超幾何関数の変換公式の導出

はじめに

　この記事ではKummerの二次変換公式
$$\F{2a}{2b}{a+b+\frac12}z=\F ab{a+b+\frac12}{4z(1-z)}$$
を筆頭とした超幾何関数の変換公式を発見的に、そして網羅的に導出していきます。
　ただ今回の記事はいつもとは違い、予め書いておいたPDFを配布する形となります。
　そのリンクはこちら$\to$ hypergeometric.pdf

　ここではPDFで紹介した変換公式をまとめておくだけにとどめておきます。

変換公式まとめ

一次変換

\begin{eqnarray*} \F abcz&=&(1-z)^{c-a-b}\F{c-a}{c-b}cz \\&=&(1-z)^{-a}\F a{c-b}c{\frac z{z-1}} =(1-z)^{-b}\F {c-a}bc{\frac z{z-1}} \\&=&\frac{\G(c)\G(c-a-b)}{\G(c-a)\G(c-b)}\F ab{a+b-c+1}{1-z} +\frac{\G(c)\G(a+b-c)}{\G(a)\G(b)}(1-z)^{c-a-b}\F{c-a}{c-b}{c-a-b+1}{1-z} \\&=&\frac{\G(c)\G(b-a)}{\G(b)\G(c-a)}(-z)^{-a}\F a{a-c+1}{a-b+1}{\frac1z} +\frac{\G(c)\G(a-b)}{\G(a)\G(c-b)}(-z)^{-b}\F b{b-c+1}{b-a+1}{\frac1z} \end{eqnarray*}

二次変換

\begin{align*} \FF ab{\frac{a+b+1}2}z &=\FF{\frac a2}{\frac b2}{\frac{a+b+1}2}{4z(1-z)}\\ &=(1-2z)\FF{\frac{a+1}2}{\frac{b+1}2}{\frac{a+b+1}2}{4z(1-z)}\\ &=(1-2z)^{-a}\FF{\frac a2}{\frac{a+1}2}{\frac{a+b+1}2}{\frac{-4z(1-z)}{(1-2z)^2}}\\ \FF a{1-a}cz &=(1-z)^{c-1}\FF{\frac{c-a}2}{\frac{c+a-1}2}c{4z(1-z)}\\ &=(1-z)^{c-1}(1-2z)\FF{\frac{c+a}2}{\frac{c-a+1}2}c{4z(1-z)}\\ &=(1-z)^{c-1}(1-2z)^{a-c}\FF{\frac{c-a}2}{\frac{c-a+1}2}c{\frac{-4z(1-z)}{(1-2z)^2}}\\ \FF ab{a-b+1}z &=(1-z)^{-a}\FF{\frac a2}{\frac{a+1}2-b}{a-b+1}{\frac{-4z}{(1-z)^2}}\\ &=(1-z)^{-a-1}(1+z)\FF{\frac{a+1}2}{\frac a2-b+1}{a-b+1}{\frac{-4z}{(1-z)^2}}\\ &=(1+z)^{-a}\FF{\frac a2}{\frac{a+1}2}{a-b+1}{\frac{4z}{(1+z)^2}}\\ &=(1+z)^{2b-a-1}(1-z)^{1-2b}\FF{\frac{a+1}2-b}{\frac a2-b+1}{a-b+1}{\frac{4z}{(1+z)^2}}\\ \FF ab{2b}z &=(1-z)^{-\frac a2}\FF{\frac a2}{b-\frac a2}{b+\frac12}{\frac{z^2}{-4(1-z)}}\\ &=(1-z)^{-\frac{a+1}2}\l(1-\frac z2\r)\FF{\frac{a+1}2}{b-\frac{a-1}2}{b+\frac12}{\frac{z^2}{-4(1-z)}}\\ &=\l(1-\frac z2\r)^{-a}\FF{\frac a2}{\frac{a+1}2}{b+\frac12}{\frac{z^2}{(2-z)^2}}\\ &=\l(1-\frac z2\r)^{a-2b}(1-z)^{b-a}\FF{b-\frac a2}{b-\frac{a-1}2}{b+\frac12}{\frac{z^2}{(2-z)^2}} \end{align*}

