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Given the function:

$$ f(n) = \lceil{a \cdot n}\rceil - a \cdot n, n \in \mathbb{N} $$

Find the minimum value of $f$ in terms of $n$, if $a$ is irrational.

We know that if $n$ was real then the minimum value of $f$ would be $0$ for $n_{k}=\frac{k}{a}, k \in \mathbb{N}$. When $n$ is natural though, the function's minimum goes to $0$ while $n$ approaches the $\frac{1}{a}$ in integer form. For example, if $a=\sqrt{2}$ then $\frac{1}{a}=0.707106...$ and when $n$ approaches $70,707,7071,...$ or $\frac{10^{k}}{a}, k \in \mathbb{N}$ then $f$ goes to $0$.

I tend to believe that $f(n) \geq \frac{1}{10n} $ but I would prefer to find a tighter lower bound, proof for the latter, or maybe even a lower bound depending on the irrational number $a$.

Thank you in advance.

Άγγελος Κρατημένος
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  • Take a look at https://en.wikipedia.org/wiki/Continued_fraction & https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem An example in relation to your example: $99^2-2\cdot 77^2=1$, so $\sqrt2\approx 99/70$ and $99-70\sqrt2=1/(99+70\sqrt2)$ hence $1/198 < 99-70\sqrt2 < 1/197$ – PM 2Ring Feb 14 '21 at 22:22

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There is a pretty standard result showing that $(\{a n\})_{n \in \mathbb{N}} \subseteq [0, 1]$ is dense if $a \in \mathbb{R} \setminus \mathbb{Q}$ (you may find it here, for example: reference). Note that $\lceil a n \rceil - a n = \lceil a n \rceil - \lfloor a n \rfloor - \{a n \} = 1 - \{a n\}.$ Since $(\{a n\})_{n \in \mathbb{N}} \subseteq [0, 1]$ is dense, we deduce that the infimum of this expression is $0.$

Actually Fritz
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