# §15.8 Transformations of Variable

## §15.8(i) Linear Transformations

All functions in this subsection and §15.8(ii) assume their principal values.

 15.8.1 $\mathbf{F}\left({a,b\atop c};z\right)=(1-z)^{-a}\mathbf{F}\left({a,c-b\atop c}% ;\frac{z}{z-1}\right)=(1-z)^{-b}\mathbf{F}\left({c-a,b\atop c};\frac{z}{z-1}% \right)=(1-z)^{c-a-b}\mathbf{F}\left({c-a,c-b\atop c};z\right),$ $|\operatorname{ph}\left(1-z\right)|<\pi$.
 15.8.2 $\displaystyle\frac{\sin\left(\pi(b-a)\right)}{\pi}\mathbf{F}\left({a,b\atop c}% ;z\right)$ $\displaystyle=\frac{(-z)^{-a}}{\Gamma\left(b\right)\Gamma\left(c-a\right)}% \mathbf{F}\left({a,a-c+1\atop a-b+1};\frac{1}{z}\right)-\frac{(-z)^{-b}}{% \Gamma\left(a\right)\Gamma\left(c-b\right)}\mathbf{F}\left({b,b-c+1\atop b-a+1% };\frac{1}{z}\right),$ $|\operatorname{ph}\left(-z\right)|<\pi$. 15.8.3 $\displaystyle\frac{\sin\left(\pi(b-a)\right)}{\pi}\mathbf{F}\left({a,b\atop c}% ;z\right)$ $\displaystyle=\frac{(1-z)^{-a}}{\Gamma\left(b\right)\Gamma\left(c-a\right)}% \mathbf{F}\left({a,c-b\atop a-b+1};\frac{1}{1-z}\right)-\frac{(1-z)^{-b}}{% \Gamma\left(a\right)\Gamma\left(c-b\right)}\mathbf{F}\left({b,c-a\atop b-a+1};% \frac{1}{1-z}\right),$ $|\operatorname{ph}\left(-z\right)|<\pi$. 15.8.4 $\displaystyle\frac{\sin\left(\pi(c-a-b)\right)}{\pi}\mathbf{F}\left({a,b\atop c% };z\right)$ $\displaystyle=\frac{1}{\Gamma\left(c-a\right)\Gamma\left(c-b\right)}\mathbf{F}% \left({a,b\atop a+b-c+1};1-z\right)-\frac{(1-z)^{c-a-b}}{\Gamma\left(a\right)% \Gamma\left(b\right)}\mathbf{F}\left({c-a,c-b\atop c-a-b+1};1-z\right),$ $|\operatorname{ph}z|<\pi$, $|\operatorname{ph}\left(1-z\right)|<\pi$. 15.8.5 $\displaystyle\frac{\sin\left(\pi(c-a-b)\right)}{\pi}\mathbf{F}\left({a,b\atop c% };z\right)$ $\displaystyle=\frac{z^{-a}}{\Gamma\left(c-a\right)\Gamma\left(c-b\right)}% \mathbf{F}\left({a,a-c+1\atop a+b-c+1};1-\frac{1}{z}\right)-\frac{(1-z)^{c-a-b% }z^{a-c}}{\Gamma\left(a\right)\Gamma\left(b\right)}\mathbf{F}\left({c-a,1-a% \atop c-a-b+1};1-\frac{1}{z}\right),$ $|\operatorname{ph}z|<\pi$, $|\operatorname{ph}\left(1-z\right)|<\pi$.

For an alternative version of the transformations (15.8.2) and (15.8.3), see (15.10.25), and for an alternative version of the transformations (15.8.4) and (15.8.5), see (15.10.21).

