Alright, I didn't know the best way to formulate my question. Basically, whilst doing some physics research, I naturally came upon the function

$$ f(x) = 2x[x] - [x]^2 $$

where I use $[x]$ as notation for the `nearest-integer function' (i.e. rounding off). Usually this function has to have a caveat of how we exactly define the value for $x \in \frac{1}{2} \mathbb Z$, but interestingly for this function it does not matter, since it turns out to be continuous! In fact, it turns out $f(x)$ is exactly given by the glued function of taking all the tangent lines of $x^2$ at integer values of $x$:

enter image description here (Note: due to properties of $x^2$, the tangent lines exactly intersect at half-integer values of $x$.)

So my question is not literally `why is it continuous?', but rather: considering it is continuous, and considering that that is not a generic property of functions which are defined in terms of nearest-integer functions, is there a better (i.e. more insightful) way of expressing $f(x)$? Relatedly, is there some part of mathematics where functions similar to these naturally arise?

  • 10
    $\begingroup$ Nice observation. However a better question would be: "which polynomials in $x$ and $[x]$ are continuous"? $\endgroup$ – goblin Dec 29 '17 at 1:06
  • 4
    $\begingroup$ This is explained by the fact that $2x(x-\frac12)-(x-\frac12)^2 = 2x(x+\frac12)-(x+\frac12)^2$; they are both equal to $x^2-\frac14$. $\endgroup$ – Rahul Dec 29 '17 at 1:12
  • 1
    $\begingroup$ Let $\,x=n+1/2\,$ for integer $\,n\,$, then $\,[x] = n + \alpha\,$ where $\,\alpha \in \{0,1\}\,$ depending on the choice of rounding for half-integers. Then $\, f(x) = 2(n+1/2)(n+\alpha) - (n+\alpha)^2 = n^2 + n + \alpha(1-\alpha) \,$ where the last term is $\,0\,$ regardless of the choice of $\,\alpha\,$ being $\,0\,$ vs. $\,1\,$. $\endgroup$ – dxiv Dec 29 '17 at 1:18
  • 4
    $\begingroup$ Is the "Computer Science" tag really relevant? $\endgroup$ – J.-E. Pin Dec 29 '17 at 13:06

Let $g(x,y)$ be any continuous function such that $g(x,-\frac12)=g(x,\frac12)$. Then $f(x)=g(x,x-[x])$ is continuous.

In particular, your function is given by $g(x,y)=x^2-y^2$. Consequently, we can write $f(x)=x^2 - (x-[x])^2$.

  • 6
    $\begingroup$ Strictly speaking, we only require $g(x,-\frac12)=g(x,\frac12)$ at half-integers $x\in\mathbb Z+\frac12$. $\endgroup$ – Rahul Dec 29 '17 at 2:03
  • $\begingroup$ Accepted! ... even though it takes a bit of the appealing mystery away ;) $\endgroup$ – Ruben Verresen Jul 29 '18 at 19:20

Call a function $f : \mathbb{R} \rightarrow \mathbb{R}$ weird-continuous iff for all $x \in \mathbb{R}$, the left and right limits of $f$ at $x$ exist, whether or not they're equal. Define the weird derivative $\Delta f$ of a function $f : \mathbb{R} \rightarrow \mathbb{R}$ to be the function $\Delta f : \mathbb{R} \rightarrow \mathbb{R}$ defined by: $$\Delta(f)(x) = (\lim_+f)(x)-(\lim_-f)(x),$$ where for example the $(\lim_+f)(x)$ denotes the limit of $f(x')$ as a $x'$ approaches $x$ from the right.

Also, for each $a \in \mathbb{R}$, write $\langle a \rangle$ for the function $\mathbb{R} \rightarrow \mathbb{R}$ defined as follows:

$$\langle a \rangle (x) = \begin{cases} 1 & x = a \\ 0 & x \neq a\end{cases}$$

For example:

  • if $H$ is the Heaviside step function, then $\Delta (H) = \langle 0\rangle$.
  • if $H$ is the Heaviside step function, then $\Delta (x \mapsto 3H(x-1)+4H(x-2)) = 3\langle 1\rangle+4\langle 2\rangle$.
  • if $f$ is continuous, then $\Delta(f) = 0$.

