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Relationship between hyperbolic arctan and logarithm

  • Thread starter sara_87
  • Start date
  • #1
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Homework Statement



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
?

Homework Equations





The Attempt at a Solution




Thank you in advance
 

Answers and Replies

  • #2
I like Serena
Homework Helper
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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]
 
  • #3
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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]
 
  • #4
I like Serena
Homework Helper
6,577
176
Yes.
 
  • #5
763
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Thanks again :)
 

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