連続q-Jacobi多項式の線形化公式5: 古典極限
前回の記事
と同様に$x:=\cos\theta$
\begin{align}
r_n(x)=r_n(x;a,b,c,d)&:=\Q43{q^{-n},abcdq^{n-1},ae^{i\theta},ae^{-i\theta}}{ab,ac,ad}{q}\\
w(x)=w(x;a,b,c,d)&:=\frac 1{\sqrt{1-x^2}}\frac{(e^{2i\theta},e^{-2i\theta};q)_{\infty}}{(ae^{i\theta},ae^{-i\theta},be^{i\theta},be^{-i\theta},ce^{i\theta},ce^{-i\theta},de^{i\theta},de^{-i\theta};q)_{\infty}}\\
h_0&:=\frac{2\pi(abcd;q)_{\infty}}{(q,ab,ac,ad,bc,bd,cd;q)_{\infty}}
\end{align}
とする. $a=q^{\frac 12},d=-q^{\frac 12}$とすると
\begin{align}
r_n(x)&=\Q43{q^{-n},-bcq^n,e^{i\theta}q^{\frac 12},e^{-i\theta}q^{\frac 12}}{bq^{\frac 12},cq^{\frac 12},-q}q\\
w(x)&=\frac 1{\sqrt{1-x^2}}\frac{(e^{2i\theta},e^{-2i\theta};q)_{\infty}}{(e^{i\theta}q^{\frac 12},e^{-i\theta}q^{\frac 12},be^{i\theta},be^{-i\theta},ce^{i\theta},ce^{-i\theta},-e^{i\theta}q^{\frac 12},-e^{-i\theta}q^{\frac 12};q)_{\infty}}\\
&=\frac 1{\sqrt{1-x^2}}\frac{(e^{2i\theta},e^{-2i\theta};q^2)_{\infty}}{(be^{i\theta},be^{-i\theta},ce^{i\theta},ce^{-i\theta};q)_{\infty}}\\
h_0&=\frac{2\pi(-bcq;q)_{\infty}}{(bc;q)_{\infty}(q^2,b^2q,c^2q;q^2)_{\infty}}
\end{align}
となる. このとき,
前回の記事
で,
\begin{align}
&\int_{-1}^1r_m(x)r_n(x)r_k(x)w(x)\,dx\\
&=h_0\frac{(-bc;q)_{m+n-k}(q,-cq^{\frac 12};q)_k(-bc;q)_{n+k-m}(c^2q^2;q^2)_{n-m}(q,bc,-bq^{\frac 12},-cq^{\frac 12};q)_n(b^2c^2;q)_{m+n+k}(q;q)_{2m}(-b/c;q)_{m+k-n}}{(q;q)_{m+n-k}(-bc,bq^{\frac 12};q)_{k}(-bc,bq^{\frac 12},cq^{\frac 12},-q;q)_{m}(c^2q;q)_{n+k-m}(q^2;q^2)_{n-m}(-bcq;q)_{m+n+k}(b^2c^2;q)_{2n}(q;q)_{m+k-n}}q^{n}(-cq^{\frac 12})^{m+k-n}\\
&\qquad\cdot\Q{10}9{c^2q^{2n-2m},cq^{n-m+1},-cq^{n-m+1},c^2,c^2q^{2n+1},-bcq^{n+k-m},-bcq^{n+k-m+1},q^{1-2m}/b^2,q^{n-m-k},q^{1+n-m-k}}{cq^{n-m},-cq^{n-m},q^{2n-2m+2},q^{1-2m},-cq^{2+n-m-k}/b,-cq^{1+n-m-k}/b,b^2c^2q^{2n+1},c^2q^{2+n+k-m},c^2q^{1+n+k-m}}{q^2;q^2}
\end{align}
が成り立つことを示した. 今回はこの古典極限を考えたいと思う. まず, $b,c$を$q^{a+\frac 12},-q^{b+\frac 12}$とすると
\begin{align}
r_n(x)&=\Q43{q^{-n},q^{a+b+n+1},e^{i\theta}q^{\frac 12},e^{-i\theta}q^{\frac 12}}{q^{a+1},-q^{b+1},-q}q
\end{align}
\begin{align}
\lim_{q\to 1}r_n(x)&=\F21{-n,a+b+n+1}{a+1}{\frac{(1-e^{i\theta})(1-e^{-i\theta})}{4}}\\
