អនុគមន៍អ៊ីពែបូលីកច្រាស់

អនុគមន៍ច្រាស់នៃអនុគមន៍អ៊ីពែបូលីកគឺជាអនុគមន៍អ៊ីពែបូលីកក្រលាផ្ទៃ។ វាគណនាក្រលាផ្ទៃនៃបំណែករបស់អ៊ីពែបូលឯកតា ${\displaystyle x^{2}-y^{2}=1\,}$ ដែលជាវិធីដូចគ្នាចំពោះអនុគមន៍ត្រីកោណមាត្រច្រាស់ ក្នុងការគណនាប្រវែងធ្នូនៃបំណែកមួយនៅលើរង្វង់ឯកតា ${\displaystyle x^{2}+y^{2}=1\,}$។ នៅក្នុងគណិតវិទ្យា ជាទូទៅវាត្រូវបានគេសរសេរកាត់ជា ${\displaystyle arsinh,arcsinh\,}$ ឬ ${\displaystyle asinh\,}$ (នៅក្នុងជំនាញវិទ្យាសាស្ត្រកុំព្យូទ័រ)។ ${\displaystyle sinh^{-1}(x),cosh^{-1}(x)\,}$ ជាដើមក៏ត្រូវបានគេប្រើប្រាស់ផងដែរ។

សញ្ញានៃប្រមាណវិធីត្រូវបានផ្តល់និយមន័យនៅក្នុងប្លង់កុំផ្លិចដោយ៖

• ${\displaystyle \operatorname {arsinh} \,x=\ln(x+{\sqrt {x^{2}+1}})}$
• ${\displaystyle \operatorname {arcosh} \,x=\ln(x+{\sqrt {x-1}}{\sqrt {x+1}})}$
• ${\displaystyle \operatorname {artanh} \,x=\ln \left({\frac {\sqrt {1-x^{2}}}{1-x}}\right)={\frac {1}{2}}\ln \left({\frac {1+x}{1-x}}\right)}$
• ${\displaystyle \operatorname {arcsch} \,x=\ln \left({\sqrt {1+{\frac {1}{x^{2}}}}}+{\frac {1}{x}}\right)}$
• ${\displaystyle \operatorname {arsech} \,x=\ln \left({\sqrt {{\frac {1}{x}}-1}}{\sqrt {{\frac {1}{x}}+1}}+{\frac {1}{x}}\right)}$
• ${\displaystyle \operatorname {arcoth} \,x={\frac {1}{2}}\ln {\frac {x+1}{x-1}}}$

អនុគមន៍អ៊ីពែបូលីកច្រាស់ក្នុងប្លង់កុំផ្លិច

${\displaystyle \operatorname {arsinh} (z)}$	${\displaystyle \operatorname {arcosh} (z)}$	${\displaystyle \operatorname {artanh} (z)}$	${\displaystyle \operatorname {arcoth} (z)}$	${\displaystyle \operatorname {arsech} (z)}$	${\displaystyle \operatorname {arcsch} (z)}$

កន្សោមស៊េរីអាចទទួលបានចំពោះអនុគមន៍ខាងលើ៖

${\displaystyle \operatorname {arsinh} \,x}$

${\displaystyle =x-\left({\frac {1}{2}}\right){\frac {x^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{5}}{5}}-\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{7}}{7}}+\cdots }$

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

${\displaystyle \operatorname {arcosh} \,x}$

${\displaystyle =\ln 2x-\left(\left({\frac {1}{2}}\right){\frac {x^{-2}}{2}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{-4}}{4}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{-6}}{6}}+\cdots \right)}$

${\displaystyle =\ln 2x-\sum _{n=1}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{-2n}}{(2n)}},\qquad x>1}$

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

${\displaystyle \operatorname {arcsch} \,x=\operatorname {arsinh} \,x^{-1}}$

${\displaystyle =x^{-1}-\left({\frac {1}{2}}\right){\frac {x^{-3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{-5}}{5}}-\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{-7}}{7}}+\cdots }$

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

${\displaystyle \operatorname {arsech} \,x=\operatorname {arcosh} \,x^{-1}}$

${\displaystyle =\ln {\frac {2}{x}}-\left(\left({\frac {1}{2}}\right){\frac {x^{2}}{2}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{4}}{4}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{6}}{6}}+\cdots \right)}$

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

${\displaystyle \operatorname {arcoth} \,x=\operatorname {artanh} \,x^{-1}}$

${\displaystyle =x^{-1}+{\frac {x^{-3}}{3}}+{\frac {x^{-5}}{5}}+{\frac {x^{-7}}{7}}+\cdots }$

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

កន្សោមអាស៊ីមតូចចំពោះ arsinh x គឺ

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

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

ឧទាហរណ៍មួយចំពោះដេរីវេ។ តាង ${\displaystyle \theta =arsinhx\,}$ គេបានៈ

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

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