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**

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

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