A function that is zero for every integer multiple of $k$

Question

I'm new to this kind of question so it may be a trivial one but I can't find a general solution:

I'm searching for a function that has a value of $0$ for every integer multiple of some fixed real $k$. In formula a function $f(x)$ such that:

$$f(x)=0\Leftarrow x=nk$$

with $n\in \Bbb Z$ and $k\in \Bbb R$

For example if $k=\pi$ then $f(x)=\sin x$ (because $\sin x=0$ iff $x=n\pi$)

And it is easy to show that if $k\in \Bbb Z$ like then $f(x)=\sin (x\pi)$

But I don't know what to do if $k$ is in general a real which is not a multiple of $\pi$. Any help ?

Answer 1

We could write the factorization of an infinite polynomial,

$$f(x) = x \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2k^2}\right),$$

in this particular case the function above is known to converge to $\sin(\pi x/k)/(\pi/k).$

Answer 2

Let r be a nonzero real number. f(x)=sin(πx/r). Now f(x)=0 iff πx/r=nπ iff x=nr

Answer 3

There are infinite such functions.

$f(x) = \frac{x}{k} - [\frac{x}{k}]$ where [w] = the greatest integer $\le$ w. ($f(x) = 0$ iff $x = nk$)

f(x) = 0 if $\frac{x}{k} \in Z$; f(x) = 37 if $\frac{x}{k} \notin Z$ is another.

$f(x) = sin(\frac{x\pi}{k})$ is probably the "nicest" and most "functionlike" though. (For instance it is continuous and differentiable which my two examples aren't.)