特殊関数論で遊ぼう

[1]収束性に関しては次の次の不等式から分かる。
$$\begin{array}{rcl}\log \prod _{n=1}^{\mathrm{\infty }}(1+\frac{6}{(n+1)(2n+9)})& =& \sum _{k=1}^{\mathrm{\infty }}\log (1+\frac{6}{(n+1)(2n+9)})\\ & \le & \sum _{n=1}^{\mathrm{\infty }}\frac{6}{(n+1)(2n+9)}\\ & \le \sum _{n=1}^{\mathrm{\infty }}\frac{6}{n^2}={\pi }^2\end{array}$$
[2]極限値に関しては
$$1+\frac{6}{(n+1)(2n+9)}=\frac{(2n+5)(n+3)}{(n+1)(2n+9)}$$
より
$$\begin{array}{rcl}\prod _{k=1}^n\frac{(2k+5)(k+3)}{(k+1)(k+9)}& =& \frac{7\cdot 9(n+2)(n+3)}{2\cdot 3(2n+7)(2n+9)}\\ & \to & \frac{21}{8}(n\to \mathrm{\infty })\end{array}$$

$$1-\frac{1}{n^2}=\frac{(n-1)(n+1)}{n^2}$$
より
$$\begin{array}{rcl}\prod _{k=2}^n(1-\frac{1}{k^2})& =& \frac{(1\cdot 2\cdot 3\cdots (n-1))(3\cdot 4\cdot 5\cdots (n+1))}{2^2{3}^2\cdots n^2}\\ & =& \frac{n+1}{2n}\\ & \to & \frac{1}{2}\end{array}$$

$$\begin{array}{rcl}\prod _{k=2}^n(1-\frac{1}{k})& =& \prod _{k=2}^n\frac{k-1}{k}\\ & =& \frac{1}{2}\frac{2}{3}\frac{n-1}{n}\\ & =& \frac{1}{n}\to 0\end{array}$$

$$\begin{array}{rcl}\log P_{2^n}& =& \sum _{k=1}^{2^n-1}\frac{1}{k}\\ & >& 1+2(\frac{1}{2}+\frac{1}{2})+\cdots 2^{n-1}(\frac{1}{2^{n-1}}+\cdots +\frac{1}{2^{n-1}})\\ & =& n\end{array}$$

$$\begin{array}{rcl}{\displaystyle \frac{1}{\mathrm{\Gamma }(z)}}& =& \underset{n\to \mathrm{\infty }}{lim}\frac{\prod _{k=0}^n(z+k)}{n^{z}n!}\\ {\displaystyle }& =& z\underset{n\to \mathrm{\infty }}{lim}{e}^{-z\log n}\prod _{k=1}^n(1+\frac{z}{k})\\ {\displaystyle }& =& z\underset{n\to \mathrm{\infty }}{lim}{e}^{(\sum _{k=1}^n\frac{1}{k}-\log n)z}\prod _{k=1}^n(1+\frac{z}{k})e^{-\frac{z}{k}}\\ {\displaystyle }& =& z{e}^{\gamma z}\prod _{k=1}^{\mathrm{\infty }}(1+\frac{z}{k})e^{-\frac{z}{k}}\end{array}$$

