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What does D = max[0; M(x)] mean?

M(x) is a function.

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Taking the maximal number amongst the parameters.

$\max\{x_1,x_2\} = \cases{x_1, \text{if }x_1 > x_2\\x_2, \text{otherwise}}$

You can define like that the maximum of any finitely many elements.

When the parameters are an infinite set of values, then it is implied that one of them is maximal (namely that there is a greatest one, unlike the set $\{-\frac{1}{n} | n\in\mathbb{N}\}$ where there is no greatest element)

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In this case, I can also think of $D$ as a function in $x$:

$D(x) = \max(0, M(x)) = \begin{cases} M(x) &\text{if}\ M(x) >0, \\ 0 &\text{otherwise}.\end{cases}$

The result is essentially $M$ "cut off" at $0$.

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  • $\begingroup$ Note that the unit step function $H(x)$ can do the job as well: $D(x)=M(x)H(M(x))$ $\endgroup$ – J. M. isn't a mathematician Dec 6 '10 at 13:42
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    $\begingroup$ Also $D(x) = (|M(x)|+M(x))/2$. $\endgroup$ – Yuval Filmus Dec 7 '10 at 0:06
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    $\begingroup$ And infinitely many other representations -- we get it, guys ;) $\endgroup$ – Raphael Dec 7 '10 at 8:52
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Its the short form of representing, a function that returns maximum among a set of given numbers.
Assume $ M(x) = \frac1x$, and $D(x) = \max \{0, M(x)\} \implies D(x) = \max \left \{0, \frac1x\right\}$

$$ D(x) = \begin{cases}\frac{1}{x} & \text{when }x > 0 \\0 & \text{if }x < 0\end{cases}$$

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