平方根であり、対数関数は複素対数である。実数の引数、例えばz = xは実数値を返すが、一定の簡素化を行うことが可能であり、例えば ${\displaystyle {\sqrt {x+1}}{\sqrt {x-1}}={\sqrt {x^{2}-1}}}$は正の平方根を使うとき、一般に真ではない。

${\displaystyle \operatorname {arsinh} (z)}$

${\displaystyle \operatorname {arcosh} (z)}$

${\displaystyle \operatorname {artanh} (z)}$

${\displaystyle \operatorname {arcoth} (z)}$

${\displaystyle \operatorname {arsech} (z)}$

${\displaystyle \operatorname {arcsch} (z)}$

z平面（複素数平面）における逆双曲線関数：平面における各点の色はその点における関数の複素数を表す。

上記の関数は次のように級数展開できる。

${\displaystyle {\begin{aligned}\operatorname {arsinh} \,x&=x-\left({\frac {1}{2}}\right){\frac {x^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{5}}{5}}-\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{7}}{7}}+\cdots \\&=\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{2n+1}}{(2n+1)}},\qquad \left|x\right|<1\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {arcosh} \,x&=\ln 2x-\left(\left({\frac {1}{2}}\right){\frac {x^{-2}}{2}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{-4}}{4}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{-6}}{6}}+\cdots \right)\\&=\ln 2x-\sum _{n=1}^{\infty }\left({\frac {(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{-2n}}{(2n)}},\qquad x>1\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {artanh} \,x&=x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+{\frac {x^{7}}{7}}+\cdots \\&=\sum _{n=0}^{\infty }{\frac {x^{2n+1}}{(2n+1)}},\qquad \left|x\right|<1\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {arcsch} \,x=\operatorname {arsinh} {\frac {1}{x}}&=x^{-1}-\left({\frac {1}{2}}\right){\frac {x^{-3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{-5}}{5}}-\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{-7}}{7}}+\cdots \\&=\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{-(2n+1)}}{(2n+1)}},\qquad \left|x\right|>1\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {arsech} \,x=\operatorname {arcosh} {\frac {1}{x}}&=\ln {\frac {2}{x}}-\left(\left({\frac {1}{2}}\right){\frac {x^{2}}{2}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{4}}{4}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{6}}{6}}+\cdots \right)\\&=\ln {\frac {2}{x}}-\sum _{n=1}^{\infty }\left({\frac {(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{2n}}{2n}},\qquad 0<x\leq 1\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {arcoth} \,x=\operatorname {artanh} {\frac {1}{x}}&=x^{-1}+{\frac {x^{-3}}{3}}+{\frac {x^{-5}}{5}}+{\frac {x^{-7}}{7}}+\cdots \\&=\sum _{n=0}^{\infty }{\frac {x^{-(2n+1)}}{(2n+1)}},\qquad \left|x\right|>1\end{aligned}}}$

またオイラーによるarctanの展開[4]の類似も成り立つ。

${\displaystyle \operatorname {artanh} \,x=\sum _{n=0}^{\infty }{\frac {2^{2n}(n!)^{2}}{(2n+1)!}}{\frac {(-1)^{n}x^{2n+1}}{(1-x^{2})^{n+1}}},\qquad \left|x\right|<{\frac {1}{\sqrt {2}}}}$

${\displaystyle (\operatorname {arsinh} \,x)^{2}=\sum _{n=0}^{\infty }{\frac {2^{2n+1}(n!)^{2}}{(2n+2)!}}(-1)^{n}x^{2n+2},\qquad \left|x\right|<1}$

arsinh x に対する漸近展開は次の式で与えられる。

${\displaystyle \operatorname {arsinh} \,x=\ln 2x+\sum \limits _{n=1}^{\infty }{\left({-1}\right)^{n-1}{\frac {\left({2n-1}\right)!!}{2n\left({2n}\right)!!}}}{\frac {1}{x^{2n}}}}$

${\displaystyle {\begin{aligned}{\frac {d}{dx}}\operatorname {arsinh} \,x&{}={\frac {1}{\sqrt {1+x^{2}}}}\\{\frac {d}{dx}}\operatorname {arcosh} \,x&{}={\frac {1}{\sqrt {x^{2}-1}}}\\{\frac {d}{dx}}\operatorname {artanh} \,x&{}={\frac {1}{1-x^{2}}}\\{\frac {d}{dx}}\operatorname {arcoth} \,x&{}={\frac {1}{1-x^{2}}}\\{\frac {d}{dx}}\operatorname {arsech} \,x&{}={\frac {-1}{x(x+1)\,{\sqrt {\frac {1-x}{1+x}}}}}\\{\frac {d}{dx}}\operatorname {arcsch} \,x&{}={\frac {-1}{x^{2}\,{\sqrt {1+{\frac {1}{x^{2}}}}}}}\\\end{aligned}}}$

