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.

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    $\begingroup$ Can't you inductively construct some sequence $g_k$ such that $\sum g_k (\sin (2 \pi x))^k$ is uniformly convergent on a neighborhood of the real axis and has rational coefficients? $\endgroup$ Commented Jun 19, 2015 at 16:11
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    $\begingroup$ @David - I must be missing something. How would you control analyticity as uniform convergence doesn't imply that the limit is real analytic. $\endgroup$
    – Ken Fan
    Commented Jun 19, 2015 at 16:55
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    $\begingroup$ @KenFan Uniform convergence on a neighbourhood of the real axis (in the complex plane) does imply analyticity in that neighbourhood. $\endgroup$ Commented Jun 19, 2015 at 17:49
  • $\begingroup$ I see. I misread "neighborhood of the real axis." I think David's comment does resolve the question. Thanks, David! $\endgroup$
    – Ken Fan
    Commented Jun 19, 2015 at 20:58

1 Answer 1


Yes, such $f$ exist, and can even be taken to be entire, and with a preassigned finite initial segment $c_0,c_1,\ldots,c_d$ of the sequence of $x^k/k!$ coefficients.

Indeed if $f(x) = \sum_{k=0}^\infty a_k \sin^k (2\pi x)$ then $f$ is entire provided $a_k \rightarrow 0$ quickly enough, say $a_k \ll 1/k!$; and the map from $\{a_k\}$ to $\{c_k\}$ is linear and invertible, with $a_k$ a linear combination of $c_0,\ldots,c_k$ and $c_k$ a linear combination of $a_0,\ldots,a_k$ for each $k$. So, solve for $a_0,\ldots,a_d$, and then inductively choose small $a_{d+1}, a_{d+2}, a_{d+3}, \ldots$ to make each of $c_{d+1}, c_{d+2}, c_{d+3}, \ldots$ rational. (One way to avoid invoking the Axiom of Choice here is to fix some enumeration of ${\bf Q}$ and then at each step with $k>d$ use the least-indexed $c_k$ that makes $|a_k| < 1/k!$.)

[I see now that David Speyer gave much the same solution in a comment (using $g_k$ for what I called $a_k$), though without bothering to make $f$ entire or preassign the start of its Taylor expansion.]


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