How to find the summation of the above trigonometric series without using desmos?

Question

How to find the summation of $$ \sin(2 + \sin(2 + \sin(2 + \cdots \infty)))? $$ I am trying this question by denoting the above summation as $S$. Therefore, $$ S = \sin(2 + \sin(2 + \sin(2 + \cdots \infty))). $$

Now, $S = \sin(2+S)$. Now I am thinking of using the Maclaurin series expansion of $\sin(2+S)$. Therefore, $$ S = (2 + S) - \frac{(2+S)^{3}}{3!} + \frac{(2+S)^{5}}{5!} - \cdots \infty. $$

But I can't understand how to find the value of $S$ further? Because this series expansion of $\sin(2+S)$ will continue until $\infty$. Please help me out with this trigonometric summation.

Answer 1

If you want to find the zero of function $$f(x)=x-\sin(2+x)$$ expand it as a Taylor series around $x=0$ $$f(x)=x-\sum_{n=0}^\infty \frac{\sin \left(\frac{\pi n}{2}+2\right)}{n!}\,x^n$$ and use power series reversion to get $$x=\frac{\csc ^2(1)}{2} t -\frac{\cot (1) \csc ^4(1)}{8} t^2 +\frac{(3+2 \cos (2)) \csc ^8(1)}{96} t^3 +O\left(t^4\right)$$ where $t=\sin(2)$.

Truncated to this level, the above gives $x=0.577266$ while the "exact" solution given by Newton method is $x=0.554196$.

Using the expansion to $O\left(t^{10}\right)$ would give $x=0.552702$.

You could have as many terms as you wish and be closer and closer to the solution.

Using my favored $1400^+$ years old ot the sine function

$$\sin(t) \simeq \frac{16 (\pi -t) t}{5 \pi ^2-4 (\pi -t) t}\qquad \text{for}\qquad 0\leq t\leq\pi$$ this leads to a cubic equation $$4 x^3+4 (8-\pi) x^2+(80+5\pi^2-24\pi) x-32 (\pi-2)=0$$ which has only one ral root. The analytical solution is obtained using the hyperbolic solution and its numerical value is $0.553977$ in a relative error of $0.04$%.

Answer 2

Use

How to solve Kepler’s equation $M=E-\epsilon\sin E$ for $E$?
Inverse of $f(x)=\sin(x)+x$

to find:

$$S=\sin(S+2)= \operatorname{hav}^{-1}\left(\operatorname I^{-1}_\frac2\pi\left(\frac32,\frac12\right)\right)-2= \pi-2+ \int_{ci-\infty}^{ci+\infty}\frac{e^{2i s}\operatorname J_s(s)}{i s (e^{\pi i s}+1)}ds$$

where inverse haversine, inverse beta regularized, and Anger J appear. The Mathematica result is shown here:

where the inverse Laplace transform is expanded and $s\to is$ is substituted.