# Logarithm of the hypergeometric function

For $F(x)={}_2F_1 (a,b;c;x)$, with $c=a+b$, $a>0$, $b>0$, it has been proved in [1] that $\log F(x)$ is convex on $(0,1)$.

I numerically checked that with a variety of $a,\ b$ values, $\log F(x)$ is not only convex, but also has a Taylor series in x consisting of strictly positive coefficients. Can this be proved?

[1] Generalized convexity and inequalities, Anderson, Vamanamurthy, Vuorinen, Journal of Mathematical Analysis and Applications, Volume 335, Issue 2, http://www.sciencedirect.com/science/article/pii/S0022247X07001825#

Here's a sketch of a proof of a stronger statement: the coefficients of the Taylor series for $\log{}_2F_1(a,b;a+b+c;x)$ are rational functions of $a$, $b$, and $c$ with positive coefficients.
To see this we first note that \begin{aligned} \frac{d\ }{dx} \log {}_2F_1(a,b;a+b+c;x) &= \frac{\displaystyle \frac{d\ }{dx}\,{}_2F_1(a,b;a+b+c;x)}{{}_2F_1(a,b;a+b+c;x)}\\[3pt] &=\frac{ab}{a+b+c}\frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b;a+b+c;x)}. \end{aligned} Then $$\begin{gathered} \frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b;a+b+c;x)} = \frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b+1;a+b+c;x)} \\ \hfill\times \frac{{}_2F_1(a,b+1;a+b+c;x)}{{}_2F_1(a,b;a+b+c;x)}.\quad \end{gathered}$$ We have continued fractions for the two quotients on the right. Let $S(x; a_1, a_2, a_3, \dots)$ denote the continued fraction $$\cfrac{1}{1-\cfrac{a_1x} {1-\cfrac{a_2x} {1-\cfrac{a_3x} {1-\ddots} }}}$$ Then $$\begin{gathered}\frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b+1;a+b+c;x)} = S \left( x;{\frac { \left( b+1 \right) \left( b+c \right) }{ \left( a +b+c+1 \right) \left( a+b+c \right) }}, \right.\hfill\\ \left. {\frac { \left( a+1 \right) \left( a+c \right) }{ \left( a+b+c+2 \right) \left( a+b+c+1 \right) }}, {\frac { \left( b+2 \right) \left( b+c+1 \right) }{ \left( a+b+c+3 \right) \left( a+b+c+2 \right) }}, \right.\\ \hfill \left. {\frac { \left( a+2 \right) \left( a+c+1 \right) }{ \left( a+b+c+4 \right) \left( a+b+c+3 \right) }},\dots \right) \end{gathered}$$ and $$\begin{gathered} \frac{{}_2F_1(a,b+1;a+b+c;x)}{{}_2F_1(a,b;a+b+c;x)} =S \left( x,{\frac {a}{a+b+c}}, {\frac { \left( b+1 \right) \left( b+c \right) }{ \left( a+b+c+1 \right) \left( a+b+c \right) }}, \right.\hfill\\ \left. {\frac { \left( a+1 \right) \left( a+c \right) }{ \left( a+b+c+2 \right) \left( a+b+c+1 \right) }}, {\frac { \left( b+2 \right) \left( b+c+1 \right) }{ \left( a+b+c+3 \right) \left( a+b+c+2 \right) }}, \right.\\ \hfill \left. {\frac { \left( a+2 \right) \left( a+c+1 \right) }{ \left( a+b+c+4 \right) \left( a+b+c+3 \right) }}, \dots\right) \end{gathered}$$ The first of these continued fractions is Gauss's well-known continued fraction, and the second can easily be derived from the first. It follows from these formulas that the coefficients of the Taylor series for $\log{}_2F_1(a,b;a+b+c;x)$ are rational functions of $a$, $b$, and $c$ with positive coefficients.