I'm looking at the series published @ Wikipedia: http://en.wikipedia.org/wiki/Inverse_hyperbolic_function(adsbygoogle = window.adsbygoogle || []).push({});

There is a series for arsinh, which I was able to derive with no problem - basically take the derivative of arsinh, which is a radical, then apply the general binomial expansion, which gives a series, from which successive derivatives can be determined, then do a Taylor series @ 0, which for every [ ( f^{(n)}/ n! ) x^{n}] term cancels out every term except the one that has the x power as 0, leaving only one term for each Taylor series term, resulting in the nice ordered series for arsinh. However, I can't seem to get something that appears to go in the direction of the series as presented @ Wikipedia.

It appears that somehow if ( 1 / x ) is used instead of ( x ), then there is some type of series that is similar to that of arsinh (i.e., in the way that the series for cos & sin are related) with a strange ln ( 2 x ) term as well. I figure that there must be some underlying identity between arsinh & arcosh, that somehow related arsinh( x ) with arcosh( 1 / x ) and ln ( 2 x ) but I can't find it anywhere.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# How to get (Taylor) series formula for arcosh?

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

Loading...

Similar Threads for Taylor series formula |
---|

I Taylor expansion of f(x+a) |

I Taylor expansions, limits and domains |

I Taylor expansion of 1/distance |

I Taylor series |

B Multiterm Taylor expansion |

**Physics Forums | Science Articles, Homework Help, Discussion**