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Solve for $\theta$:

$$x = \theta - \sin\theta$$

Is this type of isolation a matter of identities? If so, which one(s)?

gen-ℤ ready to perish
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user101434
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    That’s transcendental – InertialObserver Sep 07 '18 at 05:56
  • A closed solution for $\theta$ doesn't exist. There are numerical approaches one can use to solve such a problem, such as Newton's Method – WaveX Sep 07 '18 at 05:57
  • Are you serious? But I can create a function substituting y for θ and flip that function along y = x (a 45° diagonal through Quadrant I) and the function/curve passes the vertical line test. Doesn't that mean there is surely a way to arrive at a "θ = ..." equation? – user101434 Sep 07 '18 at 06:01
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    I don't quite see what you are doing there, but regardless of this you should ask yourself if that procedure applied to $y=x^3$ results in introducing a weird V-shaped symbol with a small three inside and then writing $x=\sqrt[3]y$. –  Sep 07 '18 at 06:09
  • @SaucyO'Path The OP means to say (I think, correct me if I'm off!) that the function appears to pass the Horizontal Line Test and to be bijective. The question then becomes, "Bijectivity implies an inverse relation. Doesn't that mean there will be a closed form solution for $\theta(x)$?" – zahbaz Sep 07 '18 at 06:17
  • And I think the answer here is no. For example, consider $y=xe^x$ on $[0,\infty)$. The solution is $W(ze^z)$ where $W$ is the Lambert W function... but do we call that closed form? It's still transcendental. – zahbaz Sep 07 '18 at 06:23
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    I applaud your confidence in deeming this simple problem as being unsolvable, but can I ask, what is it that qualifies it as being "impossible"? – user101434 Sep 07 '18 at 06:27

1 Answers1

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As said in comments, you need some numerical methods for this kind of transcendental equations.

How,we can approximate solutions to get an estimate for, say, Newton method. For example, if the solution is in the range $0 \leq \theta \leq \pi$, the maginficent approximation $$\sin(\theta) \simeq \frac{16 (\pi -\theta) \theta}{5 \pi ^2-4 (\pi -\theta) \theta}\qquad \text{for} \qquad 0\leq \theta\leq\pi$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1400$ years ago !) would lead to the cubic equation $$4\, \theta ^3-4 (x+\pi -4)\,\theta ^2+ \pi\left(4 x+5 \pi -16 \right)\,\theta-5 \pi ^2 x=0$$ which can be solved.

In the table below, I produced some values for the solution. $$\left( \begin{array}{ccc} x & \text{approximation} & \text{exact} \\ 0.0 & 0.00000 & 0.00000 \\ 0.1 & 0.84987 & 0.85375 \\ 0.2 & 1.08166 & 1.08369 \\ 0.3 & 1.24765 & 1.24852 \\ 0.4 & 1.38201 & 1.38228 \\ 0.5 & 1.49726 & 1.49730 \\ 0.6 & 1.59958 & 1.59959 \\ 0.7 & 1.69251 & 1.69259 \\ 0.8 & 1.77828 & 1.77851 \\ 0.9 & 1.85843 & 1.85881 \\ 1.0 & 1.93404 & 1.93456 \\ 1.1 & 2.00590 & 2.00655 \\ 1.2 & 2.07463 & 2.07538 \\ 1.3 & 2.14070 & 2.14151 \\ 1.4 & 2.20451 & 2.20534 \\ 1.5 & 2.26636 & 2.26717 \\ 1.6 & 2.32651 & 2.32726 \\ 1.7 & 2.38517 & 2.38584 \\ 1.8 & 2.44254 & 2.44308 \\ 1.9 & 2.49876 & 2.49915 \\ 2.0 & 2.55398 & 2.55420 \end{array} \right)$$

Another possible way would be Taylor expansion and series reversion to get $$\theta=t+\frac{t^3}{60}+\frac{t^5}{1400}+\frac{t^7}{25200}+\frac{43 t^9}{17248000}+\frac{1213 t^{11}}{7207200000}+\frac{151439 t^{13}}{12713500800000}$$ where $t=\sqrt[3]{6x}$

Claude Leibovici
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