Letting $f$ and $g$ denote weird-continuous functions, the basic results (I think) are:

Proposition 0. Suppose $f : \mathbb{R} \rightarrow \mathbb{R}$ has no removable discontinuities. Then $f$ is continuous iff $\Delta f = 0$.

Proposition 1. Additivity $\Delta(f+g) = \Delta(f)+\Delta(g)$ and $\Delta(0) = 0$.

Proposition 2. Product Rule. $\Delta (fg) = \Delta(f)g + f \Delta(g)$

(Is the product rule even true? I don't really use it below...)

Corollary to the product rule. Weird derivatives are linear with respect to continuous functions, meaning that if $f$ is continuous, then $\Delta(fg) = f\Delta(g)$.

Now think of $x$ as the identity function $\mathbb{R} \rightarrow \mathbb{R}$. Let $f = 2 x [x] - [x]^2$.

To prove that $f$ is continuous, we'll show that $\Delta(f) = 0$. We have:

$$\Delta (f) = \Delta(2 x [x] - [x]^2) = 2 x \Delta(x) - \Delta([x]^2)$$

Also: $$\Delta([x]) = \left(\sum_{n \in \mathbb{Z}+\frac{1}{2}}\langle n\rangle\right)$$

Furthermore: $$\Delta([x]^2) = \sum_{n \in \mathbb{Z}+\frac{1}{2}}((n+1/2)^2 - (n-1/2)^2)\langle n\rangle = \sum_{n \in \mathbb{Z}+\frac{1}{2}}(n+1/2+n-1/2)(n+1/2-n+1/2)\langle n\rangle = \sum_{n \in \mathbb{Z}+\frac{1}{2}}2n\langle n\rangle$$

Hence: $$\Delta (f) = 2x\left(\sum_{n \in \mathbb{Z}+\frac{1}{2}}\langle n\rangle\right)-\sum_{n \in \mathbb{Z}+\frac{1}{2}}2n\langle n\rangle = \sum_{n \in \mathbb{Z}+\frac{1}{2}}2n\langle n\rangle-\sum_{n \in \mathbb{Z}+\frac{1}{2}}2n\langle n\rangle = 0$$

So $f$ is continuous.

  • 1
    $\begingroup$ Your product rule doesn't seem to hold. Let $f(x)$ be $1$ on $(0,\infty)$ and $0$ otherwise and $g(x)=1-f(x)$. Then $fg=0$, so $\Delta(fg)(0)=0$, but the RHS of your rule is $1$. $\endgroup$ – Henning Makholm Dec 29 '17 at 2:14
  • 1
    $\begingroup$ @HenningMakholm, yes, good point. $\endgroup$ – goblin Dec 29 '17 at 2:21
  • 1
    $\begingroup$ Just a note about terminology: a function is weird-continuous (on a bounded interval) iff it is regulated, i.e. the uniform limit of step functions. This is the basic building block for the regulated integral. $\endgroup$ – Calvin Khor Dec 29 '17 at 17:51

is there a better (i.e. more insightful) way of expressing $f(x)$?

Maybe one way you could do that is by noticing that $$ f(x) = \max_{k\in\mathbb{Z}} \left( 2kx-k^2 \right). $$ Since, in any interval of fixed length, $f$ is the max of a finite number of continuous functions, then it is continuous.

is there some part of mathematics where functions similar to these naturally arise?

Personally, I have encountered this a lot in information theory, when many lower/upper bounds on communication rates are derived together, implying that the max/min of these bounds holds. This is especially true when many different linear inequalities can be naturally derived and make intuitive sense. For instance, $$ \begin{cases} R_1 \ge 2-4R_2\\ R_1 \ge 1-R_2 \end{cases} \implies R_1 \ge \max\left\{ 2-4R_2, 1-R_2 \right\}. $$ If you really want an example, off the top of my head I can think of Theorem 2 in this paper.

