Exist a function composed by simple continuous functions that $f:\Bbb R \rightarrow \Bbb N$?

Question

Im not mathematician so I dont know if this question is trivial or not. I was thinking if exist some function composed of simple continuous functions (trigonometric, exponential, powers, roots, logs, etc...) that work as Kronecker delta or the floor function that map from $\Bbb R$ to $\Bbb N$.

If the answer is a "NO" then I have a complementary subquestion: exist a limit of a function composed by simple continuous functions that make the same? What Im asking is about a function that in some limit (to 0, to infinity, to X, etc...) does the function of Kronecker delta or floor function.

I want to know if exist something similar to Kronecker delta or floor function that is defined in an analytic context, maybe with the use of limits or integration. Thanks in advance.

EDIT: sry, I didnt noticed that the function, obviously, cannot be analytic because is discrete. What I was trying to say is if exist a way to construct these type of discrete functions with functions composed of simple continuous functions not just by definition.

Sry very much for the wrong language and complication.

EDIT2: "challenge" expanded to the sign function.

Answer 1

EDIT2: I added a plot that should illustrate the convergence of $f_n^{0.3}(x)$.

EDIT: Obviously, $\sqrt[n]{x}$ does not converge towards a function that is $0$ everywhere except for $x=1$. It converges towards a function that is $1$, everywhere except for $x=0$. I've fixed this.

Referring to the part about a sequence of function converging towards the Kronecker delta (if I understood the question correctly).

Recall, that the Kronecker delta is defined as \begin{align} \delta_{ij} = \begin{cases} 1, \quad i=j \\ 0, \quad i\not = j\end{cases} \end{align} Now consider the sequence $f_n: [0,1]\rightarrow \mathbb{R}$ with \begin{align} f_n(x) = 1-\sqrt[n] x \rightarrow f(x) = \begin{cases} 1, \quad x = 0\\ 0, \quad x\not =0\end{cases} \end{align} or more similar to the Kronecker delta with $j\in [0,1]$ and again $f_n^j: [0,1]\rightarrow \mathbb{R}$ \begin{align} f_n^j(x) = 1-\sqrt[n] {(x-j)^2} \rightarrow f^j(x) = \begin{cases} 1, \quad x=j\\ 0, \quad x\not = j\,\wedge 0\leq x \leq 1\end{cases} \end{align} Since $f^j:[0,1]\rightarrow \mathbb{N}$ and $f^j(i)\equiv \delta_{ij}$ for $i,j\in [0,1]$, $f_n^j$ converges towards the Kronecker delta (on $[0,1]$).

The following plot shows $f_n^{0.3}(x)$ for different values of $n$ which converges towards $f_\infty^{0.3}(x) =f^{0.3}(x)= \delta_{x,0.3}$

And here is an extension onto $\mathbb{R}$.

Simply use $f^j_n(x) = 1-\sqrt[n]{h^j(x)}$ with $h^j(x) = 1-\frac{1}{1-(x-j)^2}$. Then $h_j(j) = 0$ and $h_j(x)\not = 0$ for $x\not = j$ and thus $f$ is defined on $\mathbb{R}$.

Here you can see a plot of $f_n^{2.5}(x)$ for $-4<x<4$.

Answer 2

You can create Fourier series, the limit of which are discontinuous (and have only natural numbers as their range). However, traditional Fourier series are periodic and I therefore suspect that covering the whole of $\mathbb N$ is rather difficult.

Since Bessel functions are orthogonal on the interval $(0, +\infty)$ it may be possible to create a series, the limit of which is $\lfloor x \rfloor$ for all $x \gt 0$.

The "sign" function, $s(x)$ can certainly be achieved by Fourier series, $F(x)$ on $(-l,l)$ for some $l \in \mathbb R$. To continue the function to infinity, we could defined the function in just three intervals $$s(x) = \begin{cases} -1 & -\infty \lt x \le -l \\ F(x) & -l \lt x \lt l \\ 1 & l \le x \lt \infty \end{cases}$$

Answer 3

I have a solution to represent floor function under continuous real functions. My idea come from periodical functions (sine, cosine, etc...) that because they are periodic they can represent naturals numbers under the expression of a continuous function (in a similar way zeta Riemann function express prime numbers by it zeros).

The problem that I had in this attempt was that periodical functions are supersymmetric for what I was searching, because they come, in some sense, from the division of a circle. In other words is the same to say that periodical functions cannot be monotonic functions.

I was trying to divide any real number into it entire part and it fractional part using the cosine with something like $\cos(2\pi x_e+2\pi x_r)$ where $x_e$ is the entire part and $x_r$ is the fractional part. The problem is that, by symmetry, the function cant differentiate between $x_r$ and $x_r+0.5$ (for $x_r \in (0,0.5)$). But what I dont see the first time is that this problem can be fixed just iterating the function at the moment where the value of the subtract have a value of zero.

In others words something like

$$f(x_{n+1})=x_n-\arccos(\cos (2\pi x_n))$$

Note that $\cos(\arccos a)=a$ (if $a \in [-1,1]$ ofc), but $\arccos(\cos a)\ne a$ for $a>\pi$ because trigonometric functions dont have real inverses.

When n reach a point then the iteration represent the floor function (the number of n to reach the equilibrium point depends on how close is $x_0$ of $\lceil x_0\rceil$). I dont know if is possible to get a explicit formula for this iteration but in any case

$$\lim_{n \to \infty}f(x_{n})=\lfloor x_0\rfloor$$

Maybe a better and more efficient way to do this using more periodic functions that just one.