Re: Problem using solve... keeps busy
- From: roberson@xxxxxxxxxxxxxxxxxx (Walter Roberson)
- Date: Thu, 31 Jul 2008 17:47:15 +0000 (UTC)
In article <g6sdfi$o79$1@xxxxxxxxxxxxxxxxxx>,
theoxarhs alex <theoxarhs@xxxxxxxxx> wrote:
I am trying to solve
[ys] = solve('L-1/2/(x^2-2*x*a+a^2+y^2-2*y*ys+ys^2+z^2)^
(1/2)*(-2*y+2*ys)+1/4*(K-1/2/(x^2-2*x*a+a^2+y^2-
2*y*ys+ys^2+z^2)^(1/2)*(-2*x+2*a))/((1/4/(x^2-2*x*a+a^2+y^2-
2*y*ys+ys^2+z^2)^(3/2)*(-2*x+2*a)^2-1/(x^2-2*x*a+a^2+y^2-
2*y*ys+ys^2+z^2)^(1/2))^2)^(1/2)/(x^2-2*x*a+a^2+y^2-
2*y*ys+ys^2+z^2)^(3/2)*(-2*x+2*a)*(-2*y+2*ys)-1/2*(K-1/2/
(x^2-2*x*a+a^2+y^2-2*y*ys+ys^2+z^2)^(1/2)*(-2*x+2*a))^2/
((1/4/(x^2-2*x*a+a^2+y^2-2*y*ys+ys^2+z^2)^(3/2)*(-2*x+2*a)
^2-1/(x^2-2*x*a+a^2+y^2-2*y*ys+ys^2+z^2)^(1/2))^2)^(3/2)*
(1/4/(x^2-2*x*a+a^2+y^2-2*y*ys+ys^2+z^2)^(3/2)*(-2*x+2*a)^2-
1/(x^2-2*x*a+a^2+y^2-2*y*ys+ys^2+z^2)^(1/2))*(-3/8/(x^2-
2*x*a+a^2+y^2-2*y*ys+ys^2+z^2)^(5/2)*(-2*x+2*a)^2*(-
2*y+2*ys)+1/2/(x^2-2*x*a+a^2+y^2-2*y*ys+ys^2+z^2)^(3/2)*(-
2*y+2*ys))=0','ys')
with no result.
The pc is kept busy for hours.
Any ideas? Is it that complex, or solve can't find a
solution?
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.
--
"Beware of bugs in the above code; I have only proved it correct,
not tried it." -- Donald Knuth
.
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