1
$\begingroup$

I'm trying to prove the following inequality:

For $x \in (0,3)$, $$ {_1F_2[1;\frac{5}{4},\frac{7}{4};\frac{-(1\cdot x)^2}{4}]}+{_1F_2[1;\frac{5}{4},\frac{7}{4};\frac{-(3\cdot x)^2}{4}]}\gt 2\cdot {_1F_2[1;\frac{5}{4},\frac{7}{4};\frac{-(5\cdot x)^2}{4}]}$$

My attempts:

I wrote Fresnel Integral Transform for Hypergeometric Functions, but it gaves me only more complicated formula to prove.

I also find some Series representations for Hypergeometric Functions. But I then tried unsuccessfully to express it and didn't find a good one to re-express it.

I also searched L.Luke's book trying to use asymptote to show it but I didn't find a good approximation. Perhaps the hypergeom can simplify with the Bessel function?

Any help would be appreciated! Thanks!

$\endgroup$
0
$\begingroup$

This can be brute-forced by using the same approach as here.

The left-hand side of the inequality will have to be expanded to order 16, and the right-hand side to order 34; the approximation errors will be bounded by the absolute values of the $x^{18}$ and the $x^{36}$ terms respectively.

Then, since everything is polynomial, a Sturm sequence can be computed to prove that the difference of the lower bound for the left-hand side and the upper bound for the right-hand side does not have zeros on $(0,3]$.

$\endgroup$
  • $\begingroup$ Thank you! Why we expand the inequality to order 16 and 34 on the LHS and RHS respectively? $\endgroup$ – user431550 Mar 10 '18 at 18:59
  • $\begingroup$ Simply because the experimentation shows that a lesser number of terms is not enough. The lower bound for the lhs diverges very fast to $-\infty$ when $x$ grows, and the upper bound for the rhs to $+\infty$. Approximations of lower orders would intersect before $x=3$. $\endgroup$ – Maxim Mar 10 '18 at 19:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.