&=\F21{-n,a+b+n+1}{a+1}{\frac{1-x}{2}}
\end{align}
となるから, 定数倍を除いてJacobi多項式に一致する. この場合の$w(x)$は
\begin{align}
\lim_{q\to 1}w(x)&=\lim_{q\to 1}\frac 1{\sqrt{1-x^2}}\frac{(e^{i\theta},-e^{i\theta},e^{-i\theta},-e^{-i\theta};q)_{\infty}}{(e^{i\theta}q^{a+\frac 12},e^{-i\theta}q^{a+\frac 12},-e^{i\theta}q^{b+\frac 12},-e^{-i\theta}q^{b+\frac 12};q)_{\infty}}\\
&=\frac 1{\sqrt{1-x^2}}(1-e^{i\theta})^{a+\frac 12}(1-e^{-i\theta})^{a+\frac 12}(1+e^{i\theta})^{b+\frac 12}(1+e^{-i\theta})^{b+\frac 12}\\
&=2^{a+b+1}(1-x)^a(1+x)^b
\end{align}
となるので定数倍を除いてJacobi多項式の重み関数に一致する. これより
\begin{align}
&2^{a+b+1}\int_{-1}^1\F21{-n,a+b+n+1}{a+1}{\frac{1-x}2}\F21{-m,a+b+m+1}{a+1}{\frac{1-x}2}\F21{-k,a+b+k+1}{a+1}{\frac{1-x}2}(1-x)^a(1+x)^b\,dx\\
&=\lim_{q\to 1}\frac{2\pi(q^{a+b+2};q)_{\infty}}{(-q^{a+b+1};q)_{\infty}(q^2,q^{2a+2},q^{2b+2};q^2)_{\infty}}\frac{(q^{a+b+1};q)_{m+n-k}(q,q^{b+1};q)_k(q^{a+b+1};q)_{n+k-m}(q^{2b+3};q^2)_{n-m}(q,-q^{a+b+1},-q^{a+1},q^{b+1};q)_n(q^{2a+2b+2};q)_{m+n+k}(q;q)_{2m}(q^{a-b};q)_{m+k-n}}{(q;q)_{m+n-k}(q^{a+b+1},q^{a+1};q)_{k}(q^{a+b+1},q^{a+1},-q^{b+1},-q;q)_{m}(q^{2b+2};q)_{n+k-m}(q^2;q^2)_{n-m}(q^{a+b+2};q)_{m+n+k}(q^{2a+2b+2};q)_{2n}(q;q)_{m+k-n}}q^{n}(q^{b+1})^{m+k-n}\\
&\qquad\cdot\Q{10}9{q^{2b+2n-2m+1},q^{b+n-m+\frac32},-q^{b+n-m+\frac 32},q^{2b+1},q^{2b+2n+2},q^{a+b+1+n+k-m},q^{a+b+2+n+k-m},q^{-2m-2a},q^{n-m-k},q^{1+n-m-k}}{q^{b+n-m+\frac 12},-q^{b+n-m+\frac 12},q^{2n-2m+2},q^{1-2m},q^{b-a+2+n-m-k},q^{b-a+1+n-m-k},q^{2a+2b+2n+3},q^{2b+3+n+k-m},q^{2b+2+n+k-m}}{q^2;q^2}\\
&=\lim_{q\to 1}\frac{2\pi(q^{a+b+1},q^{a+b+2},q^{a+b+2},q^{a+b+3};q^2)_{\infty}}{(q^{2a+2b+2},q^2,q^{2a+2},q^{2b+2};q^2)_{\infty}}\frac{4^{n-m}(a+b+1)_{m+n-k}k!(b+1)_k(a+b+1)_{n+k-m}\left(b+\frac 32\right)_{n-m}n!(b+1)_n(2a+2b+2)_{m+n+k}(2m)!(a-b)_{m+k-n}}{(m+n-k)!(a+b+1,a+1)_{k}(a+b+1,a+1)_{m}(2b+2)_{n+k-m}(n-m)!(a+b+2)_{m+n+k}(2a+2b+2)_{2n}(m+k-n)!}\\
&\qquad\cdot\F{9}8{b+n-m+\frac 12,\frac 12\left(b+n-m+\frac 12\right),b+\frac 12,b+n+1,\frac{a+b+1+n+k-m}2,\frac{a+b+2+n+k-m}2,-m-a,\frac{n-m-k}2,\frac{1+n-m-k}2}{\frac 12\left(b+n-m+\frac 12\right),1+n-m,\frac 12-m,\frac{b-a+2+n-m-k}2,\frac{b-a+1+n-m-k}2,a+b+n+\frac 32,b+\frac{3+n+k-m}2,b+\frac{2+n+k-m}2}1
