3

WolframAlpha claims that $\frac{1}{n}\sum_{k=1}^n \cos^2(2^k) \to 0$ as $n \to \infty$:

https://www.wolframalpha.com/input/?i=limit+%28sum%28%5Ccos%282%5Ek%29%5E2%2Ck%3D1%2Cn%29%29%2Fn+as+n-%3Einfinity

I doubt: By $\cos^2(2^{k+1})=(2\cos^2(2^k)-1)^2$ and Jensen's inequality applied to the convex function $x \mapsto x^2$ (or Cauchy Schwarz) $$ \frac{1}{n}\sum_{k=1}^n \cos^2(2^k)=\frac{1}{n}\sum_{k=1}^n (2\cos^2(2^{k-1})-1)^2 \ge \left(\frac{1}{n}\sum_{k=1}^n (2\cos^2(2^{k-1})-1)\right)^2 $$ $$ = \left(\frac{1}{n}\sum_{k=1}^n (2\cos^2(2^{k-1}))-1\right)^2 = \left(\frac{2}{n}\sum_{k=0}^{n-1} (\cos^2(2^{k}))-1\right)^2. $$ Now let $(n_j)$ be a sequence with $$ \frac{1}{n_j}\sum_{k=1}^{n_j} \cos^2(2^k) \to s:= \liminf_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \cos^2(2^k) \quad (j \to \infty). $$ Then $$ \frac{1}{n_j}\sum_{k=0}^{n_j-1} \cos^2(2^k) \to s \quad (j \to \infty), $$ and therefore $s \ge (2s-1)^2$, that is $4s^2-5s+1 \le 0$. Thus $s \in [1/4,1]$.

Two questions:

  1. Is this correct?
  2. Numerical experiments indicate $\frac{1}{n}\sum_{k=1}^n \cos^2(2^k) \to \frac{1}{2}$ as $n \to \infty$. Does anybody know a proof?

Thanks for any helpful comment.

Edit: I replaced $\limsup$ by $\liminf$ (same proof but more information).

Gerd
  • 7,034
  • Heuristics: \begin{align} \frac{1}{n}\sum\limits_{k = 1}^n {\cos ^2 (2^k )} & = \frac{1}{n}\int_2^n {\cos ^2 (2^x )dx} + \mathcal{O}!\left( {\frac{1}{n}} \right) = \frac{1}{{\log 2}}\frac{1}{n}\int_1^{2^n } {\frac{{\cos ^2 (t)}}{t}dt} + \mathcal{O}!\left( {\frac{1}{n}} \right)\ & = \frac{1}{{\log 2}}\frac{1}{n}\left( {\frac{1}{2}\log (2^n) + \mathcal{O}(1)} \right) + \mathcal{O}!\left( {\frac{1}{n}} \right) = \frac{1}{2} + \mathcal{O}!\left( {\frac{1}{n}} \right) \end{align} – Gary Dec 08 '21 at 13:41
  • @Gary Looks interesting. However, I don't see the first equality. For this $\sum_{k=1}^n\cos^2(2^k)- \int_2^n \cos^2(2^x)dx$ should be bounded. How can this be seen? – Gerd Dec 08 '21 at 13:58
  • That is why I wrote "heuristics". The argument is correct iff $$ \int_2^n {\cos ^2 (2^x )d\left{ x \right}} = \mathcal{O}(1) $$ as $n\to +\infty$ – Gary Dec 08 '21 at 14:01
  • 1
    I see. Thank's. I will think about it. – Gerd Dec 08 '21 at 14:03
  • The sum is also $$ \frac{1}{2} + \frac{1}{{2n}}\sum\limits_{k = 1}^n {\cos (2^{k + 1} )} $$ so it is enough to show that the cosine sum is $o(n)$. – Gary Dec 08 '21 at 14:07
  • What is the limit for degrees? The sum is periodic. – Empy2 Dec 08 '21 at 14:14
  • 3
    It is known that $2^k a !!\mod! 1$ is uniformly distributed for almost all, but not all, irrational $a$. So if it was true for $a=1/\pi$ then we would have $$ \frac{1}{n}\sum\limits_{k = 1}^n {\cos (2^{k + 1} )} = \frac{1}{n}\sum\limits_{k = 1}^n {\cos \left( {2\pi \left{ {\frac{{2^k }}{\pi }} \right}} \right)} \to \int_0^1 {\cos (2\pi x)dx} = 0. $$ – Gary Dec 08 '21 at 14:17
  • @Empy2 ... For degrees, $\frac{1}{n}\sum_{k=1}^n\cos(2^k\pi/180)^2$ : the answer is $1/2$. Start with two terms $\cos(\pi/90)^2+\cos(\pi/45)^2$, then repeat a period of $12$ terms with average value $1/2$. So limit is $1/2$. – GEdgar Dec 09 '21 at 20:08

0 Answers0