I am reading "space filling curve" by Hans Sagan. On page 17, on equation (2.3.11) in the equation, a function sgn is used.
What is sgn?
To put it into context, in the book it says $h_n=$sgn$(n)[(n-1)+i]$
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$\begingroup$ It is a slightly abbreviated form of "sign" as in positive or negative. en.wikipedia.org/wiki/Sign_function $\endgroup$– CarserJun 8, 2016 at 17:23
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7$\begingroup$ The reason for abbreviation is a little clearer when calling it the signum function. $\endgroup$– Jacob BondJun 8, 2016 at 18:40
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2 Answers
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Typically, $\operatorname{sgn}$ stands for the sign function $$\operatorname{sgn}(x) = \begin{cases} 1 & x > 0 \\ 0 & x = 0 \\ -1 & x < 0 \end{cases}.$$
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5$\begingroup$ NB the definitions of $\mathop{\rm sgn}(0)$ vary between $0$ and $1$. If $0$ is a plausible argument, you should assure yourself of the definition of $\mathop{\rm sgn}$ at $0$ in your context. $\endgroup$– AlexRJun 8, 2016 at 19:54
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8$\begingroup$ @AlexR Actually, they vary between three options: (a) $\operatorname{sgn}0=0$; (b) $\operatorname{sgn}0=1$; (c) $\operatorname{sgn}0$ is undefined, i.e., $\operatorname{Dom}\operatorname{sgn} = \mathbb{R}\setminus\{0\}$. I'm in favor of (a) and I consder (b) a deadly sin. $\endgroup$– yo'Jun 8, 2016 at 21:43
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2$\begingroup$ When $\text{sgn}(x)$ is used for applications (e.g., robust statistics), it is most commonly asserted that $\text{sgn}(0) = 0$ (as far as I know). $\endgroup$ Jun 8, 2016 at 22:10
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$\begingroup$ @yo, I guess it depends on whether or not the relationship between the sign function and the unit step function needs to be maintained, among other things. $\endgroup$ Jun 9, 2016 at 1:12
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$\begingroup$ @J.M. There is of course also a simple relationship with the unit step function $H(x)$ ($H$ stands for Heaviside) to the symmetric $\operatorname{sgn}$ function: $\operatorname{sgn}x=H(x)-H(-x)$. The relation you are alluding to with the asymmetric sign function is $\operatorname{sgn}^+x=2H(x)-1$. $\endgroup$ Jun 9, 2016 at 5:10
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I will expand on Dominik's answer.
$ \operatorname{sgn} $, standing for "sign", is usually defined as,
$$\operatorname{sgn}(x) = \begin{cases} 1 & x > 0 \\ 0 & x = 0 \\ -1 & x < 0 \end{cases}.$$
but not always. Some use
$$\operatorname{sgn}(x) = \begin{cases} 1 & x > 0 \\ 0 & x < 0 \end{cases}.$$
(that is, $ \operatorname{sgn}(0) $ is undefined).
In rare cases, one might define $ \operatorname{sgn}(0) $ as one or zero, for convinience. It really depends on the situation and usecase.