\end{align}
となる. ここで,
\begin{align}
\lim_{q\to 1}\frac{2\pi(q^{a+b+1},q^{a+b+2},q^{a+b+2},q^{a+b+3};q^2)_{\infty}}{(q^{2a+2b+2},q^2,q^{2a+2},q^{2b+2};q^2)_{\infty}}&=\frac{2\pi\Gamma(a+1)\Gamma(b+1)\Gamma(a+b+1)}{\Gamma\left(\frac{a+b+1}2\right)\Gamma\left(\frac{a+b+2}2\right)^2\Gamma\left(\frac{a+b+3}2\right)}\\
&=\frac{2^{2a+2b+2}\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}
\end{align}
となる. よって, 以下を得る.
\begin{align}
&\int_{-1}^1\F21{-n,a+b+n+1}{a+1}{\frac{1-x}2}\F21{-m,a+b+m+1}{a+1}{\frac{1-x}2}\F21{-k,a+b+k+1}{a+1}{\frac{1-x}2}(1-x)^a(1+x)^b\,dx\\
&=\frac{2^{a+b+1}\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}\frac{4^{n-m}(a+b+1)_{m+n-k}k!(b+1)_k(a+b+1)_{n+k-m}\left(b+\frac 32\right)_{n-m}n!(b+1)_n(2a+2b+2)_{m+n+k}(2m)!(a-b)_{m+k-n}}{(m+n-k)!(a+b+1,a+1)_{k}(a+b+1,a+1)_{m}(2b+2)_{n+k-m}(n-m)!(a+b+2)_{m+n+k}(2a+2b+2)_{2n}(m+k-n)!}\\
&\qquad\cdot\F{9}8{b+n-m+\frac 12,\frac 12\left(b+n-m+\frac 12\right),b+\frac 12,b+n+1,\frac{a+b+1+n+k-m}2,\frac{a+b+2+n+k-m}2,-m-a,\frac{n-m-k}2,\frac{1+n-m-k}2}{\frac 12\left(b+n-m+\frac 12\right),1+n-m,\frac 12-m,\frac{b-a+2+n-m-k}2,\frac{b-a+1+n-m-k}2,a+b+n+\frac 32,b+\frac{3+n+k-m}2,b+\frac{2+n+k-m}2}1
\end{align}
これはJacobi多項式
\begin{align}
P_n^{(a,b)}(x):=\frac{(a+1)_n}{n!}\F21{-n,a+b+n+1}{a+1}{\frac{1-x}2}
\end{align}
を用いて表すと
\begin{align}
&\int_{-1}^1P_n^{(a,b)}(x)P_m^{(a,b)}(x)P_k^{(a,b)}(x)(1-x)^a(1+x)^b\,dx\\
&=\frac{2^{a+b+1}\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}\frac{4^{n-m}(a+b+1)_{m+n-k}(b+1)_k(a+b+1)_{n+k-m}\left(b+\frac 32\right)_{n-m}(a+1,b+1)_n(2a+2b+2)_{m+n+k}(2m)!(a-b)_{m+k-n}}{m!(m+n-k)!(a+b+1)_{k}(a+b+1)_{m}(2b+2)_{n+k-m}(n-m)!(a+b+2)_{m+n+k}(2a+2b+2)_{2n}(m+k-n)!}\\
&\qquad\cdot\F{9}8{b+n-m+\frac 12,\frac 12\left(b+n-m+\frac 12\right),b+\frac 12,b+n+1,\frac{a+b+1+n+k-m}2,\frac{a+b+2+n+k-m}2,-m-a,\frac{n-m-k}2,\frac{1+n-m-k}2}{\frac 12\left(b+n-m+\frac 12\right),1+n-m,\frac 12-m,\frac{b-a+2+n-m-k}2,\frac{b-a+1+n-m-k}2,a+b+n+\frac 32,b+\frac{3+n+k-m}2,b+\frac{2+n+k-m}2}1
