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Thread: Is it possible to simplify this equation?

  1. #1
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    Question Is it possible to simplify this equation?

    $$z = \frac{y}{1+C \exp(-Wx)}$$

    Is it possible to simplify this equation by a sigmoid function or something similar?
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  2. #2
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    Re: Is it possible to simplify this equation?

    If your equation is same as this one (variable names changed for simplicity!):
    a = b / [1 + c^(-de)]
    then above simplifies to:
    c^(de) = a / (b - a)
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  3. #3
    Forum Admin topsquark's Avatar
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    Re: Is it possible to simplify this equation?

    Quote Originally Posted by brianx View Post
    $$z = \frac{y}{1+C \exp(-Wx)}$$

    Is it possible to simplify this equation by a sigmoid function or something similar?
    You could do this: Let $\displaystyle C = e^{-d}$ then
    $\displaystyle z = \frac{y}{1 + C e^{-Wx}} = y \cdot \frac{1}{1 + e^{-(Wx + d)}}$

    Then let A = Wx + d which gives
    $\displaystyle z = y \cdot \frac{1}{1 + e^{-A}}$, which is your sigmoid function.

    This looks nice but there is a catch: Setting $\displaystyle C = e^{-d}$ restricts the value of C to be positive, which would not be true in general.

    -Dan
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  4. #4
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    Re: Is it possible to simplify this equation?

    I am just wondering what is meant by "simplify". You could convert the equation to spherical or cylindrical coordinates. Would that be "simpler"? You can solve for $y$ as a function of $C,x,W,z$. That would eliminate the fraction. You could solve for $x$ in terms of $C,W,y,z$. That would eliminate the exponential (but introduce a logarithm). Is that "simpler"?
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  5. #5
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    Unhappy Re: Is it possible to simplify this equation?

    ...and me who thought that exp = exponent !
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  6. #6
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    Re: Is it possible to simplify this equation?

    Quote Originally Posted by DenisB View Post
    ...and me who thought that exp = exponent !
    Same here.
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    Re: Is it possible to simplify this equation?

    Thanks Mr.Fly: great minds think alike!!
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    Re: Is it possible to simplify this equation?

    Quote Originally Posted by brianx View Post
    $$z = \frac{y}{1+C \exp(-Wx)}$$
    Is it possible to simplify this equation?
    Take your pick.
    $ \begin{align*} z& = \dfrac{y}{{1 + C{e^{ - wx}}}} \\ z &= \dfrac{{{e^{wx}}y}}{{{e^{wx}} + C}} \\z{e^{wx}} + zC& = {e^{wx}}y \\ z{e^{wx}} - y{e^{wx}}& = - Cz \\ {e^{wx}}& = \frac{{Cz}}{{y - z}} \\ x& = \dfrac{1}{w}\left( {\log (C) + \log (z) - \log (y - z)} \right) \end{align*}$
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