连续q勒让德多项式

连续勒让德多项式是一个以基本超几何函数定义的正交多项式 P n ( x | q ) = 4 ϕ 3 ( q − n q n + 1 q 1 / 4 e i θ a 1 / 4 e − i θ q − q 1 / 2 − q ; q , q ) {\displaystyle P_{n}(x|q)=\;_{4}\phi _{3}\left({\begin{matrix}q^{-n}&q^{n+1}&q^{1/4}e^{i\theta }&a^{1/4}e^{-i\theta }&\\q&-q^{1/2}&-q\end{matrix}};q,q\right)} 令连续q勒让德多项式 q->1 得勒让德多项式 lim q → 1 P n ( x | q ) = P n ( x ) {\displaystyle \lim _{q\to 1}P_{n}(x|q)=P_{n}(x)} 验证5阶连续q勒让德多项式→勒让德多项式 lim q → 1 P 5 ( x | q ) = P 5 ( x ) {\displaystyle \lim _{q\to 1}P_{5}(x|q)=P_{5}(x)} 由定义， P 5 ( x | q ) = 1 + ( 1 − q − 5 ) ( 1 − q − 4 ) ( 1 − q − 3 ) ( 1 − q 6 ) ( 1 − q 7 ) ( 1 − q 8 ) ( 1 − q 4 ( x + i 1 − x 2 ) ) ( 1 − q 5 / 4 ( x + i 1 − x 2 ) ) ( 1 − q 9 / 4 ( x + i 1 − x 2 ) ) ( 1 − q 4 x + i 1 − x 2 ) ( 1 − q 5 / 4 x + i 1 − x 2 ) ( 1 − q 9 / 4 x + i 1 − x 2 ) q 3 ( 1 − q ) − 2 ( 1 − q 2 ) − 2 ( 1 − q 3 ) − 2 ( 1 + q ) − 1 ( 1 + q 3 / 2 ) − 1 ( 1 + q 5 / 2 ) − 1 ( 1 + q ) − 1 ( 1 + q 2 ) − 1 ( 1 + q 3 ) − 1 + ( 1 − q − 5 ) ( 1 − q − 4 ) ( 1 − q − 3 ) ( 1 − q − 2 ) ( 1 − q 6 ) ( 1 − q 7 ) ( 1 − q 8 ) ( 1 − q 9 ) ( 1 − q 4 ( x + i 1 − x 2 ) ) ( 1 − q 5 / 4 ( x + i 1 − x 2 ) ) ( 1 − q 9 / 4 ( x + i 1 − x 2 ) ) ( 1 − q 13 4 ( x + i 1 − x 2 ) ) ( 1 − q 4 x + i 1 − x 2 ) ( 1 − q 5 / 4 x + i 1 − x 2 ) ( 1 − q 9 / 4 x + i 1 − x 2 ) ( 1 − q 13 4 ( x + i 1 − x 2 ) − 1 ) q 4 ( 1 − q ) − 2 ( 1 − q 2 ) − 2 ( 1 − q 3 ) − 2 ( 1 − q 4 ) − 2 ( 1 + q ) − 1 ( 1 + q 3 / 2 ) − 1 ( 1 + q 5 / 2 ) − 1 ( 1 + q 7 / 2 ) − 1 ( 1 + q ) − 1 ( 1 + q 2 ) − 1 ( 1 + q 3 ) − 1 ( 1 + q 4 ) − 1 + ( 1 − q − 5 ) ( 1 − q − 4 ) ( 1 − q − 3 ) ( 1 − q − 2 ) ( 1 − q − 1 ) ( 1 − q 6 ) ( 1 − q 7 ) ( 1 − q 8 ) ( 1 − q 9 ) ( 1 − q 10 ) ( 1 − q 4 ( x + i 1 − x 2 ) ) ( 1 − q 5 / 4 ( x + i 1 − x 2 ) ) ( 1 − q 9 / 4 ( x + i 1 − x 2 ) ) ( 1 − q 13 4 ( x + i 1 − x 2 ) ) ( 1 − q 17 4 ( x + i 1 − x 2 ) ) ( 1 − q 4 x + i 1 − x 2 ) ( 1 − q 5 / 4 x + i 1 − x 2 ) ( 1 − q 9 / 4 x + i 1 − x 2 ) ( 1 − q 13 4 ( x + i 1 − x 2 ) − 1 ) ( 1 − q 17 4 ( x + i 1 − x 2 ) − 1 ) q 5 ( 1 − q ) − 2 ( 1 − q 2 ) − 2 ( 1 − q 3 ) − 2 ( 1 − q 4 ) − 2 ( 1 − q 5 ) − 2 ( 1 + q ) − 1 ( 1 + q 3 / 2 ) − 1 ( 1 + q 5 / 2 ) − 1 ( 1 + q 7 / 2 ) − 1 ( 1 + q 9 / 2 ) − 1 ( 1 + q ) − 1 ( 1 + q 2 ) − 1 ( 1 + q 3 ) − 1 ( 1 + q 4 ) − 1 ( 1 + q 5 ) − 1 − q 5 / 4 ( 1 − q ) 2 ( 1 + q ) ( 1 + q ) ( x + i 1 − x 