§15.10 Hypergeometric Differential Equation

§15.10(i) Fundamental Solutions

 15.10.1 $z(1-z)\frac{{\mathrm{d}}^{2}w}{{\mathrm{d}z}^{2}}+\left(c-(a+b+1)z\right)\frac% {\mathrm{d}w}{\mathrm{d}z}-abw=0.$

This is the hypergeometric differential equation. It has regular singularities at $z=0,1,\infty$, with corresponding exponent pairs $\{0,1-c\}$, $\{0,c-a-b\}$, $\{a,b\}$, respectively. When none of the exponent pairs differ by an integer, that is, when none of $c$, $c-a-b$, $a-b$ is an integer, we have the following pairs $f_{1}(z)$, $f_{2}(z)$ of fundamental solutions. They are also numerically satisfactory (§2.7(iv)) in the neighborhood of the corresponding singularity.

Singularity $z=0$

 15.10.2 $\displaystyle f_{1}(z)$ $\displaystyle=\mathop{F\/}\nolimits\!\left({a,b\atop c};z\right),$ $\displaystyle f_{2}(z)$ $\displaystyle=z^{1-c}\mathop{F\/}\nolimits\!\left({a-c+1,b-c+1\atop 2-c};z% \right),$
 15.10.3 $\mathop{\mathscr{W}\/}\nolimits\left\{f_{1}(z),f_{2}(z)\right\}=(1-c)z^{-c}(1-% z)^{c-a-b-1}.$

Singularity $z=1$

 15.10.4 $\displaystyle f_{1}(z)$ $\displaystyle=\mathop{F\/}\nolimits\!\left({a,b\atop a+b+1-c};1-z\right),$ $\displaystyle f_{2}(z)$ $\displaystyle=(1-z)^{c-a-b}\mathop{F\/}\nolimits\!\left({c-a,c-b\atop c-a-b+1}% ;1-z\right),$
 15.10.5 $\mathop{\mathscr{W}\/}\nolimits\left\{f_{1}(z),f_{2}(z)\right\}=(a+b-c)z^{-c}(% 1-z)^{c-a-b-1}.$

Singularity $z=\infty$

 15.10.6 $\displaystyle f_{1}(z)$ $\displaystyle=z^{-a}\mathop{F\/}\nolimits\!\left({a,a-c+1\atop a-b+1};\frac{1}% {z}\right),$ $\displaystyle f_{2}(z)$ $\displaystyle=z^{-b}\mathop{F\/}\nolimits\!\left({b,b-c+1\atop b-a+1};\frac{1}% {z}\right),$
 15.10.7 $\mathop{\mathscr{W}\/}\nolimits\left\{f_{1}(z),f_{2}(z)\right\}=(a-b)z^{-c}(z-% 1)^{c-a-b-1}.$

(a) If $c$ equals $n=1,2,3,\dots$, and $a=1,2,\dots,n-1$, then fundamental solutions in the neighborhood of $z=0$ are given by (15.10.2) with the interpretation (15.2.5) for $f_{2}(z)$.

(b) If $c$ equals $n=1,2,3,\dots$, and $a\neq 1,2,\dots,n-1$, then fundamental solutions in the neighborhood of $z=0$ are given by $\mathop{F\/}\nolimits\!\left(a,b;n;z\right)$ and

