# Is the square of Heaviside function equal to Heaviside?

1. Mar 17, 2006

### snooper007

Is the square of Heaviside function equal to Heaviside?
$$H(t-t')\times H(t-t')=H(t-t')?$$

2. Mar 17, 2006

### matt grime

What are ther definitions? It gives away the answer.

3. Mar 17, 2006

### mathman

Any function which has its range 0 and 1 and nothing else will be equal to its square. This is because x2=x has exactly 2 solutions, 0 and 1.

4. Mar 17, 2006

### snooper007

Thank mathman and Matt Grime for your help.
I was perplexed at this problem for several days.

5. Mar 18, 2006

### HallsofIvy

As I said in this same thread in the homework help section,
http://planetmath.org/encyclopedia/H...eFunction.html [Broken]
defines the Heaviside function by
H(x)= 0 if x< 0, 1/2 if x= 0, and 1 if x> 0.

With that definition, $H^2(x) \ne H(x)$ because
$$(H(0))^2= \frac{1}{4}\ne \frac{1}{2}= H(0)$$.

Last edited by a moderator: May 2, 2017
6. Mar 18, 2006

### mathman

This is merely a quibble. The value of the Heaviside function at 0 is irrelevant in any application.

7. Mar 18, 2006

### HallsofIvy

But not irrelevant to the question of whether H2(x)= H(x)!!

8. Mar 19, 2006

### mathman

Quibble continued. We can define H(0) to be 0 or 1 (or anything else in between), since it doesn't affect its properties. If you want H(x)=H(x)2 for all values of x, let H(0)=0 or 1.

9. Mar 20, 2006

### HallsofIvy

And you are asserting that whether or not H(x)= H2(x) is not one of its properties?

10. Mar 20, 2006

### Staff: Mentor

Halls stated the "usual" definition of $$H$$
where $$H(0) = 1/2$$

Is there another definition you want to use?

11. Mar 20, 2006

### mathman

Final quibble. I prefer a definition that H(x) is the integral of a delta function at 0. In that case, H is undefined at 0. If you want left continuity, H(0)=0, right continuity gives H(0)=1.

12. Sep 23, 2010

### wlk

No. $$H(t-t')\times H(t-t')=R(t-t')$$

Here R is the ramp function