  • 1
    $\begingroup$ Given that the max of a countable family of continuous functions isn't always continuous, how is the continuity of your $f$ automatic? $\endgroup$ – Ben Millwood Dec 30 '17 at 16:26
  • 1
    $\begingroup$ I suppose you can easily demonstrate that locally only finitely many terms of the sequence are relevant. $\endgroup$ – Ben Millwood Dec 30 '17 at 16:31
  • $\begingroup$ @BenMillwood You are correct, I was too hand-wavy there. Fixed, thanks! $\endgroup$ – jadhachem Dec 31 '17 at 4:54

The difference between the rounding function (piecewise constant) and $x$ is a triangle wave, i.e. piecewise linear, periodic, ranging in $[-\frac12,\frac12)$.

If you square it, you get a continuous, piecewise quadratic function. This is enough to explain your observation.

enter image description here


Given a $k+1$-times differentiable function $g$ such that for every integer $n$, $$\int_n^{n+1} g^{(k+1)}(t)(n+0.5-t)\,dt=0$$ we have that the following function is continous $$f(x) = g([x]) + \frac{(x-[x])^1}{1!}g'([x])+\frac{(x-[x])^2}{2!}g''([x])+\cdots+\frac{(x-[x])^k}{k!}g^{(k)}([x])$$

(In your case, $g(x) = x^2$ and $k=1$)

Observe that the function you've defined is equal to a bunch of tangent lines (or linear approximations) "stuck together". If you look at the graph, you'll see that at integer values, $f(x)$ is tangent to $x^2$. You can verify that this characterisation of your function $f$ is true because the function is locally linear, and equal to $x^2$ at integer values.

This leads to the question of: For which functions $g$ is it true that the linear approximation centered at $n\in \mathbb Z$ is equal to the linear approximation centered at $n+1$, at $x=n+0.5$. We will determine precisely when this is the case by using Taylor's theorem with the integral form of the remainder term.

So at $a=n$ and $x=n+0.5$, we have that the first order Taylor approximation is: $$g(n+0.5) =g(n) + 0.5g'(n) + \int_n^{n+0.5}g''(t)(n+0.5-t)\,dt$$ And at $a=n+1$ and $x=n+0.5$, we have that: $$g(n+0.5) = g(n+1) - 0.5g'(n+1) + \int_{n+1}^{n+0.5} g''(t)(n+0.5-t)\,dt$$ The two linear approximations centered at $a=n$ and $a=n+1$ agree at $x=n+0.5$ precisely when their remainder terms are the same, i.e.: $$\int_n^{n+0.5}g''(t)(n+0.5-t)\,dt =\int_{n+1}^{n+0.5} g''(t)(n+0.5-t)\,dt$$ which we can rearrange to $$\int_n^{n+1} g''(t)(n+0.5-t)\,dt=0$$ In particular, this is true whenever $g''(t)$ is symmetric around $n+0.5$.

From $g$, we can define a function $f$ which is equal to the tangent lines of $g$ stuck together, as follows: $$f(x) = (x - [x]) g'([x]) + g([x]) $$


Particularly I don't like rounding functions and I think $f$ could be expressed in a more appropriate form for derivation and integration.

So an alternative way to express $f$ would be as a series of functions as follows:

Let $\mathcal{X}_A$ the indicator function of the set $A$, $I_n = [ n - \frac{1}{2}, n + \frac{1}{2} [$ and $f_n(x) = (2nx-n^2) \mathcal{X}_{I_n}(x)$. Thus we can write $f$ as $$ f(x) = \sum_{n \in \mathbb{Z}} f_n(x) $$

Which is immediately absolutely convergent, but more importantly, uniformly convergent on bounded sets. This means that derivation commutes with the summation on open bounded sets of $\mathbb{R} \setminus (\mathbb{Z} + \frac{1}{2})$ and that integration commutes with the summation on compact sets of $\mathbb{R}$.

However I don't have knowledge about where similar functions naturally arises in mathematics.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.