 15.10.8 $\mathop{F\/}\nolimits\!\left({a,b\atop n};z\right)\mathop{\ln\/}\nolimits z-% \sum_{k=1}^{n-1}\frac{(n-1)!(k-1)!}{(n-k-1)!{\left(1-a\right)_{k}}{\left(1-b% \right)_{k}}}(-z)^{-k}\\ +\sum_{k=0}^{\infty}\frac{{\left(a\right)_{k}}{\left(b\right)_{k}}}{{\left(n% \right)_{k}}k!}z^{k}\left(\mathop{\psi\/}\nolimits\!\left(a+k\right)+\mathop{% \psi\/}\nolimits\!\left(b+k\right)-\mathop{\psi\/}\nolimits\!\left(1+k\right)-% \mathop{\psi\/}\nolimits\!\left(n+k\right)\right),$ $a,b\neq n-1,n-2,\dots,0,-1,-2,\dots$, Symbols: $\mathop{F\/}\nolimits\!\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $\mathop{F\/}\nolimits\!\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $=\mathop{{{}_{2}F_{1}}\/}\nolimits\!\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, ${\left(\NVar{a}\right)_{\NVar{n}}}$: Pochhammer’s symbol (or shifted factorial), $\mathop{\psi\/}\nolimits\!\left(\NVar{z}\right)$: psi (or digamma) function, $!$: factorial (as in $n!$), $\mathop{\ln\/}\nolimits\NVar{z}$: principal branch of logarithm function, $n$: integer, $z$: complex variable, $a$: real or complex parameter, $b$: real or complex parameter, $k$: integer and $m$: integer A&S Ref: 15.5.19. (Compared with (15.10.8) this result has a multiple of $\mathop{F\/}\nolimits\!\left(a,b;n;z\right)$ added to the right-hand side. It also contains an error: the conditions on $a$ and $b$ should be $a,b\neq 1,2,\dots,m$.) Referenced by: 15.10.8 Permalink: http://dlmf.nist.gov/15.10.E8 Encodings: TeX, pMML, png See also: Annotations for 15.10(i)

or

 15.10.9 $\mathop{F\/}\nolimits\!\left({-m,b\atop n};z\right)\mathop{\ln\/}\nolimits z-% \sum_{k=1}^{n-1}\frac{(n-1)!(k-1)!}{(n-k-1)!{\left(m+1\right)_{k}}{\left(1-b% \right)_{k}}}(-z)^{-k}+\sum_{k=0}^{m}\frac{{\left(-m\right)_{k}}{\left(b\right% )_{k}}}{{\left(n\right)_{k}}k!}z^{k}\left(\mathop{\psi\/}\nolimits\!\left(1+m-% k\right)+\mathop{\psi\/}\nolimits\!\left(b+k\right)-\mathop{\psi\/}\nolimits\!% \left(1+k\right)-\mathop{\psi\/}\nolimits\!\left(n+k\right)\right)+(-1)^{m}m!% \sum_{k=m+1}^{\infty}\frac{(k-1-m)!{\left(b\right)_{k}}}{{\left(n\right)_{k}}k% !}z^{k},$ $a=-m$, $m=0,1,2,\dots$; $b\neq n-1,n-2,\dots,0,-1,-2,\dots$,

or

 15.10.10 $\mathop{F\/}\nolimits\!\left({-m,-\ell\atop n};z\right)\mathop{\ln\/}\nolimits z% -\sum_{k=1}^{n-1}\frac{(n-1)!(k-1)!}{(n-k-1)!{\left(m+1\right)_{k}}{\left(\ell% +1\right)_{k}}}(-z)^{-k}+\sum_{k=0}^{\ell}\frac{{\left(-m\right)_{k}}{\left(-% \ell\right)_{k}}}{{\left(n\right)_{k}}k!}z^{k}\left(\mathop{\psi\/}\nolimits\!% \left(1+m-k\right)+\mathop{\psi\/}\nolimits\!\left(1+\ell-k\right)-\mathop{% \psi\/}\nolimits\!\left(1+k\right)-\mathop{\psi\/}\nolimits\!\left(n+k\right)% \right)+(-1)^{\ell}\ell\,!\sum_{k=\ell+1}^{m}\frac{(k-1-\ell)!{\left(-m\right)% _{k}}}{{\left(n\right)_{k}}k!}z^{k},$ $a=-m$, $m=0,1,2,\dots$; $b=-\ell$, $\ell=0,1,2,\dots,m$.

Moreover, in (15.10.9) and (15.10.10) the symbols $a$ and $b$ are interchangeable.

(c) If the parameter $c$ in the differential equation equals $2-n=0,-1,-2,\dots$, then fundamental solutions in the neighborhood of $z=0$ are given by $z^{n-1}$ times those in (a) and (b), with $a$ and $b$ replaced throughout by $a+n-1$ and $b+n-1$, respectively.

