How to find $\sup \Pi_{i=0}^{n} (\sin(i)^2 - \frac{25}{16})$? Let $\sup,\inf,{\rm dif}$ denote resp supremum , infimum and $\rm dif$ = supremum - infimum.
Does any of the 3 below have a closed form ?
$\sup \Pi_{i=0}^{n} (\sin(i)^2 - \frac{25}{16})$
$\inf \Pi_{i=0}^{n} (\sin(i)^2 - \frac{25}{16})$
${\rm dif} \Pi_{i=0}^{n} (\sin(i)^2 - \frac{25}{16})$
 A: Here is a partial answer. Let:
$$A_n := \prod_{k=0}^{n-1} \left( \sin^2 (k)-\frac{25}{16} \right),$$
and:
$$F(x) := \ln \left( \frac{5}{4}-\sin(x) \right).$$
Then:
$$\ln (A_n) = (-1)^n \sum_{k=0}^{n-1} \left(F(k)+F(-k)\right).$$
Now, $F$ has integral $0$ (this comes from the choice of the constant $5/4$), and is analytic. In addition, $1/\pi$ has a finite irrationality measure. Hence, $F$ is a coboundary for the translation by $1$ on $\mathbb{R}_{/2\pi\mathbb{Z}}$. More precisely, we can solve the equation:
$$F(x) = g(x+1)-g(x),$$
with $g(0)=0$. Taking the Fourier coefficients, we get, for $\xi \neq 0$:
$$\hat{g} (\xi) = \frac{\hat{F} (\xi)}{e^{i\xi}-1},$$
whence:
$$g(x) = \sum_{\substack{\xi \in \mathbb{Z} \\ \xi \neq 0}} \hat{F} (\xi) \frac{e^{i\xi x}-1}{e^{i\xi}-1}.$$
Going back to our initial problem, we get:
$$\ln (A_n) = (-1)^n (g(n)+g(-n)).$$
The translation by $1$ is uniquely ergodic on $\mathbb{R}_{/2\pi\mathbb{Z}}$. Better, the translation by $(1,1)$ is uniquely ergodic on $\mathbb{R}_{/2\pi\mathbb{Z}} \times \mathbb{Z}_{/2\mathbb{Z}}$, so that we don't have to worry about the sign. Hence,
$$-\ln(\inf_n A_n) = \ln(\sup_n A_n) = \max_{x \in \mathbb{R}_{/2\pi\mathbb{Z}}} |g(x)+g(-x)| = 2 \max_{x \in \mathbb{R}_{/2\pi\mathbb{Z}}} \left| \sum_{\substack{\xi \in \mathbb{Z} \\ \xi \neq 0}} \hat{F} (\xi) \frac{1-\cos(\xi x)}{1-e^{i\xi}}\right|.$$
Disclaimer : sign errors may be hidden somewhere.