## §15.8(ii) Linear Transformations: Limiting Cases

With $m=0,1,2,\dots$, polynomial cases of (15.8.2)–(15.8.5) are given by

 15.8.6 $\displaystyle F\left({-m,b\atop c};z\right)$ $\displaystyle=\frac{{\left(b\right)_{m}}}{{\left(c\right)_{m}}}(-z)^{m}F\left(% {-m,1-c-m\atop 1-b-m};\frac{1}{z}\right)=\frac{{\left(b\right)_{m}}}{{\left(c% \right)_{m}}}(1-z)^{m}F\left({-m,c-b\atop 1-b-m};\frac{1}{1-z}\right),$ ⓘ Symbols: $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, ${\left(\NVar{a}\right)_{\NVar{n}}}$: Pochhammer’s symbol (or shifted factorial), $z$: complex variable, $b$: real or complex parameter, $c$: real or complex parameter and $m$: integer Referenced by: §15.8(ii), §15.8(ii), §18.17(vii) Permalink: http://dlmf.nist.gov/15.8.E6 Encodings: TeX, pMML, png Correction (effective with 1.1.2): Pochhammer symbols now link to their definition. See also: Annotations for §15.8(ii), §15.8 and Ch.15 15.8.7 $\displaystyle F\left({-m,b\atop c};z\right)$ $\displaystyle=\frac{{\left(c-b\right)_{m}}}{{\left(c\right)_{m}}}F\left({-m,b% \atop b-c-m+1};1-z\right)=\frac{{\left(c-b\right)_{m}}}{{\left(c\right)_{m}}}z% ^{m}F\left({-m,1-c-m\atop b-c-m+1};1-\frac{1}{z}\right),$ ⓘ Symbols: $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, ${\left(\NVar{a}\right)_{\NVar{n}}}$: Pochhammer’s symbol (or shifted factorial), $z$: complex variable, $b$: real or complex parameter, $c$: real or complex parameter and $m$: integer Referenced by: §15.19(iv), §15.8(ii), §15.8(ii), (18.35.4_5) Permalink: http://dlmf.nist.gov/15.8.E7 Encodings: TeX, pMML, png Correction (effective with 1.1.2): Pochhammer symbols now link to their definition. See also: Annotations for §15.8(ii), §15.8 and Ch.15

with the understanding that if $b=-\ell$, $\ell=0,1,2,\dots$, then $m\leq\ell$.

When $b-a$ is an integer limits are taken in (15.8.2) and (15.8.3) as follows.

If $b-a$ is a nonnegative integer, then

 15.8.8 $\mathbf{F}\left({a,a+m\atop c};z\right)=\frac{(-z)^{-a}}{\Gamma\left(a+m\right% )}\sum_{k=0}^{m-1}\frac{{\left(a\right)_{k}}(m-k-1)!}{k!\Gamma\left(c-a-k% \right)}z^{-k}+\frac{(-z)^{-a}}{\Gamma\left(a\right)}\sum_{k=0}^{\infty}\frac{% {\left(a+m\right)_{k}}}{k!(k+m)!\Gamma\left(c-a-k-m\right)}(-1)^{k}z^{-k-m}\*% \left(\ln\left(-z\right)+\psi\left(k+1\right)+\psi\left(k+m+1\right)-\psi\left% (a+k+m\right)-\psi\left(c-a-k-m\right)\right),$ $|z|>1,|\operatorname{ph}\left(-z\right)|<\pi$,
 15.8.9 $\mathbf{F}\left({a,a+m\atop c};z\right)=\frac{(1-z)^{-a}}{\Gamma\left(a+m% \right)\Gamma\left(c-a\right)}\sum_{k=0}^{m-1}\frac{{\left(a\right)_{k}}{\left% (c-a-m\right)_{k}}(m-k-1)!}{k!}(z-1)^{-k}+\frac{(-1)^{m}(1-z)^{-a-m}}{\Gamma% \left(a\right)\Gamma\left(c-a-m\right)}\sum_{k=0}^{\infty}\frac{{\left(a+m% \right)_{k}}{\left(c-a\right)_{k}}}{k!(k+m)!}(1-z)^{-k}\*(\ln\left(1-z\right)+% \psi\left(k+1\right)+\psi\left(k+m+1\right)-\psi\left(a+k+m\right)-\psi\left(c% -a+k\right)),$ $|z-1|>1,|\operatorname{ph}\left(1-z\right)|<\pi$.

In (15.8.8) when $c-a-k-m$ is a nonpositive integer $\ifrac{\psi\left(c-a-k-m\right)}{\Gamma\left(c-a-k-m\right)}$ is interpreted as $(-1)^{m+k+a-c+1}(m+k+a-c)!$. Also, if $a$ is a nonpositive integer, then (15.8.6) applies.

Alternatively, if $b-a$ is a negative integer, then we interchange $a$ and $b$ in $\mathbf{F}\left(a,b;c;z\right)$.

In a similar way, when $c-a-b$ is an integer limits are taken in (15.8.4) and (15.8.5) as follows.

