Wu Characteristic

Update: March 8, 2016: Handout for a mathtable talk on Wu characteristic. Gauss-Bonnet for multi-linear valuations deals with a number in discrete geometry. But since the number satisfies formulas which in the continuum need differential calculus, like curvature, the results can be seen in the light of quantum calculus. Here are some slides: So, why is the Wu characteristic an … ….

Barycentric refinement

A finite graph has a natural Barycentric limiting space which can serve as the geometry on which to do quantum calculus or physics. The holographic picture has universal spectral properties.

Level surfaces and Lagrange

How to define level surfaces or solve extremization problems in a graph. A comment on a recent paper on Sard.

Why quantum calculus?

Quantum calculus is easy to learn, allows experimentation with small worlds and allows to use the same notation we are used to classically.

Calculus on graphs

Calculus on graphs is a natural coordinate free frame work for discrete calculus.

Exponential Function

We have seen that $f'(x)=Df(x) = (f(x+h)-f(x))/h$ satisfies $D[x]^n = n [x]^{n-1}$.
We will often leave the constant $h$ out of the notation and use terminology like $f'(x) = Df(x)$ for the “derivative”. It makes sense not to simplify $[x]^n$ to $x^n$ since the algebra structure is different.

Define the exponential function as
$exp(x) = \sum_{k=0}^{\infty} [x]^k/k!$. It solves the equation $Df=f$. Because each of the approximating polynomials $exp_n(x) = \sum_{k=0}^{n} [x]^k/k!$ is monotone and positive also $exp(x)$ is monotone and positive for all $x$. The fixed point equation $Df=f$ reads $f(x+h) = f(x) + h f(x) = (1+h) f(x)$ so that for $h=1/n$ we have $f(x+1) = f(x+n h) = (1+h)^n f(x) = e_n f(x)$
where $e_n \to e$. Because $n \to e_n$ is monotone, we see that the exponential function $\exp(x)$ depends in a monotone manner on h and that for $h \to 0$ the graphs of $\exp(x)$ converge to the graph of $\exp(x)$ as $h \to 0$.

Since the just defined exponential function is monotone, it can be inverted on the positive real axes. Its inverse is called $\log(x)$. We can also define trigonometric functions by separating real and imaginary part of $\exp(i x) = \cos(x) + i \sin(x)$. Since $D\exp=\exp$, these functions satisfy $D\cos(x) = - \sin(x)$ and $D\sin(x) = \cos(x)$ and are so both solutions to $D^2 f = -f$.