Let $f(x)$ be a real transcendental function with algebraic coefficients. So $f(x)$ and $x$ are algebraically independent. Let $\alpha$ be a transcendental number, are the numbers $$\alpha+f(\alpha),\ \alpha f(\alpha),\ \alpha/f(\alpha)$$ transcendental? It is clear these numbers are transcendental if $f(\alpha)$ is algebraic, but what if $f(\alpha)$ is transcendental? It seems in this case these numbers are also transcendental because $f(x)$ and $x$ are algebraically independent, but I'm not able to show this. I'm sorry maybe the description above is not satisfactoty. Here is a specific version: If $\alpha$ is transcendental, then are $$\alpha+\cos(\alpha), \ \alpha\cos(\alpha),\ \alpha/\cos(\alpha)\\ \alpha+e^{\alpha},\ \alpha e^{\alpha},\ \alpha/e^{\alpha}$$ transcendental?
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4$\begingroup$ Dear 好きな人がいません: Your question keeps changing from version $n$ to version $n+1$ whenever you get a negative answer to version $n$. I start fearing that the process will never converge (let alone that doing something like this is considered, I think, unfair here on MathOverflow). If you have a new question, it could make more sense to post it on a new thread (after giving it enough of a thought). With that said, I don't understand what you're asking for in the current version of the OP: Isn't $\cos(\cdot)$ a transcendental function? $\endgroup$– Salvo TringaliJun 19, 2021 at 11:25
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1$\begingroup$ You write that $f$ being "a real transcendental function with algebraic coefficients" means, for example, that "if $f(x)$ involves $\sin$ and $\cos$, then $f(x)$ is a polynomial in $\sin x$, $\cos x$ and $x$ with nonzero algebraic coefficients". But this is not the standard definition of a real transcendental function (and, to be honest, makes little sense whatsoever): Isn't the function $f: \mathbf R \to \mathbf R \colon x \mapsto \sin^2(x) + \cos^2(x) + x$ a "polynomial in $\sin x$, $\cos x$ and $x$"? $\endgroup$– Salvo TringaliJun 19, 2021 at 11:41
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$\begingroup$ I'm sorry, that sentence is not a precise description. What I want to ask is something like this: if $\alpha$ is transcendental, then are $\alpha+\cos(\alpha)$, $\alpha\cos(\alpha)$ and $\alpha/\cos(\alpha)$ transcendental? $\endgroup$– BetaJun 19, 2021 at 14:30
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1$\begingroup$ Really, why don't you start a new thread? This is a totally different question that the original one... $\endgroup$– Salvo TringaliJun 19, 2021 at 15:20
2 Answers
Here are two counterexamples to the specific question at the end.
Let $\alpha$ be the unique real solution to $xe^x=1$. This number is also called the omega constant. It is transcendental by the Lindemann-Weierstrass Theorem which (in particular) says that if $x$ is nonzero and algebraic then $e^x$ is not algebraic. So if $\alpha$ were algebraic then LW would imply that $e^\alpha=1/\alpha$ is not algebraic, a contradiction.
Let $\alpha$ be the unique real solution to $x=\cos(x)$. This number is also called the Dottie number. It is transcendental, again by the Lindemann-Weierstrass Theorem. Indeed, suppose $\alpha$ is algebraic. Then so is $i\alpha$, and so $e^{i\alpha}$ is not algebraic by LW. But $e^{i\alpha}=\cos(\alpha)+i\sin(\alpha)=\alpha+i\sqrt{1-\alpha^2}$.
Last time I checked, the entire function $f \colon \mathbf C \to \mathbf C \colon z \mapsto a\exp(iz)$ was transcendental for every non-zero $a \in \mathbf C$. So, taking $\alpha = -a = 2\pi$ yields a negative answer for the "$\alpha + f(\alpha)$" case; and taking $\alpha = a^{-1} = 2\pi$ yields a negative answer for the "$\alpha f(\alpha)$" case. The "$\alpha - f(\alpha)$" and "$\alpha/f(\alpha)$" cases are trivial variants of the "$\alpha + f(\alpha)$" and "$\alpha f(\alpha)$" cases, resp.
Edit. To address the question in the first comment under this answer, the function $f \colon \mathbf R \to \mathbf R \colon x \mapsto a \cos x$ is transcendental for every non-zero $a \in \bf R$, with $f(2\pi) = a$. So again, taking $\alpha = -a = 2\pi$ yields a negative answer for the "$\alpha + f(\alpha)$" case; and taking $\alpha = a^{-1} = 2\pi$ yields a negative answer for the "$\alpha f(\alpha)$" case.
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$\begingroup$ @Tringali, thanks so much! I edit the question for real $f(x)$ because I realize I'm dealing with real functions. So how do you think the real case? Or is there any reference on this? $\endgroup$– BetaJun 19, 2021 at 9:32
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$\begingroup$ I think this is not a question particularly suited to MathOverflow. In any case, there you have your answer. Also, I don't think it was a good idea to add in the "α−f(α)" case: It sheds a bad light on the OP, as it's a trivial variant of the "α+f(α)" case. $\endgroup$ Jun 19, 2021 at 9:57
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$\begingroup$ Ok, Tringali, after your helpful examples I now clarify the statement of the question. Please forgive my fault. $\endgroup$– BetaJun 19, 2021 at 10:18