Mixing
Is sigmoid function distributive
Sigmoid function is defined as $frac{1}{1+e^{-(x+y)}}$.
Is there a property for sigmoid such that
$frac{1}{1+e^{-(x+y)}}=$ sigmoid$(x)$ {some operation} sigmoid$(y)$ ?
Edit:
According to this link it is not. However, is there any way that I can approximate them such that they are almost equal to the original one.
Edit 2:
If I use taylor series, how can I make it distributive?
functions
asked Nov 13 '18 at 16:32
pufflespuffles
669
add a comment |
Sigmoid function is defined as $frac{1}{1+e^{-(x+y)}}$.
Is there a property for sigmoid such that
$frac{1}{1+e^{-(x+y)}}=$ sigmoid$(x)$ {some operation} sigmoid$(y)$ ?
Edit:
According to this link it is not. However, is there any way that I can approximate them such that they are almost equal to the original one.
Edit 2:
If I use taylor series, how can I make it distributive?
functions
asked Nov 13 '18 at 16:32
pufflespuffles
669
add a comment |
Sigmoid function is defined as $frac{1}{1+e^{-(x+y)}}$.
Is there a property for sigmoid such that
$frac{1}{1+e^{-(x+y)}}=$ sigmoid$(x)$ {some operation} sigmoid$(y)$ ?
Edit:
According to this link it is not. However, is there any way that I can approximate them such that they are almost equal to the original one.
Edit 2:
If I use taylor series, how can I make it distributive?
functions
asked Nov 13 '18 at 16:32
pufflespuffles
669
Sigmoid function is defined as $frac{1}{1+e^{-(x+y)}}$.
Is there a property for sigmoid such that
$frac{1}{1+e^{-(x+y)}}=$ sigmoid$(x)$ {some operation} sigmoid$(y)$ ?
Edit:
According to this link it is not. However, is there any way that I can approximate them such that they are almost equal to the original one.
Edit 2:
If I use taylor series, how can I make it distributive?
functions
functions
asked Nov 13 '18 at 16:32
pufflespuffles
669
asked Nov 13 '18 at 16:32
pufflespuffles
669
edited Nov 23 '18 at 14:45
puffles
asked Nov 13 '18 at 16:32
pufflespuffles
669
asked Nov 13 '18 at 16:32
pufflespuffles
669
asked Nov 13 '18 at 16:32
pufflespuffles
669
669
add a comment |
add a comment |
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