Khai triển biểu thức Hàm hyperbolic ngược

Biểu thức sau đây có thể được khai triển:

arsinh ⁡ x = x − ( 1 2 ) x 3 3 + ( 1 ⋅ 3 2 ⋅ 4 ) x 5 5 − ( 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ) x 7 7 ± ⋯ = ∑ n = 0 ∞ ( ( − 1 ) n ( 2 n ) ! 2 2 n ( n ! ) 2 ) x 2 n + 1 2 n + 1 , | x | < 1 {\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}}\pm \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}}} arcosh ⁡ x = ln ⁡ ( 2 x ) − ( ( 1 2 ) x − 2 2 + ( 1 ⋅ 3 2 ⋅ 4 ) x − 4 4 + ( 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ) x − 6 6 + ⋯ ) = ln ⁡ ( 2 x ) − ∑ n = 1 ∞ ( ( 2 n ) ! 2 2 n ( n ! ) 2 ) x − 2 n 2 n , | x | > 1 {\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 \left|x\right|>1\end{aligned}}} artanh ⁡ x = x + x 3 3 + x 5 5 + x 7 7 + ⋯ = ∑ n = 0 ∞ x 2 n + 1 2 n + 1 , | x | < 1 {\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}}} arcsch ⁡ x = arsinh ⁡ 1 x = x − 1 − ( 1 2 ) x − 3 3 + ( 1 ⋅ 3 2 ⋅ 4 ) x − 5 5 − ( 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ) x − 7 7 ± ⋯ = ∑ n = 0 ∞ ( ( − 1 ) n ( 2 n ) ! 2 2 n ( n ! ) 2 ) x − ( 2 n + 1 ) 2 n + 1 , | x | > 1 {\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}}\pm \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}}} arsech ⁡ x = arcosh ⁡ 1 x = ln ⁡ 2 x − ( ( 1 2 ) x 2 2 + ( 1 ⋅ 3 2 ⋅ 4 ) x 4 4 + ( 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ) x 6 6 + ⋯ ) = ln ⁡ 2 x − ∑ n = 1 ∞ ( ( 2 n ) ! 2 2 n ( n ! ) 2 ) x 2 n 2 n , 0 < x ≤ 1 {\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}}} arcoth ⁡ x = artanh ⁡ 1 x = x − 1 + x − 3 3 + x − 5 5 + x − 7 7 + ⋯ = ∑ n = 0 ∞ x − ( 2 n + 1 ) 2 n + 1 , | x | > 1 {\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}}}

Khai triển tiệm cận cho hàm arsinh x được cho bởi

arsinh ⁡ x = ln ⁡ ( 2 x ) + ∑ n = 1 ∞ ( − 1 ) n − 1 ( 2 n − 1 ) ! ! 2 n ( 2 n ) ! ! 1 x 2 n {\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}}}}


Tài liệu tham khảo

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