If $c-a-b$ is a nonnegative integer, then

 15.8.10 $\mathbf{F}\left({a,b\atop a+b+m};z\right)=\frac{1}{\Gamma\left(a+m\right)% \Gamma\left(b+m\right)}\sum_{k=0}^{m-1}\frac{{\left(a\right)_{k}}{\left(b% \right)_{k}}(m-k-1)!}{k!}(z-1)^{k}-\frac{(z-1)^{m}}{\Gamma\left(a\right)\Gamma% \left(b\right)}\sum_{k=0}^{\infty}\frac{{\left(a+m\right)_{k}}{\left(b+m\right% )_{k}}}{k!(k+m)!}(1-z)^{k}\*\left(\ln\left(1-z\right)-\psi\left(k+1\right)-% \psi\left(k+m+1\right)+\psi\left(a+k+m\right)+\psi\left(b+k+m\right)\right),$ $|z-1|<1,|\operatorname{ph}\left(1-z\right)|<\pi$,
 15.8.11 $\mathbf{F}\left({a,b\atop a+b+m};z\right)=\frac{z^{-a}}{\Gamma\left(a+m\right)% }\sum_{k=0}^{m-1}\frac{{\left(a\right)_{k}}(m-k-1)!}{k!\Gamma\left(b+m-k\right% )}\left(1-\frac{1}{z}\right)^{k}-\frac{z^{-a}}{\Gamma\left(a\right)}\sum_{k=0}% ^{\infty}\frac{{\left(a+m\right)_{k}}}{k!(k+m)!\Gamma\left(b-k\right)}(-1)^{k}% \left(1-\frac{1}{z}\right)^{k+m}\*\left(\ln\left(\frac{1-z}{z}\right)-\psi% \left(k+1\right)-\psi\left(k+m+1\right)+\psi\left(a+k+m\right)+\psi\left(b-k% \right)\right),$ $\Re z>\tfrac{1}{2},|\operatorname{ph}z|<\pi,|\operatorname{ph}\left(1-z\right)% |<\pi$.

In (15.8.11) when $b-k$ is a nonpositive integer, $\ifrac{\psi\left(b-k\right)}{\Gamma\left(b-k\right)}$ is interpreted as $(-1)^{k-b+1}(k-b)!$. Also, if $a$ or $b$ or both are nonpositive integers, then (15.8.7) applies.

Lastly, if $c-a-b$ is a negative integer, then we first apply the transformation

 15.8.12 $\mathbf{F}\left(a,b;a+b-m;z\right)=(1-z)^{-m}\mathbf{F}\left(\tilde{a},\tilde{% b};\tilde{a}+\tilde{b}+m;z\right),$ $\tilde{a}=a-m,\tilde{b}=b-m$.

A quadratic transformation relates two hypergeometric functions, with the variable in one a quadratic function of the variable in the other, possibly combined with a fractional linear transformation.

A necessary and sufficient condition that there exists a quadratic transformation is that at least one of the equations shown in Table 15.8.1 is satisfied.

The hypergeometric functions that correspond to Groups 1 and 2 have $z$ as variable. The hypergeometric functions that correspond to Groups 3 and 4 have a nonlinear function of $z$ as variable. The transformation formulas between two hypergeometric functions in Group 2, or two hypergeometric functions in Group 3, are the linear transformations (15.8.1).

In the equations that follow in this subsection all functions take their principal values.

### Group 1 $\longrightarrow$ Group 3

 15.8.13 $\displaystyle F\left({a,b\atop 2b};z\right)$ $\displaystyle=\left(1-\tfrac{1}{2}z\right)^{-a}F\left({\tfrac{1}{2}a,\tfrac{1}% {2}a+\tfrac{1}{2}\atop b+\tfrac{1}{2}};\left(\frac{z}{2-z}\right)^{2}\right),$ $|\operatorname{ph}\left(1-z\right)|<\pi$, 15.8.14 $\displaystyle F\left({a,b\atop 2b};z\right)$ $\displaystyle=\left(1-z\right)^{-\ifrac{a}{2}}F\left({\tfrac{1}{2}a,b-\tfrac{1% }{2}a\atop b+\tfrac{1}{2}};\frac{z^{2}}{4z-4}\right),$ $|\operatorname{ph}\left(1-z\right)|<\pi$.

