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Is the square of Heaviside function equal to Heaviside?

  1. Mar 17, 2006 #1
    Is the square of Heaviside function equal to Heaviside?
    [tex]H(t-t')\times H(t-t')=H(t-t')? [/tex]
    Please help me on the above equation.
     
  2. jcsd
  3. Mar 17, 2006 #2

    matt grime

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    What are ther definitions? It gives away the answer.
     
  4. Mar 17, 2006 #3

    mathman

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    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.
     
  5. Mar 17, 2006 #4
    Thank mathman and Matt Grime for your help.
    I was perplexed at this problem for several days.
     
  6. Mar 18, 2006 #5

    HallsofIvy

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    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, [itex]H^2(x) \ne H(x)[/itex] because
    [tex](H(0))^2= \frac{1}{4}\ne \frac{1}{2}= H(0)[/tex].
     
  7. Mar 18, 2006 #6

    mathman

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    This is merely a quibble. The value of the Heaviside function at 0 is irrelevant in any application.
     
  8. Mar 18, 2006 #7

    HallsofIvy

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    But not irrelevant to the question of whether H2(x)= H(x)!!
     
  9. Mar 19, 2006 #8

    mathman

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    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.
     
  10. Mar 20, 2006 #9

    HallsofIvy

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    And you are asserting that whether or not H(x)= H2(x) is not one of its properties?
     
  11. Mar 20, 2006 #10

    jim mcnamara

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

    Is there another definition you want to use?
     
  12. Mar 20, 2006 #11

    mathman

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    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.
     
  13. Sep 23, 2010 #12

    wlk

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    No. [tex]H(t-t')\times H(t-t')=R(t-t') [/tex]

    Here R is the ramp function
     
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