2
$\begingroup$

Let $$u_n=\prod_{k=1}^{n-1}\cos^2(k)$$ then $$\frac1n \ln(u_n) = \frac1n\sum_{k=0}^{n-1} \ln(\cos^2(k)) \underset{n\to\infty}\longrightarrow \frac1{2\pi} \int_0^{2\pi} \ln(\cos^2(x))\,{\rm d}x = -\ln(4)$$ We deduce that $$\ln(u_n)\sim -n\ln(4)$$ . I think that the sequence $v_n=\ln(4^n u_n))$ have not limit , So there is no constant c such that $u_n \sim c\,4^{-n}$. Can we find better then $$ \prod_{k=1}^{n-1}\cos^2(k)\sim 4^{-n}e^{o(n)}$$ . What is $\limsup 4^n u_n$ and $\liminf 4^n u_n$ ?

additional comments

wolfram does not confirm that $\liminf 4^n u_n =0$enter image description here

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ You should link to this: mathoverflow.net/questions/383866 $\endgroup$
    – François Brunault
    Commented Feb 14, 2021 at 14:30
  • 2
    $\begingroup$ You need to be careful with the limit in your second display. If $k\bmod\pi$ gets very close to $\pi/2$ for some $k\in\{1,\dotsc,n-1\}$, then $\ln(u_n)/n$ can be much smaller than $-\ln(4)$. This is why, in my response to your previous question, I only stated $\prod_{k=1}^{n-1}\cos^2(k)\leq 4^{-(1+o(1))n}$ instead of $\prod_{k=1}^{n-1}\cos^2(k)=4^{-(1+o(1))n}$. Note that $f(x):=\ln(\cos^2(x))$ is not Riemann integrable on $[0,\pi]$, because it is not bounded there. Instead, $\int_0^\pi f$ exists as an improper Riemann integral, i.e. as $\lim_{h\to 0+}(\int_0^{\pi/2-h} f+\int_{\pi/2+h}^\pi f)$. $\endgroup$
    – GH from MO
    Commented Feb 15, 2021 at 5:36
  • 1
    $\begingroup$ Thanks GH I didn't realize that f is not Riemann integrable on $I = [0,2 \ pi]$ $\endgroup$
    – Paul
    Commented Feb 15, 2021 at 13:43

1 Answer 1

Reset to default
5
$\begingroup$

I presume that exchanging the cosine by the sine will not matter for the large-$n$ behavior of the product, so let me consider $$a_n^2=4^n\prod_{k=1}^{n}\sin^2 k=\left(\prod_{k=1}^n\left|1-e^{k\alpha \pi i}\right|\right)^2\;\;\text{with}\;\;\alpha=2/\pi.$$ The convergence of $a_n$ was determined in this MO posting from five years ago, with lim inf $a_n$ equal to zero and lim sup $a_n$ equal to infinity for irrational $\alpha$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you Carlo Beenakker $u_n=4^n \prod_{k=1}^{n-1}\cos^2(k)=4\left(\prod_{k=1}^{n-1}\left|1+e^{2k i}\right|\right)^2$I'm trying to see how to apply the result of your link if we replace the - with the + $\endgroup$
    – Paul
    Commented Feb 14, 2021 at 21:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .