Talk:Inverse hyperbolic function

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square root for artanh is wrong surely, it should be over the whole quotient

No, the square root is correct. You are forgetting that the square root is of 1-x^2, not 1+x. Since 1-x^2 factorizes, it works out: ln sqrt(1-x^2)/(1-x)=ln sqrt(1-x)sqrt(1+x)/(1-x)=ln sqrt(1+x)/sqrt(1-x)=1/2 ln (1+x)/(1-x). However, an actual issue: the 'arsinh', etc. names are much less common the arc ones. I found a mathematician on the web stating that he'd never encountered this notation. 'ar' may be more correct etymologically, but math notation doesn't work that way, it works based on what people actually use. A place like wikipedia records mathematical usage of these terms: it should use the common notation rather than spreading more confusion by having people looking up the functions be confused about whether they're thinking of the same thing. —Preceding unsigned comment added by 71.182.182.215 (talk) 09:39, 2 December 2007 (UTC)

[edit] x--> z?

Should we change x to z since all of this holds for the inverse hyperbolic functions of a complex variable? futurebird (talk) 04:12, 30 November 2007 (UTC)

[edit] acosh

Shouldn't $\operatorname{arcosh}\, x = \ln(x + \sqrt{x-1}\sqrt{x+1})$ be written as $\operatorname{arcosh}\, x = \ln(x + \sqrt{x^2-1})$ ? (Asech too.) MagiMaster (talk) 05:07, 16 February 2008 (UTC)

No. That gives wrong branch cuts. Fredrik Johansson 15:45, 20 March 2008 (UTC)

For example, with $x = - 10 - 10 i$ , $\sqrt{x^2-1}\approx 10 +10i$ , while $\sqrt{x-1}\sqrt{x+1}\approx -10-10i$ , i.e. the argument shifts by $π$ . Works with $x\in\mathbb{R}$ , though.(212.247.11.156 (talk) 17:42, 29 May 2008 (UTC))

[edit] arcoth

The artile states the derivative of both artanh and arcoth as 1/(1-x**2). Is that a typo? —Preceding unsigned comment added by 88.112.61.116 (talk) 09:49, 30 May 2008 (UTC)

It is correct. Since $\mathrm{coth}(y) = \frac{1}{\mathrm{tanh}(y)}$ then $\mathrm{acoth}(x) = \mathrm{atanh}\left( \frac{1}{x}\right)$ and

$\begin{align}\frac{d}{dx}\mathrm{acoth}(x) & = \frac{d}{dx}\mathrm{atanh}\left( \frac{1}{x}\right) \\ & = \frac{1}{1 - \frac{1}{x^{2}}} \left( - \frac{1}{x^{2}}\right) \\ & = \frac{1}{1 - x^{2}} \end{align}$

It is a bit surprising though, so it would be good to think of a nice way to make it clear in the article that this isn't a mistake. --catslash (talk) 15:01, 30 May 2008 (UTC)