[1]
$$\log \mathrm{\Gamma }(z)=-\gamma z+\log z+\sum _{k=1}^{\mathrm{\infty }}(\frac{z}{k}-\log (1+\frac{z}{k}))$$
両辺を$z$で微分して
$$\frac{{\mathrm{\Gamma }}^{{}^{\prime }}(z)}{\mathrm{\Gamma }(z)}=-\gamma -\frac{1}{z}-\sum _{k=1}^{\mathrm{\infty }}(\frac{1}{z+k}-\frac{1}{k})$$
[2]
$$\begin{array}{rcl}{\displaystyle 2\frac{{\mathrm{\Gamma }}^{{}^{\prime }}(2z)}{\mathrm{\Gamma }(2z)}-\frac{{\mathrm{\Gamma }}^{{}^{\prime }}(z)}{\mathrm{\Gamma }(z)}-\frac{{\mathrm{\Gamma }}^{{}^{\prime }}(z+\frac{1}{2})}{\mathrm{\Gamma }(z+\frac{1}{2})}}& =& -2\gamma +\gamma +\gamma -\frac{1}{z}+\frac{1}{z}+\frac{1}{z+\frac{1}{2}}-2\sum _{k=1}^{\mathrm{\infty }}(\frac{1}{2z+k}-\frac{1}{k})+\sum _{k=1}^{\mathrm{\infty }}(\frac{1}{z+k}-\frac{1}{k})+\sum _{k=1}^{\mathrm{\infty }}(\frac{1}{z+k+\frac{1}{2}}-\frac{1}{k})\\ & =& \frac{2}{2z+1}+\underset{n\to \mathrm{\infty }}{lim}[-\sum _{k=1}^{2n}\frac{2}{2z+k}+2H_{2n}+\sum _{k=1}^n\frac{2}{2z+2k}-H_n+\sum _{k=1}^n\frac{2}{2z+2k+1}-H_n]\\ & =& \frac{2}{2z+1}+\underset{n\to \mathrm{\infty }}{lim}[-\sum _{k=1}^{2n}\frac{2}{2z+k}+\sum _{k=2}^{2n+1}\frac{2}{2z+k}+2H_{2n}-2H_n]\\ & =& \frac{2}{2z+1}+\underset{n\to \mathrm{\infty }}{lim}[\frac{2}{2z+2n+1}-\frac{2}{2z+1}+2H_{2n}-2H_n]\\ & =& \underset{n\to \mathrm{\infty }}{lim}(\frac{2}{2z+2n+1}+2H_{2n}-2H_n)\\ & =& \underset{n\to \mathrm{\infty }}{lim}(H_{2n}-2H_n)\\ & =& \underset{n\to \mathrm{\infty }}{lim}(2(H_{2n}-\log 2n)-2(H_n-\log n)+2\log 2n-2\log n)\\ & =& 2\underset{n\to \mathrm{\infty }}{lim}(\gamma -\gamma +\log 2+\log n-\log n)\\ & =& 2\log n\end{array}$$

問題6で$z=\frac{1}{2}$を代入すると
$$2\frac{{\mathrm{\Gamma }}^{{}^{\prime }}(1)}{\mathrm{\Gamma }(1)}-\frac{{\mathrm{\Gamma }}^{{}^{\prime }}(\frac{1}{2})}{\mathrm{\Gamma }(\frac{1}{2})}-\frac{{\mathrm{\Gamma }}^{{}^{\prime }}(1)}{\mathrm{\Gamma }(1)}=2\log 2$$
ゆえに次式を得る。
$$\begin{array}{rcl}{\mathrm{\Gamma }}^{{}^{\prime }}(\frac{1}{2})& =& (\frac{{\mathrm{\Gamma }}^{{}^{\prime }}(1)}{\mathrm{\Gamma }(1)}-2\log 2)\mathrm{\Gamma }(\frac{1}{2})\\ & =& -(\gamma +2\log 2)\sqrt{\pi }\end{array}$$

$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{z-1}dt& =& \underset{n\to \mathrm{\infty }}{lim}{\int }_0^{\mathrm{\infty }}(1+\frac{t}{n}{)}^{-n}{t}^{z-1}dt\\ & =& \underset{n\to \mathrm{\infty }}{lim}{n}^z{\int }_0^1(1-u{)}^{-n}{u}^{z-1}du\\ & =& \underset{n\to \mathrm{\infty }}{lim}\frac{n^{z}n!}{z(z+1)\cdots (z+n)}\\ & =& \mathrm{\Gamma }(z)\end{array}$$

$$\begin{array}{rcl}\mathrm{\Gamma }(s)\mathrm{\Gamma }(t)& =& {\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{x}^{s-1}{y}^{t-1}{e}^{-(x+y)}dxdy\\ & =& {\int }_0^{\mathrm{\infty }}{\int }_x^{\mathrm{\infty }}{x}^{s-1}(y-x{)}^{t-1}{e}^{-y}dydx\\ & =& {\int }_0^{\mathrm{\infty }}{\int }_0^y{x}^{s-1}(y-x{)}^{t-1}{e}^{-y}dxdy\\ & =& {\int }_0^{\mathrm{\infty }}dy{y}^{t-1}{e}^{-y}{\int }_0^y{x}^{s-1}(1-\frac{x}{y}{)}^{t-1}dx\\ & =& {\int }_0^{\mathrm{\infty }}dy{y}^{s+t-1}{e}^{-y}{\int }_0^1{x}^{s-1}(1-x{)}^{t-1}dx\\ & =& \mathrm{\Gamma }(x+y)B(x,y)\end{array}$$

