Dirac function series expansion

Question

Dirac function is generalized function. However, is it possible to use taylor expansion on it? i.e., is it possible to write $$\delta(x)=\sum_{n=0}^{\infty}A_{n}x^{n}$$?

I know that it can have Fourier transform and can be represented as Legendre polynomial, but I would like to know if it can written in terms of Taylor and corresponding $A_{n}$.

Answer 1

The Taylor series for Dirac delta function: $$ \delta(x) = \sum_{n=0}^{\infty} A_n x^n. $$ At any point $x \neq 0, ~ A_n \equiv 0$ for all $n \ge 0$. So, the series could be useful for the case when $x = 0$. Let's write down using straightforward definition of Taylor series. $$ \delta(x) = \sum_{n=0}^{\infty} \frac{\delta^{(n)}(0)}{n!}x^n. $$ For Dirac function the following relation is true (try to prove it!): $x\delta^{(n)}(x) = -n \delta^{(n-1)}(x)$, so using this recurrent relation one finds: $$ \delta(x) = \sum_{n=0}^{\infty} \frac{\delta^{(n)}(0)}{n!}x^n = \delta(0) + \sum_{n=1}^{\infty} \underbrace{(-1)^n\left[\lim_{x\to 0} \frac{\delta(x)}{x^n}\right]}_{A_n} x^n $$ However, I don't see how this is going to help you, since you have $\delta(0)$ as a summand which is itself an infinity.

Another approach that comes to mind (and potentially could be useful) is to try to use the limit definition of Dirac function, namely $$ \delta(x) = \lim_{a \to 0}\delta_a(x), $$ where $\delta_a(x) = \frac{1}{a\sqrt{\pi}}{e^{-x^2/a^2}}$. From here you can expand this smooth function in Taylor series and get the Taylor for $\delta(x)$ as a limit.

Answer 2

Whether you use the limit approach suggested by @pointguardo or anything else, you need to provide the identity: $$ \int_{-\infty}^\infty \delta(x) dx = 1. $$ If $\delta$ admits a Taylor expansion, then $$ \sum_{n = 0}^\infty A_n \int_{-\infty}^\infty x^n dx = 1. $$ Because of the symmetry of $(-\infty, \infty)$ w.r.t. $x = 0$, integrals with odd powers vanish and we have $$ 2\sum_{n = 0}^\infty A_{2n} \int_0^\infty x^{2n} dx = 1. $$ The integral is nothing else but the Mellin transform of $1$, which is equal to $$ \int_0^\infty x^{2n} dx = \delta(2n). $$ Thus, we arrive at $$ 2 \sum_{n = 0}^\infty A_{2n} \delta(2n) = 1. $$ When $n \neq 0$, $\delta(2n) = 0$. Therefore, the only non-zero term is $$ 2 A_0 \delta(0) = 1, $$ which does not hold for any $A_0$.

Thus, the answe is: no.

Answer 3

Probably it is not possible to find a Dirac delta function representing Taylor series in the real field however, as Jacques Hadamard once said, things change when you complexify i.e. transport to the complex domain the problem you are dealing with. Considering $\delta(x)$, $x\in\mathbb{R}$ as an analytic functional or as a residue current, it is possible to find its representative as a very simple Laurent series and an associated real Taylor series with complex coefficients, which overcomes the issues raised by the very basic observation by Asatur Khurshudyan.

