# Is the root of $x=\cos(x)$ a transcendental number?

This question struck me when thinking about the fixed point of $x=\cos(x)$ being "obviously" not an algebraic number (unlike something like $\sqrt{2}$, see this question).

If so, how would one prove this?

I don't see how something like a simple application of the Lindemann-Weierstrass Theorem would work in this case.

I understand that the mere fact that it is a root of a so-called transcendental equation doesn't amount to very much, since $x=\tan(x)$ has at least one root (0) which is algebraic.

• Doesn't Lindemann-Weierstrass prove that $\cos a$ is transcendental when $a$ is non-zero and algebraic? If $a=\cos a$, then $a\neq 0$ and if $a$ is algebraic, then $\cos a=a$ is transcendental, giving a contradiction. The only option is then that $a=\cos a$ is transcendental. Oct 4, 2014 at 18:11
• @Hayden Thanks! Of course $\cos(a)=(1/2)(\exp(ia)+\exp(-ia))$. If you promote your comment to an answer I'll accept it. Oct 5, 2014 at 16:50

The Lindemann-Weierstrass Theorem states that given $$\alpha_1,\ldots, \alpha_n$$ algebraic numbers that are linearly independent over $$\mathbb{Q}$$, then $$e^{\alpha_1},\ldots, e^{\alpha_n}$$ are algebraically independent over $$\mathbb{Q}$$ (i.e. for any rational function $$P(z_1,\ldots, z_n)$$ with algebraic coefficients it follows that $$P(e^{\alpha_1},\ldots, e^{\alpha_n})\neq 0$$). As a result, with $$n=1$$ we find that if $$\alpha$$ is algebraic that is non-zero, then $$e^\alpha$$ is transcendental.
Now, suppose that $$a$$ is algebraic; then $$ia$$ is also algebraic because $$i$$ is algebraic and the set of all algebraic numbers is a field. Next, consider the rational function $$P(z)=\frac{z^2-(2\cos a) z + 1}{2z}$$ and assume for the sake of a contradiction that $$\cos a$$ is algebraic. $$P(z)$$ is not identically zero and has algebraic coefficients, but $$P(e^{ia})=0$$ by the definition $$\cos a=\frac{e^{ia}+e^{-ia}}{2}$$, contradicting the assumption that $$a$$ is algebraic by Lindemann-Weierstrass.
Thus, we find that if $$a$$ is a non-zero algebraic number, then $$\cos a$$ is transcendental. As a result, if $$a$$ is a complex number such that $$a=\cos a$$, then if $$a$$ is algebraic, $$\cos a=a$$ is transcendental, giving a contradiction. Since $$a\neq 0$$ because $$\cos 0 = 1$$, we find that $$a$$ is transcendental.