ある整数$m\le x<m+1$が成り立つので以下の計算ができる。
$$\begin{array}{rcl}{\int }_0^{x}P(x)dx& =& {\int }_m^{x}P(x)dx\\ & =& {\int }_0^{x-m}P(x)dx\\ & =& {\int }_0^{x-m}(x-\frac{1}{2})dx\\ & =& \frac{1}{2}(x-m{)}^2-\frac{1}{2}(x-m)\\ & =& \frac{1}{2}(x-m-\frac{1}{2}{)}^2-\frac{1}{8}\\ & =& \frac{1}{2}(\lfloor x\rfloor -\frac{1}{2}{)}^2-\frac{1}{8}\\ & =& \frac{1}{2}{P}^2(x)-\frac{1}{8}\end{array}$$

$$\begin{array}{rcl}G(x,w)& =& e^{-aw}(1+w{)}^x\\ & =& \sum _{m=0}^{\mathrm{\infty }}\frac{(-a{)}^m{w}^m}{m!}\sum _{n=0}^{\mathrm{\infty }}\left(\begin{array}{c}x\\ n\end{array}\right)w^n\\ & =& \sum _{m=0}^{\mathrm{\infty }}\sum _{n=0}^{\mathrm{\infty }}\frac{(-a{)}^m}{m!}\left(\begin{array}{c}x\\ n\end{array}\right)w^{m+n}\\ & =& \sum _{N=0}^{\mathrm{\infty }}\sum _{n=0}^N\frac{(-a{)}^{N-n}}{(N-n)!}\left(\begin{array}{c}x\\ n\end{array}\right)w^N\end{array}$$

$$a^{x}G(x,v)G(x,w)=e^{-a(v+w)}[a(1+v)(1+w){]}^x$$
より
$$\begin{array}{rcl}\sum _{k=0}^{\mathrm{\infty }}\frac{a^{k}G(k,v)G(k,w)}{k!}& =& e^{-a(v+w)}{e}^{a(1+v)(1+w)}\\ & =& e^a{e}^{vw}\\ & =& \sum _{n=0}^{\mathrm{\infty }}\frac{e^a{a}^n(vw{)}^n}{n!}\\ & =& \sum _{k=0}^{\mathrm{\infty }}\frac{a^k}{k!}\sum _{m,n=0}^{\mathrm{\infty }}{P}_m(k)P_n(k)v^m{w}^n\\ & =& \sum _{m,n=0}^{\mathrm{\infty }}\sum _{k=0}^{\mathrm{\infty }}{P}_m(k)P_n(k)\frac{a^k}{k!}{v}^m{w}^n\end{array}$$
より$v^m{w}^n$の係数比較により証明が完了する。

[1]
$$\begin{array}{rcl}x{T}_n(x)& =& \cos n\theta \cos \theta \\ & =& \frac{1}{2}(\cos (n+1)\theta +\cos (n-1)\theta )\\ & =& \frac{T_{n+1}(x)+T_{n-1}(x)}{2}\end{array}$$
より次の漸化式を得る。
$$T_{n+1}(x)=2x{T}_n(x)-T_n(x)$$
$T_0(x)=1,T_1(x)=x$より$0,1$の時は$T_n(x)$は$x$の多項式。
そこで、$0,1,...,n$まで$T_k(x)$は$x$の$k$次多項式とする。
すると先の漸化式より、$T_{n+1}(x)$は$n+1$次多項式であることが示された。
[2]直交性
$\theta =\arccos x$と置くと、次の様に積分は書き換えれる。
$${\int }_0^{\pi }\cos m\theta \cos n\theta d\theta =\frac{1}{2}{\int }_0^{\pi }(\cos (m+n)\theta +\cos (m-n)\theta )d\theta$$
ゆえに、
[1]$m\ne n$の場合は与えられた積分は0
[2]$m=n=0$の場合$\pi$
[3]$m=n\ne 0$の場合は$\frac{\pi }{2}$

[1]
$$U_n(x)=x{U}_{n-1}(x)+T_n(x)$$
$U_0(x)=1$より$0,1,..,n-1$まで$U_k(x)$は$x$の$k$次多項式とすると、漸化式より$U_n(x)$は$n$次の多項式。
[2]
$x=\cos \theta$とおくと
$$\begin{array}{rcl}(givenintegral)& =& {\int }_0^{\pi }\sin (m+1)\theta \sin (n+1)\theta d\theta \\ & =& \frac{\pi }{2}{\delta }_{mn}\end{array}$$

