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fix LaTeX for $\gg$
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edit approved May 12, 2022 at 8:35
Z. M
edit approved May 12, 2022 at 8:35
Z. M
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Let $$ u(T)=\sum_{n = 0}^\infty a_nT^n$$ be a formal power series over a field $K$. Then why does $u(T)$ lie in $K(T)$ (i.e. is the Taylor expansion of a rational function) if and only if there is an $N > 0$ such that the Hankel determinants $$\det(a_{i + j + M}){_{0 \le i, j \le N}} = \det \begin{pmatrix} a_M & a_{M + 1} & \ldots a_{M + N} \\ a_{M + 1} & a_{M + 2} & & \\ \vdots & & & \\ a_{M + N} & \ldots & & a_{M + 2N} \end{pmatrix}$$vanish for all $M >> 0$$M\gg0$?

Let $$ u(T)=\sum_{n = 0}^\infty a_nT^n$$ be a formal power series over a field $K$. Then why does $u(T)$ lie in $K(T)$ (i.e. is the Taylor expansion of a rational function) if and only if there is an $N > 0$ such that the Hankel determinants $$\det(a_{i + j + M}){_{0 \le i, j \le N}} = \det \begin{pmatrix} a_M & a_{M + 1} & \ldots a_{M + N} \\ a_{M + 1} & a_{M + 2} & & \\ \vdots & & & \\ a_{M + N} & \ldots & & a_{M + 2N} \end{pmatrix}$$vanish for all $M >> 0$?

Let $$ u(T)=\sum_{n = 0}^\infty a_nT^n$$ be a formal power series over a field $K$. Then why does $u(T)$ lie in $K(T)$ (i.e. is the Taylor expansion of a rational function) if and only if there is an $N > 0$ such that the Hankel determinants $$\det(a_{i + j + M}){_{0 \le i, j \le N}} = \det \begin{pmatrix} a_M & a_{M + 1} & \ldots a_{M + N} \\ a_{M + 1} & a_{M + 2} & & \\ \vdots & & & \\ a_{M + N} & \ldots & & a_{M + 2N} \end{pmatrix}$$vanish for all $M\gg0$?

Latex corrected
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edit approved Sep 10, 2016 at 20:59
BigM
edit approved Sep 10, 2016 at 20:59
BigM
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Let$$u(T) = \sum_{n = 0}^\infty a_nT^n$$be $$ u(T)=\sum_{n = 0}^\infty a_nT^n$$ be a formal power series over a field $K$. Then why does $u(T)$ lie in $K(T)$ (i.e. is the Taylor expansion of a rational function) if and only if there is an $N > 0$ such that the Hankel determinants$$\det(a_{i + j + M)}{_{0 \le i, j \le N}} = \det \begin{pmatrix} a_M & a_{M + 1} & \ldots a_{M + N} \\ a_{M + 1} & a_{M + 2} & & \\ \vdots & & & \\ a_{M + N} & \ldots & & a_{M + 2N} \end{pmatrix}$$ $$\det(a_{i + j + M}){_{0 \le i, j \le N}} = \det \begin{pmatrix} a_M & a_{M + 1} & \ldots a_{M + N} \\ a_{M + 1} & a_{M + 2} & & \\ \vdots & & & \\ a_{M + N} & \ldots & & a_{M + 2N} \end{pmatrix}$$vanish for all $M >> 0$?

Let$$u(T) = \sum_{n = 0}^\infty a_nT^n$$be a formal power series over a field $K$. Then why does $u(T)$ lie in $K(T)$ (i.e. is the Taylor expansion of a rational function) if and only if there is an $N > 0$ such that the Hankel determinants$$\det(a_{i + j + M)}{_{0 \le i, j \le N}} = \det \begin{pmatrix} a_M & a_{M + 1} & \ldots a_{M + N} \\ a_{M + 1} & a_{M + 2} & & \\ \vdots & & & \\ a_{M + N} & \ldots & & a_{M + 2N} \end{pmatrix}$$vanish for all $M >> 0$?

Let $$ u(T)=\sum_{n = 0}^\infty a_nT^n$$ be a formal power series over a field $K$. Then why does $u(T)$ lie in $K(T)$ (i.e. is the Taylor expansion of a rational function) if and only if there is an $N > 0$ such that the Hankel determinants $$\det(a_{i + j + M}){_{0 \le i, j \le N}} = \det \begin{pmatrix} a_M & a_{M + 1} & \ldots a_{M + N} \\ a_{M + 1} & a_{M + 2} & & \\ \vdots & & & \\ a_{M + N} & \ldots & & a_{M + 2N} \end{pmatrix}$$vanish for all $M >> 0$?

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user97565

Formal power series is Taylor expansion of rational function iff Hankel determinants vanish?

Let$$u(T) = \sum_{n = 0}^\infty a_nT^n$$be a formal power series over a field $K$. Then why does $u(T)$ lie in $K(T)$ (i.e. is the Taylor expansion of a rational function) if and only if there is an $N > 0$ such that the Hankel determinants$$\det(a_{i + j + M)}{_{0 \le i, j \le N}} = \det \begin{pmatrix} a_M & a_{M + 1} & \ldots a_{M + N} \\ a_{M + 1} & a_{M + 2} & & \\ \vdots & & & \\ a_{M + N} & \ldots & & a_{M + 2N} \end{pmatrix}$$vanish for all $M >> 0$?