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Math Help - hyperbolic tangent

  1. #1
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    hyperbolic tangent

    Hi,

    Two things regarding hyperbolic tangent (tanh(x)):

    1) What is the inverse function of tanh(x)?

    Apparently it is NOT artanh(x).

    I need to know how to "undo" a tanh(x) function. Please help.


    2) In my grapher, artanh(1) does not exist (is undefined, invalid or whatever).

    Yet when I plug in artanh(sin(x)^2+cos(x)^2), which should be the same, I get weird splotches.

    Is the grapher getting confused, or does there really exist a difference between 1 and sin(x)^2+cos(x)^2 vis-a-vis artanh(x)?


    Thanks
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by rainer View Post
    Hi,

    Two things regarding hyperbolic tangent (tanh(x)):

    1) What is the inverse function of tanh(x)?

    Apparently it is NOT artanh(x).

    I need to know how to "undo" a tanh(x) function. Please help.


    2) In my grapher, artanh(1) does not exist (is undefined, invalid or whatever).

    Yet when I plug in artanh(sin(x)^2+cos(x)^2), which should be the same, I get weird splotches.

    Is the grapher getting confused, or does there really exist a difference between 1 and sin(x)^2+cos(x)^2 vis-a-vis artanh(x)?


    Thanks
    \tanh (x) = \frac{e^x - e^{-x}}{e^x + e^{-x} }

    to find the inverse let

    u = \tanh ^{-1} x

    \tanh(u) = x

    \frac{e^u - e^{-u}}{e^u+e^{-u}}=x

    simplify

    e^u - e^{-u} = xe^u + xe^{-u}

    e^{2u} -1 = xe^{2u} + x

    e^{2u}(x-1) +x+1=0 let e^{u} = t

    t^2 (x-1) + x+1 = 0

    t = \frac{\mp\sqrt{-4(x-1)(x+1)}}{2(x-1)}

    t = \frac{\mp 2 \sqrt{(1-x)(1+x)}}{2(x-1)}

    t =\mp \sqrt{\frac{(1-x)(1+x)}{(1-x)(1-x)}}

    t=\mp \sqrt{\frac{1+x}{1-x}}

    but t=exp(u) so

    e^u = \mp \sqrt{\frac{1+x}{1-x}}

    u=\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right)

    \tanh ^{-1}(x)=\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right)
    Last edited by Amer; September 25th 2009 at 11:49 AM.
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  3. #3
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    Amer-

    Your demonstration is very useful to me. Many thanks.

    An important question remains, however: Is artanh(x) the inverse function of tanh(x)?

    The wikipedia definition of inverse function is:

    "If is a function from A to B then an inverse function for is a function in the opposite direction, from B to A, with the property that a round trip (a composition) from A to B to A (or from B to A to B) returns each element of the initial set to itself."


    By this definition, for x=1, artanh(x) is NOT the inverse function of
    tanh(x).

    Perhaps this simply means that tanh(x) is not invertible?



    Thanks
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by rainer View Post
    Amer-

    Your demonstration is very useful to me. Many thanks.

    An important question remains, however: Is artanh(x) the inverse function of tanh(x)?

    The wikipedia definition of inverse function is:

    "If ƒ is a function from A to B then an inverse function for ƒ is a function in the opposite direction, from B to A, with the property that a round trip (a composition) from A to B to A (or from B to A to B) returns each element of the initial set to itself."


    By this definition, for x=1, artanh(x) is NOT the inverse function of
    tanh(x).

    Perhaps this simply means that tanh(x) is not invertible?



    Thanks
    ok I see let take

    \tanh ^{-1}(1) = \frac{1}{2} \ln \left(\frac{1+1}{1-1} \right) = \infty

    now

    \tanh (\infty) =\lim_{x\rightarrow \infty}\frac{e^x - e^{-x} }{e^x + e^{-x} }=\lim _{x\rightarrow \infty} \frac{e^{2x}-1}{e^{2x}+1}=1

    that's it

    or

    I will prove that \tanh ^{-1} (\tanh (x)) =x

    \tanh ^{-1}(\tanh) = \frac{1}{2} \ln\left(\frac{1+\dfrac{e^x - e^{-x} }{e^x + e^{-x} } }{1-\dfrac{e^x - e^{-x} }{e^x + e^{-x} }}\right)

    \tanh ^{-1}(\tanh) = \frac{1}{2} \ln\left(\frac{e^x + e^{-x} + e^{x} - e^{-x} }{e^x + e^{-x} -(e^x -e^{-x})}\right)

    \tanh ^{-1}(\tanh) = \frac{1}{2} \ln\left(\frac{2e^x}{2e^{-x}}\right)

    \tanh ^{-1}(\tanh) = \frac{1}{2} (\ln (e^x) - \ln(e^{-x}))=2x/2=x
    Last edited by Amer; September 26th 2009 at 10:19 AM.
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