[1]
$$\begin{array}{rcl}\frac{\partial F}{\partial w}& =& 2(x-w)F\\ & =& -\frac{\partial F}{\partial x}\end{array}$$
[2]
$$\begin{array}{rcl}G(x,w)& =& e^{2xw-w^2}\\ & =& e^{x^2}F(x,w)\\ & =& e^x\sum _{n=0}^{\mathrm{\infty }}\frac{{\partial }^{n}F}{\partial w^n}{|}_{w=0}\frac{w^n}{n!}\\ & =& \sum _{n=0}^{\mathrm{\infty }}(-1{)}^n{e}^x{D}^n{e}^{-x^2}\frac{w^n}{n!}\end{array}$$
[3]
$$\begin{array}{rcl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}G(x,v)G(x,w)e^{-x^2}dx& =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2+2(v+w)x-(v^2+w^2)}dx\\ & =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-(x-v-w{)}^2+2vw}dx\\ & =& \sqrt{\pi }{e}^{2vw}\end{array}$$
[4]
$$\begin{array}{rcl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}G(x,v)G(x,w)dx& =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\sum _{m,n}^{\mathrm{\infty }}{H}_m(x)H_n(x)\frac{v^m{w}^n}{m!n!}{e}^{-x^2}dx\\ & =& \sum _{m,n=0}^{\mathrm{\infty }}\frac{v^m{w}^n}{m!n!}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{H}_m(x)H_n(x)e^{-x^2}dx\\ & =& \sqrt{\pi }\sum _{n=0}^{\mathrm{\infty }}{2}^n\frac{(vw{)}^n}{n!}\end{array}$$
より$v^m{w}^n$の係数を比較して
$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{H}_m(x)H_n(x)e^{-x^2}dx=\sqrt{\pi }{2}^{n}n!{\delta }_{mn}$$

[1]$n=1$の場合は積の微分に他ならない
[2]そこで$1,2,..,n$まで成り立つとしよう。すると$n+1$の場合は次の様になる。
$$\begin{array}{rcl}\frac{d^{n+1}}{d{x}^{n+1}}f(x)g(x)& =& \frac{d}{dx}\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)f^{(k)}(x)g^{(n-k)}(x)\\ & =& f^{(0)}(x)g^{(n+1)}(x)+\sum _{k=1}^n(\left(\begin{array}{c}n\\ k\end{array}\right)+\left(\begin{array}{c}n\\ k-1\end{array}\right))f^{(k)}(x)g^{(n+1-k)}(x)+f^{(n+1)}(x)g^{(0)}(x)\\ & =& \sum _{k=0}^{n+1}\left(\begin{array}{c}n+1\\ k\end{array}\right)f^{(k)}(x)g^{(n+1-k)}(x)\end{array}$$
[別証明]
"R"ならば$f(x)$を微分、"U"なら$g(x)$を微分に対応するとすると$f^k(x)g^{n-k}(x)$の係数は$k$個のRと$n-k$個のUの並びの個数になるので$\left(\begin{array}{c}n\\ k\end{array}\right)$

[1]
$$\begin{array}{rcl}{D}^{n+1}{e}^{-x^2}& =& -2D^{n}x{e}^{-x^2}\\ & =& -2x{D}^n{e}^{-x^2}-2n{D}^{n-1}{e}^{-x^2}\end{array}$$
よって
$$\begin{array}{rcl}{e}^{-x^2}{H}_{n+1}(x)& =& (-1{)}^{n+1}{D}^{n+1}{e}^{-x^2}\\ & =& 2x(-1{)}^n{D}^n{e}^{-x^2}-2n(-1{)}^{n-1}{D}^{n-1}{e}^{-x^2}\end{array}$$
より
$$H_{n+1}(x)=2x{H}_n(x)-2n{H}_{n-1}(x)$$
[2]
$$\begin{array}{rcl}{H}_n^{{}^{\prime }}(x)& =& 2x{H}_n(x)-H_{n+1}(x)\end{array}$$
ゆえに[1]の漸化式と組み合わせると次式を得る。
$$H_n^{{}^{\prime }}(x)=2n{H}_{n-1}(x)$$
[3]
$$\begin{array}{rcl}{y}^{{}^{″}}& =& 2n(2x{H}_{n-1}(x)-H_n(x))\\ & =& 4nx{H}_{n-1}(x)-2H_n(x)\\ & =& 2x{y}^{{}^{\prime }}-2y\end{array}$$

