Math Forum / Math Software / MATLAB / July 2008

If you take the formula without the = 0, and you map() the simplify
function onto it, then you get a noticeably more compact expression.
However, if you then simplify() the result, you get a *large*
expression which is the sum of terms similar to

C * y^M * ys^N * x^P * a^Q * z^R *
sqrt( (y^2 - 2*y*ys + ys^2 + z^2)^2 /
         (x^2 - 2*x*a + a^2 + y^2 - 2*y*ys + ys^2 + z^2)^3 )

where C is a positive or negative multiple of 4 (e.g, +12, -36)
and M, N, P, Q, R are varying non-negative integral exponents,
and the rest of the expression appears to be the same for every
term. For example,

+24 * y * x^2 * ys^2 * z^2 * (the expression)
-36 * y^2 * x^2 * ys * a^2 * (the expression)

Then at the end of the sum, there are a few lines of other terms
that do not follow any immediately discernable pattern,
and the whole sum is divided by

2 * (x^2 - 2*x*a + a^2 + y^2 - 2*y*ys + ys^2 + z^2)^(5/2) *
(that same expression)

Clearly this is not an equation that is easily solved for!

If you use algsubs() to let
(x^2 - 2*x*a + a^2 + y^2 - 2*y*ys + ys^2 + z^2) be represented by T
and
(y^2 - 2*y*ys + ys^2 + z^2) be represented by U
and you simplify() after substitution (applying the
assumption that T and U are both nonnegative, which they
have to be as they are the sum of squares) and then collect()
on [T,U] then the expression you get back is fairly compact.
If you then substitute back the T and U values into that and
solve() the result for ys, then in a small number of seconds
you get a polynomial RootOf()... a fairly long one.
The RootOf() involves up to K^4, L^2, a^8, x^8, y^10, z^10,
and the free parameter _Z^10. There is thus likely to be
no exact symbolic solution to the RootOf() unless some
special conditions hold such as {L = 0, K=sqrt(2), a=x}
which nullifies the polynomial.

Substituting K=sqrt(2), a=x into the original equation
reduces it to a quadratic whose values are undefined at L = +/- 2
but has a solution ys = y at L=0.