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 $f$

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
    $\begingroup$ At least $$f(k) = \pi - \frac{\pi}{16} k^{2} - \frac{3 \pi}{128} k^{4} - \frac{27 \pi}{2048} k^{6} - \frac{575 \pi}{65536} k^{8} + \operatorname{O} \bigl(k^{10}\bigr),$$ as $k \to 0+$, so it is decreasing near $k=0$. $\endgroup$
    – GEdgar
    Commented Jul 6, 2012 at 15:31
  • 3
    $\begingroup$ Note that your function can also be expressed in terms of the elliptic nome: $$k\,K(k)\,\frac{1-q(k)}{2\sqrt{q(k)}}$$ $\endgroup$ Commented Jul 8, 2012 at 14:04
  • 6
    $\begingroup$ The following comment was posted by Henry Cohn on meta.MO: It's definitely possible to prove that your function is decreasing by an ugly and unilluminating calculation that shows that the derivative is nonpositive everywhere. Specifically, near $k=0$ you can compute the Taylor series expansion and bound the error. For larger $k$, you can check the values of the derivative at a bunch of points and verify that there are no sign changes in between by bounding the second derivative. So if you just need this result to get a rigorous proof of some theorem, then it will be doable... $\endgroup$ Commented Jul 12, 2012 at 6:15
  • 4
    $\begingroup$ On the other hand, much more seems to be true. Specifically, all the derivatives seem to be negative, not just the first derivative. You can see this in the Taylor series expansion, which has all negative coefficients beyond the constant term (well, it's an even function, so the odd terms vanish, but the even terms all have negative coefficients). And the terms are pretty nice: the coefficient of $k^{2i}$ seems to be pi times a rational number with denominator dividing $16^i$. $\endgroup$ Commented Jul 12, 2012 at 6:16
  • 3
    $\begingroup$ I don't know how to prove any of this, but it's more remarkable than just being a decreasing function, and all this suggests that there should be a nice way of understanding this function. Plenty of functions are decreasing for no especially good reason, but this sort of absolute monotonicity is much less common. $\endgroup$ Commented Jul 12, 2012 at 6:16

2 Answers 2


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

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


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

  • $\begingroup$ Thank you. I have absolutely no idea why once you divide $f$ by $\pi$ and make the change of variable $m=(k/4)^2$, all the coefficients seem to be negative integers... That's very interesting! $\endgroup$ Commented Jul 13, 2012 at 14:42
  • $\begingroup$ Note that the last formula can also be found in the wikipedia link for elliptic integrals given in the question. $\endgroup$ Commented Jul 19, 2012 at 19:13

See the developments here. It seems all that is left is (reasonable) numerical work.


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