Can a number have both a periodic an a non-periodic representation in a non-integer base?

Question

Fix an algebraic number $\beta$ and consider a complex number $\alpha$ which admits multiple representations in base $\beta$. If one representation of $\alpha$ is ultimately periodic, must every other representation of $\alpha$ be ultimately periodic?

Bonus question: Does this depend on the choice of the set of digits?

Edit: John Bentin showed that this is false in general, even with $\beta \in \Bbb{Q}$. Is this true at least for $\beta$ algebraic integer?

Edit 2: This question came up when trying to solve this problem of mine.

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All I know so far is that if $\beta$ is a Pisot integer, i.e. if it is a real algebraic integer greater than $1$ with every conjugate lying in the unit circle, then $\alpha \in \Bbb{Q}(\beta)$ if and only if it has an ultimately periodic expansion (which is stronger than my question).

Answer 1

We have $2=\frac23+\frac49+\cdots+(\frac23)^k+\cdots$, which is clearly periodic. Let us write, alternatively, $$2=1+\dfrac23+\dfrac{8}{27}+\dfrac{512}{19683}+\cdots,$$or$$1=\dfrac23+\dfrac{8}{27}+\dfrac{512}{19683}+\cdots=\sum_{k=1}^{\infty}\dfrac{2^k}{3^k}a_k,$$ where $a_1=1,a_2=0,a_3=1,a_4=\cdots=a_{8}=0,a_{9}=1,$ and generally $a_n=1$ if $$1-\sum_{k=1}^{n-1}\dfrac{2^k}{3^k}a_k>\dfrac{2^n}{3^n},$$ with $a_n=0$ otherwise. A proof that $\sum_{k=1}^{\infty}\left(\frac23\right)\!^ka_k=1$ is appended below. We will show that the sequence $(a_1,a_2,...)$ cannot be eventually periodic. To do this, suppose the contrary. Then there are integers $l\geqslant0$ and $m\geqslant1$ such that$$1-\left(\dfrac23+\dfrac{8}{27}+\dfrac{512}{19683}+\cdots+\dfrac{2^l}{3^l}\right) =\dfrac{2^{l+1}}{3^{l+1}}\sum_{k=1}^{m}\dfrac{2^{k-1}}{3^{k-1}}a_{l+k}\sum_{k=0}^\infty\dfrac{2^{mk}}{3^{mk}}$$ $$\qquad\qquad\qquad\quad\qquad\qquad\quad=\dfrac{2^{l+1}p}{3^l(3^m-2^m)},$$ where $p$ is an integer. Now the fraction on the LHS, in its lowest terms, has an odd numerator, while the fraction on the RHS has an even numerator: a contradiction.

Edit: To prove that $\sum_{k=1}^{\infty}\left(\frac23\right)\!^ka_k=1,$ we show by induction that $$0<1-\sum_{k=1}^{n-1}\dfrac{2^k}{3^k}a_k<2\frac{2^n}{3^n}\quad(n=1,2,..).$$This is true for $n=1,$ since $0<1<\frac43.$ Now suppose that it has been established for $n=j,$ namely $$0<1-\sum_{k=1}^{j-1}\dfrac{2^k}{3^k}a_k<2\frac{2^j}{3^j}.$$(Case 1) If $$1-\sum_{k=1}^{j-1}\frac{2^k}{3^k}a_k>\frac{2^j}{3^j},$$then $a_j=1,$ and $$0<1-\sum_{k=1}^j\frac{2^k}{3^k}a_k=1-\sum_{k=1}^{j-1}\dfrac{2^k}{3^k}a_k-\frac{2^j}{3^j}<2\frac{2^j}{3^j}-\frac{2^j}{3^j}=\frac{2^j}{3^j}<2\frac{2^{j+1}}{3^{j+1}}.$$(Case 2) If $$1-\sum_{k=1}^{j-1}\frac{2^k}{3^k}a_k<\frac{2^j}{3^j},$$then $a_j=0,$ and $$0<1-\sum_{k=1}^j\frac{2^k}{3^k}a_k=1-\sum_{k=1}^{j-1}\frac{2^k}{3^k}a_k<\frac{2^j}{3^j}<2\frac{2^{j+1}}{3^{j+1}}.$$This completes the inductive step.

Answer 2

The following counterexample answers the supplementary question added in the edit. Choose the algebraic integer $\sqrt2$ as the base. Then we may expand $1$ in the periodic form $$1=\frac12+\frac14+\cdots+\left(\frac{0}{(\sqrt2)^{2k-1}}+\frac{1}{(\sqrt2)^{2k}}\right)+\cdots.$$But we can also express it in a more irregular expansion:$$1=\frac1{\sqrt2}+\frac14+\frac1{32}+\frac1{64\sqrt2}+\cdots+\frac{a_k}{\sqrt2^k}+\cdots,$$where $a_k=1$ if $$1-\sum_{j=1}^{k-1}\frac{a_j}{\sqrt2^j}>\frac1{\sqrt2^k},$$ with $a_k=0$ otherwise, for $k=1,2,\dots\,$. To show that the irregular expansion is aperiodic, assume the contrary: namely that there exist integers $l\geqslant0$ and $m\geqslant1$ satisfying $$1=\sum_{k=1}^l\frac{a_k}{\sqrt2^k}+\frac1{\sqrt2^{l+1}}\sum_{k=1}^m\frac{a_{l+k}}{\sqrt2^k}\frac1{1-1/\sqrt2^{m}}.$$We may ensure that $l$ is positive by including, if necessary, the first cycle of $m$ terms of the periodic tail in the initial (pre-periodic) length-$l$ segment of the expansion. Since the terminal factor $1/(1-1/\sqrt2^{m})$ is either a positive rational number or the sum of a positive rational number and a positive rational multiple of $\sqrt2,$ the whole of the RHS of the above sum is of this form. Moreover it is not rational---because the initial segment (of length $l$) includes the first term $1/\sqrt2,$ and so the overall coefficient of $\sqrt2$ is at least $\frac12$. But the LHS is rational: a contradiction.