If $H(x)$ is the Heaviside step function, what is $H(-x)$? Is it $-H(x)$ or does $H(-x) = \left\{\begin{array}{ll} 1 & x < 0 \\ 1/2 & x = 0 \\ 0 & x > 0 \end{array}\right. \hspace{5ex}?$
Negated argument of the Heaviside Step Function
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1@t.b. thanks. There is always one more gimmick to know. – 2012-03-20
1 Answers
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. We find $\begin{align} H(-x) &= [-x>0] + \frac{1}{2}[-x=0] \\ &= [x<0] + \frac{1}{2}[x=0] \\ &= [x\leq 0] - \frac{1}{2}[x=0] \\ &= [\neg(x>0)] - \frac{1}{2}[x=0] \\ &= 1-[x>0] - \frac{1}{2}[x=0] \\ &= 1-H(x). \end{align}$
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0@J.M.: I think the LaTeX problem was math in \textrm{...}. – 2012-04-03