Let $x$ be an algebraic number. Must $\arctan(x)/\pi$ have finite irrationality measure? Are there any useful upper bounds?

  • $\begingroup$ What if $x=\sqrt{2}$? $\endgroup$
    – markvs
    Mar 10, 2021 at 2:22
  • $\begingroup$ I don't know. Is there a reason $\sqrt{2}$ is special here? $\endgroup$ Mar 10, 2021 at 4:29
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    $\begingroup$ Yes. The number has finite irrationality measure. A bound is obtained from Baker-Wustholtz theorem, but probably very big. en.wikipedia.org/wiki/Baker%27s_theorem $\endgroup$ Mar 10, 2021 at 5:08
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    $\begingroup$ Yes they are. The bound will depend on the degree of the algebraic number x. $\endgroup$ Mar 10, 2021 at 5:40
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    $\begingroup$ @SungjinKim: perhaps you can convert your comment into an answer? Could you explain how does it follow from Baker's Theorem? $\endgroup$
    – markvs
    Mar 10, 2021 at 6:48

1 Answer 1


Let $\alpha=\frac{1+xi}{\sqrt{1+x^2}}$. There are some cases $(\arctan x) /\pi$ is rational. For example, $x=1, \sqrt3$. In these cases, $(\arctan x)/\pi$ has the irrationality measure $1$. These occur precisely when $\alpha$ is a root of unity.

Since $x$ is algebraic, so is $\alpha$. Then $\arctan x = \arg \alpha = (\log \alpha)/i$. Using $\log(-1)=i\pi$, we have $(\arctan x)/\pi= (\log \alpha)/\log (-1)$. Here, we need a fixed determination of logarithms of complex numbers.

Suppose that $\alpha$ is not a root of unity. We must have that $(\arctan x)/\pi$ is irrational. That is, there is no nonzero rational solutions to $\beta_1 \log \alpha + \beta_2 \log(-1) = 0$. By Gelfond-Schneider theorem, $(\log \alpha)/\log(-1)$ is transcendental. Thus, the irrationality measure of $(\arctan x)/\pi$ is at least $2$. Note that the transcendence of $(\arctan x)/\pi$ also follows from the argument below.

For the upper bound of the irrationality measure, we apply Baker-Wustholz theorem. The logarithmic form $$L=\beta_1 \log\alpha + \beta_2 \log (-1)$$ is nonvanishing for any pair of integers $(\beta_1,\beta_2)\neq (0,0)$. Let $n=2$ and let $d$ be the degree of $\alpha$ over $\mathbb{Q}$. Then the theorem yields $$ \log |L| > -C(2,d)h'(\alpha)h'(-1)h'(L) $$ where $C(2,d)= 905969664 \ d^4\log(4d)$, $h'(-1)=1$, $h'(\alpha)$ is a modified height of $\alpha$ defined by $$ h'(\alpha)=\frac1d \max(h(\alpha), |\log \alpha|, 1) $$ and $h(\alpha)$ is the logarithmic Weil height of $\alpha$. We may use $h'(L)=\log (\max(|\beta_1|/b, |\beta_2|/b))+1$ although the definition is slightly different in the note. Also, $b=\gcd (\beta_1,\beta_2)$.

Hence, there are positive numbers $A(d,\alpha)$, $B(d,\alpha)$ such that for any integer $p$ and any positive integer $q$, $$ \left|\frac{\log \alpha}{\log(-1)}- \frac pq \right|> \frac{A(d,\alpha)}{q^{B(d,\alpha)}}. $$

As we see above, the number $B(d,\alpha)$ is quite huge, but it gives an upper bound of the irrationality measure of $(\arctan x)/\pi$.


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