# Formula:KLS:14.02:09

$\displaystyle {\displaystyle -\left(1-q^{-x}\right)\left(1-\gamma\delta q^{x+1}\right)\qRacah{n}@@{\mu(x)}{\alpha}{\beta}{\gamma}{\delta}{q} {}=A_n\qRacah{n+1}@@{\mu(x)}{\alpha}{\beta}{\gamma}{\delta}{q}-\left(A_n+C_n\right)\qRacah{n}@@{\mu(x)}{\alpha}{\beta}{\gamma}{\delta}{q}+C_n\qRacah{n-1}@@{\mu(x)}{\alpha}{\beta}{\gamma}{\delta}{q} }$

## Substitution(s)

$\displaystyle {\displaystyle \mu(x)=q^{-x}+\gamma\delta q^{x+1} =\lambda(x)=q^{-x}+cq^{x-N} =q^{-x}+q^{x+\gamma+\delta+1} =2a\cos@@{\theta}}$ &

$\displaystyle {\displaystyle \lambda(x)=x(x+\gamma+\delta+1)}$ &
$\displaystyle {\displaystyle C_n=\frac{q(1-q^n)(1-\beta q^n)(\gamma-\alpha\beta q^n)(\delta-\alpha q^n)} {(1-\alpha\beta q^{2n})(1-\alpha\beta q^{2n+1})}}$ &
$\displaystyle {\displaystyle A_n=\frac{(1-\alpha q^{n+1})(1-\alpha\beta q^{n+1})(1-\beta\delta q^{n+1})(1-\gamma q^{n+1})} {(1-\alpha\beta q^{2n+1})(1-\alpha\beta q^{2n+2})}}$ &
$\displaystyle {\displaystyle \mu(x):=q^{-x}+\gamma\delta q^{x+1}}$ &
$\displaystyle {\displaystyle \mu(x)=q^{-x}+\gamma\delta q^{x+1} =\lambda(x)=q^{-x}+cq^{x-N} =q^{-x}+q^{x+\gamma+\delta+1} =2a\cos@@{\theta}}$ &

$\displaystyle {\displaystyle \lambda(x)=x(x+\gamma+\delta+1)}$

## Proof

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

## Symbols List

& : logical and
$\displaystyle {\displaystyle R_{n}}$  : $\displaystyle {\displaystyle q}$ -Racah polynomial : http://dlmf.nist.gov/18.28#E19
$\displaystyle {\displaystyle \mathrm{cos}}$  : cosine function : http://dlmf.nist.gov/4.14#E2