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Math Help - Hyperbolic functions

  1. #1
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    Hyperbolic functions

    Hi,

    I'm having problem with this question:

    If y = \cosh(x) for x > 0, find x in terms of y and hence show that \cosh^{-1}x = \ln (x + \sqrt{x^2-1}). State the domain and range of the function \cosh^{-1}x.

    Currently I have x in terms of y as x = \frac{2}{e^y + e^{-y}} I am unsure if this is as far as it can go. This part: \cosh^{-1}x = \ln (x + \sqrt{x^2-1}) is where I am more confused.

    Thankyou for any help. It is greatly appreciated
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  2. #2
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    re: Hyperbolic functions

    Hello there,

    I did this pretty quickly but I think this will hopefully help you,

    Given
    y=cosh(x) where x\geq0 [1]
    then
    y=(e^{x}-e^{-x})/2
    so
    e^{x}-e^{-x}=2y
    then multiplying all by exp(x) gives
    e^{x}=\frac{2y\pm\sqrt{4y^2-4}}{2} (quadratic formula)
    thus
    x = ln(y\pm\sqrt{y^2-1})
    and since y = cosh(x) = cosh( ln( y .... ) ) then
    cosh^{-1}(y) = ln(y\pm\sqrt{y^2 -1} )

    Now, ln(a) is a real valued function where a must be positive and real, so y +/- sqrt(y^2 - 1) must be positive and real; thus y^2 - 1 >= 0 => y >= 1 (which is true since by [1] that x >= 0 and so cosh(x) >= 1), now for y +/- sqrt(y^2 - 1) to be positive then we must take y + sqrt(y^2 - 1)

    So
    cosh^{-1}(y) = ln(y+\sqrt{y^2 -1} )
    which (for letting y be x) is what was wanted.

    The domain for arccosh(y) would then be such that y + sqrt(y^2 - 1) >= ie y > 1 (since ln(0) is also undefined) and the range would then be all positive real numbers (from 0 to +infinity) which is the range of ln for domain greater than 1.





    Last edited by rain; November 4th 2009 at 09:45 AM. Reason: +/- where + should have been
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  3. #3
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    Quote Originally Posted by rain View Post
    Hello there,

    I did this pretty quickly but I think this will hopefully help you,
    Unfortunately I am not brilliant at maths, hence why I come here for help. I prefer chemistry
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  4. #4
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    Quote Originally Posted by Beard View Post
    Unfortunately I am not brilliant at maths, hence why I come here for help. I prefer chemistry
    I'd love to pick up on chemistry at some point - you're very lucky
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  5. #5
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    Quote Originally Posted by rain View Post
    Hello there,

    then multiplying all by exp(x) gives
    e^{x}=\frac{2y\pm\sqrt{4y^2-4}}{2} (quadratic formula)
    thus
    x = ln(y\pm\sqrt{y^2-1})
    and since y = cosh(x) = cosh( ln( y .... ) ) then
    cosh^{-1}(y) = ln(y\pm\sqrt{y^2 -1} )

    Now, ln(a) is a real valued function where a must be positive and real, so y +/- sqrt(y^2 - 1) must be positive and real; thus y^2 - 1 >= 0 => y >= 1 (which is true since by [1] that x >= 0 and so cosh(x) >= 1), now for y +/- sqrt(y^2 - 1) to be positive then we must take y + sqrt(y^2 - 1)

    So
    cosh^{-1}(y) = ln(y\pm\sqrt{y^2 -1} )
    which (for letting y be x) is what was wanted.

    The domain for arccosh(y) would then be such that y + sqrt(y^2 - 1) >= ie y > 1 (since ln(0) is also undefined) and the range would then be all positive real numbers (from 0 to +infinity) which is the range of ln for domain greater than 1.
    Sorry but I don't understand how the quadratic was formed when you multiply by exp(x) as you say
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  6. #6
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    My fault, I should have made this clearer,

    Given that you can get down to this equation,
    e^{x} -e^{-x} = 2y
    Then multiplying all the terms (both on the left hand side and the right hand side) by e^{x} gives
    e^{x}(e^{x} -e^{-x}) = e^{x}2y
    Then, by expanding the bracket out, you get
    e^{x}e^{x} -e^{-x}e^{x} = e^{x}2y
    Now, by standard rules of indices, a^ba^c=a^{b+c} and a^0 = 1 we get e^{-x}e^{x} = e^{-x+x} = e^{0} = 1
    Then e^{x}e^{x} -1 = e^{x}2y which can be written as
    (e^{x})^2 -e^{x}2y -1 = 0 [1] by collecting the e^{x} terms and bringing the 2y term to the left hand side
    Now, say you let W be e^{x}, just to show this is like a quadratic, you can rewrite [1] as W^2 -W2y -1 = 0 which is in the general formula for solving a quadratic equation, with roots at W=\frac{(2y)\pm\sqrt{(2y)^2-4(-1)}}{2}<br />
=\frac{2y\pm\sqrt{4y^2+4}}{2}<br />
=\frac{2y\pm\sqrt{4(y^2+1)}}{2}

    =\frac{2y\pm2\sqrt{y^2+1}}{2} (since \sqrt{4} = 2)

    =y\pm\sqrt{y^2+1} (by dividing by 2)

    thus W=e^{x}=y\pm\sqrt{y^2+1}

    and taking the log of both sides gives

    x = log(y\pm\sqrt{y^2+1}) and the rest carries on above

    *The formula for solving a quadratic equation in general is as follows in case you can't remember,
    for ax^2 +bx +c = 0
    then x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}
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  7. #7
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    Quote Originally Posted by rain View Post
    My fault, I should have made this clearer,

    Given that you can get down to this equation,
    e^{x} -e^{-x} = 2y
    Then multiplying all the terms (both on the left hand side and the right hand side) by e^{x} gives
    e^{x}(e^{x} -e^{-x}) = e^{x}2y
    Then, by expanding the bracket out, you get
    e^{x}e^{x} -e^{-x}e^{x} = e^{x}2y
    Now, by standard rules of indices, a^ba^c=a^{b+c} and a^0 = 1 we get e^{-x}e^{x} = e^{-x+x} = e^{0} = 1
    Then e^{x}e^{x} -1 = e^{x}2y which can be written as
    (e^{x})^2 -e^{x}2y -1 = 0 [1] by collecting the e^{x} terms and bringing the 2y term to the left hand side
    Now, say you let W be e^{x}, just to show this is like a quadratic, you can rewrite [1] as W^2 -W2y -1 = 0 which is in the general formula for solving a quadratic equation, with roots at W=\frac{(2y)\pm\sqrt{(2y)^2-4(-1)}}{2}<br />
=\frac{2y\pm\sqrt{4y^2+4}}{2}<br />
=\frac{2y\pm\sqrt{4(y^2+1)}}{2}

    =\frac{2y\pm2\sqrt{y^2+1}}{2} (since \sqrt{4} = 2)

    =y\pm\sqrt{y^2+1} (by dividing by 2)

    thus W=e^{x}=y\pm\sqrt{y^2+1}

    and taking the log of both sides gives

    x = log(y\pm\sqrt{y^2+1}) and the rest carries on above

    *The formula for solving a quadratic equation in general is as follows in case you can't remember,
    for ax^2 +bx +c = 0
    then x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}

    That made it a load clearer thanks again
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