My problem

Alright, so this is my dilemma and it all boils down to the very troublesome function floor ($\lfloor x \rfloor$) and piecewise constant functions in general ($\lfloor f(x) \rfloor$). Everyone knows what the indefinite integral is. Just in case it is not well known this is what it is:

$$F(x) = \int_{c}^{x} f(t)dt$$

The indefinite integral is an accumulation function with some base value that it starts out with. It varies by an arbitrary constant.

Now let me define a second operator that we will refer to as an implied integral ($\int^*$). It is best defined by example:

$\int^* \lfloor x \rfloor dx = x \lfloor x \rfloor + C(x)$

$\int^* x \lfloor x \rfloor dx = \frac 12 x^2 \lfloor x \rfloor + C(x)$

$\int^* \lfloor f(x) \rfloor dx = x \lfloor f(x) \rfloor + C(x)$

Basically, it is an indefinite integral that varies be piecewise constant functions (C(x)) and treats peicewise constant functions as constants.

Now,there is ambiguousness with the term "antiderivative" that I recently stumbled upon. The antiderivative seems to refer to both of these operations. For instance, see my question about the bug in matlab. The answer there claims that the implied integral is a valid antiderivative. I remember from elementary calculus that antiderivatives vary by constants. However, I am willing to give the benefit of the doubt since that user is a professor and probably knows way more than me.

Why is it important?

Well, while the area function is a set-in-stone idea differential equations rely upon use of the antiderivative and not the indefinite integral. Therefore, depending on which one the antiderivative follows... the result will change the solutions to differential equations. Specifically: It will decide whether or not their solutions must be continuous or if they are allowed to be discontinuous. A more concrete example is that we would no longer be able to just solve homogenous linear equations with constant coefficients. We would be able solve one's with piecewise constant coefficients trivially (it is possible to solve them for continuous solutions by doing things).

To sum it all up:

Do antiderivatives vary by constant functions or piecewise constant functions?

It truly is an important question.


marked as duplicate by The Great Duck, Community Jan 16 '17 at 19:39

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  • $\begingroup$ Can you clarify what it means for a function $f\colon \mathbb{R}\to\mathbb{R}$ to vary by another function? $\endgroup$ – user259242 Nov 11 '16 at 5:04
  • $\begingroup$ @user259242 O.O I'm not quite sure I know how to define it. In fact, I'm sure it really has a tangible definition. Like how there is the set of all antiderivatives and they all vary between each other by BLA (I dare not say constant as that's in question here). I'm not quite sure there is a rigorous sense of "variance" within a set of functions. In fact, that's probably worthy of a question in its own right. $\endgroup$ – The Great Duck Nov 11 '16 at 5:07
  • $\begingroup$ @user259242 seeing your edit. f(x) varying by another function C(x) means that the set of solutions has an arbitrary set of functions "C(x)" by which the composition f(x) + c(x) will also be in the set of solutions. Once again though. I'm referring to the concept of basic calculus in reference to infinite antiderivatives. I don't know the rigorous meaning as only implied calculus would only care about "vary by", and that is still being fleshed out (a lot), because regular calculus only uses the term once for one set of c: constants. Or at least, so I believe. $\endgroup$ – The Great Duck Nov 11 '16 at 5:10
  • $\begingroup$ I guess you are trying to express the fact that for any pair of antiderivatives $F$ and $F'$ of a function $f\colon \mathbb{R}\to\mathbb{R}$ there exists an element $c\in\mathbb{R}$ such that $F - F' = c$? $\endgroup$ – user259242 Nov 11 '16 at 5:10
  • 2
    $\begingroup$ How have I missed the point? Your question is, verbatim, "Do antiderivatives vary by constant functions or piecewise constant functions?". I have explained that two antiderivatives of the same function cannot differ by a non-constant function. $\endgroup$ – user259242 Nov 11 '16 at 5:29

Let $f\colon \mathbb{R}\to\mathbb{R}$ be a differentiable function. If $F$ and $F'$ are antiderivatives of $f$ and $F - F' = C$ where $C\colon\mathbb{R}\to\mathbb{R}$ is a function, then $C$ is constant.

We have $\frac{d}{dx} (F - F') = \frac{d}{dx} C$ and since $\frac{d}{dx}(F - F') = \frac{d}{dx} F - \frac{d}{dx} F' = f - f =0$, it follows that $\frac{d}{dx} C = 0$ and thus $C$ is a constant function.


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