# hyperbolic functions and Gauss hypergeometric series

If the Gauss hypergeometric function $F(1, 3/2, 5/2; z^2) = 3 [\tanh^{-1} (z) –z]/z^3$
what is the corresponding result for $F(1,15/8,23/8;z^8)$?

• MVS, I threw in a little formatting; please check to see that I didn't change any meanings. – Gerry Myerson Jun 6 '13 at 5:36
• Is there a reason to expect a closed form? – Noam D. Elkies Jun 6 '13 at 15:52
• Thanks for the formatting. The reason for expecting the simplification is that one of the Kummer's 24 relations seems applicable here.Further,the function F(1,1/8,9/8;1-z) seems approximately equal to z^(-1/8).e.g F(1,1/8,9/8;0.21) = 1.026 while (1-0.21)^(-1/8) = 1.029.Hence the anticipation that for both the hypergeometric functions there may be simple representations such as polynomials or hyperbolic functions. – M V S Jun 7 '13 at 7:54

There is a special form for hypergeometric functions of the form $$F(1,\beta,1+\beta,z) = \beta z^{-\beta}B(z,\beta,0)$$ when $\beta = p/q$ is a rational number (where $B$ is the incomplete beta function).
Write $$\frac{z^\beta}{\beta}F(1,\beta,1+\beta,z) = \sum_{k\geq0} \frac{z^{\beta+k}}{\beta+k} = q\sum_{k\geq0}\frac{(z^{1/q})^{p+k q}}{p+k q} = q\sum_{k\geq0}\frac{(z^{1/q})^k}{k}[k\bmod q=p].$$
The point is that you can use $$\log(1-x) = -\sum_{k\geq1} \frac{x^k}{k}$$ and $$\sum_{0\leq l < q} \zeta^{a l} = q[a\bmod q=0], \qquad \zeta=e^{2\pi i/q}, \quad a\in\mathbb{Z}$$ to write $$\sum_{0\leq l < q} -\zeta^{-l p}\log(1-\zeta^l x) = \sum_{k\geq1}\sum_{0\leq l < q} \frac{\zeta^{-lp+lk}x^k}{k} = \sum_{k\geq1} \frac{x^k}{k}q[k\bmod q=p].$$
So the closed form is $$\frac{z^\beta}{\beta}F(1,\beta,1+\beta,z) = \sum_{0\leq l < q}-\zeta^{-lp}\log(1-\zeta^l z^{1/q}), \qquad \beta=\frac pq, \quad p,q\in\mathbb{Z}.$$
I would be interested to know how this form should be generalized to irrational $\beta$.