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Finding the analytic expression for arcsinh

  1. Jan 24, 2017 #1
    1. The problem statement, all variables and given/known data
    Given that ##\sinh x = \frac{e^x-e^{-x}}{2}##, find an expression for ##arcsinh x##

    2. Relevant equations


    3. The attempt at a solution
    We can proceed by the normal procedure for finding inverses of well-defined functions, solve ##x = \frac{e^y - e^{-y}}{2}## for y. After doing some algebra and using the quadratic formula, we find that ##y = \log (x \pm \sqrt{x^2 + 1})##. How do I know which root to take? It would seem that there are two inverse functions of sinhx
     
  2. jcsd
  3. Jan 24, 2017 #2

    ehild

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    Homework Helper
    Gold Member

    What is the argument of the logarithm if you take the negative sign in ##y = \log (x \pm \sqrt{x^2 + 1})##?
     
  4. Jan 24, 2017 #3

    Stephen Tashi

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    Science Advisor

    The argument of ##log(.)## can't be a negative number.
     
  5. Jan 24, 2017 #4
    So since ##e^y## is always positive, and ##x - \sqrt{x^2 + 1}## is negative for all x, do we consider it an extraneous "solution", and thus take the other root? Why does ##x - \sqrt{x^2 + 1}## show up if it can't actually be a solution?
     
  6. Jan 25, 2017 #5

    Mark44

    Staff: Mentor

    If you had ##u^2 - 2yu - 1 = 0## and your goal was to solve for u, you would get ##u = y \pm \sqrt{y^2 + 1}##. Here there is no problem with u being negative.

    However, in your equation, you have ##e^{2x} - 2ye^x - 1 = 0##. Solving for ##e^x## gives the same solution as for u, above. This time around, ##e^x## must be nonnegative, so we have to discard the solution ##y - \sqrt{y^2 + 1}##. You weren't solving an actual quadratic equation -- instead, you were solving one that was only quadratic in form. Since ##u = e^x##, u can only take on nonnegative values.
     
  7. Jan 25, 2017 #6

    Stephen Tashi

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    Science Advisor

    Because our beloved algebraic manipulations are not completely reliable. We do them without keeping track of the assumptions we use when we apply them.

    To repeat what Mark44 said:

    From ## 2x = e^y + \frac{1}{e^y} ## we can multiply both sides by ##e^y## provided we assume ##e^y## is a number. From the properties of the function ##e^y##, we must assume ##e^y## is a positive number.

    Later where our work leads to the conclusion: ## e^y = x + \sqrt{x^2 + 1} ## or ## e^y = x - \sqrt{x^2 + 1}## this work takes place under the assumption ##e^y > 0##. So if we were using the appropriate words to keep track of our assumptions, we would see the alternative ##e^y = x - \sqrt{x^2 + 1}## is not viable.

    It's similar to evaluating an expression in proposition logic like ## A \land (B \lor \lnot A)## to imply ## A \land B##. When we do the algebraic manipulations we forget the "##A \land##" prefixes the work that leads to ##(B \lor \lnot A)## and then we wonder why the alternative ## \lnot A## doesn't work.
     
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