Dirac delta an even function?

[sirona]

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:
$$\int\delta(x-h)\;f(x)\;dx=f(h)$$
Now I've been reading some articles and I came across expressions like:
$$\int\delta(h-x)\;f(x)\;dx=f(h)$$
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
$$\delta(x)=\delta(-x)$$ means for example
$$\delta(5)=\delta(-5)$$ 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!

 

[Hurkyl]

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, $\varphi$ is a distribution, and both g(x) and g(f(x)) f'(x) are test functions, then

$$\int_{-\infty}^{+\infty} \varphi(x) g(x) \, dx = \int_{-\infty}^{+\infty} \varphi(f(x)) g(f(x)) f'(x) \, dx$$

This equation can be taken as the very definition of $\varphi(f(x))$ -- 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 $\varphi = \delta$.



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.

 

[Preno]

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.

 

[sirona]

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
$$\delta (x-h) = \delta (-x+h)$$
or just give a number
$$\delta (x-5) = \delta (-x+5)$$
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!

 

[CellCoree]

Hello,

Thank you for your question. The Dirac delta function is indeed an even function. This means that it satisfies the property f(x) = f(-x) for all values of x. In the case of the Dirac delta function, this property can be understood by looking at its definition.

The Dirac delta function is defined as a distribution, rather than a traditional function. It is often written as \delta(x), but this is just a notation and not an actual function. The Dirac delta function is defined as follows:

\int_{-\infty}^{\infty}\delta(x)\;f(x)\;dx=f(0)

This means that when we integrate the Dirac delta function with any other function f(x), we get the value of f(0). In other words, the Dirac delta function acts as a "spike" at the point x=0. It is zero everywhere else.

To understand why the Dirac delta function is an even function, we can look at the shifting property that you mentioned. When we shift the Dirac delta function by a value h, we get:

\int\delta(x-h)\;f(x)\;dx=f(h)

This means that the spike now occurs at the point x=h, and the value of the integral is f(h). However, this is equivalent to shifting the function f(x) by the same amount in the opposite direction, i.e. -h. So we can also write:

\int\delta(h-x)\;f(x)\;dx=f(h)

This shows that the Dirac delta function is indeed an even function, since it does not change when we replace x with -x.

I hope this explanation helps clarify why the Dirac delta function is an even function. Keep in mind that it is not a traditional function, and its properties may seem counterintuitive at first. But it is a useful tool in many areas of science and mathematics, and its evenness is an important aspect of its behavior. Please let me know if you have any further questions.