Editing Sign function (section)

== Properties in mathematical analysis ==

=== Discontinuity at zero ===
[[File:Discontinuity of the sign function at 0.svg|thumb|300px|The sign function is not [[continuous function | continuous]] at <math>x=0</math>.]]

Although the sign function takes the value {{math|&minus;1}} when <math>x</math> is negative, the ringed point {{math|(0, &minus;1)}} in the plot of <math>\sgn x</math> indicates that this is not the case when <math>x=0</math>. Instead, the value jumps abruptly to the solid point at {{math|(0, 0)}} where <math>\sgn(0)=0</math>. There is then a similar jump to <math>\sgn(x)=+1</math> when <math>x</math> is positive. Either jump demonstrates visually that the sign function <math>\sgn x</math> is discontinuous at zero, even though it is continuous at any point where <math>x</math> is either positive or negative.

These observations are confirmed by any of the various equivalent formal definitions of [[Continuous function|continuity]] in [[mathematical analysis]]. A function <math>f(x)</math>, such as <math>\sgn(x),</math> is continuous at a point <math>x=a</math> if the value <math>f(a)</math> can be approximated arbitrarily closely by the [[sequence]] of values <math>f(a_1),f(a_2),f(a_3),\dots,</math> where the <math>a_n</math> make up any infinite sequence which becomes arbitrarily close to <math>a</math> as <math>n</math> becomes sufficiently large. In the notation of mathematical [[Limit of a sequence|limit]]s, continuity of <math>f</math> at <math>a</math> requires that <math>f(a_n) \to f(a)</math> as <math>n \to \infty</math> for any sequence <math>\left(a_n\right)_{n=1}^\infty</math> for which <math>a_n \to a.</math> The arrow symbol can be read to mean ''approaches'', or ''tends to'', and it applies to the sequence as a whole.

This criterion fails for the sign function at <math>a=0</math>. For example, we can choose <math>a_n</math> to be the sequence <math>1,\tfrac{1}{2},\tfrac{1}{3},\tfrac{1}{4},\dots,</math> which tends towards zero as <math>n</math> increases towards infinity. In this case, <math>a_n \to a</math> as required, but <math>\sgn(a)=0</math> and <math>\sgn(a_n)=+1</math> for each <math>n,</math> so that <math>\sgn(a_n) \to 1 \neq \sgn(a)</math>. This counterexample confirms more formally the discontinuity of <math>\sgn x</math> at zero that is visible in the plot.

Despite the sign function having a very simple form, the step change at zero causes difficulties for traditional [[calculus]] techniques, which are quite stringent in their requirements. Continuity is a frequent constraint. One solution can be to approximate the sign function by a smooth continuous function; others might involve less stringent approaches that build on classical methods to accommodate larger classes of function.

=== Smooth approximations and limits ===
The signum function can be given as a number of different (pointwise) limits:
<math display="block">\begin{align}
\sgn x &= \lim_{n\to\infty}\frac{1-2^{-nx}}{1+2^{-nx}}\\
       &= \lim_{n\to\infty}\frac{2}{\pi}\operatorname{arctan}(nx)\\
       &= \lim_{n\to\infty}\tanh(nx)\\
       &= \lim_{\varepsilon\to 0} \frac{x}{\sqrt{x^2 + \varepsilon^2}}.
\end{align}</math>
Here, <math>\tanh</math> is the [[hyperbolic tangent]], and <math>\operatorname{arctan}</math> is the [[arctan|inverse tangent]].  The last of these is the derivative of <math>\sqrt{x^2+\varepsilon ^2}</math>. This is inspired from the fact that the above is exactly equal for all nonzero <math>x</math> if <math>\varepsilon=0</math>, and has the advantage of simple generalization to higher-dimensional analogues of the sign function (for example, the partial derivatives of <math>\sqrt{x^2+y^2}</math>).

See ''{{section link|Heaviside step function#Analytic approximations}}''.

