# Explicit examples of formal power series which is not rational functions?

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 $$k$$ ?

(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 $$\{0,1\}$$.

• 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? May 26, 2022 at 15:51

A good class of examples is algebraic power series that are not rational. For example, consider the Catalan number generating function $$c(x) = \frac{1-\sqrt{1-4x}}{2x}= \sum_{n=0}^\infty \frac{1}{n+1}\binom{2n}{n}x^n$$ which is defined over every field. It satisfies $$c(x) = 1+xc(x)^2$$, and it's not hard to show from this that $$c(x)$$ is not rational.

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.

• 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? May 26, 2022 at 16:04
• @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. May 26, 2022 at 16:08
• Oh yes you're right, I made a mistake on that. May 26, 2022 at 16:09