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The nature of the dirac delta function

  1. Sep 7, 2013 #1
    From what I can tell, it seems that 1/x + δ(x) = 1/x because if we think of both 1/x and the dirac delta function as the following peicewise functions:

    1/x = 1/x for x < 0
    1/x = undefined for x = 0
    1/x = 1/x for x > 0

    δ(x) = 0 for x < 0
    δ(x) = undefined for x = 0
    δ(x) = 0 for x > 0

    then for any part of 1/x that is defined, 0 is being added to it. Is it okay to ignore the fact that the function is being changed at x = 0 because it was already undefined?
  2. jcsd
  3. Sep 7, 2013 #2
    This won't work, because your definition of the dirac delta "function" is heuristical, and if you "rigorize" it, your equation won't work (if you actually to try to make it rigorous, you'll face another problem, namely that 1/x is not locally integrable, and you'll have to look at what is called the principal value of 1/x, but you can replace 1/x by 1/x², your argument remains the same, in the proper context, you'll see that it does not work).
  4. Sep 7, 2013 #3
    I don't understand-Am I thinking of the dirac delta function in the wrong way? could you explain?
  5. Sep 7, 2013 #4
    The dirac delta function is not a function. It is often presented as the function satisfying f(x)=0 for x non zero, f(0)=∞, and such that ∫f=1.
    Of course no such function exists, and its actual definition is more sophisticated.
    It is not really a bad way to picture that "function" in your head, but it's not usable as such. You have to use the actual definition, or just use the properties commonly attributed to this function (the fact that is is neutral for the convolution, that its fourier transoform is 1 etc...), all this properties can of course be proved by using the rigorous definition of δ.
  6. Sep 7, 2013 #5
    Okay, so if I change my question to "does 1/x + 0/x = 1/x", would the answer be any different?
  7. Sep 7, 2013 #6
    Well, 0/x is just the zero function over ℝ-{0}, so... your equation is tantamount to 1/x=1/x over ℝ-{0}, which is of course correct.
  8. Sep 7, 2013 #7
    So it doesn't matter that I am adding undefined to undefined at x = 0?
  9. Sep 7, 2013 #8
    Your equation has no meaning at x=0.
    You have equality of your two sides for all x non zero. There's not much more that can be said.
  10. Sep 7, 2013 #9
    okay, thanks.
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