Why can't $a=\theta-\sin(\theta)$ be solved for $\theta$ in terms of $a$ in closed form? Context:
I had taken an interest in alchemical symbols. Many of the ancient drawings are understandably crude, given the tools available at the time. In spite of their rough appearance, I imagined that the symbols were designed from geometric principles. I tried to think up some plausible rules the artists might have been aiming for, and redraw them using the tools we have today.
The equation in this question is a generalized version of one that came out of that work.
It stood out to me strongly enough that I wanted to ask another question about this class of equations. You can view the question that sparked all of this HERE.
Question:
Why can't an equation like
$$a=\theta-\sin(\theta),$$
where $a\in\mathbb R$, be solved for $θ$ in terms of $a$ in closed form?
It looks like a simple equation of elementary functions. I tried to invert the elementary functions in the equation, i.a. by applying $\arcsin$, but I failed.
Although I can't solve it, I don't have an intuition for why this can't be done.
Edit 1:
I was not familiar with the concept of a transcendental equation prior to posting.
Edit 2:
Added some introductory context.
 A: From my perspective, the key here is that generically equations can't be solved. When we learn math, we're spoiled because assignments ask us to solve things, so we get used to seeing solvable things. But a generic fifth-order or higher polynomial doesn't have a closed-form solution in terms of standard functions.
You can also think of this as a process of combining and inverting functions. If I know how to invert two functions, $f(x), g(x)$ can I figure out how to invert $f(x)+g(x)$, which is a special case of a class of functions $h(f(x),g(x))$? In general no, I can't. The cases in which I can, (for example, that $f$ and $g$ are both quadratic in $x$ and $h$ is just addition) are the special cases.
A: It really depends what you mean by "solving".
If you write $f(\theta)= \theta - \sin(\theta)$, then since $f'(\theta) = 1-\cos(\theta) \geq 0$ it is clear that $f$ is a monotone function. So in particular $f$ is injective (it has flat-tangent points, but isolated ones) and, therefore, invertible.
Let $f^{-1}$ be its inverse: then the solution is $\theta=f^{-1}(a)$. Done!
...actually, there is a problem. It turns out that we cannot express this $f^{-1}$ as a sum, product, quotient or composition of elementary functions (polynomials, $\sin(x), \cos(x), e^x, \log(x)$). So, essentially, this is what we mean by the "can't be solved" wording.
But... the equation can totally be solved by computers, be approximated by polynomial series to whatever precision you wish, etc.
