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 (obtained with Maple) :

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...
 A: The following is a strengthening of juan's conclusion. We need to show that the logarithmic derivative of
$$ f(q):=\left(\frac{1}{\sqrt{q}}-\sqrt{q}\right) \vartheta_2^2(q) $$
is negative. Multiplying the logarithmic derivative by $q$, this means that
$$ -\frac{q}{1-q}-\sum_{n=1}^\infty\frac{8nq^{4n}}{1-q^{4n}}+\sum_{n=1}^\infty
\frac{4nq^{2n}}{1+q^{2n}} < 0. $$
For $0 < q < 1$ we have
$$ \sum_{n=1}^\infty\frac{8nq^{4n}}{1-q^{4n}} > \sum_{n=1}^\infty 8nq^{4n} = \frac{8q^4}{(1-q^4)^2} $$
and
$$ \sum_{n=1}^\infty\frac{4nq^{2n}}{1+q^{2n}} < \sum_{n=1}^\infty 4nq^{2n} = \frac{4q^2}{(1-q^2)^2}, $$
hence it suffices to show that
$$ -\frac{q}{1-q}-\frac{8q^4}{(1-q^4)^2}+\frac{4q^2}{(1-q^2)^2} < 0. $$
This holds for $0 < q < 0.37795$, hence in this range we are done.
Remark: One can generate larger ranges by keeping the first few terms in the sums, and estimating the tail similarly as above. The sums converge uniformly on any interval $[0,1-\epsilon]$, hence with a complementary argument as outlined by juan for $q\in[1-\epsilon,1]$, the above strategy should indeed work. All that is left now is numerical work, namely specifying the $\epsilon>0$ and the number of terms to be kept in the above sums.
A: Without the $\pi$, we have integer coefficients, paired $+$ and $-$ ...
$$
\frac{1}{4}\Bigl(\frac{1}{\sqrt{q}}-\sqrt{q}\Bigr) \vartheta_2^2 =
1 - q + 2   q^{2} - 2   q^{3} + q^{4} - q^{5} + 2   q^{6} - 2   q^{7} + 2   q^{8} - 2   q^{9} + 3   q^{12} - 3    
q^{13} + 2   q^{14} - 2   q^{15} + 2   q^{18} 
- 2   q^{19} + 2   q^{20} - 2   q^{21} + 2   q^{22} - 2    
q^{23} + q^{24} - q^{25} + 2   q^{26} - 2   q^{27} + 2   q^{30} - 2   q^{31} + 4   q^{32} - 4   q^{33} + 2  q^{36} - 2   q^{37} + \operatorname{O} \bigl(q^{40}\bigr)
$$
A: This is more of a long comment to the remark in GH's answer. I did some calculations regarding the expression
$$-\frac{q}{1-q} - \sum_{n=1}^{\infty} \frac{8nq^{4n}}{1-q^{4n}} + \sum_{n=1}^{\infty}\frac{4nq^{2n}}{1+q^{2n}}.$$
Keeping the first terms in the two series and estimating the remaining tail as in GH's answer, we obtain the following expression (if I didn't make any mistake..) :
$$-\frac{q}{1-q} - \sum_{n=1}^{k-1} \frac{8nq^{4n}}{1-q^{4n}} + \sum_{n=1}^{k-1}\frac{4nq^{2n}}{1+q^{2n}} + \frac{8(k-1)q^{4(k+1)}-8kq^{4k}}{(1-q^4)^2} + \frac{4kq^{2k}-4(k-1)q^{2(k+1)}}{(1-q^2)^2}.$$
Given small $\epsilon$, we want to find $k$ such that the above is negative on $(0,1-\epsilon)$.
With $k=1$, we get GH's result that the function is negative for $q \in (0,0.37795)$. With $k=8$, Maple gives that the above is negative for $q \in (0, 0.78177)$. Large values of $k$ take longer to solve explicitely, but plotting the function with $k=30$ gives a negative function for $q \in (0,\alpha)$ where $\alpha>0.9$.
