# Negated argument of the Heaviside Step Function

If $H(x)$ is the Heaviside step function, what is $H(-x)$? Is it $-H(x)$ or does $$H(-x) = \left\{\begin{array}{ll} 1 & x < 0 \\ 1/2 & x = 0 \\ 0 & x > 0 \end{array}\right. \hspace{5ex}?$$

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$1 - H(x)$ labhvljaehbgvlzd – Will Jagy Mar 20 '12 at 19:12
@WillJagy: Thanks. May I ask what the letters following $1-H(x)$ mean? – alex b Mar 20 '12 at 19:14
A comment box demands a minimum of 15 characters, I put in blanks but it refused. Let me see if pairs of braces work, I think I have seen that used to make up the 15 {}{}{}{}{}{}{} edit: evidently they do not disappear – Will Jagy Mar 20 '12 at 19:20
@Will: enclose them in dollar signs: ${}{}{}{}$ then they do disappear. – t.b. Mar 20 '12 at 20:29
@t.b. thanks. There is always one more gimmick to know. – Will Jagy Mar 20 '12 at 21:05

If we use the convention $$H(x) = \begin{cases} 0 & x < 0 \\ 1/2 & x = 0 \\ 1 & x > 0 \end{cases}$$ it is straightforward to work out that $$H(-x) = 1-H(x),$$ which is equivalent to your second guess. Notice that $H(x)$ is not an odd function.
Addendum: In terms of the Iverson bracket, $$[P] = \begin{cases} 1, & \textrm{If }P\textrm{ is true} \\ 0, & \textrm{otherwise}, \end{cases}$$ the Heaviside step function is $$H(x) = [x>0]+\frac{1}{2}[x=0].$$ There are different conventions for $H(0)$. Here we choose $H(0) = 1/2$. The properties of the Iverson bracket we will exploit here are $[\neg P] = 1-[P]$ and $[x<a]+[x=a] = [x\leq a]$. We find \begin{align} H(-x) &= [-x>0] + \frac{1}{2}[-x=0] \\ &= [x<0] + \frac{1}{2}[x=0] \\ &= [x\leq 0] - \frac{1}{2}[x=0] \\ &= [\neg(x>0)] - \frac{1}{2}[x=0] \\ &= 1-[x>0] - \frac{1}{2}[x=0] \\ &= 1-H(x). \end{align}
I quite like the identity $H(x)=\dfrac{x+|x|}{2x}$ myself... thus, $\dfrac{-x+|-x|}{2(-x)}=\dfrac{x-|x|}{2x}=\dfrac{2x-(|x|+x)}{2x}=1-\dfrac{|x|+x}‌​{2x}$ – J. M. Apr 3 '12 at 18:39
(As for the $\LaTeX$, I'm not sure why. At least \begin{cases} does...) – J. M. Apr 3 '12 at 18:41