Not signed in (Sign In)

Vanilla 1.1.9 is a product of Lussumo. More Information: Documentation, Community Support.

    I asked the following question on math.stackexchange :

    So far it received 17 upvotes but remains unanswered. I was thinking of asking it on mathoverflow too. Would it be acceptable? Perhaps it will result in interesting references and comments...

    Thank you,

    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.

    • CommentAuthorHenry Cohn
    • CommentTimeJul 11th 2012

    What Fran├žois 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.

    Dear Henry, I reposted your nice observation in the comment thread on math.SE.
    • CommentAuthorHenry Cohn
    • CommentTimeJul 12th 2012


    @Henry Cohn : 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.
    Now that the bounty expired, I asked the question here on MO :