\end{align}
となる. Jacobi多項式の直交性
\begin{align}
\int_{-1}^1P_n^{(a,b)}(x)P_m^{(a,b)}(x)(1-x)^a(1+x)^b\,dx&=\frac{2^{a+b+1}}{2n+a+b+1}\frac{\Gamma(n+a+1)\Gamma(n+b+1)}{\Gamma(n+a+b+1)n!}\delta_{m,n}
\end{align}
を用いるとJacobi多項式の線形化公式
\begin{align}
P_n^{(a,b)}(x)P_m^{(a,b)}(x)=\sum_{0\leq k}b_kP_k^{(a,b)}(x)
\end{align}
の係数$b_k$を
\begin{align}
b_k&=\frac{2n+a+b+1}{a+b+1}\frac{4^{n-m}k!(a+b+1)_{m+n-k}(a+b+1)_{n+k-m}\left(b+\frac 32\right)_{n-m}(a+1,b+1)_n(2a+2b+2)_{m+n+k}(2m)!(a-b)_{m+k-n}}{m!(m+n-k)!(a+b+1)_{m}(a+1)_k(2b+2)_{n+k-m}(n-m)!(a+b+2)_{m+n+k}(2a+2b+2)_{2n}(m+k-n)!}\\
&\qquad\cdot\F{9}8{b+n-m+\frac 12,\frac 12\left(b+n-m+\frac 12\right),b+\frac 12,b+n+1,\frac{a+b+1+n+k-m}2,\frac{a+b+2+n+k-m}2,-m-a,\frac{n-m-k}2,\frac{1+n-m-k}2}{\frac 12\left(b+n-m+\frac 12\right),1+n-m,\frac 12-m,\frac{b-a+2+n-m-k}2,\frac{b-a+1+n-m-k}2,a+b+n+\frac 32,b+\frac{3+n+k-m}2,b+\frac{2+n+k-m}2}1
\end{align}
と表すことができる. これは
\begin{align}
b_{k+n-m}&=\frac{2n+2k-2m+a+b+1}{a+b+1}\frac{4^{n-m}(n-m+k)!(a+b+1)_{2m-k}(a+b+1)_{2n-2m+k}\left(b+\frac 32\right)_{n-m}(a+1,b+1)_n(2a+2b+2)_{2n+k}(2m)!(a-b)_{k}}{m!(2m-k)!(a+b+1)_{m}(a+1)_{n-m+k}(2b+2)_{2n-2m+k}(n-m)!(a+b+2)_{2n+k}(2a+2b+2)_{2n}k!}\\
&\qquad\cdot\F{9}8{b+n-m+\frac 12,\frac 12\left(b+n-m+\frac 12\right),b+\frac 12,b+n+1,\frac{a+b+k+1}2+n-m,\frac{a+b+k+2}2+n-m,-m-a,-\frac{k}2,\frac{1-k}2}{\frac 12\left(b+n-m+\frac 12\right),1+n-m,\frac 12-m,\frac{b-a+2-k}2,\frac{b-a+1-k}2,a+b+n+\frac 32,b+n-m+\frac{k+3}2,b+n-m+\frac{k+2}2}1\\
&=\frac{2n+2k-2m+a+b+1}{a+b+1}\frac{(a+b+1)_{2m}(a+b+1)_{2n-2m}(a+1,b+1)_n}{m!(a+b+1)_{m}(a+1,b+1)_{n-m}(a+b+2)_{2n}}\\
&\qquad\cdot\frac{\left(n-m+1,a+b+2n-2m+1,2a+2b+2n+2,a-b,-2m\right)_k}{k!\left(a+n-m+1,2b+2n-2m+2,a+b+2n+2,-2m-a-b\right)_k}\\
&\qquad\cdot\F{9}8{b+n-m+\frac 12,\frac 12\left(b+n-m+\frac 12\right),b+\frac 12,b+n+1,\frac{a+b+k+1}2+n-m,\frac{a+b+k+2}2+n-m,-m-a,-\frac{k}2,\frac{1-k}2}{\frac 12\left(b+n-m+\frac 12\right),1+n-m,\frac 12-m,\frac{b-a+2-k}2,\frac{b-a+1-k}2,a+b+n+\frac 32,b+n-m+\frac{k+3}2,b+n-m+\frac{k+2}2}1
\end{align}
と書き換えることもできる.
