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Let $\varrho$ be an irrational number, with irrationality measure $\mu(\varrho)$. I define the function f as

$$f(\varrho)=\liminf_{p,q\in\mathbb{Z},q\to\infty}\left(q^{\mu(\varrho)}\left|\varrho-\frac{p}{q}\right|\right)$$

So $f(\varrho)$ is the smallest number so that $\left|\varrho-\frac{p}{q}\right|<\frac{f(\varrho)}{q^{\mu(\varrho)}}$ has infinitely many solutions.

If $\mu(\varrho)=2$ then by Hurwitz's theorem $f(\varrho)\leq\frac{1}{\sqrt{5}}$ and this limit is sharp because $f\left(\frac{1+\sqrt{5}}{2}\right)=\frac{1}{\sqrt{5}}$.

Are there any other numbers for which f is known? For example, $\sqrt{2}$, $\pi$, $e$, etc.

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To real numbers $x$ and $y$ are said to be equivalent if there exist four integers $a,b,c,d$ such that

$$\begin{cases} ad-bc=\pm 1 \\ y=\frac{ax+b}{cx+d},\end{cases}$$

which exactly means that $x$ and $y$ are in the same orbit under $\mathrm{PGL}(2,\mathbb Z)$. We can also show that $x$ and $y$ are equivalent if, and only if, $x$ and $y$ have the same development in continuous fraction after a given rank.

We know that if $x$ is equivalent to

$$\frac {1+\sqrt 5}2,$$

then

$$f(x)=\frac 1{\sqrt 5}.$$

We also know that if $y\in \{\text{numbers equivalent to } \sqrt 2\}$, then

$$f(y)=\frac 1{\sqrt 8}.$$

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