楕円有理関数
L 1 ( ξ ) = ξ {\displaystyle L_{1}(\xi )=\xi \,}
L 2 ( ξ ) = 1 + t 1 − t = ( ξ + ξ 2 − 1 ) 2 {\displaystyle L_{2}(\xi )={\frac {1+t}{1-t}}=\left(\xi +{\sqrt {\xi ^{2}-1}}\right)^{2}}
L 3 ( ξ ) = ξ 3 ( 1 − x p 2 ξ 2 − x p 2 ) 2 {\displaystyle L_{3}(\xi )=\xi ^{3}\left({\frac {1-x_{p}^{2}}{\xi ^{2}-x_{p}^{2}}}\right)^{2}}
L 4 ( ξ ) = ( ξ + ( ξ 2 − 1 ) 1 / 4 ) 4 ( ξ + ξ 2 − 1 ) 2 {\displaystyle L_{4}(\xi )=\left({\sqrt {\xi }}+(\xi ^{2}-1)^{1/4}\right)^{4}\left(\xi +{\sqrt {\xi ^{2}-1}}\right)^{2}}
L 6 ( ξ ) = L 3 ( L 2 ( ξ ) ) {\displaystyle L_{6}(\xi )=L_{3}(L_{2}(\xi ))\,} etc.
n を次数とすると、零点は全部で n個あり、j を零点の番号とすると対応する零点は x n , j {\displaystyle x_{n,j}} と表せる。
x 1 , 1 = 0 {\displaystyle x_{1,1}=0\,}
x 2 , 1 = ξ 1 − t {\displaystyle x_{2,1}=\xi {\sqrt {1-t}}\,}