Calculus without limits # Discrete Cauchy Integral Formula

The following video helped a bit to get organized and to make adjustments. The discrete complex plane is $\mathbb{Z}^2$, the Shannon product. One can see this as the Gaussian integers, where two are connected if their difference has arithmetic norm 1 or 2. One can also equip the Cartesian product with the taxi metric $l^{\infty}$ and connect two points if the distance is 1. Now define translation operators Uf(z)=f(z-1) and Vf(z) = f(z-i) and $\partial_x f = [U^* f], \partial_y f = [V^*,f]$ and $D_z f = d/dz f = f'(z)=(\partial_x - i \partial_y) f/2$ as well as $d/d\overline{z} f = (\partial_x + i \partial_y) f/2$. The reason for the commutator notation in the crossed product algebra is that we have a deformation where the Leibniz rule holds. Define $[z]^n = (xU+iy V)^n$. This produces polynomials for which the rule $d/dz [z]^n = n [z]^{n-1}$ holds. By using the real Taylor series expansion in two variables in new variables $z,\overline{z}$ rather than $x,y$, we get the Taylor series in the complex for analytic functions for which $\partial/\partial \overline{z} f = 0$. Now define $\latex exp(z) = (1+1)^x (1+i)^y$ and its inverse $\log$ and the 1-form $(1/z) dz = d \log(z) = d/dz \log(z) dz$. We also have an operation which maps a 0-form f, an analytic function, to a 1-form f dz which is a function attached to edges. The analyticity condition assures that f dz is closed (unlike other frame works we do not assume that it is exact). Indeed, the Cauchy Riemann difference equations $u_x = v_y, u_y = -v_x$ are equivalent to $d/d\overline{z} f = 0$ and also equivalent to the contour integral along a square plaquette to be zero (the diagonal values of the 1-form can be filled in so that for all triangles we get a zero contour. This means that f dz is a closed 1-form in the graph theoretical sense meaning that the curl is zero. Now, by verifying two computations for the contour which is the unit circle of 0, we immediately get the Cauchy integral formula $(8 i)^{-1} \int_C f(z)/(z-a) \; dz = f(a)$ for all analytic functions. So, in these few lines we have verified (a few basic finite computations fill the details) a frame work in which the exactly same formulas hold as in the continuum. The only thing different is that the $2\pi i$ in the continuum is replaced by $8 i$ in the discrete as the unit circle in the discrete has length 8 and not $2\pi$.