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Math Help - A nasty "solve for y(x)"

  1. #1
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    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.

    I'm having a hard time doing this - please help!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Heirot View Post
    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.

    I'm having a hard time doing this - please help!
    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
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  3. #3
    Junior Member
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    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.
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