Consider the function $_2F_1(5.5, 1, 5;-|x|^2]$ for $x\in \mathbb{R}^n.$ I want to show that this function is positive. I checked that it does not have any roots so can I conclude the inequality by using continuity in $x$ of the function $_2F_1(5.5, 1, 5;-|x|^2]$?
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$\begingroup$ Probably the reason your question was closed is that it is strangely formulated. The space $\mathbb R^n$ plays no role, you are just considering negative values of the variable. Also, if you really have a proof that there are no zeroes, then as you say the positivity follows immediately. Anyway, your question is answered by basic facts on hypergeometric functions such as Euler's integral representation (I give a version of that below). But since hypergeometric functions are not in general taught even to PhD students in mathematics, I do think the question is appropriate for MathOverflow. $\endgroup$– Hjalmar RosengrenCommented Sep 27, 2021 at 9:23
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You have the integral representation $${}_2F_1\left(\begin{matrix}11/2,1\\5\end{matrix};-x\right)=\frac 4{x^4}\int_0^x\frac{(x-t)^3}{(1+t)^{11/2}}\,dt,$$ which follows by expanding $1/(1+t)^{11/2}$ using the binomial theorem and integrating termwise. This should prove the positivity.
Note that this integral is a remainder term in the Taylor expansion of $1/(1+x)^{3/2}$. So your series can be summed explicitly as $1/(1+x)^{3/2}$ minus a Taylor polynomial. This is also easy to see by writing $$\frac{(11/2)_k}{(5)_k}=\frac{4!}{(3/2)_4}\frac{(3/2)_{k+4}}{(k+4)!}.$$ Your series is essentially just the binomial series for $1/(1+x)^{3/2}$ minus the first four terms. More explicitly, $${}_2F_1\left(\begin{matrix}11/2,1\\5\end{matrix};-x\right)=\frac{128}{315x^4}\left(\frac 1{(1+x)^{3/2}}-1+\frac 32\,x-\frac{15}8\,x^2+\frac{35}{16}\,x^3\right).$$ Perhaps this fact is also useful to you.
answered Sep 26, 2021 at 6:16