### Group 2 $\longrightarrow$ Group 3

 15.8.15 $\displaystyle F\left({a,b\atop a-b+1};z\right)$ $\displaystyle=(1+z)^{-a}F\left({\frac{1}{2}a,\frac{1}{2}a+\frac{1}{2}\atop a-b% +1};\frac{4z}{(1+z)^{2}}\right),$ $|z|<1$, 15.8.16 $\displaystyle F\left({a,b\atop a-b+1};z\right)$ $\displaystyle=(1-z)^{-a}F\left({\frac{1}{2}a,\frac{1}{2}a-b+\frac{1}{2}\atop a% -b+1};\frac{-4z}{(1-z)^{2}}\right),$ $|z|<1$.
 15.8.17 $\displaystyle F\left({a,b\atop\frac{1}{2}(a+b+1)};z\right)$ $\displaystyle=(1-2z)^{-a}F\left({\frac{1}{2}a,\frac{1}{2}a+\frac{1}{2}\atop% \frac{1}{2}(a+b+1)};\frac{4z(z-1)}{(1-2z)^{2}}\right),$ $\Re z<\frac{1}{2}$, 15.8.18 $\displaystyle F\left({a,b\atop\frac{1}{2}(a+b+1)};z\right)$ $\displaystyle=F\left({\frac{1}{2}a,\frac{1}{2}b\atop\frac{1}{2}(a+b+1)};4z(1-z% )\right),$ $\Re z<\frac{1}{2}$.
 15.8.19 $\displaystyle F\left({a,1-a\atop c};z\right)$ $\displaystyle=(1-2z)^{1-a-c}(1-z)^{c-1}F\left({\frac{1}{2}(a+c),\frac{1}{2}(a+% c-1)\atop c};\frac{4z(z-1)}{(1-2z)^{2}}\right),$ $\Re z<\frac{1}{2}$, 15.8.20 $\displaystyle F\left({a,1-a\atop c};z\right)$ $\displaystyle=(1-z)^{c-1}F\left({\frac{1}{2}(c-a),\frac{1}{2}(a+c-1)\atop c};4% z(1-z)\right),$ $\Re z<\frac{1}{2}$.

### Group 2 $\longrightarrow$ Group 1

 15.8.21 $\displaystyle F\left({a,b\atop a-b+1};z\right)$ $\displaystyle=\left(1+\sqrt{z}\right)^{-2a}F\left({a,a-b+\tfrac{1}{2}\atop 2a-% 2b+1};\frac{4\sqrt{z}}{(1+\sqrt{z})^{2}}\right),$ $|\operatorname{ph}z|<\pi$, $|z|<1$. 15.8.22 $\displaystyle F\left({a,b\atop\tfrac{1}{2}(a+b+1)};z\right)$ $\displaystyle=\left(\frac{\sqrt{1-z^{-1}}-1}{\sqrt{1-z^{-1}}+1}\right)^{a}F% \left({a,\tfrac{1}{2}(a+b)\atop a+b};\frac{4\sqrt{1-z^{-1}}}{\left(\sqrt{1-z^{% -1}}+1\right)^{2}}\right),$ $|\operatorname{ph}\left(-z\right)|<\pi$, $\Re z<\frac{1}{2}$.
 15.8.23 $F\left({a,1-a\atop c};z\right)=\left(\sqrt{1-z^{-1}}-1\right)^{1-a}\left(\sqrt% {1-z^{-1}}+1\right)^{a-2c+1}\left(1-z^{-1}\right)^{c-1}F\left({c-a,c-\tfrac{1}% {2}\atop 2c-1};\frac{4\sqrt{1-z^{-1}}}{\left(\sqrt{1-z^{-1}}+1\right)^{2}}% \right),$ $|\operatorname{ph}\left(-z\right)|<\pi$, $\Re z<\frac{1}{2}$.

