Does there exist a nonconstant, real analytic function $f \colon \mathbb{R} \to \mathbb{R}$ such that $f$ is periodic with period 1 and whose Maclaurin coefficients are all rational?

(The function $\sin x$ satisfies all the conditions except that its period isn't 1.)

I can't recall exactly what I was thinking about when this question occurred to me and my primary motivation now is curiosity. Note that if the answer is no, a proof that doesn't use the irrationality of $\pi$ would be an alternative way to prove irrationality of $\pi$ because of the function $\sin(2\pi x)$.

Also, WLOG, we may assume $f(0)=0$ and if we define the vector $c = (c_k)$ where $f(x) = \sum\limits_{k = 1}^{\infty} \frac{c_k}{k!}x^k$ and let $v = (1, 1/2!, 1/3!, 1/4!, \dots)$, then by comparing the derivatives of $f$ at 0 and 1, we get that for all nonnegative integers $n$, $$v \cdot C^n c = 0,$$ where $C$ shifts the components of a vector one space to the left.

neighbourhoodof the real axis (in the complex plane) does imply analyticity in that neighbourhood. $\endgroup$