Is $y(x)=\frac{1}{2}M\left[1-\cos\left(\frac{\pi}{M}x\right)\right]$ an integer

Question

Let ${\rm y}\left(x\right) = \frac{1}{2}M\left[1-\cos\left(\frac{\pi}{M}x\right)\right]$. Is ${\rm y}\left(x\right)$ an integer for each $x = 1,2,\ldots,M$ when $M\to\infty$ ?.

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ For fixed $x$, $$ \mbox{when}\ M \to \infty\,,\quad {1 \over 2}\,M\bracks{1 - \cos\pars{{\pi \over M}x}} \sim {1 \over 2}\,\pars{{1 \over 2}\,{\pi^{2}x^{2} \over M^{2}}} = {\pi^{2}x^{2} \over 4}\,{1 \over M^{2}} \to 0 $$

Answer 2

Putting $\displaystyle\frac \pi Mx=2h\implies M=\frac{\pi x}{2h}$ and as $M\to\infty, h\to0$

$$\lim_{M\to\infty}M\left[1-\cos\left(\frac{\pi}{M}x\right)\right]=\frac{\pi x}2\lim_{h\to0}\frac{1-\cos2h}h$$

$$=\frac{\pi x}2\lim_{h\to0}\frac{2\sin^2h}h =\pi x\left(\lim_{h\to0}\frac{\sin h}h\right)^2\cdot \lim_{h\to0}h=0$$ for any finite $x$