(d) If $a+b+1-c$ equals $n=1,2,3,\dots$, or $2-n=0,-1,-2,\dots$, then fundamental solutions in the neighborhood of $z=1$ are given by those in (a), (b), and (c) with $z$ replaced by $1-z$.

(e) Finally, if $a-b+1$ equals $n=1,2,3,\dots$, or $2-n=0,-1,-2,\dots$, then fundamental solutions in the neighborhood of $z=\infty$ are given by $z^{-a}$ times those in (a), (b), and (c) with $b$ and $z$ replaced by $a-c+1$ and $\ifrac{1}{z}$, respectively.

§15.10(ii) Kummer’s 24 Solutions and Connection Formulas

The three pairs of fundamental solutions given by (15.10.2), (15.10.4), and (15.10.6) can be transformed into 18 other solutions by means of (15.8.1), leading to a total of 24 solutions known as Kummer’s solutions.

 15.10.11 $\displaystyle w_{1}(z)$ $\displaystyle=\mathop{F\/}\nolimits\!\left({a,b\atop c};z\right)=(1-z)^{c-a-b}% \mathop{F\/}\nolimits\!\left({c-a,c-b\atop c};z\right)=(1-z)^{-a}\mathop{F\/}% \nolimits\!\left({a,c-b\atop c};\frac{z}{z-1}\right)=(1-z)^{-b}\mathop{F\/}% \nolimits\!\left({c-a,b\atop c};\frac{z}{z-1}\right).$ 15.10.12 $\displaystyle w_{2}(z)$ $\displaystyle={z^{1-c}}\mathop{F\/}\nolimits\!\left({a-c+1,b-c+1\atop 2-c};z\right)$ $\displaystyle={z^{1-c}(1-z)^{c-a-b}}\*\mathop{F\/}\nolimits\!\left({1-a,1-b% \atop 2-c};z\right)$ $\displaystyle={z^{1-c}(1-z)^{c-a-1}}\*\mathop{F\/}\nolimits\!\left({a-c+1,1-b% \atop 2-c};\frac{z}{z-1}\right)$ $\displaystyle={z^{1-c}(1-z)^{c-b-1}}\*\mathop{F\/}\nolimits\!\left({1-a,b-c+1% \atop 2-c};\frac{z}{z-1}\right).$ 15.10.13 $\displaystyle w_{3}(z)$ $\displaystyle=\mathop{F\/}\nolimits\!\left({a,b\atop a+b-c+1};1-z\right)=z^{1-% c}\mathop{F\/}\nolimits\!\left({a-c+1,b-c+1\atop a+b-c+1};1-z\right)=z^{-a}% \mathop{F\/}\nolimits\!\left({a,a-c+1\atop a+b-c+1};1-\frac{1}{z}\right)=z^{-b% }\mathop{F\/}\nolimits\!\left({b,b-c+1\atop a+b-c+1};1-\frac{1}{z}\right).$ 15.10.14 $\displaystyle w_{4}(z)$ $\displaystyle=(1-z)^{c-a-b}\mathop{F\/}\nolimits\!\left({c-a,c-b\atop c-a-b+1}% ;1-z\right)$ $\displaystyle=z^{1-c}(1-z)^{c-a-b}\mathop{F\/}\nolimits\!\left({1-a,1-b\atop c% -a-b+1};1-z\right)$ $\displaystyle=z^{a-c}(1-z)^{c-a-b}\mathop{F\/}\nolimits\!\left({1-a,c-a\atop c% -a-b+1};1-\frac{1}{z}\right)$ $\displaystyle=z^{b-c}(1-z)^{c-a-b}\mathop{F\/}\nolimits\!\left({1-b,c-b\atop c% -a-b+1};1-\frac{1}{z}\right).$ 15.10.15 $\displaystyle w_{5}(z)$ $\displaystyle=e^{a\pi\mathrm{i}}z^{-a}\*\mathop{F\/}\nolimits\!\left({a,a-c+1% \atop a-b+1};\frac{1}{z}\right)$ $\displaystyle=e^{(c-b)\pi\mathrm{i}}z^{b-c}(1-z)^{c-a-b}\*\mathop{F\/}% \nolimits\!\left({1-b,c-b\atop a-b+1};\frac{1}{z}\right)$ $\displaystyle=(1-z)^{-a}\mathop{F\/}\nolimits\!\left({a,c-b\atop a-b+1};\frac{% 1}{1-z}\right)$ $\displaystyle=e^{(c-1)\pi\mathrm{i}}z^{1-c}(1-z)^{c-a-1}\*\mathop{F\/}% \nolimits\!\left({1-b,a-c+1\atop a-b+1};\frac{1}{1-z}\right).$ 15.10.16 $\displaystyle w_{6}(z)$ $\displaystyle=e^{b\pi\mathrm{i}}z^{-b}\mathop{F\/}\nolimits\!\left({b,b-c+1% \atop b-a+1};\frac{1}{z}\right)$ $\displaystyle=e^{(c-a)\pi\mathrm{i}}z^{a-c}(1-z)^{c-a-b}\*\mathop{F\/}% \nolimits\!\left({1-a,c-a\atop b-a+1};\frac{1}{z}\right)$ $\displaystyle=(1-z)^{-b}\mathop{F\/}\nolimits\!\left({b,c-a\atop b-a+1};\frac{% 1}{1-z}\right)$ $\displaystyle=e^{(c-1)\pi\mathrm{i}}z^{1-c}(1-z)^{c-b-1}\*\mathop{F\/}% \nolimits\!\left({1-a,b-c+1\atop b-a+1};\frac{1}{1-z}\right).$

