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Math Help - problem regards hyperbolic function

  1. #1
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    problem regards hyperbolic function

    Hi I have been having a problem regards this question. I spent some time trying to sit through it but to no result.

    The Question is

    Show that

    arccosech x = log((1+squareroot(1+x^2)) for x>0
    log((1-squareroot(1+x^2)) for x<0

    Justify the sign of any square roots.


    So this is what I did

    I assume that y=arccosech(x) thereby cosech(y)=x

    1/(sinh(y)) = 1/(1/2*(e^y)-(e^-y))
    2/x=e^y - e^-y
    In(2/x)=In(e^y)-In(e^-y)
    In(2/x)=2y
    2arccosechx=In(2/x)


    this is totally wrong but I do not know how to continue or to handle problem such as this one. Can any one give me a hint and direction about solving this problem?

    Best Regards
    Junks
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  2. #2
    Junior Member FelixFelicis28's Avatar
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    Re: problem regards hyperbolic function

    Quote Originally Posted by junkwisch View Post
    Hi I have been having a problem regards this question. I spent some time trying to sit through it but to no result.

    The Question is

    Show that

    arccosech x = log((1+squareroot(1+x^2)) for x>0
    log((1-squareroot(1+x^2)) for x<0

    Justify the sign of any square roots.


    So this is what I did

    I assume that y=arccosech(x) thereby cosech(y)=x

    1/(sinh(y)) = 1/(1/2*(e^y)-(e^-y))
    I think there's a mistake and it should be:

    \displaystyle \text{csch}^{-1} x = \begin{cases} \ln \left(1 + \sqrt{1 + x^2}\right) - \ln x \quad \quad \text{if} \ x > 0 \\ \ln \left(1 - \sqrt{1 + x^2}\right) - \ln x \quad \quad \text{if} \ x < 0 \end{cases}

    Now, consider:

    \displaystyle \begin{aligned} y = \text{csch}^{-1} x \implies x = \text{csch} y = \frac{1}{\sinh y} & = \frac{2}{e^y - e^{-y}} \\ \implies x\left(e^y - \frac{1}{e^y}\right) & = 2 \end{aligned}

    Let u = e^y and you'll get a quadratic in u. Solve for u, substitute u = e^y again and take natural logs.

    Oh, and I felt the need to point this out:

    2/x=e^y - e^-y
    In(2/x)=In(e^y)-In(e^-y)
    In(2/x)=2y
    2arccosechx=In(2/x)


    this is totally wrong but I do not know how to continue or to handle problem such as this one. Can any one give me a hint and direction about solving this problem?

    Best Regards
    Junks
    If you have a = b - c and you take natural logs of both sides, you take the log of the entire side i.e. \ln a = \ln (b - c) and as you know, in general \ln (x - y) \neq \ln x - \ln y.
    Last edited by FelixFelicis28; September 5th 2013 at 06:02 AM.
    Thanks from junkwisch
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  3. #3
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    Re: problem regards hyperbolic function

    thank you for your reply Felix, yes I did made a mistake. I forgot to put -In X into the equation. Just one small question though, so I subbed in U=e^y as you suggested, I did some algebra and got the equation to become u^2 - 2u/x -1 =0 I use quadratic formula for u, (-b plus of minus squareroot of (b^2 -4ac))/2a. The problem is I got U = (2/x +1) and U=-1, which is wrong. Do you know what I did wrong?
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  4. #4
    Junior Member FelixFelicis28's Avatar
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    Re: problem regards hyperbolic function

    Quote Originally Posted by junkwisch View Post
    thank you for your reply Felix, yes I did made a mistake. I forgot to put -In X into the equation. Just one small question though, so I subbed in U=e^y as you suggested, I did some algebra and got the equation to become u^2 - 2u/x -1 =0 I use quadratic formula for u, (-b plus of minus squareroot of (b^2 -4ac))/2a. The problem is I got U = (2/x +1) and U=-1, which is wrong. Do you know what I did wrong?
    Hmmm, you've done something incorrectly with the quadratic formula. I personally would've re-written u^2 - \tfrac{2}{x} u - 1 = 0 as x u^2 - 2u - x = 0 by multiplying through by x to make things easier but it shouldn't matter.

    \displaystyle \begin{aligned} u^2 - \tfrac{2}{x} u - 1 = 0 \implies u & = \frac{\tfrac{2}{x} \pm \sqrt{\left( \tfrac{2}{x} \right)^2 + 4}}{2} \\ & = \frac{1}{x} \pm \sqrt{\frac{1}{x^2} + 1} \\ & = \frac{1 + \sqrt{1 + x^2}}{x} \end{aligned}

    And then replace u with e^y and bob's your uncle.
    Last edited by FelixFelicis28; September 8th 2013 at 05:34 AM.
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