8
$\begingroup$

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.

Improve this question
$\endgroup$
4
  • 13
    $\begingroup$ 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
  • 1
    $\begingroup$ @David - I must be missing something. How would you control analyticity as uniform convergence doesn't imply that the limit is real analytic. $\endgroup$ Commented Jun 19, 2015 at 16:55
  • 1
    $\begingroup$ @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! $\endgroup$ Commented Jun 19, 2015 at 20:58

1 Answer 1

Reset to default
12
$\begingroup$

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.]

Improve this answer
$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.