Function with roots at $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ...$?

Question

I am searching for a function $~f : \mathbb{R}\rightarrow \mathbb{R}$ with $$ f\left(\frac{1}{k}\right) = 0 ~\forall k\in\mathbb{N}. $$ This function has to be smooth!

First approach: I have tried $f(x) = \sin(\pi/x)$. This works, however, this function is not smooth (as far as I am concerned?).

Context: The overall aim is to find two functions $f,g$ such that $f\neq g$ but $f(1/k) = g(1/k) ~\forall k\in\mathbb{N}$.

EDIT: Smooth means $f\in C^\infty$. $~f$ has to be at least defined on $(0,1]$.

Answer 1

Let $g(x)=\exp(-\frac1{x-1})$ for $x>1$, and $g(x)=0$ for $x\le 1$. This automatically satisfies $g(1/k)=0$, since $1/k\le 1$. For a proof that $g$ is smooth, see

https://en.wikipedia.org/wiki/Non-analytic_smooth_function#The_function_is_smooth

Note that my function $g$ is a shift of Wikipedia's function $f$, i.e. $f(x)=g(x+1)$.

Answer 2

What about $x^4\sin(\pi/x)$? This should work.

Answer 3

Using the function $e^{-1/x^2}$ for $x \ne 0$, and $0$ if $x = 0$, we get a function which is $C^{\infty}$ and supported on $[0, \infty)$. Multiplying this by a shift and flip of the function, we can easily construct a function which is $C^{\infty}$ and supported only on $[0, 1]$, giving a bump function.

Summing translates and dilates of thus localized to the intervals $[1/2, 1]$, $[1/3, 1/2]$, $[1/4, 1/3]$ and so on gives an example of a function whose zeros in $[0, 1]$ are only at the points $1, 1/2, 1/3, ...$.