I was reading the paper ImageNet Classification with Deep Convolutional Neural Networks and in section 3 were they explain the architecture of their Convolutional Neural Network they explain how they preferred using:

non-saturating nonlinearity $f(x) = max(0, x). $

because it was faster to train. In that paper they seem to refer to saturating nonlinearities as the more traditional functions used in CNNs, the sigmoid and the hyperbolic tangent functions (i.e. $f(x) = tanh(x)$ and $f(x) = \frac{1}{1 + e^{-x}} = (1 + e^{-x})^{-1}$ as saturating).

Why do they refer to these functions as "saturating" or "non-saturating"? In what sense are these function "saturating" or "non-saturating"? What do those terms mean in the context of convolutional neural networks? Are they used in other areas of machine learning (and statistics)?



A saturating activation function squeezes the input.


  • $f$ is non-saturating iff $ (|\lim_{z\to-\infty} f(z)| = +\infty) \vee (|\lim_{z\to+\infty} f(z)| = +\infty) $
  • $f$ is saturating iff $f$ is not non-saturating.

These definitions are not specific to convolutional neural networks.


The Rectified Linear Unit (ReLU) activation function, which is defined as $f(x)=max(0,x)$ is non-saturating because $\lim_{z\to+\infty} f(z) = +\infty$:

enter image description here

The sigmoid activation function, which is defined as $f(x) = \frac{1}{1 + e^{-x}}$ is saturating, because it squashes real numbers to range between $[0,1]$:

enter image description here

The tanh (hyperbolic tangent) activation function is saturating as it squashes real numbers to range between $[-1,1]$:

enter image description here

(figures are from CS231n, MIT License)

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    $\begingroup$ ah, nice makes sense! I know this wasn't my original question, but what is that property important in the context of ML and CNNs? $\endgroup$ – Charlie Parker Sep 28 '15 at 15:43
  • $\begingroup$ For ANNs, to avoid having one unit with a large output that impacts too much the ANN's output layer. $\endgroup$ – Franck Dernoncourt Sep 28 '15 at 18:03
  • $\begingroup$ whats the difference between tan and sigmoid? both of them squash the numbers in a closed range! I dont get it, Can you elaborate this abit more? I'm kind of bad in mathematics . (by the way I'm coming from a CNN perspective ) $\endgroup$ – Rika Feb 17 '16 at 11:09
  • $\begingroup$ @FranckDernoncourt Did you mean saturating for tanh activation function? I guess there is a typo? :) $\endgroup$ – CoderSpinoza Mar 24 '16 at 4:44
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    $\begingroup$ @tenCupMaximum: To saturate means to fill up to a point where no more can be added. In the context of a saturating function, it means that after a certain point, any further increase in the function's input will no longer cause a (meaningful) increase in its output, which has (very nearly) reached its maximum value. The function at that point is "all filled up", so to speak (or saturated). $\endgroup$ – Ruben van Bergen Oct 10 '17 at 11:16

The most common activation functions are LOG and TanH. These functions have a compact range, meaning that they compress the neural response into a bounded subset of the real numbers. The LOG compresses inputs to outputs between 0 and 1, the TAN H between -1 and 1. These functions display limiting behavior at the boundaries.

At the border the gradient of the output with respect to the input ∂yj/∂xj is very small. So Gradient is small hence small steps to convergence hence longer time to converge.


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