We owe Paul Dirac two excellent mathematical jokes. I have amended them with a few lesser known variations.

A. *Square root of the Laplacian*: we want $\Delta$ to be $D^2$ for some first order differential operator (for example, because it is easier to solve first order partial differential equations than second order PDEs). Writing it out,

$$\sum_{k=1}^n \frac{\partial^2}{\partial x_k^2}=\left(\sum_{i=1}^n \gamma_i \frac{\partial}{\partial x_i}\right)\left(\sum_{j=1}^n \gamma_j \frac{\partial}{\partial x_j}\right) = \sum_{i,j}\gamma_i\gamma_j \frac{\partial^2}{\partial x_i x_j},
$$

and equating the coefficients, we get that this is indeed true if

$$D=\sum_{i=1}^n \gamma_i \frac{\partial}{\partial x_i}\quad\text{and}\quad \gamma_i\gamma_j+\gamma_j\gamma_i=2\delta_{ij}.$$ (With a $2$ on the right hand side.)

It remains to come up with the right $\gamma_i$'s. Dirac realized how to accomplish it with $4\times 4$ matrices when $n=4$; but a neat follow-up joke is to simply define them to be the elements $\gamma_1,\ldots,\gamma_n$ of

$$\mathbb{R}\langle\gamma_1,\ldots,\gamma_n\rangle/(\gamma_i\gamma_j+\gamma_j\gamma_i - 2\delta_{ij}).$$

Using symmetry considerations, it is easy to conclude that the commutator of the $n$-dimensional Laplace operator $\Delta$ and the multiplication by $r^2=x_1^2+\cdots+x_n^2$ is equal to $aE+b$, where
$$E=x_1\frac{\partial}{\partial x_1}+\cdots+x_n\frac{\partial}{\partial x_n}$$ is the Euler vector field. A boring way to confirm this and to determine the coefficients $a$ and $b$ is to expand $[\Delta,r^2]$ and simplify using the commutation relations between $x$'s and $\partial$'s. A more exciting way is to act on $x_1^\lambda$, where $\lambda$ is a formal variable:

$$[\Delta,r^2]x_1^{\lambda}=((\lambda+2)(\lambda+1)+2(n-1)-\lambda(\lambda-1))x_1^{\lambda}=(4\lambda+2n)x_1^{\lambda}.$$

Since $x_1^{\lambda}$ is an eigenvector of the Euler operator $E$ with eigenvalue $\lambda$, we conclude that

$$[\Delta,r^2]=4E+2n.$$

B. *Dirac delta function*: if we can write

$$g(x)=\int g(y)\delta(x-y)dy$$

then instead of solving an inhomogeneous linear differential equation $Lf=g$ for each $g$, we can solve the equations $Lf=\delta(x-y)$ for each real $y$, where a linear differential operator $L$ acts on the variable $x,$ and combine the answers with different $y$ weighted by $g(y)$. Clearly, there are fewer real numbers than functions, and if $L$ has constant coefficients, using translation invariance the set of right hand sides is further reduced to just one, $\delta(x)$. In this form, the joke goes back to Laplace and Poisson.

What happens if instead of the ordinary geometric series we consider a doubly infinite one? Since

$$z(\cdots + z^{-n-1} + z^{-n} + \cdots + 1 + \cdots + z^n + \cdots)=
\cdots + z^{-n} + z^{-n+1} + \cdots + z + \cdots + z^{n+1} + \cdots,$$

the expression in the parenthesis is annihilated by the multiplication by $z-1$, hence it is equal to $\delta(z-1)$. Homogenizing, we get

$$\sum_{n\in\mathbb{Z}}\left(\frac{z}{w}\right)^n=\delta(z-w)$$

This identity plays an important role in conformal field theory and the theory of vertex operator algebras.

Pushing infinite geometric series in a different direction,

$$\cdots + z^{-n-1} + z^{-n} + \cdots + 1=-\frac{z}{1-z} \quad\text{and}
\quad 1 + z + \cdots + z^n + \cdots = \frac{1}{1-z},$$

which add up to $1$. This time, the sum of doubly infinite geometric series is zero!
Thus the point $0\in\mathbb{Z}$ is the sum of all lattice points on the non-negative half-line and all points on the positive half-line:

$$0=[\ldots,-2,-1,0] + [0,1,2,\ldots] $$

A vast generalization is given by Brion's formula for the generating function for the lattice points in a convex lattice polytope $\Delta\subset\mathbb{R}^N$ with vertices $v\in{\mathbb{Z}}^N$ and closed inner vertex cones $C_v\subset\mathbb{R}^N$:

$$\sum_{P\in \Delta\cap{\mathbb{Z}}^N} z^P =
\sum_v\left(\sum_{Q\in C_v\cap{\mathbb{Z}}^N} z^Q\right),$$

where the inner sums in the right hand side need to be interpreted as rational functions in $z_1,\ldots,z_N$.

Another great joke based on infinite series is the Eilenberg swindle, but I am too exhausted by fighting the math preview to do it justice.

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