Perceptrons and Decision Boundaries I am currently studying neural networks and have been trying to reason about this for a while to no avail. 

I understand that given a perceptron(such as above) with f as a step function, any decision boundaries learnt must be a hyperplane. 
However, I'm having some difficulty seeing why this should hold true when f is something else like a sigmoid or logistic function such as below:

Can someone show me how to mathematically reach this conclusion?
 A: When you make a decision based on a smooth function $f$--say, the logistic function--your rule is typically the following:

If $f(z) > p$, classify "positive". Else, classify "negative".

Where $z = \sum_{i=1}^7 w_i x_i$ is the linear combination learned by the perceptron. Often $p$ is taken to be fixed at 0.5, so let's assume this for simplicity, although it works with any $p \in [0, 1]$.
So, fix $p=0.5$. Your decision boundary is all the points $\{\vec x \mid f(\sum_i w_ix_i) = 0.5\}$. If $f$ is the sigmoid function, then it's equal to 0.5 only when its argument is exactly zero. So your decision boundary is $\{\vec x \mid \sum_i w_ix_i = 0\}$, that is, a hyperplane perpendicular to the vector $\vec w$.

Another way of looking at this is that, to make a decision based on $f$, you have to layer a decision function on top of it. The decision function needs to take in the output of $f$, which is a number in $[0,1]$, and output a decision, which is an integer 0 or 1. The decision rule I described above corresponds to the decision function 

$g(y) = \left\{\begin{array}{cc} 0 & \text{if $y<0.5$} \\ 1 & \text{otherwise} \end{array}\right.$

From a decision-making perspective, a perceptron with function $f$ and decision rule $g$ is the same as a perceptron with decision rule $g \circ f$. But $g \circ f$ is just a step function! So it's the same as a step-function perceptron, and thus its decision boundary is a hyperplane.
