Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a question about the Heaviside step function $\theta(\xi)$, defined by $$\theta(\xi):=\begin{cases} 1, & \xi\geq0\\ 0, & \xi<0 \end{cases}$$ I need to evaluate the square of the Heaviside step function, i.e.

$$[\theta(\xi)]^2$$

So, my question is: does the relation $$[\theta(\xi)]^2=\theta(\xi)$$ holds?

share|improve this question

migrated from physics.stackexchange.com Jul 13 '13 at 6:37

This question came from our site for active researchers, academics and students of physics.

1  
Related: physics.stackexchange.com/q/61176/2451 –  Qmechanic Jul 13 '13 at 0:21
    
(homework) should not be used as a standalone tag; see tag-wiki and meta. –  Martin Sleziak Jul 15 '13 at 13:43
add comment

3 Answers

up vote 5 down vote accepted

If you take $\theta(\xi)$ as you said, (with $\theta(0)=1$): $$\theta(\xi):=\begin{cases} 1, & \xi\geq0\\ 0, & \xi<0 \end{cases}$$ It will be a regular function (not a generalized, like $\delta(x)$), and there is not any reason to think about it unusually. so the relation holds:$$[\theta(\xi)]^2=\theta(\xi)$$

For the case of $\theta(0)={1\over2}$ as other answers say, $\theta^2(0)\neq \theta(0)$.

But if you define it as : $$\theta(\xi):=\begin{cases} 1, & \xi>0\\ \text{undefined},& \xi=0\\ 0, & \xi<0 \end{cases}$$ I will show that again$[\theta(\xi)]^2=\theta(\xi)$. As a generalized function, $\theta(\xi)$ will be defined through the following relation (where $g(t)\equiv\theta(\xi)$) : $$\int_{-\infty}^{+\infty}\phi(t)g^{(n)}(t) dt=(-1)^n \int_{-\infty}^{+\infty}\phi^{(n)}g(t)dt \, \,\,\,\,\,\,**$$ where $\phi$ vanishes at $\pm \infty$.

If $g^2(t)=g(t)$, you must have: $$\int_{-\infty}^{+\infty}\phi(t) {\frac{d^n g^2(t)}{dt^n}}dt=(-1)^n\int_{-\infty}^{+\infty}\phi^{(n)}(t)g(t)dt$$ We denote ${\frac{d^n g^2(t)}{dt^n}}$ with $f(t)$ for simplicity: $$\int_{-\infty}^{+\infty}\phi(t) f(t)dt=(-1)^n\int_{-\infty}^{+\infty}\phi^{(n)}(t)g(t)dt\, \,\,\,\,\,\,***$$ The right hand side is simply equal to $(-1)^n\int_{0}^{+\infty}\phi^{(n)}(t)dt=(-1)^{n-1}\phi^{(n-1)} (0)$.

Now, we use the $**$ equation with $g(t)=\delta (t)$: $$\int_{-\infty}^{+\infty}\phi(t)\delta^{(n)}(t) dt=(-1)^{n}\phi^{(n)} (0)$$

we conclude that in the left hand side of $***$, $f(t)$ must be the $(n-1)\text{th}$ derivative of $\delta (t)$, which implies ${\frac{d^n \theta^2(\xi)}{d\xi ^n}}={\frac{d^n \theta (\xi)}{d\xi ^n}}$ or $\theta^2(\xi)=\theta(\xi)$.

share|improve this answer
    
All these calculations are not needed at all for the case when $\Theta(0)$ is undefined, from the definition you can immediately observe that $\Theta^2=\Theta$ for nonzero $\zeta$, and both are undefined at $\zeta=0$, so they are the same. –  Maxim Umansky Jul 13 '13 at 5:47
add comment

Yes. For every $\xi \in \mathbb{R}$ you can check that

$$\theta(\xi)^2 = \theta(\xi)$$

so you conclude that the functions $\theta^2$ and $\theta$ agree.

share|improve this answer
add comment

If you take this definition http://mathworld.wolfram.com/HeavisideStepFunction.html, it won´t generally be true since then $H^{2}\left(0\right)=\frac{1}{4}\neq\frac{1}{2}$. In another forum http://www.physicsforums.com/showthread.php?t=114599 the same question was asked.

share|improve this answer
1  
I agree, but I'm not using this definition ($H$). –  Anuar Jul 12 '13 at 23:11
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.