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

$P_n(x|q)={}_4{\varphi }_3(\begin{array}{ccccc}{q}^{-n}& q^{n+1}& q^{1/4}{e}^{i\theta }& a^{1/4}{e}^{-i\theta }& \\ q& -q^{1/2}& -q\end{array};q,q)$

极限关系

令连续q勒让德多项式 q->1 得勒让德多项式

$\underset{q\to 1}{\lim }{P}_n(x|q)=P_n(x)$

验证5阶连续q勒让德多项式→勒让德多项式

$\underset{q\to 1}{\lim }{P}_5(x|q)=P_5(x)$

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-\sqrt[4]{q}(x+i\sqrt{1-x^2}))(1-q^{5/4}(x+i\sqrt{1-x^2}))(1-q^{9/4}(x+i\sqrt{1-x^2}))(1-\frac{\sqrt[4]{q}}{x+i\sqrt{1-x^2}})(1-\frac{q^{5/4}}{x+i\sqrt{1-x^2}})(1-\frac{q^{9/4}}{x+i\sqrt{1-x^2}})q^3{(1-q)}^{-2}{(1-q^2)}^{-2}{(1-q^3)}^{-2}{(1+\sqrt{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-\sqrt[4]{q}(x+i\sqrt{1-x^2}))(1-q^{5/4}(x+i\sqrt{1-x^2}))(1-q^{9/4}(x+i\sqrt{1-x^2}))(1-q^{\frac{13}{4}}(x+i\sqrt{1-x^2}))(1-\frac{\sqrt[4]{q}}{x+i\sqrt{1-x^2}})(1-\frac{q^{5/4}}{x+i\sqrt{1-x^2}})(1-\frac{q^{9/4}}{x+i\sqrt{1-x^2}})(1-q^{\frac{13}{4}}{(x+i\sqrt{1-x^2})}^{-1})q^4{(1-q)}^{-2}{(1-q^2)}^{-2}{(1-q^3)}^{-2}{(1-q^4)}^{-2}{(1+\sqrt{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-\sqrt[4]{q}(x+i\sqrt{1-x^2}))(1-q^{5/4}(x+i\sqrt{1-x^2}))(1-q^{9/4}(x+i\sqrt{1-x^2}))(1-q^{\frac{13}{4}}(x+i\sqrt{1-x^2}))(1-q^{\frac{17}{4}}(x+i\sqrt{1-x^2}))(1-\frac{\sqrt[4]{q}}{x+i\sqrt{1-x^2}})(1-\frac{q^{5/4}}{x+i\sqrt{1-x^2}})(1-\frac{q^{9/4}}{x+i\sqrt{1-x^2}})(1-q^{\frac{13}{4}}{(x+i\sqrt{1-x^2})}^{-1})(1-q^{\frac{17}{4}}{(x+i\sqrt{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+\sqrt{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}-\frac{q^{5/4}}{{(1-q)}^2(1+\sqrt{q})(1+q)(x+i\sqrt{1-x^2})}-\frac{q^{5/4}(x+i\sqrt{1-x^2})}{{(1-q)}^2(1+\sqrt{q})(1+q)}+q^{\frac{29}{4}}{(1-q)}^{-2}{(1+\sqrt{q})}^{-1}{(1+q)}^{-1}{(x+i\sqrt{1-x^2})}^{-1}+q^{\frac{29}{4}}(x+i\sqrt{1-x^2}){(1-q)}^{-2}{(1+\sqrt{q})}^{-1}{(1+q)}^{-1}+q^{-\frac{15}{4}}{(1-q)}^{-2}{(1+\sqrt{q})}^{-1}{(1+q)}^{-1}{(x+i\sqrt{1-x^2})}^{-1}+(x+i\sqrt{1-x^2})q^{-\frac{15}{4}}{(1-q)}^{-2}{(1+\sqrt{q})}^{-1}{(1+q)}^{-1}-\frac{q^{9/4}}{{(1-q)}^2(1+\sqrt{q})(1+q)(x+i\sqrt{1-x^2})}-\frac{q^{9/4}(x+i\sqrt{1-x^2})}{{(1-q)}^2(1+\sqrt{q})(1+q)}+(1-q^{-5})(1-q^{-4})(1-q^6)(1-q^7)(1-\sqrt[4]{q}(x+i\sqrt{1-x^2}))(1-q^{5/4}(x+i\sqrt{1-x^2}))(1-\frac{\sqrt[4]{q}}{x+i\sqrt{1-x^2}})(1-\frac{q^{5/4}}{x+i\sqrt{1-x^2}})q^2{(1-q)}^{-2}{(1-q^2)}^{-2}{(1+\sqrt{q})}^{-1}{(1+q^{3/2})}^{-1}{(1+q)}^{-1}{(1+q^2)}^{-1}+\frac{q}{{(1-q)}^2(1+\sqrt{q})(1+q)}+\frac{q^{3/2}}{{(1-q)}^2(1+\sqrt{q})(1+q)}-\frac{q^7}{{(1-q)}^2(1+\sqrt{q})(1+q)}-\frac{q^{15/2}}{{(1-q)}^2(1+\sqrt{q})(1+q)}-\frac{1}{q^4{(1-q)}^2(1+\sqrt{q})(1+q)}-\frac{1}{q^{7/2}{(1-q)}^2(1+\sqrt{q})(1+q)}+\frac{q^2}{{(1-q)}^2(1+\sqrt{q})(1+q)}+\frac{q^{5/2}}{{(1-q)}^2(1+\sqrt{q})(1+q)}\cdots$

求 q→1 的极限值： $\underset{q\to 1}{\lim }{P}_5(x|q)=\frac{63}{8}{x}^5-\frac{35}{4}{x}^3+\frac{15}{8}x$

而5阶勒让德多项式为： $P_5(x)=\frac{63}{8}{x}^5-\frac{35}{4}{x}^3+\frac{15}{8}x$

两者显然相等，所以 $\underset{q\to 1}{\lim }{P}_5(x|q)=P_5(x)$

验证毕

图集

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