三次変換

\begin{align} \FF{3a}{a+\frac13}{2a+\frac23}x &=(1+\o^2x)^{-3a}\FF a{a+\frac13}{2a+\frac23}{\frac{3\o(\o-1)x(1-x)}{(x+\o)^3}}\\ \FF{3a}{\frac13-a}{\frac12}z &=(1-z)^{-a}\FF a{\frac16-a}{\frac12}{\frac{(9-8z)^2z}{27(1-z)}}\\ &=\l(1-\frac43z\r)^{-3a}\FF a{a+\frac13}{\frac12}{\frac{(9-8z)^2z}{(4z-3)^3}}\\ \FF{3a}{a+\frac16}{\frac12}z &=(1-z)^{-2a}\FF a{\frac16-a}{\frac12}{\frac{(9-z)^2z}{-27(1-z)^2}}\\ &=\l(1+\frac z3\r)^{-3a}\FF a{a+\frac13}{\frac12}{\frac{(z-9)^2z}{(z+3)^3}}\\ \FF{3a}{\frac13-a}{2a+\frac56}z &=(1-4z)^{-3a}\FF a{a+\frac13}{2a+\frac56}{\frac{27z}{(4z-1)^3}}\\ &=(1+8z)^{-2a}(1-z)^{-a}\FF a{a+\frac12}{2a+\frac56}{\frac{27z}{(1+8z)^2(1-z)}}\\ \FF{3a}{a+\frac16}{4a+\frac23}z &=\l(1-\frac z4\r)^{-3a}\FF a{a+\frac13}{2a+\frac56}{\frac{27z^2}{(4-z)^3}}\\ &=\l(1+\frac z8\r)^{-2a}(1-z)^{-a}\FF a{a+\frac12}{2a+\frac56}{\frac{-27z^2}{(z+8)^2(1-z)}}\\ \FF{3a}{3a+\frac12}{4a+\frac23}z &=\l(1-\frac34z\r)^{-3a}\FF a{a+\frac13}{2a+\frac56}{\frac{-27z^2(1-z)}{(3z-4)^3}}\\ &=\l(1-\frac98z\r)^{-2a}\FF a{a+\frac12}{2a+\frac56}{\frac{-27z^2(1-z)}{(9z-8)^2}}\\ \FF{3a}{3a+\frac12}{2a+\frac56}z &=(1+3z)^{-3a}\FF a{a+\frac13}{2a+\frac56}{\frac{27z(1-z)^2}{(3z+1)^3}}\\ &=(1-9z)^{-2a}\FF a{a+\frac12}{2a+\frac56}{\frac{-27z(1-z)^2}{(1-9z)^2}} \end{align}

四次変換

\begin{align} \FF{4a}{\frac12-2a}{\frac23}z &=(1-z)^{-a}\FF a{\frac16-a}{\frac23}{\frac{(8-9z)^3z}{64(1-z)}}\\ &=\l(\frac{8-36z+27z^2}8\r)^{-2a}\FF a{a+\frac12}{\frac23}{\frac{(9z-8)^3z}{(8-36z+27z^2)^2}}\\ \FF{4a}{2a+\frac16}{\frac23}z &=(1-z)^{-3a}\FF a{\frac16-a}{\frac23}{\frac{(8+z)^3z}{-64(1-z)^3}}\\ &=\l(\frac{8-20z-z^2}8\r)^{-2a}\FF a{a+\frac12}{\frac23}{\frac{(8+z)^3z}{(z^2-20z-8)^2}}\\ \FF{4a}{\frac12-2a}{2a+\frac56}z &=(1-9z)^{-3a}(1-z)^{-a}\FF a{a+\frac13}{2a+\frac56}{\frac{64z}{(9z-1)^3(1-z)}}\\ &=(1+18z-27z^2)^{-2a}\FF a{a+\frac12}{2a+\frac56}{\frac{64z}{(1+18z-27z^2)^2}}\\ \FF{4a}{2a+\frac16}{6a+\frac12}z &=\l(1-\frac z9\r)^{-3a}(1-z)^{-a}\FF a{a+\frac13}{2a+\frac56}{\frac{64z^3}{(z-9)^3(1-z)}}\\ &=\l(\frac{27-18-z^2}{27}\r)^{-2a}\FF a{a+\frac12}{2a+\frac56}{\frac{64z^3}{(z^2+18z-27)^2}}\\ \FF{4a}{4a+\frac13}{6a+\frac12}z &=\l(1-\frac89z\r)^{-3a}\FF a{a+\frac13}{2a+\frac56}{\frac{64z^3(1-z)}{(9-8z)^3}}\\ &=\l(\farc{27-36z+8z^2}{27}\r)^{-2a}\FF a{a+\frac12}{2a+\frac56}{\frac{-64z^3(1-z)}{(8z^2-36z+27)^2}}\\ \FF{4a}{4a+\frac13}{2a+\frac56}z &=(1+8z)^{-3a}\FF a{a+\frac13}{2a+\frac56}{\frac{64z(1-z)^3}{(1+8z)^3}}\\ &=(1-20z-8z^2)^{-2a}\FF a{a+\frac12}{2a+\frac56}{\frac{-64z(1-z)^3}{(1-20z-8z^2)^2}}\\ \FF{4a}{2a+\frac14}{2a+\frac34}z &=(1+z)^{-4a}\FF a{a+\frac14}{2a+\frac34}{\frac{16z(1-z)^2}{(1+z)^4}}\\ &=(1-6z+z^2)^{-2a}\FF a{a+\frac12}{2a+\frac34}{\frac{-16z(1-z)^2}{(z^2-6z+1)^2}}\\ \FF{4a}{2a+\frac14}{4a+\frac12}z &=\l(1-\frac z2\r)^{-4a}\FF a{a+\frac14}{2a+\frac34}{\frac{16z^2(1-z)}{(2-z)^4}}\\ &=\l(\frac{4-4z-z^2}4\r)^{-2a}\FF a{a+\frac12}{2a+\frac34}{\frac{-16z^2(1-z)}{(z^2+4z-4)^2}}\\ \FF{4a}{\frac12}{2a+\frac34}z &=(1-2z)^{-4a}\FF a{a+\frac14}{2a+\frac34}{\frac{-16z(1-z)}{(2z-1)^4}}\\ &=(1+4z-z^2)^{-2a}\FF a{a+\frac12}{2a+\frac34}{\frac{16z(1-z)}{(1+4z-4z^2)^2}} \end{align}

