# Thread: A nasty "solve for y(x)"

1. ## A nasty "solve for y(x)"

Given:

$\alpha+\beta=\frac{\pi}{2}+\frac{x}{2}$
$\sin(\alpha)=y$
$\tan^2(\beta)=\frac{a}{\lambda y^2}\frac{a-\lambda y^2}{a-1}$
$a=\frac{1}{2}\left(1+\lambda + \sqrt{(1+\lambda)^2-4\lambda y^2}\right)$,

solve for y(x) where $\lambda$ is a known parameter.

2. Originally Posted by Heirot
Given:

$\alpha+\beta=\frac{\pi}{2}+\frac{x}{2}$
$\sin(\alpha)=y$
$\tan^2(\beta)=\frac{a}{\lambda y^2}\frac{a-\lambda y^2}{a-1}$
$a=\frac{1}{2}\left(1+\lambda + \sqrt{(1+\lambda)^2-4\lambda y^2}\right)$,

solve for y(x) where $\lambda$ is a known parameter.

Using various substitutions, such as
$\displaystyle \beta = \frac{\pi }{2} + \frac{x}{2} - alpha$

$cos(y) = \sqrt{1 - y^2}$
and the sum of angles formula for tangent, I got this down to a 4th degree polynomial in y. This can be solved in theory, so your problem does have a solution, but (as I didn't actually solve the problem) there is a chance that there are no real solutions to this monstrosity. It took me a full 10 minutes before I even saw how to attack this and just as long to come to the realization of what needs to be done. Solving a 4th degree polynomial with constant coefficients is hard enough (I've never actually managed it, the algebra is horrendous), but with the coefficients depending on a complicated formula involving x I'm not going anywhere near this thing. I wish you the best of luck.

-Dan

3. Thanks very much for your suggestion. Anyway, I came up with these equations while trying to calculate differential cross section for a classical particle scattering of a truncated harmonic potential. If y(x) cannot be found in close form, then one cannot calculate the cross section. This would indeed be sad.