# Differentiate a hypergeometric function expression

I have the following function

$$f_\epsilon (p)=\frac{1}{2}(1-p)^\epsilon 2^\epsilon {_2}F_1(1-\epsilon,\epsilon;1+\epsilon;\frac{1-p}{2}),\qquad p\in(-1,1).$$

Here $F$ is the hypergeometric function ${_2}F_1(a,b;c;z)$, see e.g. here for a definition: http://en.wikipedia.org/wiki/Hypergeometric_function

I need the first order Taylor expansion in $\epsilon$ given by

$$f_0(p)+ \left[\frac{\partial f_\epsilon}{\partial \epsilon}(p)\right]_{\epsilon=0}\cdot\epsilon.$$

The zeroth order term $f_0(p)$ should be doable, but the first order term seems to be really hard to get. The problem is to carry out the differentiation w.r.t. $\epsilon$ in the Arguments of ${_2}F_1$. I would need to compute

$$\frac{\partial }{\partial \epsilon}{_2}F_1(1-\epsilon,\epsilon;1+\epsilon;z).$$

Any ideas how this could be possible?

I will differentiate that hypergeometric function. From its definition, write $$f(\epsilon,z) = F\left(\begin{array}{c}1-\epsilon,\epsilon\\1+\epsilon\end{array}\middle|z\right) = \sum_{n\geq0} y^n \frac{(1-\epsilon)_n}{n!}\frac{\epsilon}{n+\epsilon}.$$ Differentiating w.r.t. $\epsilon$ gives $$\partial_\epsilon f(\epsilon,z) = \sum_{n\geq0} z^n \frac{(1-\epsilon)_n}{n!}\frac{\epsilon}{n+\epsilon}\left(\frac1\epsilon - \frac1{n+\epsilon} + \psi(n+1+\epsilon) - \psi(1-\epsilon) \right),$$ where $\psi=\Gamma'/\Gamma$ is the digamma function. Now the sum splits into four different sums, and only the term corresponding to $n=0$ can possibly be singular as a function $\epsilon$ because for $n\geq1$ every summand has a well-defined limit as $\epsilon\to0$. So separate out the terms $n=0$, and take the limit $\epsilon\to0$ in those terms that are not singular: $$\partial_\epsilon f(\epsilon,z) = \left(\frac1\epsilon + \sum_{n\geq}z^n \frac{n!}{n!}\frac1n + O(\epsilon)\right) - \left(\frac1\epsilon + \epsilon\sum_{n\geq1} \frac{z^n}{n^2} + O(\epsilon)\right) + \left(\epsilon\sum_{n\geq1} \frac{z^n}{n}(\psi(n+1) - \psi(1)) + O(\epsilon^2)\right).$$ The singular terms $\frac1\epsilon$ cancel, so all that is left is $$\partial_\epsilon|_{\epsilon=0} f(\epsilon,z) = -\log(1-z).$$

• Thanks, your detailed explanation helped me a lot... – flonk Jun 29 '13 at 19:32