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Is this valid when using arctanh ln identity?

  1. Jun 1, 2010 #1

    I start with [tex] arctanh\left(\frac{A}{\sqrt{A^2-1}}\right)=\frac{1}{2}ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{1-\frac{A}{\sqrt{A^2-1}}}\right)[/tex]

    The fucntion [tex] \frac{A}{\sqrt{A^2-1}} [/tex] is real, and since A>1, it too is always greater than 1.

    Is it true that it should really be the modulus around the argument of ln? therefore I can manipulate it as follows:

    [tex] ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{1-\frac{A}{\sqrt{A^2-1}}}\right)=ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{-(1-\frac{A}{\sqrt{A^2-1}})}\right)=ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{(1-\frac{A}{\sqrt{A^2-1}})}\right)[/tex]

  2. jcsd
  3. Jun 1, 2010 #2


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    Science Advisor

    In general |tanh(x)| < 1 for x real. Therefore it is to be expected that you will have a complex x for arctanh(u) when u > 1.
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