Recognition of facial expressions - Knowledge Based Systems ...

1
If f ( net
j
) is the logistic activation function, then f ( net
j
) = and:
−net
1
j
+ e
− x −1
∂a j
d(1
+ e )
= f ′(
net
j
) =
∂net
dnet
By solving the previous equation, the result is:
d
j
+ e
dnet
− x −1
( 1 )
−x
−2
j
j
−x
e 1
=
−x
−x
1+
e 1+
e
−x
1+
e −1
1
=
− x
1+
e 1+
e
−x
1+
e 1
= ( −
−x
1+
e 1+
e
= (1 − a ) a
j
= ( −1)(1
+ e
−x
−x
1
)
1+
e
− x
)
j
e
− x
( −1)
That means that:
∂a
j
∂net
j
= ( 1−
a ) a
j
j
The first derivative of the relation is
∂E
∂
a j
and
E
p
=
1
2
i
( t − a )
i
i
2
The sum is over the output units of the network. There are two cases to be considered
for the partial derivative:
- j is an output unit,
- j is not an output unit.
If j is an output unit, the derivative can be computed simply as:
In the relation, for the case that
∂E
∂a
=
j
i
= ( t
i
( t
= −1(
t
∂
=
∂a
∂(
ti
− ai
)
− ai
)
∂a
− a )( −1)
i
i
i
1 ( t )
2
i
− ai
j
2 i
− a )
i
a
j
is not an output unit, the relation is:
j
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