MHB Can you solve for y in sin(y) - y = x√(2)?

  • Thread starter Thread starter wheepep
  • Start date Start date
Click For Summary
The equation sin(y) - y = x√(2) raises questions about the existence of a closed-form solution for y. Some suggest that replacing sin(y) with its exponential form may lead to a solution involving the Lambert W function. The discussion references the Kepler equation, indicating that this problem is well-studied in celestial mechanics. For practical purposes, iterative methods, power series, or numerical interpolation are recommended for finding solutions. The periodic nature of the function suggests its relevance in various natural phenomena.
wheepep
Messages
9
Reaction score
0
sin(y) - y = x√(2)
solve for y
 
Mathematics news on Phys.org
What makes you think a closed-form solution for y can be found?
 
Replacing [math]sin(y)[/math] by [math]\frac{e^{iy}- e^{-iy}}{2i}[/math] you might be able to get a solution in terms of the "Lambert W function" https://en.wikipedia.org/wiki/Lambert_W_function.
 
Maybe there exists a closed form solution exists, maybe not... But could I ask, what is the purpose for searching such solution?

I know, in celestial mechanics, the Kepler equation is same as this - at least after a change of variables - which means, you can take a look at literature, if you find something about it. As being so, I wouldn't use my time to kick the equation, as it is one of the most researched one in the world. Unless I wanted some sort of challenge, of course.

Anyhow, your choices for the solution will most likely be an iterative method, a power series or a Fourier series or interpolation of the numerical solution over the values of x under interest.
 
In general, periodic functions are of interest due to their frequent occurrence in natural phenomenon. As speculation, this particular function may be of interest due the times and places it occurs.
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
8
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K