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Relationship between hyperbolic arctan and logarithm

  1. Oct 29, 2011 #1
    1. The problem statement, all variables and given/known data

    The relationship between arctanh and log is:

    arctanh(x)=[itex]\frac{1}{2}log(\frac{1+x}{1-x})[/itex]
    but if i take x=1.5,
    I have:
    arctanh(1.5)=0.8047 + 1.5708i
    and
    [itex]\frac{1}{2}log(\frac{1+1.5}{1-1.5})[/itex]=0.8047 + 1.5708i
    as expected, but using the laws of logarithm, why does this not equal to:
    [itex](\frac{1}{2})\frac{1}{2}log((\frac{1+1.5}{1-1.5})^2)[/itex]=0.8047
    ?
    2. Relevant equations



    3. The attempt at a solution


    Thank you in advance
     
  2. jcsd
  3. Oct 29, 2011 #2

    I like Serena

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    Hi sara_87! :smile:

    You have applied the log to a negative number yielding a complex number.
    So implicitly you have used the complex logarithm instead of the regular logarithm.

    The complex logarithm happens to be a multivalued function:
    [tex]\log re^{i\phi} = \log r + i (\phi + 2k\pi)[/tex]

    This means you have to consider the other solutions.
    By taking the square you introduce a new solution.

    In your example there is another solution, which is the right one:
    [tex](\frac{1}{2})\frac{1}{2}log((\frac{1+1.5}{1-1.5})^2) = (\frac{1}{2})\frac{1}{2}(\log 25 + i 2\pi)=\frac{1}{2}\log 5 + i \frac \pi 2[/tex]
     
  4. Oct 29, 2011 #3
    thank you :)

    so, in general, can i say that if x is less than 1,

    [itex]arctanh(x)=\frac{1}{2}*log(\frac{1+x}{1-x})=\frac{1}{2}*\frac{1}{2}log((\frac{1+x}{1-x})^2)[/itex]

    if x is more than 1, then

    [itex]arctanh(x)=\frac{1}{2}*log(\frac{1+x}{1-x})=\frac{1}{2}*\frac{1}{2}log((\frac{1+x}{1-x})^2)+i*pi/2[/itex]
     
  5. Oct 29, 2011 #4

    I like Serena

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  6. Oct 29, 2011 #5
    Thanks again :)
     
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