Mail Archives: geda-user/2013/08/19/18:30:08
On Aug 19, 2013, at 11:15 AM, Stephen R. Besch wrote:
> Note that you can do exactly the same with curves that are defined with a quadratic. You just have to solve for x using the standard quadratic solution:
>
> x=(-b+- sqrt (b^2-4ac))/2a
>
> Just remember that you will get 2 solutions (one may be imaginary)
Mmm, if the coefficients a, b, and c are real, either both solutions are real or both are complex.
> Higher order curves - other than cubics - have no closed form solutions and intersection will need to be determined numerically.
Quartics also have general solutions in terms of radicals (square roots, cube roots, ...). Higher order equations may have such solutions, but no general formula in radicals is possible. It is, however, possible to use "root isolation" techniques to obtain general solutions in terms of less familiar "root" functionals. These are more complicated than radicals, but have similar useful properties: it is possible to know which root you're dealing with, and it is possible to rigorously reason about some properties like whether a root is real. Whether this counts as "closed form" is a matter of taste.
John Doty Noqsi Aerospace, Ltd.
http://www.noqsi.com/
jpd AT noqsi DOT com
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