From Wikipedia, the free encyclopedia
In algebra , the continuant is a multivariate polynomial representing the determinant of a tridiagonal matrix and having applications in generalized continued fractions .
Definition [ edit ] The n -th continuant
K
n
(
x
1
,
x
2
,
…
,
x
n
)
{\displaystyle K_{n}(x_{1},\;x_{2},\;\ldots ,\;x_{n})}
K
0
=
1
;
{\displaystyle K_{0}=1;\,}
K
1
(
x
1
)
=
x
1
;
{\displaystyle K_{1}(x_{1})=x_{1};\,}
K
n
(
x
1
,
x
2
,
…
,
x
n
)
=
x
n
K
n
−
1
(
x
1
,
x
2
,
…
,
x
n
−
1
)
+
K
n
−
2
(
x
1
,
x
2
,
…
,
x
n
−
2
)
.
{\displaystyle 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 [ edit ] The continuant
K
n
(
x
1
,
x
2
,
…
,
x
n
)
{\displaystyle K_{n}(x_{1},\;x_{2},\;\ldots ,\;x_{n})}
x 1 ,...,x n 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
.
{\displaystyle 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
,
…
,
x
n
)
=
K
n
(
x
n
,
…
,
x
1
)
.
{\displaystyle 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
,
…
,
x
n
)
=
det
(
x
1
1
0
⋯
0
−
1
x
2
1
⋱
⋮
0
−
1
⋱
⋱
0
⋮
⋱
⋱
⋱
1
0
⋯
0
−
1
x
n
)
.
{\displaystyle 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
,
…
,
1
)
=
F
n
+
1
{\displaystyle K_{n}(1,\;\ldots ,\;1)=F_{n+1}}
n +1)-st Fibonacci number .
K
n
(
x
1
,
…
,
x
n
)
K
n
−
1
(
x
2
,
…
,
x
n
)
=
x
1
+
K
n
−
2
(
x
3
,
…
,
x
n
)
K
n
−
1
(
x
2
,
…
,
x
n
)
.
{\displaystyle {\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:
K
n
(
x
1
,
…
,
x
n
)
K
n
−
1
(
x
2
,
…
,
x
n
)
=
[
x
1
;
x
2
,
…
,
x
n
]
=
x
1
+
1
x
2
+
1
x
3
+
…
.
{\displaystyle {\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:
(
K
n
(
x
1
,
…
,
x
n
)
K
n
−
1
(
x
1
,
…
,
x
n
−
1
)
K
n
−
1
(
x
2
,
…
,
x
n
)
K
n
−
2
(
x
2
,
…
,
x
n
−
1
)
)
=
(
x
1
1
1
0
)
×
…
×
(
x
n
1
1
0
)
{\displaystyle {\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
,
…
,
x
n
)
⋅
K
n
−
2
(
x
2
,
…
,
x
n
−
1
)
−
K
n
−
1
(
x
1
,
…
,
x
n
−
1
)
⋅
K
n
−
1
(
x
2
,
…
,
x
n
)
=
(
−
1
)
n
.
{\displaystyle 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
,
…
,
x
n
)
⋅
K
n
+
2
(
x
1
,
…
,
x
n
+
2
)
−
K
n
(
x
1
,
…
,
x
n
)
⋅
K
n
+
1
(
x
2
,
…
,
x
n
+
2
)
=
(
−
1
)
n
+
1
x
n
+
2
.
{\displaystyle 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 [ edit ] 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 −1c 1 ,...,c n −1recurrence relation becomes
K
0
=
1
;
{\displaystyle K_{0}=1;\,}
K
1
=
a
1
;
{\displaystyle K_{1}=a_{1};\,}
K
n
=
a
n
K
n
−
1
−
b
n
−
1
c
n
−
1
K
n
−
2
.
{\displaystyle K_{n}=a_{n}K_{n-1}-b_{n-1}c_{n-1}K_{n-2}.\,}
Since b r c r K only as a product b r c r b r
The generalized continuant is precisely the determinant of the tridiagonal matrix
(
a
1
b
1
0
…
0
0
c
1
a
2
b
2
…
0
0
0
c
2
a
3
…
0
0
⋮
⋮
⋮
⋱
⋮
⋮
0
0
0
…
a
n
−
1
b
n
−
1
0
0
0
…
c
n
−
1
a
n
)
.
{\displaystyle {\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.
References [ edit ] 
![{\displaystyle {\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 }}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66a69721671ddcc8d06eb6f12b42d1a281e6476f)
