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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!!

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  • $\begingroup$ 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. $\endgroup$ Mar 5 '18 at 6:46
  • $\begingroup$ 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!! $\endgroup$
    – user431550
    Mar 6 '18 at 3:51
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    $\begingroup$ In particular, we have the identity $${}_1 F_2\left({{2}\atop{\frac94,\frac{11}{4}}}\middle|-\frac{x^2}{16}\right)=\frac{105\sqrt{\pi}}{8 x^{7/2}}\left(6 f\left(\sqrt{\frac{x}{\pi}}\right)+2 x g\left(\sqrt{\frac{x}{\pi}}\right)-(x+3)\cos\frac{x}{2}-(x-3)\sin\frac{x}{2}\right)$$ $\endgroup$ Mar 6 '18 at 4:03
  • $\begingroup$ Wow! Thank you! This is a nice result! Is it obviously? I already searched some special function books but cannot obtain this result :( $\endgroup$
    – user431550
    Mar 6 '18 at 5:29
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    $\begingroup$ I had some help deriving it from Mathematica; I believe Maple's convert() should be able to do something similar. $\endgroup$ Mar 6 '18 at 5:40
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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 $\left[0, \sqrt {99/2} \,\right)$.

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  • $\begingroup$ Nice proof! Thank you so much! $\endgroup$
    – user431550
    Mar 7 '18 at 0:01
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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 $x\approx 10.9691$.

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  • $\begingroup$ 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!! $\endgroup$
    – user431550
    Mar 5 '18 at 22:03

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