I only see numerical approaches to solve this equation. Is there an analytical solution to solve $x$ as a function of $y$ for the range $(0,2 \pi)$? If there is no solution, is it possible to proof it?
How to solve $y=\frac{(x-\sin x)}{ (1-\cos x)}$
-
0What do you mean by 'solve'? Are you trying to find zeroes? – 2012-06-07
-
2I would be extremely surprised if there were. – 2012-06-07
-
1@JoeJohnson126: My guess is that he wants to solve for $x$ as a function of $y$. – 2012-06-07
-
0@ Harald Hanche-Olsen: thats right – 2012-06-07
-
0Probably no solution in elementary functions. See, for example, Kepler's equation which is known to have no solution in elementary functions: http://en.wikipedia.org/wiki/Kepler's_equation – 2012-06-07
-
0out of curiosity, what is it for? Is it Newton's method or something? I'm only asking because there are contexts (like Newton's method) where you don't need one to solve the problem. – 2012-06-07
1 Answers
For a full and complete answer you might want to look into inversion of the power series, although I have not checked the inverse function theorem conditions for this one thoroughly. Nevertheless here's my $O(y^5)$-worth take on it: $$y=\frac{\frac{x^{3}}{6}+O\left(x^{5}\right)}{\frac{x^{2}}{2}+O\left(x^{4}\right)}=\frac{2}{x^{2}}\frac{\frac{x^{3}}{6}+O\left(x^{5}\right)}{1+O\left(x^{2}\right)}=\frac{2}{x^{2}}\left(\frac{x^{3}}{6}+O\left(x^{5}\right)\right)\left(1+O\left(x^{2}\right)\right)=\frac{2}{x^{2}}\left(\frac{x^{3}}{6}+O\left(x^{5}\right)\right)=\frac{x}{3}+O\left(x^{3}\right)$$ $$x=3y+O\left(y^{3}\right)$$ $$x=\left(1-\cos x\right)y+\sin x=\frac{3y^{3}}{2}+O\left(y^{5}\right)+3y+O\left(y^{3}\right)-\frac{1}{6}\left(27y^{3}+O\left(y^{5}\right)\right)=3y-3y^{3}+O\left(y^{5}\right)$$
I am not entirely (neither meromorphically) confident about the last term at $\frac{1}{6}$ in brackets in the last line, however Grapher tells me I am not too far from truth
