3

I was graphing the function $\sin(\cos(\sin(\cos(\sin(\cos...$ when I realized it started to flatten out. This meant that this approaches a constant. Since the sine and cosine repeat, we can make a finite equation for the number that I will call $\eta$: $$\eta=\sin(\cos(\eta))$$Is this transcendental? I know that $\sin$ and $\cos$ are transcendental functions, but I am not sure that the composition of two transcendental functions gives a transcendental number.

Blue
  • 75,673
Kamal Saleh
  • 6,497
  • 5
    Note that

    $$\cos x = \frac{e^{ix} + e^{-ix}}{2} \qquad \sin x = \frac{e^{ix} - e^{-ix}}{2i}$$

    so

    $$\sin(\cos(x)) = \frac{1}{2i} \left( \exp \left( \frac{e^{ix} + e^{-ix}}{2} i\right) - \exp \left(-i \frac{e^{ix} + e^{-ix}}{2} \right) \right)$$

    By Gelfond-Schneider, for $a,b$ algebraic (with $a \not \in {0,1}$ and $b \not \in \mathbb{Q}$), $a^b$ is transcendental.

    Hence I think the key to answering the question ends up may be whether $\eta \in \mathbb{Q}$.

     – PrincessEev Nov 04 '22 at 18:39
  • 1
    Numerically, $x \approx 0.6948196907307875$. – Dan Dec 14 '22 at 19:41
  • $\exp$ and $\log$ are transcendental but their composition is not, so in general, composition of transcendental functions in not necessarily transcendental. – jjagmath Jun 07 '23 at 03:13

2 Answers2

3

The number is given to $5000$ decimals at https://oeis.org/A131691/b131691.txt

There is some discussion of the number at https://oeis.org/A131691 but no indication of which it is; rational, algebraic irrational, or transcendental. Such determinations are generally quite difficult, even for simple-looking constants. E.g., to the best of my knowledge nothing is known about the nature of $\log\log2$.

An earlier question about this same number is Converging trigonometric result

Gerry Myerson
  • 179,216
0

$\def\Re{\operatorname{Re}}$

The OP confirmed they want series expansion, despite their question. If this answer is irrelevant, it can be moved to a more relevant question.

To solve $\sin(\cos(x))=x$, center $x-\sin(\cos(x)$ at $x=\frac\pi2$:

$$f(x)=\left(x+\frac\pi2\right)-\sin\left(\cos \left(x+\frac\pi2\right)\right)-\frac\pi 2=x+\sin(\sin(x))$$

and define $h(x)=f^{-1}(x)$ for $|x|\le L=f(\pi)=\pi$ via Fourier series. Then, $h(x)$ inverts $f(x)$ with quasiperiod of $2\pi$ for $x\in\Bbb R$:

$$h(x)=\frac1{2\pi}\sum_{n\in\Bbb Z}e^{i n x}\int_{-\pi}^\pi h(t) e^{- i n t}dt=\frac1{\pi}\Re\sum_{n=1}^\infty e^{inx}\int_{-\pi}^\pi te^{-inf(t)}df(t)$$

Integration by parts, a Jacobi Anger expansion, and Bessel J gives:

$$h(x)=x-\Re\sum_{n=1}^\infty i\frac{e^{i nx}}{\pi n}\int_{-\pi}^\pi e^{-i n f(t)}dt= x+2\sum_{n=1}^\infty \sum_{m\in\Bbb Z} J_m(-n)\frac{\sin(nx)}{\pi n}\int_{-\pi}^\pi e^{-i n} \sin(-n t-m \sin(t))dt $$

Applying $J_n(x)=\frac1\pi\int_0^\pi\sin(nt-x\sin(t))dt$:

$$\bbox[3px,border: 5px groove blue]{f(x)=\sin(\sin(x))+x\implies f^{-1}(x)=x+2\sum_{m,n=1}^\infty\frac{(-1)^n+(-1)^m}{n}\sin(nx)J_m(n)J_n(m)}$$

Graphing shows the inverse function:

Therefore:

$$\bbox[3px,border: 5px groove blue]{\sin(\cos(x))=x\implies x=-2\sum_{m,n=1}^\infty\frac{(-1)^n+(-1)^m}n\sin\left(\frac{\pi n }2\right)J_m(n)J_n(m)=2\sum_{n=0}^\infty\sum_{m\in\Bbb Z}\left(\frac{J_m(4n+1)J_{4n+1}(m)}{4n+1}-\frac{J_m(4n+3)J_{4n+3}(m)}{4n+3}\right)}$$

shown here. Maybe someone can evaluate $\sum\limits_{m=-\infty}^\infty J_m(n)J_n(m)$ so we can find a single series expansion.

Тyma Gaidash
  • 12,081