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I wanted to make one of those cool infinite recursive definitions for myself, and I chose one that I thought looked cool: $x=\sqrt{\sin{x}}=\sqrt{\sin{\sqrt{\sin{\sqrt{\sin{...}}}}}}$ for no other reason than because I thought it looked cool. Using my method* of finding solutions for these; start with a number e.g. 1, take sqrt(sin(1)), take sqrt(sin(that)) alot of times using Answer button until you find a good guess, then algebraically confirm said guess; I got as far as $0.8767262154$ on my TI-83 Plus and $0.87672621539$ on the calculator you get on http://www.google.com/search?q=calc and also don't forget $0$ courtesy of Mac Grapher.app,

wolfram alpha roasted all with a massive $0.876726215395062445972118643142$

What my real question is, is how would I calculate (not just compute) or solve for all the real/complex solutions of this? Also, my complex question no pun intended is, wolframalpha also gave me $x=\frac{\sqrt{i\left(e^{-ix}-e^{ix}\right)}}{\sqrt{2}}$, how did they get this?[[What are the complex solutions of this equation?]]

*I discovered this for myself when testing that $φ=1+\frac{1}{φ}$ but am aware that it may have already occured to people who know about Mandelbrot set and higher degree 2-dimensional polynomial equations for which algebraic methods of calculation hav yet to be developed

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For the complex looking thing,

Remember that $$e^{ix}=\cos x+\sin x$$ $$e^{-ix}=\cos x - \sin x$$

So $$e^{-ix} -e^{ix}=-2\sin x $$

Hence

\begin{align} x&=\sqrt{\sin x}\\ &=\sqrt{\frac{e^{-ix} -e^{-ix}}{-2}}\\ &=\frac{\sqrt{i(e^{-ix} -e^{ix})}}{\sqrt2} \end{align}

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  • $\begingroup$ Thanks, this does answer my "complex" question! $\endgroup$
    – cmarangu
    Commented Apr 10, 2018 at 15:41
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There's no reason to believe it has a solution in closed form; most equations don't. You probably have to solve it numerically. As to the complex solutions that comes from the definition of $\sin z$ when $z$ is a complex number. Best to leave that be until you study functions of a complex variable, I would say.

A glance at the graph shows that $x^2=\sin x$ has only the two real solutions you've found.

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  • $\begingroup$ oh by the way i went ahead and started studying complex analysis, phase portraits, conformal maps and stuff for a bit, made my own phase plotter in glsl, s0 i might revisit this question with newfound knowledge $\endgroup$
    – cmarangu
    Commented Jul 20, 2020 at 2:48
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So cmarangu as for your complex solutions there are ~3.

Theres 0i.

And then theres one around 0.5231i

And then theres one around 4.3069i

And don't forget ∞i.

I found these solutions using Mac Grapher app I said y=x and found intersections with y=sqrt(sin(ix)) if ya know what I mean.

In order to find the exact values for these you may have to do some weird stuff with e^(ix) and ya.

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