# Is there a better way to calculate the Arc-Cosine-Hyperbolic

I am using $$\operatorname{arcosh} x = \operatorname{arcsinh}(\sqrt { x^2-1 } )$$ with bad results. The standard power series $$\operatorname{arcosh} x = \ln(2x) - \left( \left( \frac {1} {2} \right) \frac {x^{-2}} {2} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {x^{-4}} {4} + \left( \frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} \right) \frac {x^{-6}} {6} +\cdots \right) \\ = \ln(2x) - \sum_{n=1}^\infty \left( \frac {(2n)!} {2^{2n}(n!)^2} \right) \frac {x^{-2n}} {2n} , \qquad \left| x \right| > 1$$ is also messy for me since it wants to use $$\ln(2x)$$ and it has the slow calculation of $$x^{-6} / 6$$ and others like it. I have a $$\ln()$$ power series but it also takes a long time to calculate. Also, this one is not good $$\operatorname {arcosh} (x) = \ln \left(x + \sqrt{x^{2} - 1} \right) \qquad x \geqslant 1 \\$$ It uses an $$\ln()$$ and square root function. Calculating a square root is as bad as a $$\ln()$$. Please, no Bernoulli or Euler numbers also. They are messy in a computer system. Least way, my system. Sorry for being picky.

Is there another series to be used? Thank you.

$$Blue$$, 1st Thank you for all of your help before. Yes, I followed your process, and it made sense, but I can not see how to remove $$ln(2x)$$ from the series.

$$Jean-Claude Arbaut$$, The problem, for me only, is that 1. Within the C code, they do use the $$log$$ function built within the computer system. 2. They are also using fixed-length (20 decimal places) constants, example::$$ln2 = 6.93147180559945286227e-01;$$ I want and can go longer than this, if I need to.

$$Yves Daoust$$, Bernoulli or Euler are "messy" from the view point that I would have to carry them in a table and they are only given a certain fixed-length. Also, I have been trying to find "how do you create Bernoulli or Euler numbers". I have found nothing simple to use.

• As with the various other faster-series-computation questions you've asked, there's a Horner-like form that speeds up the calculation here. I'd hoped that walking through the process in my previous answer would be enough to help you handle other cases on your own, but if you need more guidance, I can try to provide it. Have you tried applying the Horner form to this problem? If so, can you convey where you're having difficulties? I'd like to address your specific issues instead of posting another general solution. – Blue Apr 22 at 2:40
• If you lookup how it's done in widespread libraries, the formula with log is not that uncommon. See fdlibm or fn on Netlib. You can check other libraries. – Jean-Claude Arbaut Apr 22 at 7:35
• If you want a "least way" solution, you must specify which range of values you want to support, and what accuracy. You are not picky, you are incomplete, sorry. By the way, I don't see how the Bernouilli or Euler are "messy". They are mere constants. – Yves Daoust Apr 22 at 8:10

For "small" values of $$x$$ (say $$x<3$$), you could use Puiseux-Laurent series since $$\frac{\cosh^{-1}(x)}{\sqrt{2(x-1)} }=1-\frac{x-1}{12}+\frac{3(x-1)^2}{160} -\frac{5(x-1)^3}{896} +\frac{35 (x-1)^4}{18432}-\frac{63 (x-1)^5}{90112}+\frac{231 (x-1)^6}{851968}-\frac{143 (x-1)^7}{1310720}+\frac{6435 (x-1)^8}{142606336}-\frac{12155 (x-1)^9}{637534208}+\frac{46189 (x-1)^{10}}{5637144576}-\frac{88179 (x-1)^{11}}{24696061952}+\frac{676039 (x-1)^{12}}{429496729600}-\frac{1300075 (x-1)^{13}}{1855425871872}+\frac{5014575 (x-1)^{14}}{15942918602752}+O\left((x-1)^{15}\right)$$
Have a look at $$OEIS$$ sequences $$A055786$$ and $$A091019$$ if you want more terms.
Otherwise, using $$\cosh^{-1}(x)= \log(2x)- \sum_{n=1}^\infty \left( \frac {(2n)!} {2^{2n}(n!)^2} \right) \frac {x^{-2n}} {2n}$$ let $$a_n=\left( \frac {(2n)!} {2^{2n}(n!)^2} \right) \frac {x^{-2n}} {2n}\implies a_{n+1}=\frac{n (2 n+1)}{2 (n+1)^2 }\,a_n\,x^{-2}$$ which is quite good for programming.