MitPY 12: Inverse Hyperbolic Functions

This MitPY comes from occasional commenter, Craig:

Hello folks, just a quick (probably stupid) question. Is the following step justified?

\boldsymbol{x = \ln\left(\cosh x + \sqrt{(\cosh x)^2-1}\right)}

\boldsymbol{\mbox{\bf arccosh}\, x = \ln\left(x+\sqrt{x^2 -1}\right)}


14 Replies to “MitPY 12: Inverse Hyperbolic Functions”

  1. Hi Craig. I’d say yes (but I can think of easier ways (for me, anyway) of getting \text{arccosh} (x) than deriving(?) the first identity).

    Here is how I would justify it:

    Let y = \cosh (x) be defined from the identity \displaystyle x = \ln \left( y + \sqrt{y^2 - 1} \right).

    To get the inverse, let \displaystyle x = \cosh (y) where y = \text{arccosh} (x).

    Then from the above identity:

    \displaystyle y = \ln \left( x + \sqrt{x^2 - 1} \right).

    1. Thanks John. I had to show that arccosh was actually ln(stuff) in disguise, so I tried starting with the exponential definition for cosh then try and get arccosh somehow. I got the right formula but x and arccoshx were in each others place. Just wondered if I could swap them around. And I still have to explain how I got rid of a plus/minus at some point. But I’m happy to know that the last two lines are fine as you say.

  2. arccosh(x) is only defined for x ≥ 1, so you have to add that in as a condition as well.

    Once you do that, yes, this is fine and justified, since if f(x) = g(x), then f(h(x)) = g(h(x)) (assuming everything in question exists).

          1. There are no restrictions on z but since the complex logarithm is multi-valued you need to use the Principal Value log and so there are restrictions on the values of the complex logarithm.

  3. Hi Craig!

    I think it is always worth thinking about if you are consistent with notation and variables. In the first equation, I guess you know how to argue that it makes sense for all real x.

    But is that variable x the *same* x as in the second equation? Or is it something different?


    1. Thanks Glen. In the first equation, x means the input for the cosh function. For the second equation, x means the input for the inverse cosh function. So it looks like they are different xs. I just had to get to the second equation and I got to the first so I thought I would take a shortcut and see if that shortcut made sense and that’s why I’m here. So I guess I should have changed the x in the second equation to something else. Or actually the first x to something else. Which is I think what happens in John’s notes.

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