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2 Correct typo $(2^v v)!$ for $(2^v v!)$; also reset a couple of formulas
Some further manipulation past my answer of a week ago yields a formula that should reduce the proof of the observed behavior to routine (if not entirely pleasant) estimates. Whereas $f$ is constant at $2/B(1/2,n/2)$ 2/B(\frac12,\frac{n}2)$for even$n$, in the present case of odd$n$the maximum exceeds$2/B(1/2,n/2)$2/B(\frac12,\frac{n}2)$ by a tiny amount that is very nearly $$\frac{4}{\pi} \phantom. \frac1{n+2} \frac2{n+4} \frac3{n+6} \cdots \frac{n}{3n} = \frac4\pi n! \frac{n!!}{(3n)!!} = (27+o(1))^{-n/2}$$ for large $n$. Here and later we use "$u!!$" only for positive odd $u$ to mean the product of all odd integers in $[1,u]$; that is, $u!! := u!/(2^v v)!$ v!)$where$u=2v+1$. Recall the previous notations: we take$n=2m+1$and $$g(x) = f(x+\frac\pi2) = g(-x) = -g(x+n\pi),$$ which has a finite Fourier expansion in cosines of odd multiples of$X := x/n$, namely $$f(x) = (-1)^m 2^{-n} \sum_{j=0}^n (-1)^j {n\choose j} \frac{\cos \phantom. tX}{\sin \frac{\pi t}{2n}}$$ where$t = n-2j$. Even before we use this expansion, we deduce from the original formula $$f(\beta)=\sum_{k=1}^n \sin^n\frac{-\beta+k \pi}{n}$$ that$f(\beta)-f(\beta+\pi) = 2\phantom.\sin^n (\beta/n)$, from which it follows that$g(x)$is maximized somewhere in$|x| \leq \pi/2$, but that changing the optimal$x$by a small integral multiple of$\pi$reduces$g$by a tiny amount; this explains the near-maxima I observed at$x=\pm\pi$for$2|m$, and indeed the further oscillations for both odd and even$m$that I later noticed as$n$grows further. This also suggests that in and near the interval$|x| \leq \pi/2$our function$g$should be very nearly approximated for large$n$by an even periodic function$\tilde g(x)$of period$\pi$. We next outline the derivation of such an approximation, with$\tilde g$having an explicit cosine-Fourier expansion $$\tilde g(x) = g_0 + g_1 \cos 2x + g_2 \cos 4x + g_3 \cos 6x + \cdots$$ where$g_0 = 2/B(1/2,n/2)$2/B(\frac12,\frac{n}2)$ and, for $l>0$, $$g_l = (-1)^{m+l-1} \frac4\pi \frac{n!}{2l+1} \frac{((2l-1)n)!!}{((2l+1)n)!!}$$ with the double-factorial notation defined as above. Thus $$\tilde g(x) = g_0 + (-1)^m \frac{4n!}\pi \left(\frac{n!!}{(3n)!!} \cos 2x - \frac13 \frac{(3n)!!}{(5n)!!} \cos 4x + \frac15 \frac{(5n)!!}{(7n)!!} \cos 6x - + \cdots \right).$$ For large $n$, this is maximized at $x=0$ or $x=\pm\pi/2$ according as $m$ is even or odd. Since we already know by symmetry arguments that $g'(0) = g'(\pm \pi/2) = 0$, this point or points will also be where $g$ is maximized, once it is checked that $g - \tilde g$ and its first two derivatives are even tinier there.
The key to all this is the partial-fraction expansion of the factor $1 / \sin (\pi t /2n)$ in the Fourier series of $g$, obtained by substituting $\theta = \pi t / 2n$ into $$\frac1{\sin \pi\theta} = \frac1\pi \sum_{l=-\infty}^\infty \frac{(-1)^l}{\theta-l}$$ with the conditionally convergent sum interpreted as a principal value or Cesàro limit etc. I already noted in the previous note that the main term, for $l=0$, yields the convolution of $\cos^n (x/n)$ with a symmetrical square wave, which is thus maximized at $x=0$ and almost constant near $x=0$; we identify the constant with $2/B(1/2,n/2)$ 2/B(\frac12,\frac{n}2)$using the known product formula for$\int_{-\pi/2}^{\pi/2} \cos^n X \phantom. dX$. The new observation is that each of the error terms$(-1)^l/(\theta-l)$likewise yields the convolution with a square wave of$(-1)^l \cos(2lx) \phantom. \cos^n(x/n)$. If we approximate this square wave with a constant, we get the formula for$g_l$displayed above, via the formula for the$n$-th finite difference of a function$1/(j_0-j)$. The error in this approximation is still tiny (albeit not necessarily negative) because$\cos^n (x/n)$is minuscule when$x$is within$\pi/2$of the square wave's jump at$\pm \pi n / 2$. I've checked these approximations numerically to high precision (modern computers and gp make this easy) for$n$as large as$100$or so, in both of the odd congruence classes mod$4$, and it all works as expected; for example, when$n=99$we have$f(0) - g_0 = 2.57990478176660\ldots \cdot 10^{-70}$, which almost exactly matches the main term$g_1 = (4/\pi) \phantom. 99! \phantom. 99!!/297!!$but exceeds it by$5.9110495\ldots \cdot 10^{-102}$, which is almost exactly$g_2 = (4/\pi) \phantom. 99! \phantom. 297!!/(3 \cdot 495!!)$but too large by$7.92129\ldots \cdot 10^{-120}$, which is almost exactly$g_3 = (4/\pi) \phantom. 99! \phantom. 495!!/(5 \cdot 693!!)