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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!

  • 0
    Have you tried differentiating with respect to $k$?2012-07-06
  • 1
    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 $k=0$.2012-07-06
  • 0
    @tomchuchta : Yes, but this results in a complicated expression, and it's not clear (to me at least..) that the resulting expression is negative. It seems to be the case though, again with Maple.2012-07-06
  • 1
    How did you get to this function?2012-07-06
  • 3
    Note that your function can also be expressed in terms of the [elliptic nome](http://mathworld.wolfram.com/Nome.html): $$k\,K(k)\,\frac{1-q(k)}{2\sqrt{q(k)}}$$2012-07-08
  • 0
    @J.M. : Interesting, thank you. I didn't know my function was related to the elliptic nome, that might be useful..2012-07-08
  • 0
    @Norbert : It's complicated to explain, but let's just say this function is part of a calculation I'm doing to verify the simplest non-trivial case of a conjecture I have regarding Analytic Capacity (part of my PhD thesis). I'm not hoping that someone does all the work for me, but I posted this question hoping to get some insight about the function $f(k)$ and maybe some good references about elliptic integrals (I really don't know much about them).2012-07-08
  • 0
    I'm interested, whether this problem is worth such efforts if you can check with mathematica that derivative is negative.2012-07-08
  • 0
    It's an idea only. Hopefully it works, but "ugly". Using GEdgar with error estimation, you obtain a $k_0$ such that f is decreasing on $[0,k_0]$. Then on $[k_0,1]$ calculating the asymptotic of g(k) with error estimation you obtain the negativity of g.2012-07-10
  • 6
    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...2012-07-12
  • 4
    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 $16^i$.2012-07-12
  • 3
    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.2012-07-12
  • 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

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