Let $f$ be a formal power series with every coefficient either 0 or 1 , $f(2/3)=2017/(2^{2018})$. Find the period of decimal of $f(1/2)$.
I am new to power series but i searched all properties of formal power series but could not find any way to solve the problem.
Let $d_n$ be the sequences in $\{0,1\}$ such that $$f(x)=\sum_{n=1}^\infty d_n x^n$$ If $d_n$ is a periodic sequences in $\{0,1\}$ with period $T>0$ then \begin{align} f\left(\frac 23\right)=\sum_{n=1}^\infty d_n\left(\frac 23\right)^n &=\sum_{k=0}^\infty\sum_{n=1}^T d_{kT+n}\left(\frac 23\right)^{kT+n}\\ &=\sum_{k=0}^\infty\left(\frac 23\right)^{kT}\sum_{n=1}^T d_n\left(\frac 23\right)^n\\ &=\frac{3^T}{3^T-2^T}\sum_{n=1}^T d_n\left(\frac 23\right)^n\\ &=\frac 1{3^T-2^T}\sum_{n=1}^T d_n 2^n 3^{T-n} \end{align} is a rational number with odd denominator. The same conclusion holds also for $d_n$ eventually periodic. Thus, in order to have $f(2/3)=2017/2^{2018}$, the sequences $d_n$ must to be aperiodic. Consequently, the number $$f\left(\frac 12\right)=\sum_{n=1}^\infty d_n\left(\frac 12\right)^n$$ is irrational because the $d_n$ are its binary digits. Then the period decimal of $f(1/2)$ is infinite.