The $\binom{6}{3}=20$ connection formulas for the principal branches of Kummer’s solutions are:

 15.10.17 $\displaystyle w_{3}(z)$ $\displaystyle=\frac{\mathop{\Gamma\/}\nolimits\!\left(1-c\right)\mathop{\Gamma% \/}\nolimits\!\left(a+b-c+1\right)}{\mathop{\Gamma\/}\nolimits\!\left(a-c+1% \right)\mathop{\Gamma\/}\nolimits\!\left(b-c+1\right)}w_{1}(z)+\frac{\mathop{% \Gamma\/}\nolimits\!\left(c-1\right)\mathop{\Gamma\/}\nolimits\!\left(a+b-c+1% \right)}{\mathop{\Gamma\/}\nolimits\!\left(a\right)\mathop{\Gamma\/}\nolimits% \!\left(b\right)}w_{2}(z),$ 15.10.18 $\displaystyle w_{4}(z)$ $\displaystyle=\frac{\mathop{\Gamma\/}\nolimits\!\left(1-c\right)\mathop{\Gamma% \/}\nolimits\!\left(c-a-b+1\right)}{\mathop{\Gamma\/}\nolimits\!\left(1-a% \right)\mathop{\Gamma\/}\nolimits\!\left(1-b\right)}w_{1}(z)+\frac{\mathop{% \Gamma\/}\nolimits\!\left(c-1\right)\mathop{\Gamma\/}\nolimits\!\left(c-a-b+1% \right)}{\mathop{\Gamma\/}\nolimits\!\left(c-a\right)\mathop{\Gamma\/}% \nolimits\!\left(c-b\right)}w_{2}(z),$ 15.10.19 $\displaystyle w_{5}(z)$ $\displaystyle=\frac{\mathop{\Gamma\/}\nolimits\!\left(1-c\right)\mathop{\Gamma% \/}\nolimits\!\left(a-b+1\right)}{\mathop{\Gamma\/}\nolimits\!\left(a-c+1% \right)\mathop{\Gamma\/}\nolimits\!\left(1-b\right)}w_{1}(z)+e^{(c-1)\pi% \mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(c-1\right)\mathop{\Gamma\/}% \nolimits\!\left(a-b+1\right)}{\mathop{\Gamma\/}\nolimits\!\left(a\right)% \mathop{\Gamma\/}\nolimits\!\left(c-b\right)}w_{2}(z),$ 15.10.20 $\displaystyle w_{6}(z)$ $\displaystyle=\frac{\mathop{\Gamma\/}\nolimits\!\left(1-c\right)\mathop{\Gamma% \/}\nolimits\!\left(b-a+1\right)}{\mathop{\Gamma\/}\nolimits\!\left(b-c+1% \right)\mathop{\Gamma\/}\nolimits\!\left(1-a\right)}w_{1}(z)+e^{(c-1)\pi% \mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(c-1\right)\mathop{\Gamma\/}% \nolimits\!\left(b-a+1\right)}{\mathop{\Gamma\/}\nolimits\!\left(b\right)% \mathop{\Gamma\/}\nolimits\!\left(c-a\right)}w_{2}(z).$
 15.10.21 $\displaystyle w_{1}(z)$ $\displaystyle=\frac{\mathop{\Gamma\/}\nolimits\!\left(c\right)\mathop{\Gamma\/% }\nolimits\!\left(c-a-b\right)}{\mathop{\Gamma\/}\nolimits\!\left(c-a\right)% \mathop{\Gamma\/}\nolimits\!\left(c-b\right)}w_{3}(z)+\frac{\mathop{\Gamma\/}% \nolimits\!\left(c\right)\mathop{\Gamma\/}\nolimits\!\left(a+b-c\right)}{% \mathop{\Gamma\/}\nolimits\!\left(a\right)\mathop{\Gamma\/}\nolimits\!