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Sorry for the dumb question, but I'm not involved in math. I need to reverse the following formula, to calculate $\alpha$:

$$a = b(\alpha + \sin \alpha)/c$$

So I have:

$$(\alpha + \sin \alpha)=ac/b = K$$

Since $a$, $b$, $c$ are constant, I put equal to $K$.

$\alpha$ is measured in radians. I need to find the value of $\alpha$ (in radians or degree).

Thanks to all!!

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4 Answers 4

up vote 4 down vote accepted

There is no "closed-form" solution to this equation. You can use numerical methods to solve it for any given value of $K$. If $K$ is small, you can use a series:

$$\alpha = {\frac {1}{2}}K+{\frac {1}{96}}{K}^{3}+{\frac {1}{1920}}{K}^{5}+{\frac {43}{1290240}}{K}^{7}+{\frac {223}{92897280}}{K}^{9} + \ldots $$

The error in the approximation using the terms above is less than about $2 \times 10^{-7}$ for $-1 \le K \le 1$.

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Unfortunately K is at list $300 \cdot 600$. But thanks, you remembered me the use of series! –  Tommaso Jun 14 '12 at 17:53
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@Tommaso: A similar series can be used in other ranges of $K$. If you write $\alpha = 2 n \pi + t$, $\alpha + \sin(\alpha) = 2 n \pi + t + \sin(t) = K$ if $t + \sin(t) = K - 2 n \pi$. So if $K$ is near $2 n \pi$, use my series with $K$ replaced by $K - 2 n \pi$ and then add $2 n \pi$ to the result. –  Robert Israel Jun 14 '12 at 18:00
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@Tommaso: If $K$ is that big, the approximation $\alpha \approx K$ shouldn't be too bad... –  Hans Lundmark Jun 14 '12 at 18:01
    
@RobertIsrael Thanks for the suggestion! I will try to create a code in order to get at least an approximation of the solution –  Tommaso Jun 14 '12 at 18:18
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These are what we call Transcendental Equations.

Solving these involve graphical or numerical analysis, both of which yield approximate results.

In graphical analysis, you first rearrange the equation as:

$\sin\alpha = K - \alpha $

Now, on a graph, plot the curve $y=\sin\alpha$. On the same graph, plot the straight line $y=K-\alpha$

The point at which the two intersect is basically the solution of the equation. From the $y$ coordinate of the intersection point, you can then easily calculate the value of $\alpha$ using $y=K-\alpha$

Numeric solutions involve methods like the Newton Raphson Method, Bisection Method, etc.

Here is a nice wikipedia article enlisting all such methods.

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A useful link with a useful terminology as well

http://en.wikipedia.org/wiki/Lagrange_inversion_theorem

This equation is often present in the mathematical problems of celestial mechanics

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You are trying to find the root of the equation $f(\alpha) = \sin(\alpha) + \alpha - K$. Just use Newton-Raphson to get to the solution. If your $K$ is fairly small, then initializing Newton-Raphson with $\frac{K}{2}$ should be good and if its quite large, then initializing $\alpha$ with $K$ should do.

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