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

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September 2nd 2010, 07:27 AM
problady

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
September 2nd 2010, 08:59 AM
CaptainBlack

Quote:

Originally Posted by problady

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
September 2nd 2010, 09:18 AM
problady

y=1/1+exp^-alpha*x --> alpha = gain factor
September 2nd 2010, 01:12 PM
CaptainBlack

Quote:

Originally Posted by problady

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
September 3rd 2010, 12:44 AM
problady

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 :(
September 3rd 2010, 01:45 AM
CaptainBlack

Quote:

Originally Posted by problady

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