Introduction
The sign function assigns to each real number its sign, disregarding its magnitude. The function is defined as follows:
\[sgn ( x ) = \{ - 1 & \text{if} x < 0 \\ 0 & \text{if} x = 0 \\ 1 & \text{if} x > 0 \forall x \in \mathbb{R}\]Specifically, $sgn ( x )$ returns $- 1$ for negative values, $0$ when $x = 0$, and $1$ for positive values. The function does not quantify the magnitude of $x$ but solely indicates the position of $x$ relative to zero. For example, applying the definition we have:
\[sgn ( - 7 ) = - 1 sgn ( 0 ) = 0 sgn ( 4 ) = 1\]The graph of $y = sgn ( x )$ consists of two horizontal rays and one isolated point. The ray on $y = - 1$ extends over all $x < 0$, the ray on $y = 1$ extends over all $x > 0$, and the isolated point at the origin $( 0 , 0 )$ lies on $y = 0$. The two rays approach but do not intersect the y-axis.
The sign function is classified as an odd function because it satisfies the identity:
\[sgn ( - x ) = - sgn ( x ) \forall x \in \mathbb{R}\]Properties
- Domain: $\mathbb{R}$.
- Range: $- 1 , 0 , 1$.
- The function is odd, since $sgn ( - x ) = - sgn ( x )$.
- The function has exactly one root at $x = 0$, since $sgn ( x ) = 0$ only when $x = 0$.
- The function is constant on $( - \infty , 0 )$ and on $( 0 , + \infty )$, hence increasing in the sense that it is non-decreasing over $\mathbb{R}$.
- The function has a jump discontinuity at $x = 0$; it is continuous everywhere else.
- The function is not differentiable at $x = 0$. It is differentiable, with zero derivative, at every other point.
- Limits approaching $x = 0$ from either side: \(\underset{x \rightarrow 0^{-}}{lim} sgn ( x ) & = - 1 \\ \underset{x \rightarrow 0^{+}}{lim} sgn ( x ) & = 1\)
Since the two one-sided limits differ, the two-sided the following limit does not exist: \(\underset{x \rightarrow 0}{lim} sgn ( x )\)
Limits at infinity
As $x$ moves away from the origin in either direction, the sign function stabilizes at a constant value:
\[\underset{x \rightarrow - \infty}{lim} sgn ( x ) & = - 1 \\ \underset{x \rightarrow + \infty}{lim} sgn ( x ) & = 1\]These are simply a consequence of the fact that $sgn ( x ) = - 1$ for all $x < 0$ and $sgn ( x ) = 1$ for all $x > 0$, so the function value does not change as $x$ moves further from zero.
Derivative and integral
On each of the two open half-lines where the sign function is constant, its derivative is zero:
\[\frac{d}{d x} sgn ( x ) = 0 \text{for} x \neq 0\]At $x = 0$, the derivative does not exist because the function is discontinuous there. In the sense of distributions, however, the derivative of the sign function is:
\[\frac{d}{d x} sgn ( x ) = 2 \delta ( x )\]$\delta ( x )$ is the Dirac delta, a generalised function that is zero everywhere except at the origin and integrates to one over the entire real line. This distributional identity demonstrates that the sign function exhibits a discontinuity of amplitude $2$ at the origin, as:
\(\underset{x \rightarrow 0^{-}}{lim} sgn ( x ) = - 1\) \(\underset{x \rightarrow 0^{+}}{lim} sgn ( x ) = 1\)
This amplitude explains the presence of the factor $2$ preceding the Dirac delta function.
The indefinite integral of the sign function, computed away from the origin, gives back the absolute value:
\[\int sgn ( x ) d x = | x | + c\]| This is consistent with the fact that the derivative of $ | x | $ equals $sgn ( x )$ wherever the former is differentiable. |
Relationship with the absolute value function
A direct algebraic relationship exists between the sign function and the absolute value. For any $x \neq 0$, the following identity holds:
\[sgn ( x ) = \frac{x}{| x |}\]| This result is consistent with the definition: when $x > 0$, the ratio $x / | x | = x / x = 1$. When $x < 0$, the ratio $x / | x | = x / ( - x ) = - 1$. The formula is undefined at $x = 0$, so the value $sgn ( 0 ) = 0$ is assigned separately by convention. |
Conversely, the absolute value can be expressed in terms of the sign function using the following identity:
\[| x | = x \cdot sgn ( x )\]| This identity holds for all $x \in \mathbb{R}$, including $x = 0$, where both sides are zero. Together, these two identities demonstrate that $ | x | $ and $sgn ( x )$ are complementary: the absolute value preserves magnitude and omits sign, whereas the sign function preserves sign and omits magnitude. |
Two additional identities arise from the relationship between the sign function and the absolute value. The first identity is as follows:
\[x = | x | \cdot sgn ( x )\]| This identity expresses any real number as the product of its magnitude and its sign. The second identity utilises the equality $ | x | = \sqrt{x^{2}}$, which is valid for all $x \in \mathbb{R}$, and provides an alternative representation of the sign function for $x \neq 0$: |
Relationship with the Heaviside step function
The Heaviside step function $H ( x )$ is defined as:
\[H ( x ) = \{ 0 & \text{if} x < 0 \\ \frac{1}{2} & \text{if} x = 0 \\ 1 & \text{if} x > 0\]
The sign function and the Heaviside step function are related by a simple linear transformation. Specifically:
\[sgn ( x ) = 2 H ( x ) - 1\]Equivalently, we have:
\[H ( x ) = \frac{1 + sgn ( x )}{2}\]This relationship is frequently utilised as converting between these representations can simplify calculations. The Heaviside function maps $( - \infty , 0 )$ to $0$ and $( 0 , + \infty )$ to $1$, and may therefore be interpreted as a shifted and rescaled form of the sign function.