2 ) − q 5 / 4 ( x + i 1 − x 2 ) ( 1 − q ) 2 ( 1 + q ) ( 1 + q ) + q 29 4 ( 1 − q ) − 2 ( 1 + q ) − 1 ( 1 + q ) − 1 ( x + i 1 − x 2 ) − 1 + q 29 4 ( x + i 1 − x 2 ) ( 1 − q ) − 2 ( 1 + q ) − 1 ( 1 + q ) − 1 + q − 15 4 ( 1 − q ) − 2 ( 1 + q ) − 1 ( 1 + q ) − 1 ( x + i 1 − x 2 ) − 1 + ( x + i 1 − x 2 ) q − 15 4 ( 1 − q ) − 2 ( 1 + q ) − 1 ( 1 + q ) − 1 − q 9 / 4 ( 1 − q ) 2 ( 1 + q ) ( 1 + q ) ( x + i 1 − x 2 ) − q 9 / 4 ( x + i 1 − x 2 ) ( 1 − q ) 2 ( 1 + q ) ( 1 + q ) + ( 1 − q − 5 ) ( 1 − q − 4 ) ( 1 − q 6 ) ( 1 − q 7 ) ( 1 − q 4 ( x + i 1 − x 2 ) ) ( 1 − q 5 / 4 ( x + i 1 − x 2 ) ) ( 1 − q 4 x + i 1 − x 2 ) ( 1 − q 5 / 4 x + i 1 − x 2 ) q 2 ( 1 − q ) − 2 ( 1 − q 2 ) − 2 ( 1 + q ) − 1 ( 1 + q 3 / 2 ) − 1 ( 1 + q ) − 1 ( 1 + q 2 ) − 1 + q ( 1 − q ) 2 ( 1 + q ) ( 1 + q ) + q 3 / 2 ( 1 − q ) 2 ( 1 + q ) ( 1 + q ) − q 7 ( 1 − q ) 2 ( 1 + q ) ( 1 + q ) − q 15 / 2 ( 1 − q ) 2 ( 1 + q ) ( 1 + q ) − 1 q 4 ( 1 − q ) 2 ( 1 + q ) ( 1 + q ) − 1 q 7 / 2 ( 1 − q ) 2 ( 1 + q ) ( 1 + q ) + q 2 ( 1 − q ) 2 ( 1 + q ) ( 1 + q ) + q 5 / 2 ( 1 − q ) 2 ( 1 + q ) ( 1 + q ) ⋯ {\displaystyle P_{5}(x|q)=1+\left(1-{q}^{-5}\right)\left(1-{q}^{-4}\right)\left(1-{q}^{-3}\right)\left(1-{q}^{6}\right)\left(1-{q}^{7}\right)\left(1-{q}^{8}\right)\left(1-{\sqrt[{4}]{q}}\left(x+i{\sqrt {1-{x}^{2}}}\right)\right)\left(1-{q}^{5/4}\left(x+i{\sqrt {1-{x}^{2}}}\right)\right)\left(1-{q}^{9/4}\left(x+i{\sqrt {1-{x}^{2}}}\right)\right)\left(1-{\frac {\sqrt[{4}]{q}}{x+i{\sqrt {1-{x}^{2}}}}}\right)\left(1-{\frac {{q}^{5/4}}{x+i{\sqrt {1-{x}^{2}}}}}\right)\left(1-{\frac {{q}^{9/4}}{x+i{\sqrt {1-{x}^{2}}}}}\right){q}^{3}\left(1-q\right)^{-2}\left(1-{q}^{2}\right)^{-2}\left(1-{q}^{3}\right)^{-2}\left(1+{\sqrt {q}}\right)^{-1}\left(1+{q}^{3/2}\right)^{-1}\left(1+{q}^{5/2}\right)^{-1}\left(1+q\right)^{-1}\left(1+{q}^{2}\right)^{-1}\left(1+{q}^{3}\right)^{-1}+\left(1-{q}^{-5}\right)\left(1-{q}^{-4}\right)\left(1-{q}^{-3}\right)\left(1-{q}^{-2}\right)\left(1-{q}^{6}\right)\left(1-{q}^{7}\right)\left(1-{q}^{8}\right)\left(1-{q}^{9}\right)\left(1-{\sqrt[{4}]{q}}\left(x+i{\sqrt {1-{x}^{2}}}\right)\right)\left(1-{q}^{5/4}\left(x+i{\sqrt {1-{x}^{2}}}\right)\right)\left(1-{q}^{9/4}\left(x+i{\sqrt {1-{x}^{2}}}\right)\right)\left(1-{q}^{\frac {13}{4}}\left(x+i{\sqrt {1-{x}^{2}}}\right)\right)\left(1-{\frac {\sqrt[{4}]{q}}{x+i{\sqrt {1-{x}^{2}}}}}\right)\left(1-{\frac {{q}^{5/4}}{x+i{\sqrt {1-{x}^{2}}}}}\right)\left(1-{\frac {{q}^{9/4}}{x+i{\sqrt {1-{x}^{2}}}}}\right)\left(1-{q}^{\frac {13}{4}}\left(x+i{\sqrt {1-{x}^{2}}}\right)^{-1}\right){q}^{4}\left(1-q\right)^{-2}\left(1-{q}^{2}\right)^{-2}\left(1-{q}^{3}\right)^{-2}\left(1-{q}^{4}\right)^{-2}\left(1+{\sqrt {q}}\right)^{-1}\left(1+{q}^{3/2}\right)^{-1}\left(1+{q}^{5/2}\right)^{-1}\left(1+{q}^{7/2}\right)^{-1}\left(1+q\right)^{-1}\left(1+{q}^{2}\right)^{-1}\left(1+{q}^{3}\right)^{-1}\left(1+{q}^{4}\right)^{-1}+\left(1-{q}^{-5}\right)\left(1-{q}^{-4}\right)\left(1-{q}^{-3}\right)\left(1-{q}^{-2}\right)\left(1-{q}^{-1}\right)\left(1-{q}^{6}\right)\left(1-{q}^{7}\right)\left(1-{q}^{8}\right)\left(1-{q}^{9}\right)\left(1-{q}^{10}\right)\left(1-{\sqrt[{4}]{q}}\left(x+i{\sqrt {1-{x}^{2}}}\right)\right)\left(1-{q}^{5/4}\left(x+i{\sqrt {1-{x}^{2}}}\right)\right)\left(1-{q}^{9/4}\left(x+i{\sqrt {1-{x}^{2}}}\right)\right)\left(1-{q}^{\frac {13}{4}}\left(x+i{\sqrt {1-{x}^{2}}}\right)\right)\left(1-{q}^{\frac {17}{4}}\left(x+i{\sqrt {1-{x}^{2}}}\right)\right)\left(1-{\frac {\sqrt[{4}]{q}}{x+i{\sqrt {1-{x}^{2}}}}}\right)\left(1-{\frac {{q}^{5/4}}{x+i{\sqrt {1-{x}^{2}}}}}\right)\left(1-{\frac {{q}^{9/4}}{x+i{\sqrt {1-{x}^{2}}}}}\right)\left(1-{q}^{\frac {13}{4}}\left(x+i{\sqrt {1-{x}^{2}}}\right)^{-1}\right)\left(1-{q}^{\frac {17}{4}}\left(x+i{\sqrt {1-{x}^{2}}}\right)^{-1}\right){q}^{5}\left(1-q\right)^{-2}\left(1-{q}^{2}\right)^{-2}\left(1-{q}^{3}\right)^{-2}\left(1-{q}^{4}\right)^{-2}\left(1-{q}^{5}\right)^{-2}\left(1+{\sqrt {q}}\right)^{-1}\left(1+{q}^{3/2}\right)^{-1}\left(1+{q}^{5/2}\right)^{-1}\left(1+{q}^{7/2}\right)^{-1}\left(1+{q}^{9/2}\right)^{-1}\left(1+q\right)^{-1}\left(1+{q}^{2}\right)^{-1}\left(1+{q}^{3}\right)^{-1}\left(1+{q}^{4}\right)^{-1}\left(1+{q}^{5}\right)^{-1}-{\frac {{q}^{5/4}}{\left(1-q\right)^{2}\left(1+{\sqrt {q}}\right)\left(1+q\right)\left(x+i{\sqrt {1-{x}^{2}}}\right)}}-{\frac {{q}^{5/4}\left(x+i{\sqrt {1-{x}^{2}}}\right)}{\left(1-q\right)^{2}\left(1+{\sqrt {q}}\right)\left(1+q\right)}}+{q}^{\frac {29}{4}}\left(1-q\right)^{-2}\left(1+{\sqrt {q}}\right)^{-1}\left(1+q\right)^{-1}\left(x+i{\sqrt {1-{x}^{2}}}\right)^{-1}+{q}^{\frac {29}{4}}\left(x+i{\sqrt {1-{x}^{2}}}\right)\left(1-q\right)^{-2}\left(1+{\sqrt {q}}\right)^{-1}\left(1+q\right)^{-1}+{q}^{-{\frac {15}{4}}}\left(1-q\right)^{-2}\left(1+{\sqrt {q}}\right)^{-1}\left(1+q\right)^{-1}\left(x+i{\sqrt {1-{x}^{2}}}\right)^{-1}+\left(x+i{\sqrt {1-{x}^{2}}}\right){q}^{-{\frac {15}{4}}}\left(1-q\right)^{-2}\left(1+{\sqrt {q}}\right)^{-1}\left(1+q\right)^{-1}-{\frac {{q}^{9/4}}{\left(1-q\right)^{2}\left(1+{\sqrt {q}}\right)\left(1+q\right)\left(x+i{\sqrt {1-{x}^{2}}}\right)}}-{\frac {{q}^{9/4}\left(x+i{\sqrt {1-{x}^{2}}}\right)}{\left(1-q\right)^{2}\left(1+{\sqrt {q}}\right)\left(1+q\right)}}+\left(1-{q}^{-5}\right)\left(1-{q}^{-4}\right)\left(1-{q}^{6}\right)\left(1-{q}^{7}\right)\left(1-{\sqrt[{4}]{q}}\left(x+i{\sqrt {1-{x}^{2}}}\right)\right)\left(1-{q}^{5/4}\left(x+i{\sqrt {1-{x}^{2}}}\right)\right)\left(1-{\frac {\sqrt[{4}]{q}}{x+i{\sqrt {1-{x}^{2}}}}}\right)\left(1-{\frac {{q}^{5/4}}{x+i{\sqrt {1-{x}^{2}}}}}\right){q}^{2}\left(1-q\right)^{-2}\left(1-{q}^{2}\right)^{-2}\left(1+{\sqrt {q}}\right)^{-1}\left(1+{q}^{3/2}\right)^{-1}\left(1+q\right)^{-1}\left(1+{q}^{2}\right)^{-1}+{\frac {q}{\left(1-q\right)^{2}\left(1+{\sqrt {q}}\right)\left(1+q\right)}}+{\frac {{q}^{3/2}}{\left(1-q\right)^{2}\left(1+{\sqrt {q}}\right)\left(1+q\right)}}-{\frac {{q}^{7}}{\left(1-q\right)^{2}\left(1+{\sqrt {q}}\right)\left(1+q\right)}}-{\frac {{q}^{15/2}}{\left(1-q\right)^{2}\left(1+{\sqrt {q}}\right)\left(1+q\right)}}-{\frac {1}{{q}^{4}\left(1-q\right)^{2}\left(1+{\sqrt {q}}\right)\left(1+q\right)}}-{\frac {1}{{q}^{7/2}\left(1-q\right)^{2}\left(1+{\sqrt {q}}\right)\left(1+q\right)}}+{\frac {{q}^{2}}{\left(1-q\right)^{2}\left(1+{\sqrt {q}}\right)\left(1+q\right)}}+{\frac {{q}^{5/2}}{\left(1-q\right)^{2}\left(1+{\sqrt {q}}\right)\left(1+q\right)}}\cdots } 求 q→1 的极限值： lim q → 1 P 5 ( x | q ) = 63 8 x 5 − 35 4 x 3 + 15 8 x {\displaystyle \lim _{q\to 1}P_{5}(x|q)={\frac {63}{8}}\,{x}^{5}-{\frac {35}{4}}\,{x}^{3}+{\frac {15}{8}}\,x} 而5阶勒让德多项式为： P 5 ( x ) = 63 8 x 5 − 35 4 x 3 + 15 8 x {\displaystyle P_{5}(x)={\frac {63}{8}}\,{x}^{5}-{\frac {35}{4}}\,{x}^{3}+{\frac {15}{8}}\,x} 两者显然相等，所以 lim q → 1 P 5 ( x | q ) = P 5 ( x ) {\displaystyle \lim _{q\to 1}P_{5}(x|q)=P_{5}(x)} 验证毕 Roelof Koekoek, Hypergeometric Orthogonal Polynomials and its q-Analogues, p475,Springer,2010