Formula:KLS:14.10:08: Difference between revisions

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Latest revision as of 08:37, 22 December 2019


2 ⁢ x ⁢ P ~ n ( n , β ) ⁡ ( x | q ) = A n ⁢ P ~ n ( n + 1 , β ) ⁡ ( x | q ) + [ q 1 2 ⁢ α + 1 4 + q - 1 2 ⁢ α - 1 4 - ( A n + C n ) ] ⁢ P ~ n ( n , β ) ⁡ ( x | q ) + C n ⁢ P ~ n ( n - 1 , β ) ⁡ ( x | q ) 2 𝑥 continuous-q-Jacobi-polynomial-P-tilde 𝑛 𝛽 𝑛 𝑥 𝑞 subscript 𝐴 𝑛 continuous-q-Jacobi-polynomial-P-tilde 𝑛 1 𝛽 𝑛 𝑥 𝑞 delimited-[] superscript 𝑞 1 2 𝛼 1 4 superscript 𝑞 1 2 𝛼 1 4 subscript 𝐴 𝑛 subscript 𝐶 𝑛 continuous-q-Jacobi-polynomial-P-tilde 𝑛 𝛽 𝑛 𝑥 𝑞 subscript 𝐶 𝑛 continuous-q-Jacobi-polynomial-P-tilde 𝑛 1 𝛽 𝑛 𝑥 𝑞 {\displaystyle{\displaystyle{\displaystyle 2x{\tilde{P}}^{(n,\beta)}_{n}\!% \left(x|q\right)=A_{n}{\tilde{P}}^{(n+1,\beta)}_{n}\!\left(x|q\right)+\left[q^% {\frac{1}{2}\alpha+\frac{1}{4}}+q^{-\frac{1}{2}\alpha-\frac{1}{4}}-\left(A_{n}% +C_{n}\right)\right]{\tilde{P}}^{(n,\beta)}_{n}\!\left(x|q\right){}+C_{n}{% \tilde{P}}^{(n-1,\beta)}_{n}\!\left(x|q\right)}}}

Substitution(s)

C n = q 1 2 ⁢ α + 1 4 ⁢ ( 1 - q n ) ⁢ ( 1 - q n + β ) ⁢ ( 1 + q n + 1 2 ⁢ ( α + β ) ) ⁢ ( 1 + q n + 1 2 ⁢ ( α + β + 1 ) ) ( 1 - q 2 ⁢ n + α + β ) ⁢ ( 1 - q 2 ⁢ n + α + β + 1 ) subscript 𝐶 𝑛 superscript 𝑞 1 2 𝛼 1 4 1 superscript 𝑞 𝑛 1 superscript 𝑞 𝑛 𝛽 1 superscript 𝑞 𝑛 1 2 𝛼 𝛽 1 superscript 𝑞 𝑛 1 2 𝛼 𝛽 1 1 superscript 𝑞 2 𝑛 𝛼 𝛽 1 superscript 𝑞 2 𝑛 𝛼 𝛽 1 {\displaystyle{\displaystyle{\displaystyle C_{n}=\frac{q^{\frac{1}{2}\alpha+% \frac{1}{4}}(1-q^{n})(1-q^{n+\beta})(1+q^{n+\frac{1}{2}(\alpha+\beta)})(1+q^{n% +\frac{1}{2}(\alpha+\beta+1)})}{(1-q^{2n+\alpha+\beta})(1-q^{2n+\alpha+\beta+1% })}}}} &
A n = ( 1 - q n + α + 1 ) ⁢ ( 1 - q n + α + β + 1 ) ⁢ ( 1 + q n + 1 2 ⁢ ( α + β + 1 ) ) ⁢ ( 1 + q n + 1 2 ⁢ ( α + β + 2 ) ) q 1 2 ⁢ α + 1 4 ⁢ ( 1 - q 2 ⁢ n + α + β + 1 ) ⁢ ( 1 - q 2 ⁢ n + α + β + 2 ) subscript 𝐴 𝑛 1 superscript 𝑞 𝑛 𝛼 1 1 superscript 𝑞 𝑛 𝛼 𝛽 1 1 superscript 𝑞 𝑛 1 2 𝛼 𝛽 1 1 superscript 𝑞 𝑛 1 2 𝛼 𝛽 2 superscript 𝑞 1 2 𝛼 1 4 1 superscript 𝑞 2 𝑛 𝛼 𝛽 1 1 superscript 𝑞 2 𝑛 𝛼 𝛽 2 {\displaystyle{\displaystyle{\displaystyle A_{n}=\frac{(1-q^{n+\alpha+1})(1-q^% {n+\alpha+\beta+1})(1+q^{n+\frac{1}{2}(\alpha+\beta+1)})(1+q^{n+\frac{1}{2}(% \alpha+\beta+2)})}{q^{\frac{1}{2}\alpha+\frac{1}{4}}(1-q^{2n+\alpha+\beta+1})(% 1-q^{2n+\alpha+\beta+2})}}}}


Proof

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Symbols List

& : logical and
P ~ n ( α , β ) subscript superscript ~ 𝑃 𝛼 𝛽 𝑛 {\displaystyle{\displaystyle{\displaystyle{\tilde{P}}^{(\alpha,\beta)}_{n}}}}  : normalized continuous q 𝑞 {\displaystyle{\displaystyle{\displaystyle q}}} -Jacobi polynomial P ~ ~ 𝑃 {\displaystyle{\displaystyle{\displaystyle{\tilde{P}}}}}  : http://drmf.wmflabs.org/wiki/Definition:normctsqJacobiPtilde

Bibliography

Equation in Section 14.10 of KLS.

URL links

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