### Group 2 $\longrightarrow$ Group 4

 15.8.24 $F\left({a,b\atop a-b+1};z\right)=(1-z)^{-a}\frac{\Gamma\left(a-b+1\right)% \Gamma\left(\tfrac{1}{2}\right)}{\Gamma\left(\tfrac{1}{2}a+\tfrac{1}{2}\right)% \Gamma\left(\tfrac{1}{2}a-b+1\right)}F\left({\tfrac{1}{2}a,\tfrac{1}{2}a-b+% \tfrac{1}{2}\atop\tfrac{1}{2}};\left(\frac{z+1}{z-1}\right)^{2}\right)+(1+z)(1% -z)^{-a-1}\frac{\Gamma\left(a-b+1\right)\Gamma\left(-\tfrac{1}{2}\right)}{% \Gamma\left(\tfrac{1}{2}a\right)\Gamma\left(\tfrac{1}{2}a-b+\tfrac{1}{2}\right% )}F\left({\tfrac{1}{2}a+\tfrac{1}{2},\tfrac{1}{2}a-b+1\atop\tfrac{3}{2}};\left% (\frac{z+1}{z-1}\right)^{2}\right),$ $|\operatorname{ph}\left(-z\right)|<\pi$.
 15.8.25 $F\left({a,b\atop\tfrac{1}{2}(a+b+1)};z\right)=\frac{\Gamma\left(\tfrac{1}{2}(a% +b+1)\right)\Gamma\left(\tfrac{1}{2}\right)}{\Gamma\left(\tfrac{1}{2}a+\tfrac{% 1}{2}\right)\Gamma\left(\tfrac{1}{2}b+\tfrac{1}{2}\right)}F\left({\tfrac{1}{2}% a,\tfrac{1}{2}b\atop\tfrac{1}{2}};(1-2z)^{2}\right)+(1-2z)\frac{\Gamma\left(% \tfrac{1}{2}(a+b+1)\right)\Gamma\left(-\tfrac{1}{2}\right)}{\Gamma\left(\tfrac% {1}{2}a\right)\Gamma\left(\tfrac{1}{2}b\right)}F\left({\tfrac{1}{2}a+\tfrac{1}% {2},\tfrac{1}{2}b+\tfrac{1}{2}\atop\tfrac{3}{2}};(1-2z)^{2}\right),$ $|\operatorname{ph}z|<\pi$, $|\operatorname{ph}\left(1-z\right)|<\pi$.
 15.8.26 $F\left({a,1-a\atop c};z\right)=(1-z)^{c-1}\frac{\Gamma\left(c\right)\Gamma% \left(\tfrac{1}{2}\right)}{\Gamma\left(\tfrac{1}{2}(c-a+1)\right)\Gamma\left(% \tfrac{1}{2}c+\tfrac{1}{2}a\right)}F\left({\tfrac{1}{2}c-\tfrac{1}{2}a,\tfrac{% 1}{2}c+\tfrac{1}{2}a-\tfrac{1}{2}\atop\tfrac{1}{2}};(1-2z)^{2}\right)+(1-2z)(1% -z)^{c-1}\frac{\Gamma\left(c\right)\Gamma\left(-\tfrac{1}{2}\right)}{\Gamma% \left(\tfrac{1}{2}c-\tfrac{1}{2}a\right)\Gamma\left(\tfrac{1}{2}(c+a-1)\right)% }F\left({\tfrac{1}{2}c-\tfrac{1}{2}a+\tfrac{1}{2},\tfrac{1}{2}c+\tfrac{1}{2}a% \atop\tfrac{3}{2}};(1-2z)^{2}\right),$ $|\operatorname{ph}z|<\pi$, $|\operatorname{ph}\left(1-z\right)|<\pi$.

### Group 4 $\longrightarrow$ Group 2

 15.8.27 $\frac{2\Gamma\left(\tfrac{1}{2}\right)\Gamma\left(a+b+\tfrac{1}{2}\right)}{% \Gamma\left(a+\tfrac{1}{2}\right)\Gamma\left(b+\tfrac{1}{2}\right)}F\left(a,b;% \tfrac{1}{2};z\right)=F\left(2a,2b;a+b+\tfrac{1}{2};\tfrac{1}{2}-\tfrac{1}{2}% \sqrt{z}\right)+F\left(2a,2b;a+b+\tfrac{1}{2};\tfrac{1}{2}+\tfrac{1}{2}\sqrt{z% }\right),$ $|\operatorname{ph}z|<\pi$, $|\operatorname{ph}\left(1-z\right)|<\pi$.
 15.8.28 $\frac{2\sqrt{z}\Gamma\left(-\tfrac{1}{2}\right)\Gamma\left(a+b-\tfrac{1}{2}% \right)}{\Gamma\left(a-\tfrac{1}{2}\right)\Gamma\left(b-\tfrac{1}{2}\right)}F% \left(a,b;\tfrac{3}{2};z\right)=F\left(2a-1,2b-1;a+b-\tfrac{1}{2};\tfrac{1}{2}% -\tfrac{1}{2}\sqrt{z}\right)-F\left(2a-1,2b-1;a+b-\tfrac{1}{2};\tfrac{1}{2}+% \tfrac{1}{2}\sqrt{z}\right),$ $|\operatorname{ph}z|<\pi$, $|\operatorname{ph}\left(1-z\right)|<\pi$.

When the intersection of two groups in Table 15.8.1 is not empty there exist special quadratic transformations, with only one free parameter, between two hypergeometric functions in the same group.

### Examples

$b=\tfrac{1}{3}a+\tfrac{1}{3}$, $c=2b=a-b+1$ in Groups 1 and 2.

