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It seems ELU (Exponential Linear Units) is used as an activation function for deep learning. But its' graph is very similar to the graph of $log(1+e^x)$. So why has $log(1+e^x)$ not been used as the activation functions instead of ELU?

In other words what is the advantage of ELU over $log(1+e^x)$?

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ReLU and all its variants ( except ReLU-6 ) are linear i.e $ y = x $ for values greater than or equal to 0.

This gives ReLU and specifically ELU an advantage like:

  • Linearity means that the slope does not plateau or saturate when $x$ becomes larger. Hence, the vanishing gradient problem is solved.

enter image description here

Now, the graph $ y = log( 1 + e^x ) $ isn't linear for values > 0.

For larger negative values, the graph produces values which are very close to zero. This is also found in sigmoid where larger values produce a fully saturated activation. Hence, $ y = log( 1 + e^x ) $ can raise problems which sigmoid and tanh suffer.

About ELU:

enter image description here

ELU has a log curve for all negative values which is $ y = \alpha( e^x - 1 )$. It does not produce a saturated firing for some extent but saturates for larger negative values.

See here for more information.

Hence, $ y = log( 1 + e^x ) $ is not used because of early saturation for negative values and also non linearity for values > 0. This may produce problems and even bring down some features which ReLU and variants exhibit.

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  • $\begingroup$ Thanks for your comments. But still I am not convinced. log(1+exp(x)), although it is non linear, it does not saturate for x>0. Do you say log(1+exp(x)) has early saturation for x<0 compared to ELU? $\endgroup$ – user570593 Jun 10 at 3:38
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    $\begingroup$ For all values > 0, log(1+exp(x)) is a non-linear function whereas ReLU is linear ( y = x ). For large negative values, the value of log(1+exp(x)) becomes closer to zero, which indeed is a problem suffered by sigmoid too. $\endgroup$ – Shubham Panchal Jun 10 at 3:41

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