# Continuant (mathematics)

In algebra, the continuant is a multivariate polynomial representing the determinant of a tridiagonal matrix and having applications in continued fractions.

## Definition

The n-th continuant $K_{n}(x_{1},\;x_{2},\;\ldots ,\;x_{n})$ is defined recursively by
    $K_{0}=1;\,$
    $K_{1}(x_{1})=x_{1};\,$
    $K_{n}(x_{1},\;x_{2},\;\ldots ,\;x_{n})=x_{n}K_{n-1}(x_{1},\;x_{2},\;\ldots ,\;x_{n-1})+K_{n-2}(x_{1},\;x_{2},\;\ldots ,\;x_{n-2}).\,$

## Properties

* The continuant $K_{n}(x_{1},\;x_{2},\;\ldots ,\;x_{n})$ can be computed by taking the sum of all possible products of x$_1$,...,x$_n$, in which any number of disjoint pairs of consecutive terms are deleted (Euler's rule). For example,
    $K_{5}(x_{1},\;x_{2},\;x_{3},\;x_{4},\;x_{5})=x_{1}x_{2}x_{3}x_{4}x_{5}\;+\;x_{3}x_{4}x_{5}\;+\;x_{1}x_{4}x_{5}\;+\;x_{1}x_{2}x_{5}\;+\;x_{1}x_{2}x_{3}\;+\;x_{1}\;+\;x_{3}\;+\;x_{5}.$
    It follows that continuants are invariant with respect to reversing the order of indeterminates: $K_{n}(x_{1},\;\ldots ,\;x_{n})=K_{n}(x_{n},\;\ldots ,\;x_{1}).$
* The continuant can be computed as the determinant of a tridiagonal matrix:
    $K_{n}(x_{1},\;x_{2},\;\ldots ,\;x_{n})=\det {\begin{pmatrix}x_{1}&1&0&\cdots &0\\-1&x_{2}&1&\ddots &\vdots \\0&-1&\ddots &\ddots &0\\\vdots &\ddots &\ddots &\ddots &1\\0&\cdots &0&-1&x_{n}\end{pmatrix}}.$
* $K_{n}(1,\;\ldots ,\;1)=F_{n+1}$, the (n+1)-st Fibonacci number.
* ${\frac {K_{n}(x_{1},\;\ldots ,\;x_{n})}{K_{n-1}(x_{2},\;\ldots ,\;x_{n})}}=x_{1}+{\frac {K_{n-2}(x_{3},\;\ldots ,\;x_{n})}{K_{n-1}(x_{2},\;\ldots ,\;x_{n})}}.$
* Ratios of continuants represent (convergents to) continued fractions as follows:
    ${\frac {K_{n}(x_{1},\;\ldots ,x_{n})}{K_{n-1}(x_{2},\;\ldots ,\;x_{n})}}=[x_{1};\;x_{2},\;\ldots ,\;x_{n}]=x_{1}+{\frac {1}{\displaystyle {x_{2}+{\frac {1}{x_{3}+\ldots }}}}}.$
* The following matrix identity holds:
    ${\begin{pmatrix}K_{n}(x_{1},\;\ldots ,\;x_{n})&K_{n-1}(x_{1},\;\ldots ,\;x_{n-1})\\K_{n-1}(x_{2},\;\ldots ,\;x_{n})&K_{n-2}(x_{2},\;\ldots ,\;x_{n-1})\end{pmatrix}}={\begin{pmatrix}x_{1}&1\\1&0\end{pmatrix}}\times \ldots \times {\begin{pmatrix}x_{n}&1\\1&0\end{pmatrix}}$.
** For determinants, it implies that
    $K_{n}(x_{1},\;\ldots ,\;x_{n})\cdot K_{n-2}(x_{2},\;\ldots ,\;x_{n-1})-K_{n-1}(x_{1},\;\ldots ,\;x_{n-1})\cdot K_{n-1}(x_{2},\;\ldots ,\;x_{n})=(-1)^{n}.$
** and also
    $K_{n-1}(x_{2},\;\ldots ,\;x_{n})\cdot K_{n+2}(x_{1},\;\ldots ,\;x_{n+2})-K_{n}(x_{1},\;\ldots ,\;x_{n})\cdot K_{n+1}(x_{2},\;\ldots ,\;x_{n+2})=(-1)^{n+1}x_{n+2}.$

## Generalizations

A generalized definition takes the continuant with respect to three sequences a, b and c, so that K(n) is a polynomial of a$_1$,...,a$_n$, b$_1$,...,b$_{n−1}$ and c$_1$,...,c$_{n−1}$. In this case the recurrence relation becomes
    $K_{0}=1;\,$
    $K_{1}=a_{1};\,$
    $K_{n}=a_{n}K_{n-1}-b_{n-1}c_{n-1}K_{n-2}.\,$
Since b$_r$ and c$_r$ enter into K only as a product b$_r$c$_r$ there is no loss of generality in assuming that the b$_r$ are all equal to 1.
The generalized continuant is precisely the determinant of the tridiagonal matrix
    ${\begin{pmatrix}a_{1}&b_{1}&0&\ldots &0&0\\c_{1}&a_{2}&b_{2}&\ldots &0&0\\0&c_{2}&a_{3}&\ldots &0&0\\\vdots &\vdots &\vdots &\ddots &\vdots &\vdots \\0&0&0&\ldots &a_{n-1}&b_{n-1}\\0&0&0&\ldots &c_{n-1}&a_{n}\end{pmatrix}}.$
In Muir's book the generalized continuant is simply called continuant.