Trying to define the set of alternate derivatives associated with the set of periodic functions with some rational number period

Question

I have been working on some math proofs and general theory stuff in a paper linked below and I have hit a mental block on something.

Piecewise Constant Functions in Differential and Functional Equations

On page 22 there is a definition given for the set of alternate derivatives associated with any differential fork, which is a set of functions. The set of periodic functions with a rational number period seems to meet the definition of a differential fork. So does anyone know how to write the associated alternate differential for that set aside from that definition? I'm trying to come up with a more useful way of writing it for that particular set, but I haven't been able to come up with anything for months.

I have really no clue at all how to go about doing it.

Answer 1

This is general theory and will still need some work. First we will prove a lemma needed to justify a restriction that will follow.

The set $S$ of all functions periodic with a period of $1$ is a differential fork.

Proof: First we establish that the set of all constant functions is a subset of $S$. This is observed by the fact that constant functions are periodic with any real number period.

Now we must prove that if $f \in S$ and $g \in S$ and $H(x,y)$ is some function of $x$ and $y$ then $M(x) = H(f(x),g(x)) \in S$. By the definition of periodicity we know that for all real numbers $x$ that $f(x) = f(x+1)$ and $g(x) = g(x+1)$. Therefore, substituting into $H(f(x),g(x))$ we have that $H(f(x),g(x)) = H(f(x),g(x))$ and therefore by transitivity we have that $M(x) = M(x+1)$ for all real numbers $x$. Therefore, by the definition of periodicity we have that $M(x) \in S$.

Therefore, by the definition of a differential fork we have that the set $S$ is a differential fork. []

The reason I did this is because it is particularly easier to deal with functions with period $1$ rather than arbitrary periods that are rational multiples of $1$. In fact, I would not be surprised if such an alternate differential has no closed form as a standard limit.

Here is my thinking which I will further develop into a formal definition. Consider the following graph of $\sin{x}$ (and please ignore that I forgot scale the period to $1$).

Each point on the graph $x$ has a line segment connecting it to $x+1$ and $x-1$. My thinking is that the operator I am seeking should a derivative whose value at $x$ is some (as of yet undetermined) set of linear combinations of the slope between each point and the point in the adjacent period.

In the case of sin or any periodic function this results in $0$ everywhere, which makes sense. Interestingly enough consider this graph of $x * \sin{x}$:

https://www.google.com/search?q=x*sin%28x%29&ie=utf-8&oe=utf-8

(zoom out a lot)

You should upper and lower curves that look like $x$ and $-x$. Note, the result we expect with the operator (ignoring the incompatible period) is $\sin{x}$! Taking that derivative on the upper and lower curves we would have them both evaluate to $1$ and $-1$.

This lends credence to method potentially working. Mostly, I am just working out the analytical details and whatnot it took me 2 years to formalize the other alternate derivative (for piecewise constant functions), so please don't get upset that I have not quite figured out the exact expression yet.