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While doing some research I got stuck trying to prove that the following function is decreasing

$f(k):= k K(k) \sinh \left(\frac{\pi}{2} \frac{K(\sqrt{1-k^2})}{K(k)}\right)$ for $k \in (0,1)$.

Here $K$ is the Complete elliptic integral of the first kind, defined by $K(k):= \int_{0}^{1} \frac{dt}{\sqrt{1-t^2} \sqrt{1-k^2t^2}}.$

This seems to be true, as the graph below suggests :

graph of <span class=f">

I really don't know much about elliptic integrals, so perhaps someone here can give some insight. Any relevant reference on elliptic integrals of the first kind is welcome.

Thank you, Malik

EDIT (2012-07-09) :

Using J.M.'s suggestion to rewrite the function $f(k)$ as $f(k) = kK(k) \frac{1-q(k)}{2 \sqrt{q(k)}}$ and using the derivative formulas $K'(k) = \frac{E(k)}{k(1-k^2)} - \frac{K(k)}{k},$ $q'(k)=\frac{\pi^2}{2} \frac{q(k)} { K(k)^2 (1-k^2)k}$ where $E(k)$ is the Complete elliptic integral of the second kind, I was able to calculate $f'(k)$ and reduce the problem to showing that the following function is negative for $k \in (0,1)$ :

$g(k):= 4(1-q(k))K(k)E(k) - \pi^2 (1+q(k)).$

Below is the graph of $g$ obtained with Maple :

enter image description here

EDIT (19-07-2012)

I asked the question on MathOverflow!

  • 1
    @DanPetersen: Thank you for copying Henry Cohn's comment here. For information, here is my response on meta.MO : That is a great and illuminating comment, thank you! I suspected something more was going on about this function, and I'm quite interested in finding exactly what it is. Your comment clarifies this a lot. As soon as the bounty expires, I'll post a modified version of the question on MO which will include an interrogation regarding this "absolute monotonicity" you pointed out.2012-07-12

2 Answers 2

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A few more terms for those investigating. From Maple. These coefficients are not listed in the On-line Encyclopedia of Integer Sequences.

$ \frac{4}{\pi} \sqrt{m} \;K(4 \sqrt{m}) \sinh \biggl(\frac{\pi\; K(\sqrt{1 - 16 m})}{2\;K(4 \sqrt{m})}\biggr) \\ = 1 - m - 6 m^{2} - 54 m^{3} - 575 m^{4} - 6715 m^{5} - 83134 m^{6} - 1071482 m^{7} - \\ \quad{}\quad{} 14221974 m^{8} - 193050435 m^{9} - 2667157340 m^{10} - 37378279402 m^{11} - \\ \quad{}\quad{} 530024062361 m^{12} - 7590192561912 m^{13} - \\ \quad{}\quad{}109610113457650 m^{14} - 1594344146568120 m^{15} - \\ \quad{}\quad{}23336667998911128 m^{16} - 343468859344118109 m^{17} - \\ \quad{}\quad{}5079858166426507168 m^{18} - 75457168334744888190 m^{19} - \\ \quad{}\quad{}1125223725054635766392 m^{20} + \operatorname{O} \bigl(m^{21}\bigr) $

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Who knows if this is relevant? See A002849 $ \frac{2}{\pi}K(4\sqrt{m}) = 1+4m+36m^2+400m^3+4900m^4+\dots =\sum_{n=0}^\infty \binom{2n}{n}^2m^n $

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    Note that the last formula can also be found in the wikipedia link for elliptic integrals given in the question.2012-07-19
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See the developments here. It seems all that is left is (reasonable) numerical work.