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Homework Help: Arctanh and arccoth

  1. Jul 23, 2010 #1
    How can I prove this:

    1. For all y ∈ ]-1;1[ : Arctanh y = (1/2) ln( 1+y / 1-y )

    2. For all y ∈ ]-∞;-1[ U ]1;+∞[ : Arccoth y = (1/2) ln( y+1 / y-1 )

    Can I solve it by using this:

    Arcsinh y = ln (x+square(x^2 +1))
    Arccosh y = ln (x+square (x^2 -1))
     
  2. jcsd
  3. Jul 23, 2010 #2

    Office_Shredder

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    Arctanhy is defined to be the number x such that tanh(x)=y. So to prove that arcanh(y)=(1/2) ln( 1+y / 1-y ), it's the same thing as showing that tanh((1/2) ln( 1+y / 1-y ))=y
     
  4. Jul 23, 2010 #3
    Thx, but I still don't know how I can prove that..
     
  5. Jul 23, 2010 #4

    Office_Shredder

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    The definition of tanh is in terms of sinh and cosh, which are in terms of exponentials. Surely you can calculate an exponential raised to a logarithmic power, then there's just a bunch of algebra to do. If you get stuck post how far you've gotten and we can see how to progress
     
  6. Jul 23, 2010 #5

    Mark44

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    Please do not double-post.
     
  7. Jul 23, 2010 #6
    I get :

    2 tanh (1/2 ln (1+y / 1-y) = ... = (1+y)/2 - (1-y)/2
     
  8. Jul 25, 2010 #7
    I found it! Thank you very much!
     
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