$, etc.; and likewise for$n=101$except that the maximum occurs at$\beta = \pi/2$and is approximated by an alternating sum$g_1 - g_2 + g_3 \ldots$(actually here this approximation is exact because$x=0$). 1 Some further manipulation past my answer of a week ago yields a formula that should reduce the proof of the observed behavior to routine (if not entirely pleasant) estimates. Whereas$f$is constant at$2/B(1/2,n/2)$for even$n$, in the present case of odd$n$the maximum exceeds$2/B(1/2,n/2)$by a tiny amount that is very nearly $$\frac{4}{\pi} \frac1{n+2} \frac2{n+4} \frac3{n+6} \cdots \frac{n}{3n} = \frac4\pi n! \frac{n!!}{(3n)!!} = (27+o(1))^{-n/2}$$ for large$n$. Here and later we use "$u!!$" only for positive odd$u$to mean the product of all odd integers in$[1,u]$; that is,$u!! := u!/(2^v v)!$where$u=2v+1$. Recall the previous notations: we take$n=2m+1$and $$g(x) = f(x+\frac\pi2) = g(-x) = -g(x+n\pi),$$ which has a finite Fourier expansion in cosines of odd multiples of$X := x/n$, namely $$f(x) = (-1)^m 2^{-n} \sum_{j=0}^n (-1)^j {n\choose j} \frac{\cos \phantom. tX}{\sin \frac{\pi t}{2n}}$$ where$t = n-2j$. Even before we use this expansion, we deduce from the original formula $$f(\beta)=\sum_{k=1}^n \sin^n\frac{-\beta+k \pi}{n}$$ that$f(\beta)-f(\beta+\pi) = 2\phantom.\sin^n (\beta/n)$, from which it follows that$g(x)$is maximized somewhere in$|x| \leq \pi/2$, but that changing the optimal$x$by a small integral multiple of$\pi$reduces$g$by a tiny amount; this explains the near-maxima I observed at$x=\pm\pi$for$2|m$, and indeed the further oscillations for both odd and even$m$that I later noticed as$n$grows further. This also suggests that in and near the interval$|x| \leq \pi/2$our function$g$should be very nearly approximated for large$n$by an even periodic function$\tilde g(x)$of period$\pi$. We next outline the derivation of such an approximation, with$\tilde g$having an explicit cosine-Fourier expansion $$\tilde g(x) = g_0 + g_1 \cos 2x + g_2 \cos 4x + g_3 \cos 6x + \cdots$$ where$g_0 = 2/B(1/2,n/2)$and, for$l>0$, $$g_l = (-1)^{m+l-1} \frac4\pi \frac{n!}{2l+1} \frac{((2l-1)n)!!}{((2l+1)n)!!}$$ with the double-factorial notation defined as above. Thus $$\tilde g(x) = g_0 + (-1)^m \frac{4n!}\pi \left(\frac{n!!}{(3n)!!} \cos 2x - \frac13 \frac{(3n)!!}{(5n)!!} \cos 4x + \frac15 \frac{(5n)!!}{(7n)!!} \cos 6x - + \cdots \right).$$ For large$n$, this is maximized at$x=0$or$x=\pm\pi/2$according as$m$is even or odd. Since we already know by symmetry arguments that$g'(0) = g'(\pm \pi/2) = 0$, this point or points will also be where$g$is maximized, once it is checked that$g - \tilde g$and its first two derivatives are even tinier there. The key to all this is the partial-fraction expansion of the factor$1 / \sin (\pi t /2n)$in the Fourier series of$g$, obtained by substituting$\theta = \pi t / 2n$into $$\frac1{\sin \pi\theta} = \frac1\pi \sum_{l=-\infty}^\infty \frac{(-1)^l}{\theta-l}$$ with the conditionally convergent sum interpreted as a principal value or Cesàro limit etc. I already noted in the previous note that the main term, for$l=0$, yields the convolution of$\cos^n (x/n)$with a symmetrical square wave, which is thus maximized at$x=0$and almost constant near$x=0$; we identify the constant with$2/B(1/2,n/2)$using the known product formula for$\int_{-\pi/2}^{\pi/2} \cos^n X \phantom. dX$. The new observation is that each of the error terms$(-1)^l/(\theta-l)$likewise yields the convolution with a square wave of$(-1)^l \cos(2lx) \phantom. \cos^n(x/n)$. If we approximate this square wave with a constant, we get the formula for$g_l$displayed above, via the formula for the$n$-th finite difference of a function$1/(j_0-j)$. The error in this approximation is still tiny (albeit not necessarily negative) because$\cos^n (x/n)$is minuscule when$x$is within$\pi/2$of the square wave's jump at$\pm \pi n / 2$. I've checked these approximations numerically to high precision (modern computers and gp make this easy) for$n$as large as$100$or so, in both of the odd congruence classes mod$4$, and it all works as expected; for example, when$n=99$we have$f(0) - g_0 = 2.57990478176660\ldots \cdot 10^{-70}$, which almost exactly matches the main term$g_1 = (4/\pi) \phantom. 99! \phantom. 99!!/297!!$but exceeds it by$5.9110495\ldots \cdot 10^{-102}$, which is almost exactly$g_2 = (4/\pi) \phantom. 99! \phantom. 297!!/(3 \cdot 495!!)$but too large by$7.92129\ldots \cdot 10^{-120}$, which is almost exactly$g_3 = (4/\pi) \phantom. 99! \phantom. 495!!/(5 \cdot 693!!)$, etc.; and likewise for$n=101$except that the maximum occurs at$\beta = \pi/2$and is approximated by an alternating sum$g_1 - g_2 + g_3 \ldots$(actually here this approximation is exact because$x=0\$).