3
$\begingroup$

I have a problem:

Let $a \in \mathbb{Z}$, $a \geq 3$, and set $\xi= \sum_{n=0}^\infty 10^{-a^{2n}}>0$. Then the inequality
$$\Big|\,\xi - \dfrac{x}{y}\,\Big| \leq \dfrac{1}{y^a}$$ has infinitely many solutions with $x,y \in \mathbb{Z}$, $y>0$ and $\gcd(x,y)=1$.

I would imitate the proof of Dirichlet's theorem as follow, I claim my lemma

Lemma: Let $\xi=\sum_{n=0}^\infty 10^{-a^{2n}}$, for every integer $Q \geq 2$, there are integers $x,y$ which are not both equal to $0$, such that $$|x - \xi y| \leq \dfrac{1}{Q^{a-1}}$$ with $0<y \leq Q$ and $\gcd(x,y) =1$

  • Try to prove this lemma:

Partition the interval $[0,1]$ into $Q^{a-1}$ subintervals of length $\dfrac{1}{Q^{a-1}}$. Consider $Q^{a-1}+1$ numbers $\xi-[\xi]$,..,$Q^{a-1}\xi-[Q^{a-1}\xi]$ and $1$. By the Dirichlet principle, two among these numbers must lie in the same subinterval of length $\dfrac{1}{Q^{a-1}}$. Hence we can find $x,y \in \mathbb{z}$ such that $|x-y \xi| \leq \frac{1}{Q^{a-1}}$. But now my trouble is $y$ is not smaller than $Q$.

Does anyone have other ideas?

$\endgroup$
  • 1
    $\begingroup$ It is related to Liouville's theorem, not to Dirichlet's. An idea to follow is at the beginning of the article (it could be put shorter...). $\endgroup$ – metamorphy Sep 11 at 9:49
  • 1
    $\begingroup$ It's quite related but in Liouville's number you have $|\xi -\frac{x}{y}| \leq \frac{1}{y^n}$, meanwhile my problem is $|\xi -\frac{x}{y}| \leq \frac{1}{y^a}$. It's fixed a. $\endgroup$ – Desunkid Sep 11 at 13:21
1
$\begingroup$

With regards to other ideas, just straight attack. Noting $y=10^{a^{2n}}$ then $$\sum\limits_{k=0}^{n}\frac{1}{10^{a^{2k}}}=\frac{\sum\limits_{k=0}^{n}10^{a^{2n}-a^{2k}}}{y}$$ where $x=\sum\limits_{k=0}^{n}10^{a^{2n}-a^{2k}}=\sum\limits_{k=0}^{\color{red}{n-1}}10^{a^{2n}-a^{2k}}\color{red}{+1}=10\cdot Q\color{red}{+1}$ and $\gcd(x,y)=1$. Now $$\left|\xi -\frac{x}{y}\right|= \sum\limits_{k=n+1}^{\infty}\frac{1}{10^{a^{2k}}}= \frac{1}{10^{a^{2n+2}}}\left(\sum\limits_{k=n+1}^{\infty}\frac{1}{10^{a^{2k}-a^{2n+2}}}\right)< \\ \frac{1}{10^{a^{2n+2}}}\left(\sum\limits_{k=0}^{\infty}\frac{1}{10^{k}}\right)= \frac{1}{10^{a^{2n+2}}}\left(\frac{1}{1-\frac{1}{10}}\right)=\\ \frac{1}{10^{a^{2n}\cdot a^2}}\cdot \frac{10}{9}= \frac{1}{y^{a^2}}\cdot \frac{10}{9}< \frac{1}{y^{a+1}}\cdot \frac{10}{9}=\frac{1}{y^a}\cdot \frac{10}{y\cdot9}<\frac{1}{y^a}$$ for infinitely $n$ and because $y\geq10, \forall n\geq0$ and $a^2 > a+1, \forall a\geq3$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.