2
$\begingroup$

Wikipedia mentions that the number $$a = \dfrac{\Gamma\left(\dfrac{1}{4}\right)}{\pi^{1/4}}$$ is transcendental. Since $\Gamma(1/2) = \sqrt{\pi}$ the above number $a$ seems to connected to a combination of $\Gamma(1/4)$ and $\Gamma(1/2)$. On squaring the connection is clear as we can see that $$a^{2} = \frac{\Gamma^{2}(1/4)}{\Gamma(1/2)} = B(1/4, 1/4)$$ where $B$ represents the beta function.

Are there any other known values of beta function which are transcendental? Any reference to a proof of the transcendence of number $a$ given above would also be helpful.

$\endgroup$
  • 1
    $\begingroup$ To mention the obvious: It is clear from the reflection formula that at least one of $q!$ or $(-q)!$ is transcendental, for $q\in\mathbb Q$. $\endgroup$ – Lucian Jun 28 '15 at 11:01
  • $\begingroup$ @Lucian: Nice remark to deal with certain values of $\Gamma$. $\endgroup$ – Paramanand Singh Jun 28 '15 at 11:09
5
$\begingroup$

You have the following result.

Theorem. Let $a$ and $b$ be rational numbers which are not integers and such that $a+b$ is not an integer. Then the number $$B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=\int_{0}^{1}u^{a-1}(1-u)^{b-1} \,du$$ is transcendental.

Here is, among others, an interesting reference from Michel Waldschmidt: Transcendence of Periods (see p. 6).

For example, $$ \frac{\Gamma\!\left(\dfrac{1}{3}\right)}{\pi^{1/3}}\quad \text{and}\quad \frac{\Gamma\!\left(\dfrac{1}{3}\right)}{\pi^{2/3}}$$ are transcendental (see here).

$\endgroup$

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.