Mumbai University > Computer Engineering > Sem 7 > Soft Computing

Marks: 5 Marks

Year: May 2016

Delta Training rules for bipolar continuous activation function:

The activation function in the case of bipolar continuous activation function is given by

$$f(net)=\dfrac{2}{1+exp(-net)}-1$$

We obtain

$$f' (net)=\dfrac{2exp(-net)}{[1+exp(-net)]^2}$$

An useful identify can be applied here

$$\dfrac{2exp (-net)}{[1+exp(-net)]^2} =\dfrac12 (1-0^2 )$$

Verification of identity

Letting o=f(net)

$$\dfrac12 (1-0^2 )=\dfrac12 \bigg[1-\bigg(\dfrac{1-exp(-net)}{1+exp(-net)^2}\bigg) \bigg] \\ \dfrac12 \bigg[1-\bigg(\dfrac{1-exp(-net)}{1+exp(-net)^2}\bigg) \bigg]=\dfrac{2exp(-net)}{[1+exp(-net)]^2}$$

LHS=RHS

The delta value for a bipolar continuous activation function is given by

$δ_ok=\dfrac12 (d_k-o_k )(1-o_k^2)$

Which uses the following identity for f’(net)

$$f' (net)=\dfrac12 (1-o^2 )$$

Hence, for bipolar continuous activation function $f’ (net) = ½(1-o^2)$