The nature of the dirac delta function

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Discussion Overview

The discussion revolves around the nature of the Dirac delta function and its interaction with the function 1/x, particularly in the context of mathematical rigor and definitions. Participants explore whether the addition of the Dirac delta function to 1/x can be justified and what implications arise from this interaction.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant suggests that 1/x + δ(x) = 1/x by treating both as piecewise functions, questioning the significance of the undefined point at x = 0.
  • Another participant counters that this reasoning is heuristic and that a rigorous approach reveals that 1/x is not locally integrable, complicating the equation.
  • A participant expresses confusion about the Dirac delta function and seeks clarification on its nature.
  • It is noted that the Dirac delta function is not a function in the traditional sense and that its properties are derived from a more sophisticated definition.
  • One participant rephrases the question to whether 1/x + 0/x = 1/x, leading to a discussion about the zero function and its implications.
  • Another participant clarifies that 0/x is the zero function over ℝ-{0}, affirming that the equation holds true for all non-zero x.
  • Questions arise about the meaning of adding undefined values at x = 0, with a participant stating that the equation has no meaning at that point.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the original equation involving the Dirac delta function and 1/x. There are competing views regarding the interpretation and rigor of the Dirac delta function and its mathematical implications.

Contextual Notes

Limitations include the dependence on definitions of the Dirac delta function and the local integrability of 1/x, which are not resolved in the discussion.

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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?
 
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Hi,
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).
 
I don't understand-Am I thinking of the dirac delta function in the wrong way? could you explain?
 
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.
 
Okay, so if I change my question to "does 1/x + 0/x = 1/x", would the answer be any different?
 
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.
 
So it doesn't matter that I am adding undefined to undefined at x = 0?
 
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.
 
okay, thanks.
 

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