How can the Dirac delta function be even? Watch

hhoque1890
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My understanding of the Dirac delta function is that its a function that can be visualized to be infinity at some point A and 0 at all other points.

For any function to be even that function must be symmetric with respect to the y-axis. However the Dirac delta function seems to be only symmetric with respect to the y-axis when it is centered at the origin (i.e. when A=0, Dirac delta(x)=infinity)

Hence how does it make sense for a Dirac delta function not centered at the origin to be still referred to as an even function?
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DFranklin
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No, the definition is that "it's infinite at 0", not at an arbitrary point A. (You can just use \delta(x-A) if you want the more general behaviour).
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hhoque1890
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Sorry, I realize now my definition for the Dirac delta function was slightly wrong. However even still I don't know how the general form of the Dirac delta function (giving the more generalized behavior as you say) can be visualized as even. (Please see image attached).

This is because when I visualize a Dirac delta function in this 'general form', I imagine a single sharply peaked function centered at some arbitrary value (x' . A function like this is clearly not symmetric with respect to the y-axis (unless x'=0).

However I have seen various proofs online (although I'm not sure if I fully understand them) that even in this more general form, the Dirac delta function is still even.


My end question is how can the general form of the Dirac delta function be regarded as even despite not being symmetric with respect to the y-axis for all values of x' not equal to 0?
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hhoque1890
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***The image corresponding to my previous post doesn't seem to show so here it is again:
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DFranklin
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(Original post by hhoque1890)
***The image corresponding to my previous post doesn't seem to show so here it is again:
What I told you in my previous post still holds. \delta(x) = 0 for x \neq 0. Obviously \delta(x-a) is 0 unless x = a, and it looks like this is what you're thinking about. But this isn't the delta function; it's not what we're asserting is even. (anymore than we would expect cos(x-1) to be an even function just because cos(x) is).
Last edited by DFranklin; 1 week ago
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hhoque1890
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Just to be 100% clear are you saying that:


\delta(x) is even

\delta(x-a) is not even
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DFranklin
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(Original post by hhoque1890)
Just to be 100% clear are you saying that:


\delta(x) is even

\delta(x-a) is not even
Yes. (Exception: if a = 0 the latter function is also even).
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hhoque1890
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Sorry the screenshot of the proof from the video in my previous post doesn't seem to show again. Here it is again:
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DFranklin
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(Original post by hhoque1890)
Sorry the screenshot of the proof from the video in my previous post doesn't seem to show again. Here it is again:
This isn't saying anything that disagrees with my previous posts.

\delta(x - a) = \delta(a-x) is merely repeating that the delta function \delta(x) is even. It is NOT saying that the function f defined by f(x) = \delta(x-a) is even.
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