\left(b% \right)}w_{4}(z),$ 15.10.22 $\displaystyle w_{2}(z)$ $\displaystyle=\frac{\mathop{\Gamma\/}\nolimits\!\left(2-c\right)\mathop{\Gamma% \/}\nolimits\!\left(c-a-b\right)}{\mathop{\Gamma\/}\nolimits\!\left(1-a\right)% \mathop{\Gamma\/}\nolimits\!\left(1-b\right)}w_{3}(z)+\frac{\mathop{\Gamma\/}% \nolimits\!\left(2-c\right)\mathop{\Gamma\/}\nolimits\!\left(a+b-c\right)}{% \mathop{\Gamma\/}\nolimits\!\left(a-c+1\right)\mathop{\Gamma\/}\nolimits\!% \left(b-c+1\right)}w_{4}(z),$ 15.10.23 $\displaystyle w_{5}(z)$ $\displaystyle=e^{a\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(a-b+1% \right)\mathop{\Gamma\/}\nolimits\!\left(c-a-b\right)}{\mathop{\Gamma\/}% \nolimits\!\left(1-b\right)\mathop{\Gamma\/}\nolimits\!\left(c-b\right)}w_{3}(% z)+e^{(c-b)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(a-b+1\right)% \mathop{\Gamma\/}\nolimits\!\left(a+b-c\right)}{\mathop{\Gamma\/}\nolimits\!% \left(a\right)\mathop{\Gamma\/}\nolimits\!\left(a-c+1\right)}w_{4}(z),$ 15.10.24 $\displaystyle w_{6}(z)$ $\displaystyle=e^{b\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(b-a+1% \right)\mathop{\Gamma\/}\nolimits\!\left(c-a-b\right)}{\mathop{\Gamma\/}% \nolimits\!\left(1-a\right)\mathop{\Gamma\/}\nolimits\!\left(c-a\right)}w_{3}(% z)+e^{(c-a)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(b-a+1\right)% \mathop{\Gamma\/}\nolimits\!\left(a+b-c\right)}{\mathop{\Gamma\/}\nolimits\!% \left(b\right)\mathop{\Gamma\/}\nolimits\!\left(b-c+1\right)}w_{4}(z).$
 15.10.25 $\displaystyle w_{1}(z)$ $\displaystyle=\frac{\mathop{\Gamma\/}\nolimits\!\left(c\right)\mathop{\Gamma\/% }\nolimits\!\left(b-a\right)}{\mathop{\Gamma\/}\nolimits\!\left(b\right)% \mathop{\Gamma\/}\nolimits\!\left(c-a\right)}w_{5}(z)+\frac{\mathop{\Gamma\/}% \nolimits\!\left(c\right)\mathop{\Gamma\/}\nolimits\!\left(a-b\right)}{\mathop% {\Gamma\/}\nolimits\!\left(a\right)\mathop{\Gamma\/}\nolimits\!\left(c-b\right% )}w_{6}(z),$ 15.10.26 $\displaystyle w_{2}(z)$ $\displaystyle=e^{(1-c)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(2-% c\right)\mathop{\Gamma\/}\nolimits\!\left(b-a\right)}{\mathop{\Gamma\/}% \nolimits\!\left(1-a\right)\mathop{\Gamma\/}\nolimits\!\left(b-c+1\right)}w_{5% }(z)+e^{(1-c)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(2-c\right)% \mathop{\Gamma\/}\nolimits\!\left(a-b\right)}{\mathop{\Gamma\/}\nolimits\!% \left(1-b\right)\mathop{\Gamma\/}\nolimits\!\left(a-c+1\right)}w_{6}(z),$ 15.10.27 $\displaystyle w_{3}(z)$ $\displaystyle=e^{-a\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(a+b-c% +1\right)\mathop{\Gamma\/}\nolimits\!