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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.

Sammy Black
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Syamaprasad Chakrabarti
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    There won't be a closed form solution for $S=\sin(2+S)$. But you can find numerical approximations. Is that what you're after? – Stefan Lafon Nov 15 '23 at 18:54
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    Extending the comment of @StefanLafon, you are looking for the real number $~\theta~$ such that $~\sin(2 + \theta) = \theta.~$ If you graph the two functions $~f(\theta) = \sin(2 + \theta)~$ and $~g(\theta) = \theta,~$ you can then examine their graphs to get a rough idea of where they intersect. – user2661923 Nov 15 '23 at 18:57
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    The expression $sin(2+sin(2+sin(2+...\infty)))$ does not have any meaning, because $\infty$ is not a number. I recommend learning how to write such expressions rigorously, for otherwise a lot of mistaken thoughts will hamper your learning. – Greg Martin Nov 15 '23 at 19:00
  • There seems to be oscillation between $x\approx 0.4$ and $x\approx 0.63$. Your equation is a case of the Kepler equation with has these solutions and a Mathematica closed form here – Тyma Gaidash Nov 15 '23 at 19:55
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    Extending the comment of @GregMartin, consider $~a_1 = \sin(2), ~a_{n+1} = \sin(2 + a_n).~$ Then, you are attempting to compute $~\displaystyle \lim_{n\to \infty} a_n.$ – user2661923 Nov 16 '23 at 00:12
  • How to find the intersection point@user2661923? without using desmos? – Syamaprasad Chakrabarti Nov 16 '23 at 05:15
  • How to write these expressions @GregMartin? – Syamaprasad Chakrabarti Nov 16 '23 at 05:15
  • You can't "find the intersection point", only numerical approximations of it. So the question should be reformulated in terms of accuracy and the effort to get there. Note that already the sine on the computer or pocket calculator is an approximation, so one can't expect any better for solutions of equations with the sine. Exceptions to that principle are just that, but they dominate in textbooks. // So as another example, if you set $2+S=\pi-x$, you get $\pi-2=x+\sin x$ and with the small angle approximation a starting guess at $x=\frac\pi2-1$. – Lutz Lehmann Nov 17 '23 at 09:29

2 Answers2

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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 $0.04$%.

Claude Leibovici
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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:

enter image description here

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

Тyma Gaidash
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