# Formula:KLS:14.10:101

Jump to navigation Jump to search

${\displaystyle{\displaystyle{\displaystyle{\tilde{w}}(x;1|q)P_{n}\!\left(x|q% \right)=\left(\frac{q-1}{2}\right)^{n}\frac{q^{\frac{1}{4}n^{2}}}{\left(q,-q^{% \frac{1}{2}},-q;q\right)_{n}}\left(D_{q}\right)^{n}\left[{\tilde{w}}(x;q^{% \frac{1}{2}n}|q)\right]}}}$

## Substitution(s)

${\displaystyle{\displaystyle{\displaystyle{\tilde{w}}(x;a|q):=\frac{w(x;a|q)}{% \sqrt{1-x^{2}}}}}}$ &

$\displaystyle {\displaystyle w(x;a|q)=\left|\frac{\qPochhammer{\expe^{2\iunit\theta}}{q}{\infty}}{\qPochhammer{aq^{\frac{1}{4}}\expe^{\iunit\theta} aq^{\frac{3}{4}}\expe^{\iunit\theta},-aq^{\frac{1}{4}}\expe^{\iunit\theta} -aq^{\frac{3}{4}}\expe^{\iunit\theta}}{q}{\infty}}\right|^2 =\left|\frac{\qPochhammer{\expe^{\iunit\theta},-\expe^{\iunit\theta}}{q^{\frac{1}{2}}}{\infty}} {\qPochhammer{aq^{\frac{1}{4}}\expe^{\iunit\theta},-aq^{\frac{1}{4}}\expe^{\iunit\theta}}{q^{\frac{1}{2}}}{\infty}}\right|^2 =\left|\frac{\qPochhammer{\expe^{2\iunit\theta}}{q}{\infty}}{\qPochhammer{a^2q^{\frac{1}{2}}\expe^{2\iunit\theta}}{q}{\infty}}\right|^2 =\frac{h(x,1)h(x,-1)h(x,q^{\frac{1}{2}})h(x,-q^{\frac{1}{2}})} {h(x,aq^{\frac{1}{4}})h(x,aq^{\frac{3}{4}})h(x,-aq^{\frac{1}{4}})h(x,-aq^{\frac{3}{4}})}}$ &
${\displaystyle{\displaystyle{\displaystyle h(x,\alpha):=\prod_{k=0}^{\infty}% \left(1-2\alpha xq^{k}+\alpha^{2}q^{2k}\right)=\qPochhammer{\alpha\expe^{% \iunit\theta},\alpha\expe^{-\iunit\theta}}{q}{\infty}}}}$ &

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

## Proof

We ask users to provide proof(s), reference(s) to proof(s), or further clarification on the proof(s) in this space.

## URL links

We ask users to provide relevant URL links in this space.