Timeline for Find better than $ 4^n\prod_{k=1}^{n-1}\cos^2(k)\sim e^{o(n)}$

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Feb 15, 2021 at 14:14	vote	accept	Paul
Feb 15, 2021 at 13:43	comment	added	Paul		Thanks GH I didn't realize that f is not Riemann integrable on $I = [0,2 \ pi]$
Feb 15, 2021 at 5:36	comment	added	GH from MO		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 $\lim_{h\to 0+}(\int_0^{\pi/2-h} f+\int_{\pi/2+h}^\pi f)$.
Feb 15, 2021 at 5:26	history	edited	GH from MO		edited tags; edited tags
Feb 15, 2021 at 3:23	history	edited	Paul	CC BY-SA 4.0	added 275 characters in body
Feb 14, 2021 at 15:31	answer	added	Carlo Beenakker		timeline score: 5
Feb 14, 2021 at 14:30	comment	added	François Brunault		You should link to this: mathoverflow.net/questions/383866
Feb 14, 2021 at 12:46	history	asked	Paul	CC BY-SA 4.0