\left(b-a\right)}{\mathop{\Gamma\/}% \nolimits\!\left(b\right)\mathop{\Gamma\/}\nolimits\!\left(b-c+1\right)}w_{5}(% z)+e^{-b\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(a+b-c+1\right)% \mathop{\Gamma\/}\nolimits\!\left(a-b\right)}{\mathop{\Gamma\/}\nolimits\!% \left(a\right)\mathop{\Gamma\/}\nolimits\!\left(a-c+1\right)}w_{6}(z),$ 15.10.28 $\displaystyle w_{4}(z)$ $\displaystyle=e^{(b-c)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(c-% a-b+1\right)\mathop{\Gamma\/}\nolimits\!\left(b-a\right)}{\mathop{\Gamma\/}% \nolimits\!\left(1-a\right)\mathop{\Gamma\/}\nolimits\!\left(c-a\right)}w_{5}(% z)+e^{(a-c)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(c-a-b+1\right% )\mathop{\Gamma\/}\nolimits\!\left(a-b\right)}{\mathop{\Gamma\/}\nolimits\!% \left(1-b\right)\mathop{\Gamma\/}\nolimits\!\left(c-b\right)}w_{6}(z).$
 15.10.29 $\displaystyle w_{1}(z)$ $\displaystyle=e^{b\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(c% \right)\mathop{\Gamma\/}\nolimits\!\left(a-c+1\right)}{\mathop{\Gamma\/}% \nolimits\!\left(a+b-c+1\right)\mathop{\Gamma\/}\nolimits\!\left(c-b\right)}w_% {3}(z)+e^{(b-c)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(c\right)% \mathop{\Gamma\/}\nolimits\!\left(a-c+1\right)}{\mathop{\Gamma\/}\nolimits\!% \left(b\right)\mathop{\Gamma\/}\nolimits\!\left(a-b+1\right)}w_{5}(z),$ 15.10.30 $\displaystyle w_{1}(z)$ $\displaystyle=e^{a\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(c% \right)\mathop{\Gamma\/}\nolimits\!\left(b-c+1\right)}{\mathop{\Gamma\/}% \nolimits\!\left(a+b-c+1\right)\mathop{\Gamma\/}\nolimits\!\left(c-a\right)}w_% {3}(z)+e^{(a-c)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(c\right)% \mathop{\Gamma\/}\nolimits\!\left(b-c+1\right)}{\mathop{\Gamma\/}\nolimits\!% \left(a\right)\mathop{\Gamma\/}\nolimits\!\left(b-a+1\right)}w_{6}(z),$ 15.10.31 $\displaystyle w_{2}(z)$ $\displaystyle=e^{(b-c+1)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(% 2-c\right)\mathop{\Gamma\/}\nolimits\!\left(a\right)}{\mathop{\Gamma\/}% \nolimits\!\left(a+b-c+1\right)\mathop{\Gamma\/}\nolimits\!\left(1-b\right)}w_% {3}(z)+e^{(b-c)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(2-c\right% )\mathop{\Gamma\/}\nolimits\!\left(a\right)}{\mathop{\Gamma\/}\nolimits\!\left% (a-b+1\right)\mathop{\Gamma\/}\nolimits\!