approximating a maximum function by a differentiable function Is it possible to approximate the $max\{x,y\}$ by a differentiable function?
$f(x,y)=max \{x,y\} ;\ x,y>0$
 A: You can use an approximation of max norm a.k.a. infinity norm with a p-norm of relatively high $p$.
Caveat #1: make sure $p$ isn't too high though because that can cause numerical instabilities, overflow to infinity, esp. if you use 32-bit floats
Caveat #2: norm works with the absolute values so you'll need to transform your numbers with e.g. log(exp(x)+1) (a.k.a. softplus) if you have to deal with negative values as well (and then back). Softplus quickly approximates $x$ for $x>20ish$ btw so this can be faster than it seems.
A: Yes it is. One possibility is the following: Note that $\def\abs#1{\left|#1\right|}$
$$ \max\{x,y\} = \frac 12 \bigl( x+ y + \abs{x-y}\bigr), $$
take a differentiable approximation of $\abs\cdot$, for example $\def\abe{\mathop{\rm abs}\nolimits_\epsilon}$$\abe \colon \mathbb R \to \mathbb R$ for $\epsilon > 0$ given by
$$ \abe(x) := \sqrt{x^2 + \epsilon}, \quad x \in \mathbb R $$
and define $\max_\epsilon \colon \mathbb R^2 \to \mathbb R$ by 
$$ \max\nolimits_\epsilon(x,y) := \frac 12 \bigl( x+y+\abe(x-y)\bigr). $$
Another possibility is to take a smooth mollifier $\phi_\epsilon$ and let $\max'_\epsilon :=\mathord\max * \phi_\epsilon$.
A: Another possibility is given by:
\begin{equation}
\max (x,y) \approx \ln(e^x+e^y)
\end{equation}
This approximation doesn't work well for similar values of $x$ and $y$. We can however, remedy this by introducing a scaling parameter $N$:
\begin{equation}
\max (x,y) \approx \frac1N\ln(e^{N x}+e^{N y})
\end{equation}
for large values of $N$.
A general definition is given by:
\begin{equation}
\max\limits_{x \in S} \approx \frac1N\ln(\sum\limits_{x\in S} e^{N x})
\end{equation}
Note that in practice $e^{Nx}$ will give unworkably large answers for large $N$.
