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GH from MO
GH from MO
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For the coefficients $2^n$ the equidistribution theorem fails. In fact it is easy to exhibit an irrational $a$ such that the sequence $(2^na)_{n=0}^\infty$ is not even dense in $(0,1)$ modulo $1$. For example, take an $a\in (0,1)$ whose binary expansion consists of increasing blocks of $1$'s seperated by $0$'s: $$ a=0.101101110111101111101111110\dots $$

Added. For algebraic irrationals such as $a=\sqrt{2}$, it is not even known if they are normalevery digit occurs infinitely often in a given base, let alone $(2^n a)_{n=0}^\infty$ being dense or even equidistributed in $(0,1)$ modulo $1$.

For the coefficients $2^n$ the equidistribution theorem fails. In fact it is easy to exhibit an irrational $a$ such that the sequence $(2^na)_{n=0}^\infty$ is not even dense in $(0,1)$ modulo $1$. For example, take an $a\in (0,1)$ whose binary expansion consists of increasing blocks of $1$'s seperated by $0$'s: $$ a=0.101101110111101111101111110\dots $$

Added. For algebraic irrationals such as $a=\sqrt{2}$, it is not even known if they are normal in a given base, let alone $(2^n a)_{n=0}^\infty$ being dense or even equidistributed in $(0,1)$ modulo $1$.

For the coefficients $2^n$ the equidistribution theorem fails. In fact it is easy to exhibit an irrational $a$ such that the sequence $(2^na)_{n=0}^\infty$ is not even dense in $(0,1)$ modulo $1$. For example, take an $a\in (0,1)$ whose binary expansion consists of increasing blocks of $1$'s seperated by $0$'s: $$ a=0.101101110111101111101111110\dots $$

Added. For algebraic irrationals such as $a=\sqrt{2}$, it is not even known if every digit occurs infinitely often in a given base, let alone $(2^n a)_{n=0}^\infty$ being dense or even equidistributed in $(0,1)$ modulo $1$.

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Source Link
GH from MO
GH from MO
  • 105.4k
  • 8
  • 293
  • 398

For the coefficients $2^n$ the equidistribution theorem fails. In fact it is easy to exhibit an irrational $a$ such that the sequence $(2^na)_{n=1}^\infty$$(2^na)_{n=0}^\infty$ is not even dense in $(0,1)$ modulo $1$. For example, take an $a\in (0,1)$ whose binary expansion consists of increasing blocks of $1$'s seperated by $0$'s: $$ a=0.101101110111101111101111110\dots $$

Added. For algebraic irrationals such as $a=\sqrt{2}$, it is not even known if they are normal in a given base, let alone $(2^n a)_{n=0}^\infty$ being dense or even equidistributed in $(0,1)$ modulo $1$.

For the coefficients $2^n$ the equidistribution theorem fails. In fact it is easy to exhibit an irrational $a$ such that the sequence $(2^na)_{n=1}^\infty$ is not even dense in $(0,1)$ modulo $1$. For example, take an $a\in (0,1)$ whose binary expansion consists of increasing blocks of $1$'s seperated by $0$'s: $$ a=0.101101110111101111101111110\dots $$

For the coefficients $2^n$ the equidistribution theorem fails. In fact it is easy to exhibit an irrational $a$ such that the sequence $(2^na)_{n=0}^\infty$ is not even dense in $(0,1)$ modulo $1$. For example, take an $a\in (0,1)$ whose binary expansion consists of increasing blocks of $1$'s seperated by $0$'s: $$ a=0.101101110111101111101111110\dots $$

Added. For algebraic irrationals such as $a=\sqrt{2}$, it is not even known if they are normal in a given base, let alone $(2^n a)_{n=0}^\infty$ being dense or even equidistributed in $(0,1)$ modulo $1$.

Source Link
GH from MO
GH from MO
  • 105.4k
  • 8
  • 293
  • 398

For the coefficients $2^n$ the equidistribution theorem fails. In fact it is easy to exhibit an irrational $a$ such that the sequence $(2^na)_{n=1}^\infty$ is not even dense in $(0,1)$ modulo $1$. For example, take an $a\in (0,1)$ whose binary expansion consists of increasing blocks of $1$'s seperated by $0$'s: $$ a=0.101101110111101111101111110\dots $$