3
$\begingroup$

A lemma by Kronecker states that the series $f(z):= \sum_{i=0}^{\infty} c_i z^i$ represents a rational function if and only if for every $m \gg 0$ the determinant of the following matrix is equal to zero:

\begin{bmatrix} c_0 & c_1 & \cdots & c_m \\ c_1 & c_2 & \cdots & c_{m+1}\\ \vdots & \vdots & & \vdots \\ c_m & c_{m+1} & \cdots & c_{2m} \end{bmatrix}

Question: What are alternative ways to detect this property, that is $f(z)$ being a rational function?

Added later: I am looking for algebraic/analytic characterisations, as I don't have the precise values of $c_i$ at hand. However, I know that $c_i$ are integers if that is of any help.

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ The above property is the same as the coefficients $c_i$ satisfies a linear recursion. Detecting this is done by the above determinant trick though... $\endgroup$
    – Per Alexandersson
    Commented Aug 27, 2018 at 13:30

1 Answer 1

Reset to default
5
$\begingroup$

Probably not what OP is asking for, but a method I find very successful when the coefficients are integers:

Take $c_0,c_1,c_2,\dotsc$ and plug into OEIS. If it match a rational function with a description that match the way you produced $f(z)$ you are done.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Not a very mathematical but certainly a very feasible and indeed successful method. If this had been a facebook post, I'd have added a large smiley! $\endgroup$
    – Wolfgang
    Commented Aug 27, 2018 at 13:44
  • 1
    $\begingroup$ You might also try the gfun package in Maple. $\endgroup$
    – Robert Israel
    Commented Aug 27, 2018 at 15:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .