# To show Hypergeometric Functions(1F2) positive.

I don't have experience with hypergeoemtric functions, but wish to show $$_1F_2[2; \frac{9}{4}, \frac{11}{4}; -\frac{x^2}{16}] > 0 , \text{ when } x\in(0,5).$$ I used Maple to plot the graph and it does greater than $0$ when $x\in(0,5)$. For the proof, my idea is to use asymptotic expansion to show it but I didn't get the result I wanted.

Any help would be appreciated. Thanks!!

• Your inequality can be re-expressed in terms of the Fresnel auxiliary functions: $$6 f\left(\frac{\sqrt{x}}{\sqrt{\pi}}\right)+2 x g\left(\frac{\sqrt{x}}{\sqrt{\pi}}\right)>(x+3)\cos\left(\frac{x}{2}\right)-(3-x) \sin\left(\frac{x}{2}\right)$$ You might be able to use the fact that the auxiliary functions are always positive and monotonically decreasing for nonnegative arguments. Mar 5 '18 at 6:46
• Thank you! When I plug in square root of x/π in Fresnel auxiliary functions(f and g), I don't know what the relation it has with hypergeometric fcts. Could you give me some hints that how to obtain your inequality? Sorry for my background about special functions. Thanks again!! Mar 6 '18 at 3:51
• In particular, we have the identity $${}_1 F_2\left({{2}\atop{\frac94,\frac{11}{4}}}\middle|-\frac{x^2}{16}\right)=\frac{105\sqrt{\pi}}{8 x^{7/2}}\left(6 f\left(\sqrt{\frac{x}{\pi}}\right)+2 x g\left(\sqrt{\frac{x}{\pi}}\right)-(x+3)\cos\frac{x}{2}-(x-3)\sin\frac{x}{2}\right)$$ Mar 6 '18 at 4:03
• Wow! Thank you! This is a nice result! Is it obviously? I already searched some special function books but cannot obtain this result :( Mar 6 '18 at 5:29
• I had some help deriving it from Mathematica; I believe Maple's convert() should be able to do something similar. Mar 6 '18 at 5:40

Since the argument $-x^2/16$ is negative, the expansion of the hypergeometric function at zero is an alternating series, with the ratio between the successive terms given by $$\frac {a_{k+1}} {a_k} = -\frac {k+2} {(k+1)(4k+9)(4k+11)} x^2.$$ For a fixed $x \in \left[0, \sqrt {99/2} \,\right)$, the absolute values of $a_k$ go to zero monotonically, and the error can be estimated by the first discarded term: $$\left| {_1F_2} \left(2; \frac 9 4, \frac {11} 4; -\frac {x^2} {16} \right) - 1 \right| \leq \frac {2x^2} {99},$$ which implies that the function is positive on $\left[0, \sqrt {99/2} \,\right)$.

• Nice proof! Thank you so much! Mar 7 '18 at 0:01

This is not a proof since based on numerical calculations.

If we consider the function $$f(x)=\, _1F_2\left(2;\frac{9}{4},\frac{11}{4};-\frac{x^2}{16}\right)$$ the plot reveals that it is positive up to the first root which is $\approx 10.6155$.

If we expand its as truncated series up to $O(x^{2n})$ and solve for the first positive root $x_{2n}$ using Newton method, we get $$\left( \begin{array}{cc} n & x_{2n} \\ 1 & 7.03562 \\ 3 & 9.30115 \\ 5 & 10.4850 \\ 6 & 10.6338 \\ 7 & 10.6139 \\ 8 & 10.6156 \\ 9 & 10.6155 \end{array} \right)$$ For $n=2$ and $n=4$, there is no real solution to $f(x)=0$.

What we can also do is to build Padé approximants around $x=0$. A simple one, which matches quite well the function $f(x)$ is $$g(x)=\frac{1-\frac{102272 }{7156611}x^2+\frac{139136 }{2799746235} x^4}{1+\frac{1282 }{216867}x^2+\frac{961 }{70048041}x^4 }$$ which shows a first root at $x\approx 10.9691$.

• Thanks a million! I am very interested in Padé approximation your mentioned but don't know how to obtain your g(x). Could you teach me or give me some hints? I searched a lot but only find few results about Padé approximants for generalized hypergeometric function. Thanks again!! Mar 5 '18 at 22:03