(15.8.21) becomes

 15.8.29 $F\left({a,\tfrac{1}{3}a+\tfrac{1}{3}\atop\tfrac{2}{3}a+\tfrac{2}{3}};z\right)=% \left(1+\sqrt{z}\right)^{-2a}\*F\left({a,\tfrac{2}{3}a+\tfrac{1}{6}\atop\tfrac% {4}{3}a+\tfrac{1}{3}};\frac{4\sqrt{z}}{(1+\sqrt{z})^{2}}\right).$

This is a quadratic transformation between two cases in Group 1.

We can also use (15.8.13), followed by the inverse of (15.8.15), and obtain

 15.8.30 $\left(1-\tfrac{1}{2}z\right)^{-a}F\left({\tfrac{1}{2}a,\tfrac{1}{2}a+\tfrac{1}% {2}\atop\tfrac{1}{3}a+\tfrac{5}{6}};\left(\frac{z}{2-z}\right)^{2}\right)=F% \left({a,\tfrac{1}{3}a+\tfrac{1}{3}\atop\tfrac{2}{3}a+\tfrac{2}{3}};z\right)=(% 1+z)^{-a}F\left({\tfrac{1}{2}a,\tfrac{1}{2}a+\tfrac{1}{2}\atop\tfrac{2}{3}a+% \tfrac{2}{3}};\frac{4z}{(1+z)^{2}}\right),$

which is a quadratic transformation between two cases in Group 3.

For further examples see Andrews et al. (1999, pp. 130–132 and 176–177).

## §15.8(v) Cubic Transformations

### Examples

 15.8.31 $F\left({3a,3a+\frac{1}{2}\atop 4a+\frac{2}{3}};z\right)=\left(1-\tfrac{9}{8}z% \right)^{-2a}\*F\left({a,a+\frac{1}{2}\atop 2a+\frac{5}{6}};\frac{27z^{2}(z-1)% }{(9z-8)^{2}}\right),$ $\Re z<\frac{8}{9}$.

With $\zeta={\mathrm{e}}^{\ifrac{2\pi\mathrm{i}}{3}}(1-z)/\left(z-{\mathrm{e}}^{% \ifrac{4\pi\mathrm{i}}{3}}\right)$

 15.8.32 $\frac{\left(1-z^{3}\right)^{a}}{\left(-z\right)^{3a}}\left(\frac{1}{\Gamma% \left(a+\frac{2}{3}\right)\Gamma\left(\frac{2}{3}\right)}F\left({a,a+\frac{1}{% 3}\atop\frac{2}{3}};z^{-3}\right)+\frac{{\mathrm{e}}^{\frac{1}{3}\pi\mathrm{i}% }}{z\Gamma\left(a\right)\Gamma\left(\frac{4}{3}\right)}F\left({a+\frac{1}{3},a% +\frac{2}{3}\atop\frac{4}{3}};z^{-3}\right)\right)=\frac{3^{\frac{3}{2}a+\frac% {1}{2}}{\mathrm{e}}^{\frac{1}{2}a\pi\mathrm{i}}\Gamma\left(a+\frac{1}{3}\right% )(1-\zeta)^{a}}{2\pi\Gamma\left(2a+\frac{2}{3}\right)(-\zeta)^{2a}}F\left({a+% \frac{1}{3},3a\atop 2a+\frac{2}{3}};\zeta^{-1}\right),$ $|z|>1$, $|\operatorname{ph}\left(-z\right)|<\frac{1}{3}\pi$.

### Ramanujan’s Cubic Transformation

 15.8.33 $F\left({\frac{1}{3},\frac{2}{3}\atop 1};1-\left(\frac{1-z}{1+2z}\right)^{3}% \right)=(1+2z)F\left({\frac{1}{3},\frac{2}{3}\atop 1};z^{3}\right),$

provided that $z$ lies in the intersection of the open disks $\left|z-\frac{1}{4}\pm\frac{1}{4}\sqrt{3}\mathrm{i}\right|<\frac{1}{2}\sqrt{3}$, or equivalently, $\left|\operatorname{ph}\left(\ifrac{(1-z)}{(1+2z)}\right)\right|<\pi/3$. This is used in a cubic analog of the arithmetic-geometric mean. See Borwein and Borwein (1991), and also Berndt et al. (1995).

For further examples and higher-order transformations see Goursat (1881), Watson (1910), Vidūnas (2005), and Tu and Yang (2013); see also Erdélyi et al. (1953a, pp. 67 and 113–114).