連續勒讓德多項式是一個以基本超幾何函數定義的正交多項式[1]
令連續q勒讓德多項式 q->1 得勒讓德多項式
- 驗證5階連續q勒讓德多項式→勒讓德多項式
由定義，
![{\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 }](//wikimedia.org/api/rest_v1/media/math/render/svg/7e0faba9fff78428a46eecfed01865a009ed6e9f)
求 q→1 的極限值：
而5階勒讓德多項式為：
兩者顯然相等，所以
驗證畢
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Roelof Koekoek, Hypergeometric Orthogonal Polynomials and its q-Analogues, p475,Springer,2010
![{\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 }](http://wikimedia.org/api/rest_v1/media/math/render/svg/7e0faba9fff78428a46eecfed01865a009ed6e9f)
