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 $y=\frac{(x-\sin x)}{ (1-\cos x)}$
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0out of curiosity, what is it for? Is it Newton's method or something? I'm only asking because there are conte$x$ts (like Newton's method) where $y$ou 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