Taylor series expansion of Dirac delta

[friend]

I'm trying to understand how the algebraic properties of the Dirac delta function might be passed onto the argument of the delta function.

One way to go from a function to its argument is to derive a Taylor series expansion of the function in terms of its argument. Then you are dealing with the argument and powers of it.

Normally, the Taylor series expansion for f(x) is:
$$f(x) = \sum\limits_{n = 0}^\infty {\frac{{{f^{(n)}}({x_0})}}{{n!}} \cdot {{(x - {x_0})}^n}}$$

But how does this change for a composite function? How do you expand $f(g(x))$? Do you just replace x in the sum with g(x) and x₀ with g(x₀)? Or do you differentiat f(g(x)) with respect to x in each term before setting x to x₀?

But I suspect the even harder question is how do you take the derivatives of the Dirac delta function for each term in the sum. And if you have to evaluate it at x₀, won't that make each term infinity? Is there an easy expression for ${\delta ^{(n)}}({x_0})$? Or better yet, how do we evaluate ${\delta ^{(n)}}(g(x))$? Thank you.

 

[CompuChip]

The mistake you are making is that the Dirac delta is not a function. It has some properties of a function, and is sometimes called a generalized function (better known as distribution). One of these is that you can define an n-th order derivative. But one of the things you can't do is write down a series, see for example this old PF thread.

There are different ways to define the Dirac delta. Normally, in distribution theory, it is defined by the way it acts as integration kernel, i.e. by the property $\int \delta(x) f(x) \, dx = f(0)$. However, you can write it down as the limit of a series of continuous functions, my favourite one being
$$\delta_a(x) = \frac{1}{a \sqrt{\pi}} e^{-x^2 / a^2}$$
such that $\delta(x) = \lim_{a \to 0} \delta_a(x)$
and of course you are free to expand $\delta_a(x)$ because it's as smooth as you could possibly wish.

 

[friend]

The mistake you are making is that the Dirac delta is not a function. It has some properties of a function, and is sometimes called a generalized function (better known as distribution). One of these is that you can define an n-th order derivative. But one of the things you can't do is write down a series, see for example this old PF thread.

There are different ways to define the Dirac delta. Normally, in distribution theory, it is defined by the way it acts as integration kernel, i.e. by the property $\int \delta(x) f(x) \, dx = f(0)$. However, you can write it down as the limit of a series of continuous functions, my favourite one being
$$\delta_a(x) = \frac{1}{a \sqrt{\pi}} e^{-x^2 / a^2}$$
such that $\delta(x) = \lim_{a \to 0} \delta_a(x)$
and of course you are free to expand $\delta_a(x)$ because it's as smooth as you could possibly wish.

I was thinking more in terms like this for the k^th derivative of the delta,
$${\delta ^{(k)}}[\varphi ] = {( - 1)^k}{\varphi ^{(k)}}(0)$$
that I found about a third of the way down on the wikipedia.com website for the Dirac delta. Could this be used to construct an expansion? I don't know if the $\varphi$ here can be used as the g(x) in my composition.

 

[CompuChip]

Yes, this is the definition as you would get it from distribution theory, where the Dirac delta would be defined by the way it "acts on" other functions $\phi$.
This definition is what I said earlier:
$$\int \delta(x) \phi(x) dx = \phi(0)$$

To find the k'th derivative, apply partial integration k times to
$$\int \delta^{(k)}(x) f(x) dx$$
to move it over to the test function f(x). This gives
$$\int \delta^{(k)}(x) f(x) dx = (-1)^k \int \delta(x) f^{(k)}(x) \, dx$$
which is $(-1)^k f^{(k)}(0)$ by the definition applied to $\phi = f^{(k)}$.

 

[blue_raver22]

The Dirac delta function is a mathematical construct that represents an infinitely narrow spike at a specific point, with an area of 1 under the spike. It is not a conventional function, so it cannot be expanded using a Taylor series in the traditional sense. However, there are ways to approximate the Dirac delta function using a series of functions.

One approach is to use a sequence of functions, such as a Gaussian function, that become increasingly narrow and tall as the sequence approaches the Dirac delta function. This allows for a Taylor series expansion to be applied to the sequence of functions, which can then be used to approximate the properties of the Dirac delta function.

Another approach is to use the theory of distributions, which extends the concept of a function to a broader class of objects that includes the Dirac delta function. In this framework, the Dirac delta function is defined as a linear functional that acts on a test function, and its derivatives can be calculated using integration by parts.

In either case, the algebraic properties of the Dirac delta function, such as its ability to "pass on" to the argument, can be understood through its definition and the properties of the approximating functions or distributions. It is important to note that the Dirac delta function is a tool for mathematical convenience and should be used with caution, as its properties can lead to unexpected results if not used correctly.