kitchen table math, the sequel: algebra I and optics...

Saturday, October 2, 2010

algebra I and optics...

Solve for x, given k = x + 1/x and that k is fixed.

This is a reduced and massively rearranged derivation of the lens equation, when the separation distance between an image and object is known, and the focal length of a lens known and fixed at a single value. x = do / di. (Or due to symmetry, it could just as well be the other way round.)

Mathematica kindly pointed out (as it solved the equation for me) that you could write it this way:

k = x + x-1

which resembles a quadratic equation, oooh. kinda. In real life you already know every k (= S/f -2) determines two solutions from experiment, but I was trying to decide how symmetric the ratios theoretically were.

Algebraic equation solvers tell me the solution has approximately the same form as quadratic roots, taking the form p +/- sqrt(q). Kinda. The general solution is x = 0.5 [ k +/- sqrt(k2 -4)]

Unlike most quadratics though, there is a gaping asymptote, and the function y=k rapidly changes from -inf to +inf as x passes from -c to c, where c is a number arbitrarily close to zero. When x is large, k is approximately x; when x is small, k is approximately 1/x.

I puzzle in my head how to solve it algebraically on paper -- this college student who has (up to this point) taken Calc III, linear algebra, and the mess of multiple equilibrium scenarios that tricky chemistry professors throw at you. How do you algebraically solve "inverse-order" polynomials? I mean, this equation should be simple to solve, right?

Daniel Ethier said...

Multiply both sides by x, keeping in mind that x cannot be 0. This gives you a proper quadratic equation:

kx = x^2 + 1

or

0 = x^2 - kx + 1

Using the quadratic formula, you get

x = 1/2 [k +/- sqrt(k^2 - 4)]