Vanilla 1.1.9 is a product of Lussumo. More Information: Documentation, Community Support.
1 to 7 of 7
It's probably best to wait until the bounty has expired to avoid problems. If you do post on MO, make sure to include a link to the original question.
What François says sounds reasonable. In the meantime, here are a few comments (since I don't have the reputation to post comments on MSE):
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.
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 $16^i$. 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.
Thanks!
1 to 7 of 7