Solving $z=w/2-\sin(tw)/(2t)$ for $w$

Question

Is it possible to solve $$z=\frac{w}{2}-\frac{\sin(tw)}{2t},$$ for $w$?

My first thoughts were that we would have to be careful about the domain of $f(w)$ so that the inverse was actually a function (and not a multi-valued function). But, on graphing $f(w)$ for various values of $t$, e.g. $t=1$ below, I think the domain can actually be $\mathbb{R}$.

My first attempt was to use the Lagrange inversion formula, but this produces complicated terms which grow in size. Might there be a solution in terms of the Lambert W function?

Another approach I am working on, but also seems to become quite complicated, is a recursive approach, e.g. rearranging,

$$w=2z+\frac{\sin(tw)}{t}.\tag{1}$$

Recursively substituting $(1)$ into $(1)$ gives

$$w=2z+\frac{1}{t}\sin(2tz+\sin(tw))=\cdots.$$

Expanding,

$$w=2z+\frac{1}{t}\sin(2tz)\cos(\sin(tw))+\frac{1}{t}\cos(2tz)\sin(\sin(tw))=\cdots.$$

Again, the expressions become complicated...

Answer 1

Your equation is equivalent to the Kepler equation: $$M=E-\epsilon \sin(E)$$ for a suitable choose of $E$,$M$,$\epsilon$:

$$M=2tz$$ $$E=wt$$ $$\epsilon=1$$

A solution in terms of Bessel functions $J_n$ can be built to solve the Kepler equation: $$E(M)=M+\sum_{n=1}\frac{2\sin(Mn)J_n(n\epsilon)}{n}$$

Your question is strictly related to these questions:

How to solve Kepler's equation $M=E-\varepsilon \sin E$ for $E$?

Solving $2x - \sin 2x = \pi/2$ for $0 < x < \pi/2$

EDIT From a computationl point of view, the above seen series converges more and more slowly when $\epsilon$ (the eccentricity of original Kepler problem) approaches 1, that is your case. The convergence can be improved through acceleration series techniques as Levin acceleration.

The Burniston Siewert approach

A different analytical approach to obtain a solution can be tha Burniston Siewert integral representation.

In this paper the Kepler equation is solved by an quadrature:

E. E. Burniston and C. E. Siewert, "Exact Analytical Solutions of the Transcendental Equation a sin(z)=z," SIAM Journal on Applied Mathematics, 24 (1973) 460-466 (The PDF is here)

C. E. Siewert and E. E. Burniston, "An Exact Analytical Solution of Kepler's Equation," Celestial Mechanics, 6 (1972) 294-304 (The PDF is here)

References

http://eaton.math.rpi.edu/faculty/Kovacic/Classes/Math-4210/Papers/Bessel.pdf

http://www.willbell.com/math/mc12.htm (Solving Kepler's equation over three centuries, Peter Collwell)

Answer 2

Equations like $$f(w)=\frac{w}{2}-\frac{\sin(tw)}{2t}-z=0$$ which mix polynomial and trigonometric terms do not show analytical or closed form solutions and only numerical methods can be used.

Probably, the simplest would be to use Newton method which, starting from a reasonable guess $w_0$, will update it according to $$w_{n+1}=w_n-\frac{f(w_n)}{f'(w_n)}$$ Here $$f'(w)=\frac{1}{2}-\frac{1}{2} \cos (t w)=\sin ^2\left(\frac{t w}{2}\right)$$

Now, the problem is to find a reasonable guess; for given $t,z$, you could plot the function.

Let us take $t=1$ and $z=5$; the plot shows a root close to $w=10$. So, let us start iterating from $w_0=10$. The method then generates the following estimates : $9.70419$, $9.71440$, $9.71441$ which is the solution for six significant figures.