Results 1 to 3 of 3

Math Help - LambertW

  1. #1
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1

    LambertW

    Hello all:

    I seen a problem on another site where someone was wanting to know the inverse.

    y=4+x+5e^{x-3}

    As Pka said, the derivative is always positive, so it has an inverse, but can not be found by elementary means.

    I ran it through Maple and it kicked back:

    x=y-4-LambertW(\frac{5e^{y}}{e^{7}})

    What is Lambert?. I am unfamiliar. Does anyone know how this would be done without technology?. I thought it would something interesting to look into. I did find an interesting page on Wiki
    Last edited by galactus; January 31st 2007 at 06:06 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by galactus View Post
    Hello all:

    I seen a problem on another site where someone was wanting to know the inverse.

    y=4+x+5e^{x-3}

    As Plato said, the derivative is always positive, so it has an inverse, but can not be found by elementary means.

    I ran it through Maple and it kicked back:

    x=y-4-LambertW(\frac{5e^{y}}{e^{7}})

    What is Lambert?. I am unfamiliar. Does anyone know how this would be done without technology?. I thought it would something interesting to look into. I did find an interesting page on Wiki
    The wikipedia pge is a good overview, and computational schemes.

    For real arguments > -1/e Newton-Raphson is quite a good method og evaluating it.

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    In my engineering class we were studing a hanging cable. The problem is that the equations that describe it are hyperbolic functions and hence cannot be solved through "normally". However, I have been able to reduce the problem to solving,
    ax+b=e^x for a\not = 0.
    In that case the linear-exponential equations can always be solved (I imagine). The full solution is here. http://www.mathhelpforum.com/math-he...e-problem.html
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum