# I Heaviside function

#### Sudhir Regmi

Hi everyone,
I could not figure out whether H(x-a) and H(a-x) are same or different. Please help me to understand this.

#### Replusz

Member warned to be careful of posting incorrect information
Since H(x)=-H(-x), it is true.
Hope this helps.

#### Sudhir Regmi

Since H(x)=-H(-x), it is true.
Hope this helps.
Thanks, it means H(a-x) = -H(x-a) is true.

Exactly :)

#### jasonRF

Gold Member
Since H(x)=-H(-x), it is true.
Hope this helps.
The Heaviside step function is usually define: H(x) = 0 for x<0, and H(x) = 1 for x>0. So it is not true that H(x) = -H(-x). For example, H(1) = 1 but H(-1) = 0.

Jason

#### Replusz

No. The Heaviside function is H(0)=0 and H(x<0)= -1 and H(x>1)=+1

#### Mark44

Mentor
No. The Heaviside function is H(0)=0 and H(x<0)= -1 and H(x>1)=+1
Take another look at that graph near the top of the page that you linked to. If x < 0, H(x) = 0, not -1.

#### Sudhir Regmi

Take another look at that graph near the top of the page that you linked to. If x < 0, H(x) = 0, not -1.
Hello Mark44,
What do you think about my original question, is H( x-a) equal to H( a-x)? Where x is a variable and a is a constant.

#### Mark44

Mentor
Hello Mark44,
What do you think about my original question, is H( x-a) equal to H( a-x)? Where x is a variable and a is a constant.
$H(x - a) = \begin{cases} 1 & x > a \\ 0 & x < a \\ \end{cases}$
H(a - x) = H(-(x - a)) -- this is the reflection of the graph of H(x - a) across the line x = a, so the graphs of these two functions are not the same.

#### Sudhir Regmi

Thank you Mark44, Replusz and jasonRF for responding to my question.