Following Gel'fand and Shilov (1964) (ch. II, §1.5 pp. 161-162), calling $\mathscr{A}(\mathbb{R})$ is the spaces of all entire functions $\psi(s)$, $s=x+iy\in\mathbb{C}$ such that $$ |s|^q|\psi(s)|\leq C_q e^{a|y|}\quad q\in\mathbb{N} $$ we can define $\delta(x)$ by the following formula $$ \psi(0)=\langle \delta,\psi\rangle=\frac{1}{2\pi}\left[\int\limits_{-\infty+ai}^{+\infty+ai}\frac{\psi(s)}{s}\mathrm{d}s-\int\limits_{-\infty-ai}^{+\infty-ai}\frac{\psi(s)}{s}\mathrm{d}s\right]=\frac{1}{2\pi}\int\limits_{|s|=1}\frac{\psi(s)}{s}\mathrm{d}s \quad a>0\tag{1}\label{1} $$ In this approach we can say that $$ \delta(x)=\sum_{n=-\infty}^{+\infty}A_ns^n=s^{-1}=\frac{1}{x+iy}=\sum_{n=0}^{+\infty}\frac{(-1)^n}{(iy)^{n+1}}x^n\tag{2}\label{2} $$ and already we see that the issue raised by Asatur Khurshudyan is avoided. Approaching the problem in $\mathscr{A}^\prime(\mathbb{R})$ could seem a limited way to deal with it: however, the same approach can be extended to $\mathscr{D}^\prime(\mathbb{R})$ by understanding \eqref{1} in the sense of residue currents, as done by Berenstein et al. (1993) (ch. 1,§1, pp. 1-2). Following their approach, it is possible to define the following functional: $$ I(\varphi, \epsilon)=\frac{1}{2\pi}\int\limits_{|s|=\epsilon}\frac{\varphi(s)}{s}\mathrm{d}s \qquad\forall\varphi\in\mathscr{D}(\mathbb{R}) $$ and by their proposition 1.1 (Berenstein et al. (1993), ch. 1, §1, p.2) we have that $$ \varphi(0)=\langle \delta,\varphi\rangle=\lim_{\epsilon\to 0}I(\varphi, \epsilon)\triangleq\mathrm{res}_{x=0}\frac{\varphi(s)}{s}\qquad\forall\varphi\in\mathscr{D}(\mathbb{R}) $$

[1] Berenstein, C. A.; Gay, R.; Vidras, A.; Yger, A. (1993), Residue currents and Bezout identities, Progress in Mathematics, 114, Basel–Berlin–Boston: Birkhäuser Verlag, pp. xi+158, doi:10.1007/978-3-0348-8560-7, ISBN 3-7643-2945-9 / 0-8176-2945-9 / 978-3-0348-8560-7, MR 1249478, Zbl 0802.32001.

[2] Gel'fand, I. M.; Shilov, G. E. (1964) [1958], Generalized functions. Vol. I: Properties and operations, New York–London: Academic Press, pp. xviii+423, ISBN 978-0-12-279501-5, MR 0166596, Zbl 0115.33101.

Answer 4

From my experience in physics, the Taylor expansion of the Dirac delta often appears. The concept is still obscure to me and I was hoping to find an answer to give me a better insight. We can find in numerous scientific article an expression equivalent to

$\begin{equation} \delta (\mathbf{r}-\mathbf{u}) = \delta(\mathbf{r}) - \mathbf{u}\cdot \nabla \delta(\mathbf{r}) + \frac{1}{2} \mathbf{u} \cdot \nabla \nabla \delta(\mathbf{r}) \cdot \mathbf{u} + O(3) \end{equation}$

where r is an atomic position and u is a small displacement due to a perturbation applied to the atom. So to answer your question, is it possible to represent a $\delta(x)$ as a Taylor series: Yes and it gives sensible results in physics.

Now, you can push further and ask the question : Is the $\delta(x)$ function correctly represented or is it only a trick ?

For this, I've made a python code. Note that, I've approximated the delta dirac function as a gaussian. Remember that one of the definition of a delta dirac is a gaussian with $\lim\limits_{a=0}\sigma=a$. According to me, this function seems to converge with enough terms and an infinitely small sigma. This gives the following plot and I've linked the code if you want to play with sigma to convince yourself :

def gaussian_delta():
# Define the parameters of the Gaussian function
sigma = 0.1  # standard deviation
r0 = 0.5     # center position

# Define the displacement vector
u = np.array(0.1)

# Define the x-axis range for plotting
x = np.linspace(-1, 1, 1000)

# Define the Gaussian function
G = lambda r: np.exp(-0.5*((r-r0)/sigma)**2)

# Define the first three terms of the Taylor series
term1 = lambda r: G(r)
term2 = lambda r: G(r) - u.dot(np.gradient(G(r), r))
term3 = lambda r: G(r) - u.dot(np.gradient(G(r), r)) + 0.5 * u.dot(np.gradient(np.gradient(G(r), r),r)).dot(u)

# Plot the functions
plt.plot(x, term1(x), color='b', label="Term 1")
plt.plot(x, term2(x), color='g', label="Term 1+2")
plt.plot(x, term3(x), color='y', label="Term 1+2+3")
plt.plot(x, np.zeros_like(x), 'r', label="Delta Dirac")

plt.xlim(-1,1)
plt.ylim(-0.2,1.4)

# For axvline, 0|1 correspond to the bottom|top of the graph
ymin = ((0-plt.ylim()[0])/(plt.ylim()[1]-plt.ylim()[0]))
plt.axvline(r0+u, color='r', ymin=ymin, ymax=1)

plt.legend()
plt.show()