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It is well-known that a real number is a rational if, and only, if its decimal expansion is periodic (see Wikipedia). This result is a finiteness result, as it says the decimal expansion is, in a way, not really infinite, as it consists in the repeating infinitely a finite block.

A similar finiteness result exists for Laurent series $\alpha=\sum_{i=1}^{\infty}\alpha_i T^{-i} \in K((1/T))$, which are the analogues of reals modulo 1. Namely, associate to $\alpha$ the infinite matrix $(\alpha_{(i+j)-1})_{i,j\in\mathbb{N}^2}$. Then Hankel’s theorem states that $\alpha$ is rational if and only if the matrix is of finite rank.

We show here a quantitative version of this statement. Let $\displaystyle{M_{\alpha,n}=\begin{pmatrix} \alpha_{1} & \cdots & \alpha_{n} & \\ \alpha_{2} & \cdots & \alpha_{n+1} & \\ \vdots & \vdots & \vdots & \\ \alpha_{n} & \cdots & \alpha_{2n-1} & \end{pmatrix}.}$

The rank of this matrix is related to Diophantine properties of $\alpha$.
To see this, we introduce for $\beta=\sum_{i=-\infty}^{m}\beta_it^i\in\mathbb{F}_p((\frac{1}{t}))$ the fractional part $\{\beta\}= \sum_{i=-\infty}^{-1}\beta_it^i$ and the norm $\lVert{\beta}\rVert=\lvert\{\beta\}\rvert=q^k$, where $k\leq -1$ is the largest $i\leq -1$ such that $\beta_i\neq 0$. Then we have the following result.

If the rank of $M_{\alpha,n}$ is at most $r$, then there exist a decomposition $r=i+j$ and a monic polynomial $P$ of degree $i$ such that $\lVert P\alpha\rVert. The converse holds too.

The classical theory of Hankel matrices (see Chapter X, Paragraph 10 of Gantmacher’s book) tells us that if the rank of the matrix is $r$, then there exist integers $i$ and $h$ such that $i+h=r$ and the $i$ first rows are linearly independent, each of the following $n-r$ rows is a linear combination of them,
and the minor formed of the first $i$ and last $h$ rows and columns is nonzero. Let $L_1,\ldots,L_n$ be the rows of $M_{\alpha,n}$. Write $L_m=(L_m',a_m)$ where $a_m$ is the last coefficient of $L_m$ and $L'_m$ the row of the first $n-1$ coefficients. By the above, we have a relation $L_{i+1}=\sum_{m=1}^i c_mL_m$. In fact, we shall show that for any $j=0,\ldots,n-r-1$, we have $L_{i+1+j}=\sum_{m=1}^i c_m L_{m+j}.$

This equation holds for $j=0$. So we argue by induction and assume it holds for any $j'\leq j$ for some $j and prove it for $j+1$. To start with, the displayed equation implies that $\displaystyle{L'_{i+1+j+1}=\sum_{m=1}^i c_mL'_{m+j+1}.}$
Applying the induction hypothesis iteratively, we find coefficients $c_m^{(k)}$ for $m=1,\ldots,i$ and $k\leq j+1$ such that $\displaystyle{L_{i+1+k}=\sum_{m=1}^i c_m^{(k)}L_{m}}$

and $\displaystyle{ L'_{i+1+j+1}=\sum_{m=1}^i c_m^{(j+1)}L'_{m}.}$
These coefficients satisfy the initial condition $c_m^{(0)}=c_m$ and the recurrence relations $c_m^{(k+1)}=c_{m-1}^{(k)}+c_ic_m^{(k)}$ for $m>1$ and $c_1^{(k)}=c_h^kc_1$.

On the other hand, we know that there exist coefficients $d_1,\ldots,d_i$
such that $\displaystyle{L_{i+1+j+1}=\sum_{m=1}^i d_mL_{m}.}$
Comparing the last equation to the one involving $L'$, and using the linear independence of the first $i$ lines, we find that $d_m=c_m^{(j+1)}$.

Thus $a_{i+1+j+1}=\sum_{m=1}^i c_m^{(j+1)}a_m$. But $\sum_{m=1}^i c_ma_{m+j+1}=\sum_{m=1}^i c_m^{(j+1)}a_m$ by definition of the coefficients $c_m^{(k)}$. We infer that $a_{i+1+j+1}=\sum_{m=1}^i c_ma_{m+j+1}$,
which conclude the inductive argument.

To see the connexion with Diophantine properties of $\alpha$, notice that $L_i$ is the row of the first $n$ coefficients of $\{t^{i-1}\alpha\}$. Thus the validity of  the identity for $L'$ for all $j=0,\ldots,n-r-1$ implies that $\{t^i\alpha\}=\{\sum_{m=1}^ic_mt^{m-1}\alpha\}+\beta$
where $\beta\in\mathbb{T}$ satisfies $\lvert\beta\rvert.
That is, using the polynomial $P=t^i+\sum_{m=1}^ic_mt^{m-1}$, we find that $\lVert P\alpha\rVert which concludes the proof of the direct statement. The converse is straightforward.

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