Return to Answer

deleted 12 characters in body

Source Link

edited Apr 11, 2017 at 20:16

Christian Remling

24.2k
2
48
83

I've decided to upgrade my comments and make an answer out of them, even though I'm just addressing the (easier) variant suggested by the OP at the end of the post, where we replace the golden ratio by other numbers $b$.

If $b$ can be well approximated by rationals, then it's easy to see that the sum cannot stay bounded: More precisely, suppose there are infinitely many (reduced) fractions with odd denominators $q$ such that $$ \left| \frac{b}{2} - \frac{p}{q} \right| = o(q^{-2}) $$ (the golden ratio is not of this type; I can only get $O(q^{-2})$, and in fact it is usually thought of as the number with the worst rational approximations).

The shift by $p/q$ is periodic with odd period, so gives a bias of at least $1$ to one of the two halves of the circle. Also, if I keep a distance $\gtrsim 1/q$ to the endpoints of the intervals, then a perturbation by $\le \delta/q$ of the initial point will not change anything.

Now I can just wait long enough until $nb$ is close to such an initial point, and the approximation of $b/2$ by $p/q$ will be justified during a time interval of length $kq$, for any $k\ge 1$ (because $kq\cdot o(q^{-2})\ll 1/q$). So I pick up an overall bias of at least $k$.

I've decided to upgrade my comments and make an answer out of them, even though I'm just addressing the (easier) variant suggested by the OP at the end of the post, where we replace the golden ratio by other numbers $b$.

If $b$ can be well approximated by rationals, then it's easy to see that the sum cannot stay bounded: More precisely, suppose there are infinitely many (reduced) fractions with odd denominators $q$ such that $$ \left| \frac{b}{2} - \frac{p}{q} \right| = o(q^{-2}) $$ (the golden ratio is not of this type; I can only get $O(q^{-2})$, and in fact it is usually thought of as the number with the worst rational approximations).

The shift by $p/q$ is periodic with odd period, so gives a bias of at least $1$ to one of the two halves of the circle. Also, if I keep a distance $\gtrsim 1/q$ to the endpoints of the intervals, then a perturbation by $\le \delta/q$ of the initial point will not change anything.

Now I can just wait long enough until $nb$ is close to such an initial point, and the approximation of $b/2$ by $p/q$ will be justified during a time interval of length $kq$, for any $k\ge 1$ (because $kq\cdot o(q^{-2})\ll 1/q$). So I pick up an overall bias of at least $k$.

I've decided to upgrade my comments and make an answer out of them, even though I'm just addressing the (easier) variant suggested by the OP at the end of the post, where we replace the golden ratio by other numbers $b$.

If $b$ can be well approximated by rationals, then it's easy to see that the sum cannot stay bounded: More precisely, suppose there are infinitely many (reduced) fractions with odd denominators $q$ such that $$ \left| \frac{b}{2} - \frac{p}{q} \right| = o(q^{-2}) $$ (the golden ratio is not of this type; I can only get $O(q^{-2})$, and in fact it is usually thought of as the number with the worst rational approximations).

The shift by $p/q$ is periodic with odd period, so gives a bias of at least $1$ to one of the two halves of the circle. Also, if I keep a distance $\gtrsim 1/q$ to the endpoints of the intervals, then a perturbation by $\le \delta/q$ of the initial point will not change anything.

Now I can just wait until $nb$ is close to such an initial point, and the approximation of $b/2$ by $p/q$ will be justified during a time interval of length $kq$, for any $k\ge 1$ (because $kq\cdot o(q^{-2})\ll 1/q$). So I pick up an overall bias of at least $k$.

Source Link

created Apr 11, 2017 at 20:03

Christian Remling

24.2k
2
48
83

I've decided to upgrade my comments and make an answer out of them, even though I'm just addressing the (easier) variant suggested by the OP at the end of the post, where we replace the golden ratio by other numbers $b$.

If $b$ can be well approximated by rationals, then it's easy to see that the sum cannot stay bounded: More precisely, suppose there are infinitely many (reduced) fractions with odd denominators $q$ such that $$ \left| \frac{b}{2} - \frac{p}{q} \right| = o(q^{-2}) $$ (the golden ratio is not of this type; I can only get $O(q^{-2})$, and in fact it is usually thought of as the number with the worst rational approximations).

The shift by $p/q$ is periodic with odd period, so gives a bias of at least $1$ to one of the two halves of the circle. Also, if I keep a distance $\gtrsim 1/q$ to the endpoints of the intervals, then a perturbation by $\le \delta/q$ of the initial point will not change anything.

Now I can just wait long enough until $nb$ is close to such an initial point, and the approximation of $b/2$ by $p/q$ will be justified during a time interval of length $kq$, for any $k\ge 1$ (because $kq\cdot o(q^{-2})\ll 1/q$). So I pick up an overall bias of at least $k$.