六次変換

\begin{align} \FF{6a}{2a+\frac13}{4a+\frac23}z &=(1-z+z^2)^{-3a}\FF a{a+\frac13}{2a+\frac56}{\frac{27z^2(1-z)^2}{4(z^2-z+1)^3}}\\ &=\l(\frac{2-3z-3z^2+2z^3}2\r)^{-2a} \\ &\qquad\times\FF a{a+\frac12}{2a+\frac56}{\frac{-27z^2(1-z)^2}{(2z^3-3z^2-3z+2)^2}}\\ \FF{6a}{\frac23-2a}{2a+\frac56}z &=(1-16z+16z^2)^{-3a}\FF a{a+\frac13}{2a+\frac56}{\frac{108z(1-z)}{(1-16z+16z^2)^3}}\\ &=(1+30z-96z^2+64z^3)^{-2a} \\ &\qquad\times\FF a{a+\frac12}{2a+\frac56}{\frac{-108z(1-z)}{(64z^3-96z^2+30z+1)^2}}\\ \FF{6a}{4a+\frac16}{2a+\frac56}z &=\l(\frac{16-16z+z^2}{16}\r)^{-3a}\FF a{a+\frac13}{2a+\frac56}{\frac{-108z^4(1-z)}{(z^2-16z+16)^3}}\\ &=\l(\frac{64-96z+30z^2+z^3}{64}\r)^{-2a} \\ &\qquad\times\FF a{a+\frac12}{2a+\frac56}{\frac{-108z^4(1-z)}{(z^3+30z^2-96z+64)^2}}\\ \FF{6a}{4a+\frac16}{8a+\frac13}z &=(1+14z+z^2)^{-3a}\FF a{a+\frac13}{2a+\frac56}{\frac{-108z(1-z)^4}{(1+14z+z^2)^3}}\\ &=(1-33z-33z^2+z^3)^{-2a} \\ &\qquad\times\FF a{a+\frac12}{2a+\frac56}{\frac{108z(1-z)^4}{(z^3-33z-33z+1)^2}} \end{align}

