ラウス・フルビッツの安定判別法
D 1 = a 1 , {\displaystyle D_{1}=a_{1},~} D 2 = | a 1 a 3 a 0 a 2 | , {\displaystyle D_{2}={\begin{vmatrix}a_{1}&a_{3}\\a_{0}&a_{2}\end{vmatrix}},~} D 3 = | a 1 a 3 a 5 a 0 a 2 a 4 0 a 1 a 3 | , {\displaystyle D_{3}={\begin{vmatrix}a_{1}&a_{3}&a_{5}\\a_{0}&a_{2}&a_{4}\\0&a_{1}&a_{3}\end{vmatrix}},~} ⋮ {\displaystyle \vdots } D n − 1 = | a 1 a 3 a 5 ⋯ a 2 n − 3 a 0 a 2 a 4 ⋯ a 2 n − 4 0 a 1 a 3 ⋯ a 2 n − 5 0 a 0 a 2 ⋯ a 2 n − 6 0 0 a 1 ⋯ a 2 n − 7 ⋮ ⋮ ⋱ ⋱ ⋮ 0 0 0 ⋯ a n − 1 | {\displaystyle D_{n-1}={\begin{vmatrix}a_{1}&a_{3}&a_{5}&\cdots &a_{2n-3}\\a_{0}&a_{2}&a_{4}&\cdots &a_{2n-4}\\0&a_{1}&a_{3}&\cdots &a_{2n-5}\\0&a_{0}&a_{2}&\cdots &a_{2n-6}\\0&0&a_{1}&\cdots &a_{2n-7}\\\vdots &\vdots &\ddots &\ddots &\vdots \\0&0&0&\cdots &a_{n-1}\\\end{vmatrix}}} このとき、特性方程式の次数の関係で存在しない係数は0として扱う。