Wikipedia:Reference desk/Archives/Mathematics/2014 December 19

= December 19 =

Cosine of powers series
Does $$\sum_{n=0}^\infty (-1)^n\cos(z^n)$$ have any meaning? I do know that I can identify it with the Weierstrass function with a = -1, b = z, and x = 1/π. The Weierstrass function is not differentiable anywhere but I'm going to conjecture that the function defined by this series is differentiable in some neighborhood of the origin. Am I right?--Jasper Deng (talk) 19:55, 19 December 2014 (UTC)
 * The Weierstrass function is a function of x. You've set x to a constant. I can't see how to turn anything like what you have into something that converges. Dmcq (talk) 21:57, 19 December 2014 (UTC)
 * As Dmcq indicates, the similarity to the Weierstrass function is meaningless here. Your function is divergent (possibly on every z, so it may be the empty function). As to differentiable anywhere, this seems even more impossible. —Quondum 22:47, 19 December 2014 (UTC)
 * I had a tentative proof but it didn't work. I think that's true, but proving it never converges for z>1 is an interesting exercise! :) I was considering the properties of zn so cos( zn) tended to zero as n tended to infinity. Dmcq (talk) 23:08, 19 December 2014 (UTC)
 * Yes, it looks rather challenging. But intuition says that if z exist for which it converges, such z are extremely sparse, and never at a point of continuity. Anyhow, the question of differentiability near the origin is settled: the sum diverges for |z|≤1. —Quondum 04:09, 20 December 2014 (UTC)
 * It seems like a good candidate for Cesaro summability (e.g., the series is Cesaro summable at z=0, with value 1/2). It seems rather hard to prove this in general.  (Although it may be useful first to prove that $$z^n$$ mod 2&pi; is equidistributed.)   Sławomir Biały  (talk) 14:04, 20 December 2014 (UTC)
 * The thought of taming the divergence through any of several summation techniques had occurred to me, and it seems pretty likely that it would be so summable at an infinite number of points in any interval. But the summation used to define the function would have to be defined as such. Perhaps one use as a premise that "every" z produces a series that tends to uniform distribution of zn mod 2&pi; (or rather that sum from after a certain point tends to zero in a certain sense) for large z, and see whether the problem becomes tractable. —Quondum 16:03, 20 December 2014 (UTC)
 * Yes, I should have added that for |z|&le;1, Cesaro summability follows straightforwardly from $$\cos t = 1+O(t^2).$$ It is |z|>1 that is the interesting case, where presumably one can show equidistribution of the powers of z.  This may or may not be enough, though.  Equidistribution should show that the means (notation from our Cesaro sum article) $$s_n/n$$ tend to zero, so $$s_n=o(n)$$.  But of course, this is not enough to guarantee Cesaro summability.  For that, at a minimum we would need to get this down to $$s_n=O(1)$$ (that is, bounded partial sums).  But the cosine is a fairly amicable function to deal with, so I think that can probably be arranged with a little effort (although one might need a stronger kind of ergodicity on the powers, not sure, there are lots of intricate Tauberian theorems, etc.).   Sławomir Biały  (talk) 18:39, 20 December 2014 (UTC)
 * There is also the possibility of using a stronger summation method, e.g., Abel summation. So study the asymptotics of
 * $$\sum (-1)^n e^{-n\epsilon}\cos(z^n)$$
 * as $$\epsilon\to 0$$ (e.g., Hardy–Littlewood tauberian theorem).  Sławomir Biały  (talk) 19:14, 20 December 2014 (UTC)
 * It may make more sense to first study the behaviour (and indeed determine the function) if redefined using the Cesaro sum (or any stronger sum, if desired) in the interval |z|≤1. If it turns out to be analytic, it could then be extended through analytic continuation. —Quondum 19:37, 20 December 2014 (UTC)
 * Yes, one could try that, but the series has the vibe of something that is the boundary value of an analytic function rather than something that is real analytic (hence the need for a nuanced notion of summation). But I'd happily be wrong about that.   Sławomir Biały  (talk) 19:56, 20 December 2014 (UTC)
 * Interestingly, $$\lim_{\epsilon \downarrow 0} \sum_{n=0}^{\inf} (-1)^n e^{-n\epsilon} \cos(nz)=\tfrac{1}{2}$$ (for real z, at least). Perhaps not surprising, but I'm wondering whether many other simple functions as the parameter to $$\cos$$ might not also give this constant result. —Quondum 15:56, 22 December 2014 (UTC)