\left(b-c+1\right)}w_{5}(z),$ 15.10.32 $\displaystyle w_{2}(z)$ $\displaystyle=e^{(a-c+1)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(% 2-c\right)\mathop{\Gamma\/}\nolimits\!\left(b\right)}{\mathop{\Gamma\/}% \nolimits\!\left(a+b-c+1\right)\mathop{\Gamma\/}\nolimits\!\left(1-a\right)}w_% {3}(z)+e^{(a-c)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(2-c\right% )\mathop{\Gamma\/}\nolimits\!\left(b\right)}{\mathop{\Gamma\/}\nolimits\!\left% (b-a+1\right)\mathop{\Gamma\/}\nolimits\!\left(a-c+1\right)}w_{6}(z).$
 15.10.33 $\displaystyle w_{1}(z)$ $\displaystyle=e^{(c-a)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(c% \right)\mathop{\Gamma\/}\nolimits\!\left(1-b\right)}{\mathop{\Gamma\/}% \nolimits\!\left(a\right)\mathop{\Gamma\/}\nolimits\!\left(c-a-b+1\right)}w_{4% }(z)+e^{-a\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(c\right)% \mathop{\Gamma\/}\nolimits\!\left(1-b\right)}{\mathop{\Gamma\/}\nolimits\!% \left(a-b+1\right)\mathop{\Gamma\/}\nolimits\!\left(c-a\right)}w_{5}(z),$ 15.10.34 $\displaystyle w_{1}(z)$ $\displaystyle=e^{(c-b)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(c% \right)\mathop{\Gamma\/}\nolimits\!\left(1-a\right)}{\mathop{\Gamma\/}% \nolimits\!\left(b\right)\mathop{\Gamma\/}\nolimits\!\left(c-a-b+1\right)}w_{4% }(z)+e^{-b\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(c\right)% \mathop{\Gamma\/}\nolimits\!\left(1-a\right)}{\mathop{\Gamma\/}\nolimits\!% \left(b-a+1\right)\mathop{\Gamma\/}\nolimits\!\left(c-b\right)}w_{6}(z),$ 15.10.35 $\displaystyle w_{2}(z)$ $\displaystyle=e^{(1-a)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(2-% c\right)\mathop{\Gamma\/}\nolimits\!\left(c-b\right)}{\mathop{\Gamma\/}% \nolimits\!\left(a-c+1\right)\mathop{\Gamma\/}\nolimits\!\left(c-a-b+1\right)}% w_{4}(z)+e^{-a\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(2-c\right)% \mathop{\Gamma\/}\nolimits\!\left(c-b\right)}{\mathop{\Gamma\/}\nolimits\!% \left(a-b+1\right)\mathop{\Gamma\/}\nolimits\!\left(1-a\right)}w_{5}(z),$ 15.10.36 $\displaystyle w_{2}(z)$ $\displaystyle=e^{(1-b)\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(2-% c\right)\mathop{\Gamma\/}\nolimits\!\left(c-a\right)}{\mathop{\Gamma\/}% \nolimits\!\left(b-c+1\right)\mathop{\Gamma\/}\nolimits\!\left(c-a-b+1\right)}% w_{4}(z)+e^{-b\pi\mathrm{i}}\frac{\mathop{\Gamma\/}\nolimits\!\left(2-c\right)% \mathop{\Gamma\/}\nolimits\!\left(c-a\right)}{\mathop{\Gamma\/}\nolimits\!% \left(b-a+1\right)\mathop{\Gamma\/}\nolimits\!\left(1-b\right)}w_{6}(z).$