អនុគមន៍ artanh
អនុគមន៍ច្រាស់នៃអនុគមន៍អ៊ីពែបូលីកគឺជាអនុគមន៍អ៊ីពែបូលីកក្រលាផ្ទៃ។ វាគណនាក្រលាផ្ទៃនៃបំណែករបស់អ៊ីពែបូលឯកតា
ដែលជាវិធីដូចគ្នាចំពោះអនុគមន៍ត្រីកោណមាត្រច្រាស់ ក្នុងការគណនាប្រវែងធ្នូនៃបំណែកមួយនៅលើរង្វង់ឯកតា
។ នៅក្នុងគណិតវិទ្យា ជាទូទៅវាត្រូវបានគេសរសេរកាត់ជា
ឬ
(នៅក្នុងជំនាញវិទ្យាសាស្ត្រកុំព្យូទ័រ)។
ជាដើមក៏ត្រូវបានគេប្រើប្រាស់ផងដែរ។
សញ្ញានៃប្រមាណវិធីត្រូវបានផ្តល់និយមន័យនៅក្នុងប្លង់កុំផ្លិចដោយ៖
![{\displaystyle \operatorname {arsinh} \,x=\ln(x+{\sqrt {x^{2}+1}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b5a940f68091896efb3c640e005836039af0193)
![{\displaystyle \operatorname {arcosh} \,x=\ln(x+{\sqrt {x-1}}{\sqrt {x+1}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f65d5d0842e197ab161f8c765ca7edf0b77dcbc9)
![{\displaystyle \operatorname {artanh} \,x=\ln \left({\frac {\sqrt {1-x^{2}}}{1-x}}\right)={\frac {1}{2}}\ln \left({\frac {1+x}{1-x}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e145b67709abd59cf19315aaaaee51e9d86a1043)
![{\displaystyle \operatorname {arcsch} \,x=\ln \left({\sqrt {1+{\frac {1}{x^{2}}}}}+{\frac {1}{x}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0985123c671436e13edb807c1a46eead73250622)
![{\displaystyle \operatorname {arsech} \,x=\ln \left({\sqrt {{\frac {1}{x}}-1}}{\sqrt {{\frac {1}{x}}+1}}+{\frac {1}{x}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c923adf336fe2636acac8fb1b38e872e732533b8)
![{\displaystyle \operatorname {arcoth} \,x={\frac {1}{2}}\ln {\frac {x+1}{x-1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79467f032f6b5cb8950e20483a82758a7decbab2)
អនុគមន៍អ៊ីពែបូលីកច្រាស់ក្នុងប្លង់កុំផ្លិច
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កន្សោមស៊េរីអាចទទួលបានចំពោះអនុគមន៍ខាងលើ៖
![{\displaystyle =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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/7306dbcf947ca114605b7c9ac1d90fad12e7a123)
![{\displaystyle =\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d548a52ebc547d2663ece8b454ae9165c54c1cba)
![{\displaystyle =\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/87f2e040edd7ba424a27c7a7b62090afb2e03dc6)
![{\displaystyle =\ln 2x-\sum _{n=1}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{-2n}}{(2n)}},\qquad x>1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/619a6112dee56bb382e60e82d93c98171ee158d0)
![{\displaystyle \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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/042385ac6d62082a4ce68602d475b601d45068c4)
![{\displaystyle =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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c239e579e423dd7f76809b535a3f0413d20f8587)
![{\displaystyle =\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a559ff704836a9fdcfdb8812d5f102a70635f59e)
![{\displaystyle =\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44c227e49955b475b8d15dfe231dda2de950df6c)
![{\displaystyle =\ln {\frac {2}{x}}-\sum _{n=1}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{2n}}{2n}},\qquad 0<x\leq 1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/588a3bf00dae48b336c9583df3c09b644506e93f)
![{\displaystyle =x^{-1}+{\frac {x^{-3}}{3}}+{\frac {x^{-5}}{5}}+{\frac {x^{-7}}{7}}+\cdots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/7cef86af717cecf1dada61c4f0835a038db53816)
![{\displaystyle =\sum _{n=0}^{\infty }{\frac {x^{-(2n+1)}}{(2n+1)}},\qquad \left|x\right|>1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02a381055841fd8ee3c8f5a00fcd51fbec4a3f13)
កន្សោមអាស៊ីមតូចចំពោះ arsinh x គឺ
![{\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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e6ff85ab18e02ce0b9f828449f91b40beb7c6e6)
ដេរីវេនៃអនុគមន៍អ៊ីពែបូលីកច្រាស់
[កែប្រែ]
![{\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 {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ad371657f6a59a54752017197743c510b8c68e9)
ឧទាហរណ៍មួយចំពោះដេរីវេ។ តាង
គេបានៈ
![{\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}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7d56ae857b6e2927eee07f6cc12e7fcec483d90)