# hypergeometric closed form for z=1/4,-1/3

There exist the linear identities for the 2f1 hypergeometric function where z is either -1, 1, or 1/2 using the quadratic transdormations it is easy to derive new identities in terms of gamma functions for z = -1/8,8/9 and 1/9

I have seen some identities with z=1/4 and z=-1/3 but there is no obvious transformation from the linear to the quadratic that yield these. So how do you derive them using the existing linear, quadratic transformation?

this problem i assure you is extremely difficult

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This is too obscure. What identities do you mean? There is an infinite number of hypergeometric identities, derived by a wide variety of methods, so this post is completely vacuous. –  Igor Rivin Feb 27 '11 at 3:35
more specifically this is the identity i want proven functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/… as you can see the z value is 1/4 you can verify it by checking the differential equation or integral, but i would like to know what combination of the linear and quadratic transformations give this. –  mathers Feb 27 '11 at 6:33
Ah, OK. That clarifies things. Maybe you should edit the OP to include this... –  Igor Rivin Feb 28 '11 at 3:33
I don't think it's possible to derive these identities from quadratic transformations. However, it is possible to derive identities with $z=1/4$ from cubic transformations, though I don't know if the specific identity that the questioner asks about can be proved this way.
Here's an example. In the cubic transformation \begin{multline*}{}_{2}F_{1}\left({{3a,1/3-a}\atop 2a+5/6}\Bigm |z\right)\\ =\left( 1-z \right) ^{-a} \left( 1+8z \right) ^{-2a} {}_{2}F_{ 1}\left({{a,a+1/2}\atop 2a+5/6}\Bigm |{\frac {27 z}{ \left(1-z \right) \left( 1+8z \right) ^{2}}}\right) \end{multline*} we can set $z=1/4$ to get \begin{align*} {}_{2}F_{1}\left({{3a,1/3-a}\atop 2a+5/6}\Bigm |1/4\right)&= \left( {\frac {4}{27}} \right) ^{a} {}_{2} F_{1}\left({{a,a+1/2}\atop 2a+5/6}\Bigm |1\right)\\ &= \frac{2\pi}{\sqrt3} \left( {\frac {4}{27}} \right) ^{a}{\frac {\Gamma \left( 2a+5/6 \right) }{\Gamma \left( 2/3 \right) \Gamma \left( a+5/6 \right) \Gamma \left( a+1/3 \right) }}. \end{align*}