連續q勒讓德多項式

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

$$P_n(x|q)=\;_{4}\phi_3\left(\begin{matrix} q^{-n} & q^{n+1} & q^{1/4}e^{i\theta} & a^{1/4}e^{-i\theta} & \\ q & -q^{1/2} &-q \end{matrix} ; q,q \right)$$

## 極限關係

令連續q勒讓德多項式 q->1 得勒讓德多項式

$$\lim_{q \to 1}P_{n}(x|q)=P_{n}(x)$$

驗證5階連續q勒讓德多項式→勒讓德多項式

$$\lim_{q \to 1}P_{5}(x|q)=P_{5}(x)$$

由定義。

$$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) <pre> \left( 1-{q}^{8} \right) \left( 1-\sqrt [4]{q} \left( x+i\sqrt {1-{x </pre> }^{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}} <pre> \right) \right) \left( 1-{\frac {\sqrt [4]{q}}{x+i\sqrt {1-{x}^{2}} </pre> }} \right) \left( 1-{\frac {{q}^{5/4}}{x+i\sqrt {1-{x}^{2}}}} <pre> \right) \left( 1-{\frac {{q}^{9/4}}{x+i\sqrt {1-{x}^{2}}}} \right) { </pre> 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 <pre> \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 </pre> } \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}} <pre> \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-{ </pre> 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}}}} <pre> \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) ^{ </pre> -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} <pre> \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}^ </pre> {3} \right) ^{-1} \left( 1+{q}^{4} \right) ^{-1}+ \left( 1-{q}^{-5} <pre> \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 </pre> } \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} <pre> \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 { </pre> 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}}} <pre> \left( x+i\sqrt {1-{x}^{2}} \right) ^{-1} \right) \left( 1-{q}^{{ </pre> \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} <pre> \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) ^{- </pre> 1} \left( 1+{q}^{2} \right) ^{-1} \left( 1+{q}^{3} \right) ^{-1} <pre> \left( 1+{q}^{4} \right) ^{-1} \left( 1+{q}^{5} \right) ^{-1}-{\frac </pre> {{q}^{5/4}}{ \left( 1-q \right) ^{2} \left( 1+\sqrt {q} \right) <pre> \left( 1+q \right) \left( x+i\sqrt {1-{x}^{2}} \right) }}-{\frac {{q </pre> }^{5/4} \left( x+i\sqrt {1-{x}^{2}} \right) }{ \left( 1-q \right) ^{2} <pre> \left( 1+\sqrt {q} \right) \left( 1+q \right) }}+{q}^{{\frac {29}{4} </pre> }} \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} <pre> \right) \left( 1+q \right) \left( x+i\sqrt {1-{x}^{2}} \right) }}-{ </pre> \frac {{q}^{9/4} \left( x+i\sqrt {1-{x}^{2}} \right) }{ \left( 1-q <pre> \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 </pre> } \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} <pre> \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} </pre> +{\frac {q}{ \left( 1-q \right) ^{2} \left( 1+\sqrt {q} \right) <pre> \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+ </pre> \sqrt {q} \right) \left( 1+q \right) }}-{\frac {1}{{q}^{4} \left( 1-q <pre> \right) ^{2} \left( 1+\sqrt {q} \right) \left( 1+q \right) }}-{ </pre> \frac {1}{{q}^{7/2} \left( 1-q \right) ^{2} \left( 1+\sqrt {q} <pre> \right) \left( 1+q \right) }}+{\frac {{q}^{2}}{ \left( 1-q \right) ^ </pre> {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 <pre> \right) }}\cdots$$

求 q→1 的極限值：

$$\lim_{q \to 1}P_{5}(x|q)={\frac {63}{8}}\,{x}^{5}-{\frac {35}{4}}\,{x}^{3}+{\frac {15}{8}}\,x$$

而5階勒讓德多項式為：

$$P_5(x)={\frac {63}{8}}\,{x}^{5}-{\frac {35}{4}}\,{x}^{3}+{\frac {15}{8}}\,x$$

兩者顯然相等，所以$$\lim_{q \to 1}P_{5}(x|q)=P_{5}(x)$$

驗證畢

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