対称的な表示
前回の記事
で, 以下の対称的な表示を示した.
\begin{align}
&\int_{-1}^1r_m(x)r_n(x)r_k(x)w(x)\,dx\\
&=h_0\frac{(-bq^{\frac 12},q;q)_k(-bq^{\frac 12},q;q)_{m}(-bq^{\frac 12},q;q)_n(-bc;q)_{m+n-k}(-bc;q)_{m+k-n}(-bc;q)_{n+k-m}}{(-bc,cq^{\frac 12};q)_{k}(-bc,cq^{\frac 12};q)_{m}(-bc,cq^{\frac 12};q)_{n}(-bcq;q)_{m+n+k}}(-cq^{\frac 52})^{m+n+k}q^{\binom{m+n+k}2}\\
&\qquad\cdot\sum_{0\leq j}\frac{(1-q^{4j+1-2m-2n-2k}/b^2)(c^2q;q^2)_j(b^2c^2;q^2)_{m+n+k-j}(-b/c,b^2/q;q)_{m+n+k-2j}}{(1-q/b^2)(q^2;q^2)_{j-m}(q^2;q^2)_{j-n}(q^2;q^2)_{j-k}(b^2q;q^2)_{m+n-j}(b^2q;q^2)_{m+k-j}(b^2q;q^2)_{n+k-j}(-bc,q;q)_{m+n+k-2j}}\left(\frac{q^{-2m-2n-2k-1}}{c^2}\right)^jq^{4\binom j2}
\end{align}
先ほどのようにこの古典極限を考えると
\begin{align}
&\int_{-1}^1\F21{-m,a+b+m+1}{a+1}{\frac{1-x}2}\F21{-n,a+b+n+1}{a+1}{\frac{1-x}2}\F21{-k,a+b+k+1}{a+1}{\frac{1-x}2}(1-x)^a(1+x)^b\,dx\\
&=\frac{2^{a+b+1}\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}\frac{m!n!k!(a+b+1)_{m+n-k}(a+b+1)_{m+k-n}(a+b+1)_{n+k-m}}{(a+b+1)_k(a+b+1)_m(a+b+1)_n(a+b+2)_{m+n+k}}\\
&\qquad\cdot\sum_{0\leq j}\frac{(4j+1-2m-2n-2k-2a)(b+1)_j(a+b+1)_{m+n+k-j}(a-b,2a)_{m+n+k-2j}}{(1-2a)(j-m)!(j-n)!(j-k)!(a+1)_{m+n-j}(a+1)_{m+k-j}(a+1)_{n+k-j}(a+b+1)_{m+n+k-2j}(m+n+k-2j)!}
\end{align}
を得る. Jacobi多項式を用いて書き換えると
\begin{align}
&\int_{-1}^1P_m^{(a,b)}(x)P_n^{(a,b)}(x)P_k^{(a,b)}(x)(1-x)^a(1+x)^b\,dx\\
&=\frac{2^{a+b+1}\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}\frac{(a+1)_m(a+1)_n(a+1)_k(a+b+1)_{m+n-k}(a+b+1)_{m+k-n}(a+b+1)_{n+k-m}}{(a+b+1)_k(a+b+1)_m(a+b+1)_n(a+b+2)_{m+n+k}}\\
&\qquad\cdot\sum_{0\leq j}\frac{(4j+1-2m-2n-2k-2a)(b+1)_j(a+b+1)_{m+n+k-j}(a-b,2a)_{m+n+k-2j}}{(1-2a)(j-m)!(j-n)!(j-k)!(a+1)_{m+n-j}(a+1)_{m+k-j}(a+1)_{n+k-j}(a+b+1)_{m+n+k-2j}(m+n+k-2j)!}
\end{align}
となる.
