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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 $\mathbb{R}$.

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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...

pshmath0
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    Again, it is interesting to observe that nested/iterates $sin$ or $cos$ can be simplified through the Bessel functions:

    http://en.wikipedia.org/wiki/Jacobi%E2%80%93Anger_expansion

    suggesting that your expansion should become equivalent to the Bessel one.

     – giorgiomugnaini Apr 04 '15 at 08:57

2 Answers2

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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)

metamorphy
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giorgiomugnaini
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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.

Claude Leibovici
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    So don't you consider the Bessel series an analytical solution? – giorgiomugnaini Apr 04 '15 at 08:53
  • For sure, it is (even a nice) analytical solution ! I was thinking more about closed forms and tried to show how simple can be the problem from a purely numerical point of view. By the way, $+1$ since, thanks to you, I refreshed my (too old) memory ! Cheers :-) – Claude Leibovici Apr 04 '15 at 08:58
  • I agree, It isn't a "closed form" in the sense of terms of "finite combination of elementary functions". But it is not only a mere formal solution, it is an analytical formula that we can numerically evaluate, moreover we can write an arbitrary term of the solution series without cumbersome algebraic manipulations, if we admit the employement of well-known functions as Bessel functions. In conclusion I think that the problem of transcendental equation solving is only the problem of finding the most "comfortable" representation capable of numerical evalutions. – giorgiomugnaini Apr 04 '15 at 11:29
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    @giorgiomugnaini. I totally agree with you, be sure ! Cheers :-) – Claude Leibovici Apr 04 '15 at 11:38