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Compute the error using the sigmoid function in the backflow

I have a quick question regarding a return sale. I consider the following:

http://www4.rgu.ac.uk/files/chapter3%20-%20bp.pdf

This article talks about calculating neuron error as

Error = Exit (i) * (1 - Exit (i)) * (Target (i) - Exit (i))

I put part of the equation that I do not understand in bold. The document says that because of the sigmoid function, output (i) * is required (1 - output (i)) , but I still do not understand why this would be inconvenient.

What would be wrong when using

Error = abs(Output(i) - Target(i))

?

Is the error function independent of the activation / transfer function of neurons?

+5
source
2

, , , .

, ( )

.

:

E = -(target - output)^2

dE/dinput = dE/doutput * doutput/dinput

doutput/dinput:

output = sigmoid (input)

doutput/dinput = output * (1 - output)    (derivative of sigmoid function)

:

dE/dinput = 2 * (target - output) * output * (1 - output)
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