The fact that $R_n(\omega)=1$ means that $\omega$ belongs to $\Delta_{k,n}$ for some odd integer $k$. That is, $k\leqslant2^n\omega\lt k+1$ and $k=2\ell+\color{red}{1}$ for some integer $\ell$. Hence $2^n\omega=2\ell+\color{red}{1}+s$ with $0\leqslant s\lt 1$. Since $0\leqslant k\lt2^n$, $0\leqslant\ell\lt2^{n-1}$ and $\ell$ is a binary integer $\ell=\sum\limits_{i=0}^{n-2}\ell_i2^i$ with $\ell_i=0$ or $\ell_i=1$. Thus,
$$
\omega=\frac2{2^n}\sum\limits_{i=0}^{n-2}\ell_i2^i+\frac{\color{red}{1}}{2^n}+\frac{s}{2^n}=\sum\limits_{i=1}^{n-1}\frac{\ell_{n-i-1}}{2^i}+\frac{\color{red}{1}}{2^n}+\frac{s}{2^n}.
$$
This is the binary expansion of $\omega$, up to and including its $n$th bit, and this $n$th bit is $\color{red}{1}$.

To solve the case $R_n(\omega)=-1$, rewrite the whole paragraph replacing each $\color{red}{1}$ by $\color{blue}{0}$.

The Rademacher functions for counting binary digits
Frontal impact vehicle is there fan motor touches the oil cooler,oil cooler slight impact is there, in coolant tank oil and coolant mixed, what are the reasons to mix the coolant and oil in this situation. 


