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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
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@t.b. thanks. There is always one more gimmick to know. –  Will Jagy Mar 20 '12 at 21:05
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1 Answer

up vote 3 down vote accepted

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 $$\begin{align} H(-x) &= [-x>0] + \frac{1}{2}[-x=0] \\ &= [x<0] + \frac{1}{2}[x=0] \\ &= [x\leq 0] - \frac{1}{2}[x=0] \\ &= [\neg(x>0)] - \frac{1}{2}[x=0] \\ &= 1-[x>0] - \frac{1}{2}[x=0] \\ &= 1-H(x). \end{align}$$

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@J.M.: Thanks for the help. Why doesn't that LaTeX work? –  user26872 Apr 3 '12 at 18:37
    
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
    
@J.M.: Very nice. By the way, I first learned of the Iverson bracket reading one of your answers. They're great! Maybe I'll try and get cases to work. –  user26872 Apr 3 '12 at 18:43
    
I'm happy that you are now using Iverson brackets because of something I wrote. Please do spread the word. ;) –  J. M. Apr 3 '12 at 18:47
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