# Formula:KLS:14.10:14

${\displaystyle{\displaystyle{\displaystyle(1-q)^{2}D_{q}\left[{\tilde{w}}(x;q^% {\alpha+1},q^{\beta+1}|q)D_{q}y(x)\right]+\lambda_{n}{\tilde{w}}(x;q^{\alpha},% q^{\beta}|q)y(x)=0}}}$

## Substitution(s)

${\displaystyle{\displaystyle{\displaystyle\lambda_{n}=4q^{-n+1}(1-q^{n})(1-q^{% n+\alpha+\beta+1})}}}$ &

${\displaystyle{\displaystyle{\displaystyle{\tilde{w}}(x;q^{\alpha},q^{\beta}|q% ):=\frac{w(x;q^{\alpha},q^{\beta}|q)}{\sqrt{1-x^{2}}}}}}$ &
${\displaystyle{\displaystyle{\displaystyle y(x)=\ctsqJacobi{\alpha}{\beta}{n}@% {x}{q}}}}$ &
$\displaystyle {\displaystyle w(x):=w(x;q^{\alpha},q^{\beta}|q) =\left|\frac{\qPochhammer{\expe^{2\iunit\theta}}{q}{\infty}} {\qPochhammer{q^{\frac{1}{2}\alpha+\frac{1}{4}}\expe^{\iunit\theta},q^{\frac{1}{2}\alpha+\frac{3}{4}}\expe^{\iunit\theta} -q^{\frac{1}{2}\beta+\frac{1}{4}}\expe^{\iunit\theta},-q^{\frac{1}{2}\beta+\frac{3}{4}}\expe^{\iunit\theta}}{q}{\infty}}\right|^2 =\left|\frac{\qPochhammer{\expe^{\iunit\theta},-\expe^{\iunit\theta}}{q^{\frac{1}{2}}}{\infty}}{\qPochhammer{q^{\frac{1}{2}\alpha+\frac{1}{4}}\expe^{\iunit\theta} -q^{\frac{1}{2}\beta+\frac{1}{4}}\expe^{\iunit\theta}}{q^{\frac{1}{2}}}{\infty}}\right|^2 =\frac{h(x,1)h(x,-1)h(x,q^{\frac{1}{2}})h(x,-q^{\frac{1}{2}})} {h(x,q^{\frac{1}{2}\alpha+\frac{1}{4}})h(x,q^{\frac{1}{2}\alpha+\frac{3}{4}}) h(x,-q^{\frac{1}{2}\beta+\frac{1}{4}})h(x,-q^{\frac{1}{2}\beta+\frac{3}{4}})}}$ &
${\displaystyle{\displaystyle{\displaystyle h(x,\alpha):=\prod_{k=0}^{\infty}% \left(1-2\alpha xq^{k}+\alpha^{2}q^{2k}\right)=\qPochhammer{\alpha e^{i\theta}% ,\alpha e^{-i\theta}}{q}{\infty}}}}$ &

${\displaystyle{\displaystyle{\displaystyle x=\cos\theta}}}$

## Proof

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