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[This answer has been heavily edited in response to a long chain of comments on Eric Stucky's answer.]

I came up with these few theorems and I am curious whether or not my hypothesis are true. I don't really have a method for proving these, mind you. I'm more considering these as things I've noticed to be true. I just don't know whether they are always true.

Some of these are definitions to which I will append "def." to the end of the title. Think of these as a given. It's like how the integral is defined abstractly as area. I'm defining terms.

Jump Series Theorem def.

A function $f$ [consisting of floor] is said to have a jump series if there is a series $\Sigma=\sum_{n=1}^\infty a_nH_n$ of scaled step functions such that $f+\Sigma$ no longer has any discontinuity attributable to floor. The sequence of scaling factors $a_n$ is called the jump sequence for the function.

Definition of Implied Derivative def.

We define the evaluation of the implied derivative of a function $y$ at the point $x$ as the set of values $y^{\to}(x) = \{y | y = \lim_{h \to 0^+} \frac {y(x+h) - f(h)}{h} \lor y = \lim_{h \to 0^-} \frac {y(x+h) - f(h)}{h}\}$. It is in this sense that the implied derivative has the potential of being a multi-valued operator.

Jump Series Resolution Theorem

[The original wording is preserved; I have a formalism but it assumes that having a jump function is equivalent to being piecewise continuous, which I'm not sure is true.]

The integral of a floor-based function is floor treated as a constant minus the appropriate jump series of the integral.

Integration Continuity Theorem

The (ordinary) indefinite integral of any function $f$ having a jump series must be continuous (or fixably discontinuous) on the domain of $f$. However, this can vary for the implied indefinite integral.

Calculus Jump Theorem def.

[Again, the original wording is preserved; the meaning of this phrase has not been properly illuminated]

Any form of jumping within a function representable as a floor function can be considered as a portion of the constant of the implied integration for floor.

Jump Location Theorem

For any function $a(x)$ with a jump series, the function $\lfloor a(x)\rfloor$, has points of discontinuity precisely at those $x$ such that $a(x) - [a(x)] = 0$.

Composite Floor Function Jump Sequence Theorem One

If $a$ is a function such that $a(x) - \lfloor a(x) \rfloor = 0$ when $x - \lfloor x \rfloor = 0$, then the jump sequence of the composite function $f(x - \lfloor a\rfloor)$ is a constant value $f(0) - f(1)$.

Implied Differential Equations Conjecture

Any continuous solution to an equation constructed with implied derivatives of various orders is the same as the solution to the similarly constructed differential equation.

user64742
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    These sound interesting, but I think you need to explain several of these a bit more. – Paul Jan 13 '16 at 00:04
  • Your concepts of the floor function are like having what's called a piecewise continuous function, a function that is continuous on a number of open intervals. – Bob Krueger Jan 13 '16 at 00:27
  • I am not sure I understand the formulation of your Integration Continuity Theorem, but perhaps you mean the result on Riemann integral which is shown here. – Martin Sleziak Jan 15 '16 at 06:05
  • I've edited your question based on our discussion in the comments. Please feel free to re-edit it if it is not what you intended. Also, please help me keep this page clean by deleting all of your comments on my answer except the ones you feel will benefit future readers. – Eric Stucky Feb 17 '16 at 05:54
  • FWIW: I am almost certain that having a jump series is a weaker condition than being piecewise continuous because we permit $\Sigma$ to converge conditionally... – Eric Stucky Feb 17 '16 at 05:56
  • @TheGreatDuck: Whoops, sorry for not seeing this until now! I'm certainly interested in taking a look; you can see my user page for contact information. (To delete your own comments, you can mouse over the pencil at the end; when you get close enough there will be an X that shows up; click that and you can delete it.) – Eric Stucky Feb 26 '16 at 23:43
  • I haven't looked at your work in a while, and I don't think I'm likely to do so for at least another couple weeks: it's finals season. After that, I expect winter "break" to be kind of nuts but I think it would be good for me to start looking at it again for maybe an hour a day. I'll keep in touch (and if not, you should ping me :P) – Eric Stucky Dec 04 '16 at 18:07
  • Do not delete the most part of your question. – Did Jan 25 '17 at 01:30

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My preliminary thoughts:

[After the edit(s) on February 17, this answer is out of date.]

The Jump Series Theorem is definitely false, but since you gave it as a definition, we will consider the class of functions on which it is true. I won't say so explicitly in future lines, but from here on, I mean true/false with respect to this class.

The Jump Series Resolution Theorem is ill-defined, but the most reasonable interpretation that I can think of is true by definition.

The Integration Continuity Theorem is true by the theorem cited in the comments.

The Calculus Jump Theorem is, as best I can tell, meaningless because the floor function does not imply any sort of integration is happening; much less a constant of integration; even less a portion of such a constant, which is not a term I'm familiar with.

The Implied Integration and Differentiation Theorem is definitely false, but again I will from now on assume that we are only looking at the class of functions for which it is true.

The Jump Location Theorem is unambiguously true.

The Composite Floor Function Jump Sequence Theorem One is possible, but you don't explain what a jump sequence is. Based on the way you use it, it does not make sense for it to be a jump series, nor does it make sense to be the set of domain values where the function jumps. So I'm at a loss here.

Eric Stucky
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  • what is the definition of a "jump series", or at least what would be an example of ? – reuns Feb 12 '16 at 08:11
  • for $x \ge 0$ the (distributional) derivative of $\lfloor x \rfloor$ is $\sum_{n = 0}^\infty \delta(x-n)$ so (I agree there is a problem with the negative side), that works for any piecewise constant function – reuns Feb 12 '16 at 08:25
  • I don't ever remember you mentioning the $f(x)/f(x)$ thing; I think I would say you are considering formal functions so that you don't have to worry with domain issues; fine. However, I, like others before me, don't see why this means you can treat the floor function as a constant-- I suspect that this is because your understanding of 'constant' is different than mine somehow. I think I understand what you mean by a "jump sequence" now, but now I am confused as to how this is different than a jump series. Is the only difference that the latter takes $x$-values into consideration? – Eric Stucky Feb 13 '16 at 06:57
  • @TheGreatDuck [comment 2]: I see. The phrase you are looking for is 'piecewise constant'. But the reason $f(x)=\lfloor x \rfloor$ does not have defined derivative at the integers is because it contains some indeterminate form $p(x)/p(x)$ (with $p$ a polynomial), but this isn't true. Rather, it's because at an integer, the one-sided limits of the difference quotient, from the right and from the left, are different—but each is well-defined. – Eric Stucky Feb 14 '16 at 20:49
  • It may be worth noting that based on your examples, it looks like you are doing something closely related to the analysis of "locally $L^p$ functions", but working with a more restricted equivalence relation that guarantees the existence of some more fine-grained data (namely, a jump series). Somewhat less abstractly, you are considering any function which is defined at codiscrete subset of $\Bbb R$ to be equivalent to its continuous extensions, if any exist. – Eric Stucky Feb 16 '16 at 10:55
  • @TheGreatDuck: Sorry if I wasn't clear: I only say these words so that you can have something to look up. It's pretty hard to google vague math ideas so some terminology in the bag can help you to find neighboring things. – Eric Stucky Feb 26 '16 at 23:40