無理変換

\begin{align*} \FF ab{a+b+\frac12}z &=\FF{2a}{2b}{a+b+\frac12}{\frac{1-\sqrt{1-z}}2}\\ &=\l(\frac{1+\sqrt{1-z}}2\r)^{\frac12-a-b}\FF{b-a+\frac12}{a-b+\frac12}{a+b+\frac12}{\frac{1-\sqrt{1-z}}2}\\ &=(\sqrt{1-z}+\sqrt{-z})^{-2a}\FF{2a}{a+b}{2(a+b)}{\frac{2\sqrt{-z}}{\sqrt{1-z}+\sqrt{-z}}}\\ &=\l(\frac{1+\sqrt{1-z}}2\r)^{-2a}\FF{2a}{a-b+\frac12}{a+b+\frac12}{\frac{\sqrt{1-z}-1}{\sqrt{1-z}+1}}\\ \FF ab{a+b-\frac12}z &=\frac1{\sqrt{1-z}}\FF{2a-1}{2b-1}{a+b-\frac12}{\frac{1-\sqrt{1-z}}2}\\ &=\frac1{\sqrt{1-z}}\l(\frac{1+\sqrt{1-z}}2\r)^{\frac32-a-b}\FF{b-a+\frac12}{a-b+\frac12}{a+b-\frac12}{\frac{1-\sqrt{1-z}}2}\\ &=\frac1{\sqrt{1-z}}(\sqrt{1-z}+\sqrt{-z})^{1-2a}\FF{2a-1}{a+b-1}{2(a+b-1)}{\frac{2\sqrt{-z}}{\sqrt{1-z}+\sqrt{-z}}}\\ &=\frac1{\sqrt{1-z}}\l(\frac{1+\sqrt{1-z}}2\r)^{1-2a}\FF{2a-1}{a-b+\frac12}{a+b-\frac12}{\frac{\sqrt{1-z}-1}{\sqrt{1-z}+1}}\\ \FF a{a+\frac12}cz &=(1-z)^{-a}\FF{2a}{2(c-a)-1}c{\frac{\sqrt{1-z}-1}{2\sqrt{1-z}}}\\ &=(1-z)^{\frac{c-2a-1}2}\l(\frac{1+\sqrt{1-z}}2\r)^{1-c}\FF{c-2a}{c+2a+1}c{\frac{\sqrt{1-z}-1}{2\sqrt{1-z}}}\\ &=(1+\sqrt z)^{-2a}\FF{2a}{c-\frac12}{2c-1}{\frac{2\sqrt z}{1+\sqrt z}}\\ &=(1+\sqrt z)^{\frac12-c}(1-\sqrt z)^{c-2a-\frac12}\FF{2(c-a)-1}{c-\frac12}{2c-1}{\frac{2\sqrt z}{1+\sqrt z}}\\ &=\l(\frac{1+\sqrt{1-z}}2\r)^{-2a}\FF{2a}{2a-c+1}c{\frac{1-\sqrt{1-z}}{1+\sqrt{1-z}}}\\ &=\l(\frac{1+\sqrt{1-z}}2\r)^{2a-2c+1}(1-z)^{c-2a-\frac12}\FF{c-2a}{2(c-a)-1}c{\frac{1-\sqrt{1-z}}{1+\sqrt{1-z}}} \end{align*}

特殊な無理変換

\begin{align} \FF ab{2b}z &=(1-z)^{-\frac a2}\FF a{2b-a}{b+\frac12}{\frac{(1-\sqrt{1-z})^2}{-4\sqrt{1-z}}}\\ &=(1-z)^{\frac14+\frac{b-a}2}\l(\frac{1+\sqrt{1-z}}2\r)^{1-2b}\FF{b-a+\frac12}{a-b+\frac12}{b+\frac12}{\frac{(1-\sqrt{1-z})^2}{-4\sqrt{1-z}}}\\ &=\l(\frac{1+\sqrt{1-z}}2\r)^{-2a}\FF a{a-b+\frac12}{b+\frac12}{\l(\frac{1-\sqrt{1-z}}{1+\sqrt{1-z}}\r)^2}\\ &=\l(\frac{1+\sqrt{1-z}}2\r)^{2a-4b}(1-z)^{b-a}\FF{2b-a}{a-b+\frac12}{b+\frac12}{\l(\frac{1-\sqrt{1-z}}{1+\sqrt{1-z}}\r)^2}\\ \FF ab{a-b+1}z &=(1+\sqrt z)^{-2a}\FF a{a-b+\frac12}{2(a-b)+1}{\frac{4\sqrt z}{(1+\sqrt z)^2}}\\ &=(1+\sqrt z)^{4b-2a-2}(1-z)^{1-2b}\FF{a-2b+1}{a-b+\frac12}{2(a-b)+1}{\frac{4\sqrt z}{(1+\sqrt z)^2}}\\ \FF ab{\frac{a+b+1}2}z &=(\sqrt{1-z}+\sqrt{-z})^{-2a}\FF a{\frac{a+b}2}{a+b}{\frac{4\sqrt{z(z-1)}}{(\sqrt{1-z}+\sqrt{-z})^2}}\\ \FF a{1-a}cz &=(\sqrt{1-z}+\sqrt{-z})^{2a-2c}(1-z)^{c-1}\FF{c-a}{c-\frac12}{2c-1}{\frac{4\sqrt{z(z-1)}}{(\sqrt{1-z}+\sqrt{-z})^2}} \end{align}