Functions Mapping Integers to Zero?

Question

I am looking for functions such that:

$z∈$ Z ⇔ $f(z)=0$

That is to say, functions that map from Z to the zero set.

One example is $f(z)=\sin(πz)$.

EDIT: To narrow the possible group of functions, the following conditions must be met:

1. $f(z)$ must be a continuous, differentiable function (it may not be a piecewise-defined function)
2. The function must be expressible in terms of other elementary, special or algebraic functions.

Specifically, I am looking for functions that are easy to analyze and manipulate. The function should not be trivial. This was a question posed to me by my teacher; sine was the example function. I am looking for further examples.

Answer 1

With the added restrictions, the problem is no longer too wide open.

You already have $f(x) = \sin(\pi x)$, of course. In fact, anything of the form $f(x) = \sin(n\pi x),$ where $n$ is an integer, will do the trick.

You can combine several of these functions together to obtain different repeating patterns. Try $f(x) = \sin(\pi x) + \sin(3\pi x)$, for example.

These functions so far are all odd functions, meaning you can rotate them $180$ degrees around the origin and end up with the same function. In a sense, these functions have "equal" parts above and below the $x$-axis. But you can have a function that never goes below the $x$-axis at all, such as $f(x) = \sin^2(\pi x)$. You can also have a function that has larger excursions above the axis than below, such as $f(x) = \frac12 + \sin\left(\pi \left(2x - \frac16\right)\right)$.

An example of a non-periodic function would be $f(x) = \sin(\pi x^2)$. You can also try $f(x) = \sin\left(\frac12\pi (x^2 + x)\right)$.

As suggested in comments, once you have a function that satisfies the requirements, you can multiply it by a non-trivial, continuous, differentiable function to obtain a new function that satisfies the requirements. For example, $x \sin(\pi x)$ or $x^2 \sin(\pi x)$ both have larger and larger oscillations as you go farther from $x= 0$. But $\frac{1}{1+x^2}\sin(\pi x)$ has smaller and smaller oscillations as you go farther from $x = 0$, and $e^x \sin(\pi x)$ shrinks in one direction and grows in the other.

Functions such as $\sin^2(\pi x)$, which not only are zero at each integer value of $x$ but also have zero derivative at each such point, can be multiplied by functions that are discontinuous (but bounded) at the integers and still produce a continuous, differentiable function. But you might find that introducing the discontinuous factor makes your product function a little harder to work with.

It's a bit challenging to think of any such function that is not somehow based on sinusoidal functions (sine or cosine), because we want the building blocks of the functions to be selected from a fairly limited set, and there are not a lot of functions in that set that keep returning to zero. One example is $$ f(x) = 16 \left(x - \lfloor x \rfloor\right)^2 \left(x - \lfloor x \rfloor - 1\right)^2, $$

which looks almost as if it might be sinusoidal, but is not, or $$ f(x) = 16 \left(x - \lfloor x \rfloor\right)^4 \left(x - \lfloor x \rfloor - 1\right)^2, $$ which looks even stranger. But while these functions satisfy the criteria (including continuity and differentiability), the discontinuous building blocks might make them harder to work with than you would like.

If you want to have some fun, start putting these formulas into a decent graphing calculator or one of the on-line graphing calculators such as desmos.com. Mix and match, multiply functions by each other, and so forth, and see what you can do.

Answer 2

As an obvious example, $$f_\gamma(z)=\inf_{k\in\Bbb Z}|z-k|^\gamma$$ for $\gamma>0$.

This question is too broad.

Answer 3

Here's three:

$$ f(z) = z - [z]\\ f(z) = z - \lfloor z\rfloor\\ f(z) = z - \lceil z \rceil $$