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Math Help - show that derivative of sigmoid function can be defined recursively in terms o itself

  1. #1
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    Unhappy show that derivative of sigmoid function can be defined recursively in terms o itself

    I will be very very happy if someone has knowledge to explain this in a right way... and understandable for me! I know to solve simple derivatives but this is something that I really dont have a clue! Please explain what this sentence "recursively in terms o itself" means as well
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by problady View Post
    I will be very very happy if someone has knowledge to explain this in a right way... and understandable for me! I know to solve simple derivatives but this is something that I really dont have a clue! Please explain what this sentence "recursively in terms o itself" means as well
    What sigmoid function are we discussing?

    CB
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    y=1/1+exp^-alpha*x --> alpha = gain factor
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    Quote Originally Posted by problady View Post
    y=1/1+exp^-alpha*x --> alpha = gain factor
    y=\dfrac{1}{1+e^{-\alpha x}}

    \dfrac{dy}{dx}=\dfrac{-\alpha e^{-\alpha x}}{(1+e^{-\alpha x})^2}=-\alpha e^{-\alpha x}y^2=-\alpha \dfrac{1-y}{y}y^2=-\alpha (1-y)y

    which is a variant of the logistic equation.

    CB
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  5. #5
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    Dear CaptainBlack can you explain this example a bit more! I would like to be able to generalize this example in other ones... in this case I cannot do that, since I dont understand what you have done and why
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    Quote Originally Posted by problady View Post
    Dear CaptainBlack can you explain this example a bit more! I would like to be able to generalize this example in other ones... in this case I cannot do that, since I dont understand what you have done and why
    You differentiate with respect to x and then use the definition of the function to remove all reference to x by replacing terms that contain x by equivalent ones in terms of y only.

    here we use:

     \dfrac{1}{1+e^{-\alpha x}}=y

    and

     e^{-\alpha x}=\dfrac{1-y}{y}




    CB
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