Formula:KLS:14.19:12

From DRMF
Jump to navigation Jump to search


( 1 - q ) 2 D q [ w ~ ( x ; q α + 1 | q ) D q y ( x ) ] + 4 q - n + 1 ( 1 - q n ) w ~ ( x ; q α | q ) y ( x ) = 0 superscript 1 𝑞 2 subscript 𝐷 𝑞 delimited-[] ~ 𝑤 𝑥 conditional superscript 𝑞 𝛼 1 𝑞 subscript 𝐷 𝑞 𝑦 𝑥 4 superscript 𝑞 𝑛 1 1 superscript 𝑞 𝑛 ~ 𝑤 𝑥 conditional superscript 𝑞 𝛼 𝑞 𝑦 𝑥 0 {\displaystyle{\displaystyle{\displaystyle(1-q)^{2}D_{q}\left[{\tilde{w}}(x;q^% {\alpha+1}|q)D_{q}y(x)\right]+4q^{-n+1}(1-q^{n}){\tilde{w}}(x;q^{\alpha}|q)y(x% )=0}}}

Substitution(s)

w ~ ( x ; q α | q ) := w ( x ; q α | q ) 1 - x 2 assign ~ 𝑤 𝑥 conditional superscript 𝑞 𝛼 𝑞 𝑤 𝑥 conditional superscript 𝑞 𝛼 𝑞 1 superscript 𝑥 2 {\displaystyle{\displaystyle{\displaystyle{\tilde{w}}(x;q^{\alpha}|q):=\frac{w% (x;q^{\alpha}|q)}{\sqrt{1-x^{2}}}}}} &

y ( x ) = P n ( α ) ( x | q ) 𝑦 𝑥 continuous-q-Laguerre-polynomial-P 𝛼 𝑛 𝑥 𝑞 {\displaystyle{\displaystyle{\displaystyle y(x)=P^{(\alpha)}_{n}\!\left(x|q% \right)}}} &
w ( x ) := w ( x ; q α | q ) = | ( e 2 i θ ; q ) ( q 1 2 α + 1 4 e i θ q 1 2 α + 3 4 e i θ ; q ) | 2 = | ( e i θ , - e i θ ; q 1 2 ) ( q 1 2 α + 1 4 e i θ ; q 1 2 ) | 2 = h ( x , 1 ) h ( x , - 1 ) h ( x , q 1 2 ) h ( x , - q 1 2 ) h ( x , q 1 2 α + 1 4 ) h ( x , q 1 2 α + 3 4 ) assign 𝑤 𝑥 𝑤 𝑥 conditional superscript 𝑞 𝛼 𝑞 superscript q-Pochhammer-symbol 2 imaginary-unit 𝜃 𝑞 q-Pochhammer-symbol superscript 𝑞 1 2 𝛼 1 4 imaginary-unit 𝜃 superscript 𝑞 1 2 𝛼 3 4 imaginary-unit 𝜃 𝑞 2 superscript q-Pochhammer-symbol imaginary-unit 𝜃 imaginary-unit 𝜃 superscript 𝑞 1 2 q-Pochhammer-symbol superscript 𝑞 1 2 𝛼 1 4 imaginary-unit 𝜃 superscript 𝑞 1 2 2 𝑥 1 𝑥 1 𝑥 superscript 𝑞 1 2 𝑥 superscript 𝑞 1 2 𝑥 superscript 𝑞 1 2 𝛼 1 4 𝑥 superscript 𝑞 1 2 𝛼 3 4 {\displaystyle{\displaystyle{\displaystyle w(x):=w(x;q^{\alpha}|q)=\left|\frac% {\left({\mathrm{e}^{2\mathrm{i}\theta}};q\right)_{\infty}}{\left(q^{\frac{1}{2% }\alpha+\frac{1}{4}}{\mathrm{e}^{\mathrm{i}\theta}}q^{\frac{1}{2}\alpha+\frac{% 3}{4}}{\mathrm{e}^{\mathrm{i}\theta}};q\right)_{\infty}}\right|^{2}=\left|% \frac{\left({\mathrm{e}^{\mathrm{i}\theta}},-{\mathrm{e}^{\mathrm{i}\theta}};q% ^{\frac{1}{2}}\right)_{\infty}}{\left(q^{\frac{1}{2}\alpha+\frac{1}{4}}{% \mathrm{e}^{\mathrm{i}\theta}};q^{\frac{1}{2}}\right)_{\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 , α ) := k = 0 ( 1 - 2 α x q k + α 2 q 2 k ) = ( α e i θ , α e - i θ ; q ) assign 𝑥 𝛼 superscript subscript product 𝑘 0 1 2 𝛼 𝑥 superscript 𝑞 𝑘 superscript 𝛼 2 superscript 𝑞 2 𝑘 q-Pochhammer-symbol 𝛼 imaginary-unit 𝜃 𝛼 imaginary-unit 𝜃 𝑞 {\displaystyle{\displaystyle{\displaystyle h(x,\alpha):=\prod_{k=0}^{\infty}% \left(1-2\alpha xq^{k}+\alpha^{2}q^{2k}\right)=\left(\alpha{\mathrm{e}^{% \mathrm{i}\theta}},\alpha{\mathrm{e}^{-\mathrm{i}\theta}};q\right)_{\infty}}}} &

x = cos θ 𝑥 𝜃 {\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.

Symbols List

& : logical and
P α ( n ) subscript superscript 𝑃 𝑛 𝛼 {\displaystyle{\displaystyle{\displaystyle P^{(n)}_{\alpha}}}}  : continuous q 𝑞 {\displaystyle{\displaystyle{\displaystyle q}}} -Laguerre polynomial : http://drmf.wmflabs.org/wiki/Definition:ctsqLaguerre
( a ; q ) n subscript 𝑎 𝑞 𝑛 {\displaystyle{\displaystyle{\displaystyle(a;q)_{n}}}}  : q 𝑞 {\displaystyle{\displaystyle{\displaystyle q}}} -Pochhammer symbol : http://dlmf.nist.gov/5.18#i http://dlmf.nist.gov/17.2#SS1.p1
e e {\displaystyle{\displaystyle{\displaystyle\mathrm{e}}}}  : the base of the natural logarithm : http://dlmf.nist.gov/4.2.E11
i i {\displaystyle{\displaystyle{\displaystyle\mathrm{i}}}}  : imaginary unit : http://dlmf.nist.gov/1.9.i
Π Π {\displaystyle{\displaystyle{\displaystyle\Pi}}}  : product : http://drmf.wmflabs.org/wiki/Definition:prod
cos cos {\displaystyle{\displaystyle{\displaystyle\mathrm{cos}}}}  : cosine function : http://dlmf.nist.gov/4.14#E2

Bibliography

Equation in Section 14.19 of KLS.

URL links

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