小q拉盖尔多项式

**小 q 拉蓋爾多項式**是一個以基本超幾何函數定義的正交多項式

　
: $\displaystyle p_{n}(x;a|q)={}_{2}\phi _{1}(q^{-n},0;aq;q,qx)={\frac {1}{(a^{-1}q^{-n};q)_{n}}}{}_{2}\phi _{0}(q^{-n},x^{-1};;q,x/a)$

## 極限關係

**大 q 拉蓋爾多項式→小 q 拉蓋爾多項式**

在大 q 拉蓋爾多項式中，令 $x\to bqx$ , 並令 $b\to \infty $ 即得小 q 拉蓋爾多項式

$$
\lim _{b\to \infty }P_{n}(bqx;a,b;q)=p_{n}(x;a|q)
$$

仿射 Q 克拉夫楚克多項式→ 小 q 拉蓋爾多項式：

 $\lim _{a\to 1}=K_{n}^{aff}(q^{x-N};p,N|q)=p_{n}(q^{x};p,q)$ 令小 q 拉蓋爾多項式 $a=q^{a}$ $x=(1-q)*x$ , 然後令 q→1
即得拉蓋爾多項式

$$
\lim _{q\to 1}P_{a}(1-q)x;q^{a}|q)={\frac {L_{n}^{(a)}(x)}{L_{n}^{(a)}(0)}}
$$

**驗證 9 階小 q 拉蓋爾多項式→拉蓋爾多項式**

作上述代換，

 
 $P_{a}(1-q)x;q^{a}|q)=1+{\frac {qx}{1-{q}^{\alpha }q}}-{\frac {x}{{q}^{8}\left(1-{q}^{\alpha }q\right)}}$
 $+\left(1-{q}^{-9}\right)\left(1-{q}^{-8}\right){q}^{2}\left(1-q\right){x}^{2}\left(1-{q}^{2}\right)^{-1}\left(1-{q}^{\alpha }q\right)^{-1}\left(1-{q}^{\alpha }{q}^{2}\right)^{-1}$ $+\left(1-{q}^{-9}\right)\left(1-{q}^{-8}\right)\left(1-{q}^{-7}\right){q}^{3}\left(1-q\right)^{2}{x}^{3}\left(1-{q}^{2}\right)^{-1}$ $\left(1-{q}^{3}\right)^{-1}\left(1-{q}^{\alpha }q\right)^{-1}\left(1-{q}^{\alpha }{q}^{2}\right)^{-1}\left(1-{q}^{\alpha }{q}^{3}\right)^{-1}+\cdots $

 

 

 
求 q→1 極限得

令 a=3, 得 $(1-{\frac {9}{4}}x+{\frac {9}{5}}{x}^{2}-{\frac {7}{10}}{x}^{3}+{\frac {3}{20}}{x}^{4}-{\frac {3}{160}}{x}^{5}+{\frac {1}{720}}{x}^{6}-{\frac {1}{16800}}{x}^{7}+{\frac {1}{739200}}{x}^{8}-{\frac {1}{79833600}}{x}^{9})$

另一方面

 ${\frac {L_{n}^{(3)}(x)}{L_{n}^{(3)}(0)}}$
= $(1-{\frac {9}{4}}x+{\frac {9}{5}}{x}^{2}-{\frac {7}{10}}{x}^{3}+{\frac {3}{20}}{x}^{4}-{\frac {3}{160}}{x}^{5}+{\frac {1}{720}}{x}^{6}-{\frac {1}{16800}}{x}^{7}+{\frac {1}{739200}}{x}^{8}-{\frac {1}{79833600}}{x}^{9})$

二者顯然相等 QED

## 圖集

## 參考文獻

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