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Decomposing Heavy Side Step Function into even and odd components

  1. Sep 3, 2014 #1

    grandpa2390

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    1. The problem statement, all variables and given/known data

    Decompose H(x) into even and odd components

    2. Relevant equations

    e(x) = (f(x)+f(-x))/2
    o(x) = (f(x)-f(-x))/2

    3. The attempt at a solution

    i said f(x)=1 and f(-x)=0 and I got e(x)=1/2 and o(x)=1/2. but this isn't true. e(x)=1/2 but o(x) = 1/2 when x>0 and -1/2 when x<0. what am I doing wrong?
     
  2. jcsd
  3. Sep 3, 2014 #2

    vela

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    ##H(x)=1## and ##H(-x)=0## only hold if ##x>0##, so your conclusions about ##e(x)## and ##o(x)## are only valid for positive ##x##.
     
  4. Sep 3, 2014 #3

    grandpa2390

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    can you give me more information. like tell me the correct the way to evaluate if x<0
     
  5. Sep 3, 2014 #4

    vela

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    No, because I'd essentially be doing the problem for you, which is against the forum rules, and I think it would really help you more if you thought about it for a while and identified where your attempt is failing. Look for assumptions you're using that you're not aware of, etc.
     
  6. Sep 3, 2014 #5

    grandpa2390

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    it doesn't help. If you don't want to explain how to do piecewise functions, can you help me by linking me to an explanation of how to do piece-wise functions, splitting them into even and odd.

    My problem is not that I can not do other problems. I can do e^x. I have no clue how to do this. My textbook does not explain a method for piecewise functions. Don't solve the problem for me, but can you please advise me on how to treat it theoretically.

    the only way I can make the this work, is if I plug in 1 for f(-x) in the o(x) function. and 0 for F(x) when x<0. This gives me the correct answer, but I don't understand why I would do that. What is the concept that causes me to do this.

    the only thing I can think, is that I have to know that it is negative before-hand, and work backwards. And say that for x<0, my odd function would have to be negative. so I flip my original equation over the y axis and f(-x)=1

    btw I have been looking at this for 4 days now. And I can't find any explanation anywheres. This is the only reason I signed up for the forum. because I'm not getting it.
     
    Last edited: Sep 3, 2014
  7. Sep 3, 2014 #6

    vanhees71

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    It's indeed true that, if we give more hints, it's like solving the problem. But one thing should be said, it's the Heaviside step function. It's named after Oliver Heaviside, one of the most remarkable geniuses in the history of electromagnetism. He developed nearly all modern mathematical tools, including the vector notation we still use today and the application of Laplace transformation to solve differential equations:

    http://en.wikipedia.org/wiki/Oliver_Heaviside
     
  8. Sep 3, 2014 #7

    grandpa2390

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    for x<0 am I supposed to multiply the O(x) by -1? why don't I do this for even equation?
    if I split e^x into a piecewise, why don't I have to do anything for the negative side?

    edit: I take that back, perhaps I would have to do it for e^x for the odd equation?

    edit: no that doesn't work.

    I don't understand. why I have to do this for heavy side function, a piecewise function, and only on the odd part and not on any other function. or even part.
     
    Last edited: Sep 3, 2014
  9. Sep 3, 2014 #8

    grandpa2390

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    or maybe when x<0 the equations flip?
    e(x)= (f(-x)+f(x))/2 (0+1)/2 = 1/2 x<0
    o(x) = (f(-x)-f(x))/2 (0-1)/2 = -1/2 x<0

    my textbook by bracewell just gives a short blurb and the equations I posted. and my instructor just reads the textbook.

    edit: that doesn't work for e^x
     
    Last edited: Sep 3, 2014
  10. Sep 3, 2014 #9

    grandpa2390

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    can't you do an example with another piecewise problem and I can look at how you did it? Is an actual explanation of this process some atomic secret that can only be learned through self-discovery? I am not looking for the answer. I know the answer. My instructor knows that I have the answer. I just need an explanation on why my answer is the answer. so that I can attack the other problems in my hw. it has nothing to do with my hw. she refuses to give me an explanation unless everyone else in the class is struggling with it. Chances are I am the only idiot who doesn't understand this.

    That must be what it is. Well when I finally figure this process out, I am going to sell the secret to Iran and I'll retire.
     
  11. Sep 3, 2014 #10

    vanhees71

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    Ok, one little hint. Just think about the cases [itex]x>0[/itex] and [itex]x<0[/itex] and consider the two functions
    [tex]e(x)=\frac{1}{2}[H(x)+H(-x)], \quad o(x)=\frac{1}{2} [H(x)-H(-x)].[/tex]
    If I say more, I nearly solve the problem, which is strictly forbidden by forum rules, and that's really for your best!
     
  12. Sep 3, 2014 #11

    grandpa2390

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    well maybe H(x)=H(-x) for e(x)
    and for o(x), H(-x) = -H(x)

    so for x<0

    e(x)=(H(-x)=H(x)) so (H(x) + H(-x))/2 x<0

    o(x) = (H(-x) - H(x)) = (-H(x) -H(-x))/2 x<0
     
  13. Sep 3, 2014 #12

    grandpa2390

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    Well I give up. Apparently I'm the only person in the world who considers the equations themselves an insufficient explanation. Obviously I am too stupid to do this. I'll have to drop the class if yahoo answers can't help me.
     
  14. Sep 3, 2014 #13

    grandpa2390

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    all I can see is that I get 1/2 and I am supposed to attach the negative as I see fit.
     
  15. Sep 3, 2014 #14

    vela

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    In your argument, you set f(x)=H(x) and plugged it into the definition. Then you said f(x)=1 and f(x)=0. This is fine if, say, x=1 because H(1)=1 and H(-1)=0. But what about when x=-2? It's not true that f(x) = H(-2) = 1 or f(-x) = H(+2) = 0, right? So there's a problem with simply saying f(x)=1 and f(-x)=0. You need to be more careful. If you assume x>0, then you can say f(x)=1 and f(-x)=0, and everything you wrote is fine, but your conclusions are then valid only for x>0.

    What if you now assume x<0? What changes?
     
  16. Sep 3, 2014 #15

    grandpa2390

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    the x's flip.
    f(x) x<0 = (f(-x)-(f(x))/2 = -1/2

    I said this earlier, so why doesn't this work for e^x, x<0? what am I doing wrong here? by the way this is one is not a homework problem. I just found this example online and so I know the solution.
    if I say e^x when x<0
    I plug in (e^-x - e^x)/2 I get the opposite sign of what I should get.
     
  17. Sep 3, 2014 #16

    vela

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    No, the x's don't flip. You're changing the definition of o(x) if you do that, which is why it doesn't work when you try that with ex.

    If x<0, what does f(x)=H(x) equal?
     
  18. Sep 3, 2014 #17

    grandpa2390

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    if x<0 then H(x)=0, and I thought that f(x)=H(x) so that f(x) would also equal 0. but you are saying that that is not true if x<0.
     
  19. Sep 3, 2014 #18

    grandpa2390

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    nevermind, a physics tutor explained to me.

    what we're actually doing is find the mean for any equation just for x>0. whether it be 1/2 for H(x) or cosh(x) and sinh(x) for e^x, and then we reflect the graph across the axis. for the even function we reflect across the x and for the odd function we reflect across the y and then the x axis and that is our graph.
     
  20. Sep 3, 2014 #19

    Orodruin

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    If x < 0, what is -x?
     
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