连续q勒让德多项式

令连续q勒让德多项式 q->1 得勒让德多项式

${\displaystyle \lim _{q\to 1}P_{n}(x|q)=P_{n}(x)}$

验证5阶连续q勒让德多项式→勒让德多项式

${\displaystyle \lim _{q\to 1}P_{5}(x|q)=P_{5}(x)}$

由定义， ${\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： ${\displaystyle \lim _{q\to 1}P_{5}(x|q)={\frac {63}{8}}\,{x}^{5}-{\frac {35}{4}}\,{x}^{3}+{\frac {15}{8}}\,x}$

而5阶勒让德多项式为： ${\displaystyle P_{5}(x)={\frac {63}{8}}\,{x}^{5}-{\frac {35}{4}}\,{x}^{3}+{\frac {15}{8}}\,x}$

两者显然相等，所以 ${\displaystyle \lim _{q\to 1}P_{5}(x|q)=P_{5}(x)}$

验证毕

1. Roelof Koekoek, Hypergeometric Orthogonal Polynomials and its q-Analogues, p475,Springer,2010