Dirac delta an even function?

  • Thread starter sirona
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Hi this is my first post here so I'm sorry if my question seems trivial.
I haven't worked a lot with the dirac delta function before, so i always thought that the shifting property would only work as:
[tex]\int\delta(x-h)\;f(x)\;dx=f(h)[/tex]
Now I've been reading some articles and I came across expressions like:
[tex]\int\delta(h-x)\;f(x)\;dx=f(h)[/tex]
which didn't make sense to me so I checked on the internet and saw that the delta function is supposed to be an even function.
Now I know it is not entirely a true function, but still the only description I know is that it's zero everywhere except for x. If I can give values to x other than zero, than
[tex]\delta(x)=\delta(-x)[/tex] means for example
[tex]\delta(5)=\delta(-5)[/tex] which doesn't make sense to me.
So I'll be very glad if someone can explain to me why it's an even function.
Thanks!
 

Answers and Replies

  • #2
Hurkyl
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To a great extent, the kind of object the dirac delta function is plays well with changes of variable. The simplest case is when f is an invertible, differentiable, increasing function, [itex]\varphi[/itex] is a distribution, and both g(x) and g(f(x)) f'(x) are test functions, then

[tex]\int_{-\infty}^{+\infty} \varphi(x) g(x) \, dx = \int_{-\infty}^{+\infty} \varphi(f(x)) g(f(x)) f'(x) \, dx[/tex]

This equation can be taken as the very definition of [itex]\varphi(f(x))[/itex] -- although some effort must be taken to show this definition is well-defined.

For other f, things are a little trickier. f(x) = -x isn't so bad, since it's still differentiable and invertible. Other f simply won't make sense all, such as when f(x) = 0 in the case where [itex]\varphi = \delta[/itex].



I should point out that there is an ambiguity in notation. When we have expressions involving a variable, such as sin(x) or x + 3, sometimes we mean for that expression to represent the (undetermined) real number that is the result of the computation involving the (undetermined) real number x. Other times, we mean for that expression to represent a function of a single variable, which we have named x.

The latter kind of notation is often applied to distributions like the Dirac delta, since it is rather convenient to write.
 
  • #3
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Now I know it is not entirely a true function, but still the only description I know is that it's zero everywhere except for x.
x being which number, exactly?

The delta function is, roughly, zero everywhere except for 0, and as such is obviously even - delta(0) = +infinity (metaphorically speaking), delta(non-zero) = 0. To say that it's zero everywhere except for "x" doesn't even make sense.
 
  • #4
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yes ok i see your point.
no need to be hostile, what i meant to ask was why it's an even function in general. so i could also phrase it like
[tex]\delta (x-h) = \delta (-x+h)[/tex]
or just give a number
[tex]\delta (x-5) = \delta (-x+5)[/tex]
In this case I just felt like the "spike" formed by the delta function would be on the other side of the axis, but the distribution argument makes sense.
Thanks for the replies!
 

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