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Simplfying Inverse Hyperbolic Cosine

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Simplify the following expression:

    [tex] arccosh \left(\frac{1}{\sqrt{1 - x^2}}\right) \forall x ∈ (-1, 1)[/tex]


    2. Relevant equations

    [tex] cosh(u) = \left(\frac{1}{\sqrt{1 - tanh^{2}u}}\right) u ∈ ℝ [/tex]


    3. The attempt at a solution

    [tex] x = tanhu ∴ u = arctanhx [/tex]

    [tex] u ∈ (arctanh(-1), arctanh(1)) [/tex]

    [tex] arccosh(coshu) = u = arctanhx [/tex]

    but that gives me the interval with infinities in it which is what's confusing me.

    I have the solution but I don't understand it particularly the first line.

    4. Solution

    Suppose that [tex] x ∈ [0, 1) [/tex] (Why are they supposing this?)

    and that [tex] u = arctanhx ≥ 0 [/tex]

    so
    [tex] cosh(u) = \left(\frac{1}{\sqrt{1 - tanh^{2}u}}\right) = \frac{1}{\sqrt{1 - x^2}} [/tex] so therefore

    [tex] arccosh \left(\frac{1}{\sqrt{1 - x^2}}\right) = arccosh(coshu) [/tex]

    but [tex] arccosh(coshu) = u [/tex] as long as [tex] u ≥ 0 [/tex] which is the case here

    From where [tex] arccosh \left(\frac{1}{\sqrt{1 - x^2}}\right) = u = arctanh(x) [/tex]

    Since [tex] arccosh \left(\frac{1}{\sqrt{1 - x^2}}\right) [/tex] is an even function (another big ? here, don't know what an even function is) in [tex] x ∈ (-1, 1) [/tex] we obtain [tex] \forall x ∈ (-1, 1), arccosh \left(\frac{1}{\sqrt{1 - x^2}}\right) = arctanh(|x|) [/tex]


    I appreciate the help!
     
  2. jcsd
  3. Oct 21, 2012 #2
    Just as a hint for the last part, an even function is one such that f(-x) = f(x), meaning that the function is symmetric across the y-axis. If a function is even, that means we can solve for it over the interval (0, L) and we've implicitly solved it over (-L, L) since it is the same function on both sides of the axis.
     
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