実数xに対して、

${\displaystyle {\begin{aligned}{\frac {d}{dx}}\operatorname {arsech} \,x&{}={\frac {\mp 1}{x\,{\sqrt {1-x^{2}}}}};\qquad \Re \{x\}\gtrless 0\\{\frac {d}{dx}}\operatorname {arcsch} \,x&{}={\frac {\mp 1}{x\,{\sqrt {1+x^{2}}}}};\qquad \Re \{x\}\gtrless 0\end{aligned}}}$

微分法の例：θ = arsinh xとおくと、

${\displaystyle {\frac {d\,\operatorname {arsinh} \,x}{dx}}={\frac {d\theta }{d\sinh \theta }}={\frac {1}{\cosh \theta }}={\frac {1}{\sqrt {1+\sinh ^{2}\theta }}}={\frac {1}{\sqrt {1+x^{2}}}}}$

${\displaystyle {\begin{aligned}\sinh(\operatorname {arcosh} x)&={\sqrt {x^{2}-1}}&{\text{for }}|x|>1\\\sinh(\operatorname {artanh} x)&={\frac {x}{\sqrt {1-x^{2}}}}&{\text{for }}|x|<1\\\cosh(\operatorname {arsinh} x)&={\sqrt {1+x^{2}}}\\\cosh(\operatorname {artanh} x)&={\frac {1}{\sqrt {1-x^{2}}}}&{\text{for }}|x|<1\\\tanh(\operatorname {arsinh} x)&={\frac {x}{\sqrt {1+x^{2}}}}\\\tanh(\operatorname {arcosh} x)&={\frac {\sqrt {x^{2}-1}}{x}}&{\text{for }}|x|>1\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {arsinh} u\pm \operatorname {arsinh} v&=\operatorname {arsinh} \left(u{\sqrt {1+v^{2}}}\pm v{\sqrt {1+u^{2}}}\right)\\\operatorname {arcosh} u\pm \operatorname {arcosh} v&=\operatorname {arcosh} \left(uv\pm {\sqrt {\left(u^{2}-1\right)\left(v^{2}-1\right)}}\right)\\\operatorname {artanh} u\pm \operatorname {artanh} v&=\operatorname {artanh} \left({\frac {u\pm v}{1\pm uv}}\right)\\\operatorname {arsinh} u+\operatorname {arcosh} v&=\operatorname {arsinh} \left(uv+{\sqrt {\left(1+u^{2}\right)\left(v^{2}-1\right)}}\right)\\&=\operatorname {arcosh} \left(v{\sqrt {1+u^{2}}}+u{\sqrt {v^{2}-1}}\right)\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {arcosh} (2x^{2}-1)&=2\operatorname {arcosh} (x)&{\text{ for }}x\geq 1\\\operatorname {arcosh} (8x^{4}-8x^{2}+1)&=4\operatorname {arcosh} (x)&{\text{ for }}x\geq 1\\\operatorname {arcosh} (2x^{2}+1)&=2\operatorname {arsinh} (x)&{\text{ for }}x\geq 0\\\operatorname {arcosh} (8x^{4}+8x^{2}+1)&=4\operatorname {arsinh} (x)&{\text{ for }}x\geq 0\end{aligned}}}$

• グーデルマン関数
• 逆双曲線関数の原始関数の一覧
• 関数一覧

• Herbert Busemann and Paul J. Kelly (1953) Projective Geometry and Projective Metrics, page 207, Academic Press.

• Weisstein, Eric W. "Inverse hyperbolic functions". MathWorld (英語).
• area functions - PlanetMath.（英語）
• Hazewinkel, Michiel, ed. (2001), "Inverse hyperbolic functions", Encyclopaedia of Mathematics, Springer, ISBN 978-1-55608-010-4。
• Hazewinkel, Michiel, ed. (2001), "Area-function", Encyclopaedia of Mathematics, Springer, ISBN 978-1-55608-010-4。
• Inverse hyperbolic functions - University College London Department of Mathematics
• 2.22 逆双曲線関数 - 同志社大学・近藤弘一