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Is $-\frac{\pi^5}{2\sqrt2}\times\cot(\frac{\pi}{\sqrt2})$ irrational? I have an idea, and this is the last question for this idea. Pi, and square root of 2 is irrational numbers, but these expression, i don't know. Please, help! Thanks for the answers!

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Kristóf Lőrincz
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  • Related question https://math.stackexchange.com/questions/3295902/is-pi4-sum-k-0-infty-frac-zeta2k2k-irrational ... and please use MathJax – Robert Z Jul 17 '19 at 18:02
  • Yes, this is also my question. But i don't now, this expression is an irrational number. I'd like a mathematical proof for this. – Kristóf Lőrincz Jul 17 '19 at 18:04
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    It is almost certainly irrational, but I doubt that it is provably so in the current state of the art. We can't even prove that $\pi + e$ is irrational. – Robert Israel Jul 17 '19 at 18:10
  • @RobertIsrael Perhaps surprisingly, while what you say is true, we do know that things like $\pi$ and $e^\pi$ are algebraically independent. The number in the question is not quite of the right form for this result to be applied, but it's very similar, and I suppose answer might be known. – Wojowu Jul 17 '19 at 19:25

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COMMENT.- It is known from Lindemann's paper in which he proved the transcendance of $\pi$ that, among other results, the value in $x$ of the trigonometric function $\cot(x)$ is trascendental if $x$ is non-zero algebraic. But if $x$ is trascendental then $\cot (x)$ can be algebraic (even rational), or transcendental.

On the other hand, the product of two transcendental numbers is not necessarily transcendental (for example until today it is ignored if $e\times \pi$ is algebraic or transcendental).

If $\cot\left(\dfrac{\pi}{\sqrt2}\right)\approx-0.7612229$ is algebraic then we can say that the proposed number is transcendental but if $\cot\left(\dfrac{\pi}{\sqrt2}\right)$ is transcendental then we cannot answer.

The true problem is to know the nature of $\cot\left(\dfrac{\pi}{\sqrt2}\right)$ and this is not known as far as I know.

Piquito
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    Transcendence of $\cot(\pi/\sqrt{2})$ follows from that of $e^{i\pi/\sqrt{2}}$, which is immediate from Gelfond-Schneider (since this is a value of $(-1)^{1/\sqrt{2}}$). For the number in the question, if it wasn't for the $i$, it would follow from results of Neserenko, but I don't know whether $\pi, e^{i\sqrt{2}\pi}$ are algebraically independent. – Wojowu Jul 17 '19 at 19:20
  • So, similar to the transcendance of $\pi$ by Lindemann. Well, thus we cannot answer to the problem. – Piquito Jul 17 '19 at 19:25