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.

Fundamental Theorem

Let denote the discrete derivative of a continuous function f on the real line. In this post, I assume that all functions are continuous of have compact support. They are zero outside some large interval. No smoothness is required of course. Here is the simplest version of the fundamental theorem of calculus in a discrete setting: Theorem: . Proof. Choose … ….

Critical points

Assume f is a continuous function of one real variable. Lets call a point p a critical point of f if Df(p)=0 where Df(x) = f(x+1)-f(x) is the discrete derivative of f. As in classical calculus, a point p is called a local maximum of f, if there exists an open neighborhood U of a, such that for all x … ….