Heaviside Function: Clarifying H(x-a) & H(a-x)

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Discussion Overview

The discussion centers around the properties of the Heaviside step function, specifically the relationship between H(x-a) and H(a-x). Participants explore whether these two expressions are equivalent or different, delving into definitions and characteristics of the Heaviside function.

Discussion Character

  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant questions whether H(x-a) and H(a-x) are the same or different.
  • Some participants assert that H(a-x) = -H(x-a) based on the property H(x) = -H(-x).
  • Another participant challenges this assertion, stating that the Heaviside function is defined as H(x) = 0 for x < 0 and H(x) = 1 for x > 0, implying that the earlier claim is incorrect.
  • Further clarification is provided that H(0) = 0, H(x < 0) = 0, and H(x > 0) = 1, contradicting the earlier claims about negative values.
  • A participant explains that H(a-x) can be viewed as the reflection of H(x-a) across the line x = a, suggesting that their graphs are not the same.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between H(x-a) and H(a-x). There is no consensus on whether they are equivalent, with multiple competing interpretations of the Heaviside function's properties.

Contextual Notes

There are unresolved definitions and interpretations of the Heaviside function, particularly regarding its values at and around zero, which contribute to the disagreement in the discussion.

Sudhir Regmi
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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.
 
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Since H(x)=-H(-x), it is true.
Hope this helps.
 
Replusz said:
Since H(x)=-H(-x), it is true.
Hope this helps.
Thanks, it means H(a-x) = -H(x-a) is true.
 
Exactly :)
 
Replusz said:
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
 
No. The Heaviside function is H(0)=0 and H(x<0)= -1 and H(x>1)=+1
 
Mark44 said:
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.
 
  • #10
Sudhir Regmi said:
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.
 
  • #11
Thank you Mark44, Replusz and jasonRF for responding to my question.
 

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