[1]
最初の微分方程式は次の計算により証明できる。
$$\begin{array}{rcl}\frac{\partial H}{\partial w}& =& \frac{x-w}{(1-2xw+w^2{)}^{\frac{3}{2}}}\\ & =& \frac{x-w}{1-2xw+w^2}H\end{array}$$
この微分方程式に$H(x,w)=\sum _{n=0}^{\mathrm{\infty }}{P}_n(x)w^n$を代入すると以下の式を得る。
$$\begin{array}{r}{P}_1(x)-x{P}_0(x)+\sum _{n=1}^{\mathrm{\infty }}((n+1)P_{n+1}(x)-(2n+1)P_n(x)+n{P}_{n-1}(x))w^n=0\end{array}$$
$w^n$の係数を比較することで、所要の漸化式を得る。
[2]
$A_v=\frac{1+v^2}{2v},A_w=\frac{1+w^2}{2w}$とおくと次の様に計算できる
$$\begin{array}{rcl}{\int }_{-1}^{1}H(x,v)H(x,w)dx& =& {\int }_{-1}^1\frac{dx}{((1-2vx+v^2)(1-2wx+w^2){)}^{\frac{1}{2}}}\\ & =& \frac{1}{2(vw{)}^{\frac{1}{2}}}{\int }_{-1}^1\frac{dx}{((A_v-x)(A_w-x){)}^{\frac{1}{2}}}\\ & =& \frac{1}{2(vw{)}^{\frac{1}{2}}}[-2\log ((A_v-x{)}^{\frac{1}{2}})+(A_w-x{)}^{\frac{1}{2}}{]}_{-1}^1\\ & =& \frac{1}{(vw{)}^{\frac{1}{2}}}\log \frac{(A_v+1{)}^{\frac{1}{2}}+(A_w+1{)}^{\frac{1}{2}}}{(A_v-1{)}^{\frac{1}{2}}+(A_w-1{)}^{\frac{1}{2}}}\\ & =& \frac{1}{(vw{)}^{\frac{1}{2}}}\log \frac{\frac{v+1}{(2v{)}^{\frac{1}{2}}}+\frac{w+1}{(2w{)}^{\frac{1}{2}}}}{\frac{1-v}{(2v{)}^{\frac{1}{2}}}+\frac{1-w}{(2w{)}^{\frac{1}{2}}}}\\ & =& \frac{1}{(vw{)}^{\frac{1}{2}}}\frac{w^{\frac{1}{2}}(v+1)+v^{\frac{1}{2}}(w+1)}{w^{\frac{1}{2}}(1-v)+v^{\frac{1}{2}}(1-w)}\\ & =& \frac{1}{(vw{)}^{\frac{1}{2}}}\log \frac{v^{\frac{1}{2}}+w^{\frac{1}{2}}+(vw{)}^{\frac{1}{2}}(v^{\frac{1}{2}}+w^{\frac{1}{2}})}{v^{\frac{1}{2}}+w^{\frac{1}{2}}-(vw{)}^{\frac{1}{2}}(v^{\frac{1}{2}}+w^{\frac{1}{2}})}\\ & =& \frac{1}{(vw{)}^{\frac{1}{2}}}\log \frac{1+(vw{)}^{\frac{1}{2}}}{1-(vw{)}^{\frac{1}{2}}}\\ & =& \sum _{n=1}^{\mathrm{\infty }}\frac{(vw{)}^n}{2n-1}\\ & =& {\int }_{-1}^1\sum _{m,n=0}^{\mathrm{\infty }}{P}_m(x)P_n(x)v^m{w}^{n}x\\ & =& \sum _{m,n=0}^{\mathrm{\infty }}{v}^m{w}^n{\int }_{-1}^1{P}_m(x)P_n(x)dx\end{array}$$
を得る。
ゆえに$v^m{w}^n$の係数を比較して次式を得る。
$${\int }_{-1}^1{P}_m(x)P_n(x)dx=\frac{1}{2n-1}{\delta }_{mn}$$