# approximating a maximum function by a differentiable function

Is it possible to approximate the $max\{x,y\}$ by a differentiable function? $f(x,y)=max \{x,y\} ;\ x,y>0$

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$\max'_\epsilon :=\mathord\max * \phi_\epsilon$. • (+1), good advice. One smoothing I stumbled across recently is$|x| \approx x\operatorname{erf}(ax)$with$a \gg 1$. Taking$a=5$works well enough. – Antonio Vargas Oct 8 '13 at 22:37 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$N$. • How could deal with overflow and underflow in case of large numbers multiplied by large scaling parameter N? – Feras Oct 22 '17 at 17:49 • I don't know. That's why I said it is unworkable for large N. – Angelorf Oct 23 '17 at 22:34 • I'm testing another function$max(x,y) = pow((pow(x,n) + pow(y,n), 1/n)$it is doesn't have overflow but in one case when x or y is zero so I'm adding this small number 1e-5 to avoid it. Do you have any idea ? – Feras Oct 24 '17 at 16:46 • That doesn't work for negative numbers. Why avoid zero?$0^x = 0$(except$0^0$, but$n \neq 0\$ anyway) – Angelorf Oct 25 '17 at 19:42