Is the square of Heaviside function equal to Heaviside?

[snooper007]

Is the square of Heaviside function equal to Heaviside?
H(t-t')\times H(t-t')=H(t-t')?
Please help me on the above equation.

 

[matt grime]

What are ther definitions? It gives away the answer.

 

[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.

 

[snooper007]

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

 

[HallsofIvy]

Please don't multi-post!

As I said in this same thread in the homework help section,
http://planetmath.org/encyclopedia/H...eFunction.html
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).

 

[mathman]

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

 

[HallsofIvy]

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

 

[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.

 

[HallsofIvy]

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.

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

 

[jim mcnamara]

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

Is there another definition you want to use?

 

[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.

 

[wlk]

Is the square of Heaviside function equal to Heaviside?
H(t-t&#039;)\times H(t-t&#039;)=H(t-t&#039;)?
Please help me on the above equation.

No. H(t-t&#039;)\times H(t-t&#039;)=R(t-t&#039;)

Here R is the ramp function