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Hi,

I asked some time ago the following question on math.stackexchange, but I ask it here too since it remains unanswered.

The question concerns a function I encountered during research :

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

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

More specifically, my question is the following :

Is $f$ decreasing on $(0,1)$?

This seems to be true, as the graph below suggests (obtained with Maple) :

graph of $f$

In fact, as remarked by Henry Cohn, much more seems to be true : all the derivatives of $f$ seem to be negative. This can be seen by looking at the Taylor series expansion of $f$ (see the link to math.stackexchange). The Taylor series expansion seems to have all negative coefficients (except the constant term), and the coefficient of $k^{2j}$ seems to be $\pi$ times a rational number with denominator dividing $16^j$...

Any comment or relevant reference is welcome.

Thank you, Malik

EDIT (20-07-2012) It was remarked by J.M. on M.SE that $f$ can be written as $$f(k)=kK(k)\frac{1-q(k)}{2\sqrt{q(k)}},$$ where $q(k)$ is the Elliptic nome. Maybe this is useful...
show/hide this revision's text 2 added 236 characters in body

Hi,

I asked some time ago the following question on math.stackexchange, but I ask it here too since it remains unanswered.

The question concerns a function I encountered during research :

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

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

More specifically, my question is the following :

Is $f$ decreasing on $(0,1)$?

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

graph of $f$

In fact, as remarked by Henry Cohn, much more seems to be true : all the derivatives of $f$ seem to be negative. This can be seen by looking at the Taylor series expansion of $f$ (see the link to math.stackexchange). The Taylor series expansion seems to have all negative coefficients (except the constant term), and the coefficient of $k^{2j}$ seems to be $\pi$ times a rational number with denominator dividing $16^j$...

Any comment or relevant reference is welcome.

Thank you, Malik

EDIT (20-07-2012) It was remarked by J.M. on M.SE that $f$ can be written as $$f(k)=kK(k)\frac{1-q(k)}{2\sqrt{q(k)}},$$ where $q(k)$ is the Elliptic nome. Maybe this is useful...
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Is the following function decreasing on $(0,1)$?

Hi,

I asked some time ago the following question on math.stackexchange, but I ask it here too since it remains unanswered.

The question concerns a function I encountered during research :

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

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

More specifically, my question is the following :

Is $f$ decreasing on $(0,1)$?

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

graph of $f$

In fact, as remarked by Henry Cohn, much more seems to be true : all the derivatives of $f$ seem to be negative. This can be seen by looking at the Taylor series expansion of $f$ (see the link to math.stackexchange). The Taylor series expansion seems to have all negative coefficients (except the constant term), and the coefficient of $k^{2j}$ seems to be $\pi$ times a rational number with denominator dividing $16^j$...

Any comment or relevant reference is welcome.

Thank you, Malik