4
$\begingroup$

This MO question https://mathoverflow.net/questions/249541/formal-power-series-is-taylor-expansion-of-rational-function-iff-hankel-determin states that if $k$ is a field and $k[[T]]$ is the power series ring over $k$ then $u(T)\in k[[T]]$ is the power series of some rational function if and only if a condition on Hankel determinant holds. However, I could not think of any explicit power series which does not satisfy this condition.
I know that if $k=\mathbb{R},\mathbb{C}$, there are a lot of examples of power series which are not rational functions. But are there any explicit examples of power series which work for an arbitrary field $k$ ?

$\endgroup$
0

2 Answers 2

5
$\begingroup$

(1) A lacunary series like $\sum_{j=0}^{\infty} T^{j!}$ is not a rational function. You will get many Hankel determinants with $1$ on the anti-diagonal (upper right to lower left) and $0$ everywhere else.

(2) If $k$ is a countable field, than there are countably many rational functions, but uncountably many power series $\sum_{j=0}^\infty a_j T^j$ with coefficients in $\{0,1\}$.

$\endgroup$
1
  • $\begingroup$ I just saw the answer of Lubin to this question link. He said that $\sum x^{2^n}$ is not rational over $Q$ by using some specific feature of $Q$, but can I generalize this to general field because the coefficients are only 0 and 1 and they satisfy the criterion on Hankel determinants? $\endgroup$
    – jlidm
    May 26 at 15:51
2
$\begingroup$

It's easiest to use condition (2):

There is a finite sequence $q_0,\ldots, q_N$, not all zero, such that for all sufficiently large $m$,

$$a_m q_N + a_{m+1} q_{N-1} + \cdots + a_{m + N}q_0 = 0.$$

In other words, the coefficients must obey a linear recurrence (with a finite number of exceptions.)

If you choose coefficients "sufficiently random" you would not expect this to hold. For an explicit counterexample, you can take $f(T) = \sum_{n=0}^\infty a_n T^n$ where $a_n = 1$ if $n = 2^m$ for some $m$, otherwise $a_n = 0$. Then you will have arbitrarily long "gaps" in the sequence, with only coefficients equal to $0$. If the recurrence relation is obeyed, then any $N$ consecutive zeros $a_m, \ldots, a_{m+N-1}$ would mean that all coefficients after must be $0$, since you can solve for $a_{m+N}$ as a linear combination of $a_{m}, a_{m+1}, \ldots, a_{m+N-1}$, so that $a_{m+N} = 0$, and so on infinitum. Then we conclude $f(T)$ must be a polynomial, which we know it is not. So no such recurrence can hold.

$\endgroup$
3
  • $\begingroup$ Thanks a lot. I think I got what you mean. So in the same way, any power series $\sum x^{k_n}$, where $k_n$ tends to infinity is an example, right? $\endgroup$
    – jlidm
    May 26 at 16:04
  • $\begingroup$ @jlidm Well no, for example $k_n = n$ would not work. But if $k_{n+1} - k_n \rightarrow \infty$ (so we will have arbitrarily long gaps between nonzero coefficients) then it will work. $\endgroup$ May 26 at 16:08
  • $\begingroup$ Oh yes you're right, I made a mistake on that. $\endgroup$
    – jlidm
    May 26 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.