=== Differentiation ===

The signum function <math>\sgn x</math> is [[Differentiable function|differentiable]] everywhere except when <math>x=0.</math> Its [[derivative]] is zero when <math>x</math> is non-zero:
<math display="block"> \frac{\text{d}\, (\sgn x)}{\text{d}x} = 0 \qquad \text{for } x \ne 0\,.</math>

This follows from the differentiability of any [[constant function]], for which the derivative is always zero on its domain of definition. The signum <math>\sgn x</math> acts as a constant function when it is restricted to the negative [[Interval (mathematics)#Definitions and terminology|open region]] <math>x<0,</math> where it equals {{math|−1}}. It can similarly be regarded as a constant function within the positive open region <math>x>0,</math> where the corresponding constant is {{math|+1}}. Although these are two different constant functions, their derivative is equal to zero in each case.

It is not possible to define a classical derivative at <math>x=0</math>, because there is a discontinuity there.

Although it is not differentiable at <math>x=0</math> in the ordinary sense, under the generalized notion of differentiation in [[distribution (mathematics)|distribution theory]], 
the derivative of the signum function is two times the [[Dirac delta function]]. This can be demonstrated using the identity <ref>{{MathWorld |title=Sign |id=Sign}}</ref>
<math display="block"> \sgn x = 2 H(x) - 1 \,,</math>
where <math>H(x)</math> is the [[Heaviside step function]] using the standard <math>H(0)=\frac{1}{2}</math> formalism.
Using this identity, it is easy to derive the distributional derivative:<ref>{{MathWorld |title=Heaviside Step Function |id=HeavisideStepFunction}}</ref>
<math display="block"> \frac{\text{d}\sgn x}{\text{d}x} = 2 \frac{\text{d} H(x)}{\text{d}x} = 2\delta(x) \,.</math>

=== Integration ===

The signum function has a [[definite integral]] between any pair of finite values {{mvar|a}} and {{mvar|b}}, even when the interval of integration includes zero. The resulting integral for {{mvar|a}} and {{mvar|b}} is then equal to the difference between their absolute values:
<math display="block"> \int_a^b (\sgn x) \, \text{d}x = |b| - |a| \,.</math>

In fact, the signum function is the derivative of the absolute value function, except where there is an abrupt change in [[slope|gradient]] at zero:
<math display="block"> \frac{\text{d} |x|}{\text{d}x} =  \sgn x \qquad \text{for } x \ne 0\,.</math>

We can understand this as before by considering the definition of the absolute value <math>|x|</math> on the separate regions <math>x>0</math> and <math>x<0.</math> For example, the absolute value function is identical to <math>x</math> in the region <math>x>0,</math> whose derivative is the constant value {{math|+1}}, which equals the value of <math>\sgn x</math> there.

Because the absolute value is a [[convex function]], there is at least one [[subderivative]] at every point, including at the origin. Everywhere except zero, the resulting [[subdifferential]] consists of a single value, equal to the value of the sign function. In contrast, there are many subderivatives at zero, with just one of them taking the value <math>\sgn(0) = 0</math>. A subderivative value {{math|0}} occurs here because the absolute value function is at a minimum. The full family of valid subderivatives at zero constitutes the subdifferential interval <math>[-1,1]</math>, which might be thought of informally as "filling in" the graph of the sign function with a vertical line through the origin, making it continuous as a two dimensional curve.

In integration theory, the signum function is a [[weak derivative]] of the absolute value function. Weak derivatives are equivalent if they are equal [[almost everywhere]], making them impervious to isolated anomalies at a single point. This includes the change in gradient of the absolute value function at zero, which prohibits there being a classical derivative.

=== Fourier transform ===
The [[Fourier transform]] of the signum function is<ref>{{cite journal|last1=Burrows|first1=B. L.|last2=Colwell|first2=D. J.|title=The Fourier transform of the unit step function|journal=International Journal of Mathematical Education in Science and Technology|date=1990|volume=21|issue=4|pages=629–635|doi=10.1080/0020739900210418}}</ref>
<math display="block">PV\int_{-\infty}^\infty (\sgn x) e^{-ikx}\text{d}x = \frac{2}{ik} \qquad \text{for } k \ne 0,</math>
where <math>PV</math> means taking the [[Cauchy principal value]].