Re: better than a quartic?

From: John R Ramsden (jr_at_adslate.com)
Date: 07/04/04 

Date: Sun, 04 Jul 2004 14:33:26 +0100

On 3 Jul 2004 07:17:07 -0700, Carlo_02@libero.it (Carlo) wrote:

> I have to solve this set of equations:
>
> t5 = t1 + (t1-t2)^2/( (t1-t3)*(Pa-1) )
>
> t1 = t5 + (t5-t4)^1/( (t5-t3)*(Pb-1) )
>
> t1 and t5 are unknown. t2, t3, t4, Pa, Pb are known.
> In practice, these properties are also true:
> 0 < t2 < t3 < t4
> 0 < Pa < 1
> 0 < Pb < 1
>
> With some manipulation, I can turn the equations above into a quartic,
> and solve it numerically (I know there is also a symbolic solutions to
> a quartic, however, the coefficients of this quartic are already a
> mess, never mind the symbolic quartic solution...). Once I get t1 or
> t5 this way, I can find the other easily by applying one of the above.
>
> Unfortunately, with the numbers I have to play with, the quartic is
> relatively a mess.

Probably the best that can be done algebraically is try and solve
for the unknowns in terms of a parameter that satisfies a simpler
quartic than maybe the one you found.

For typing convenience I've relabelled your variables and unknowns
as follows:

  x, y, a, b, c, d, e := t5, t1, t2, t4, Pa - 1, Pb - 1, t3

so your equations become:

  x - y = (x - b)^2 / (d.(x - e)) = (y - a)^2 / (c.(y - e))

taking:

  u, v, A, B := (x - b)/d, (y - a)/c, (b - e)/d, (a - e)/d

these become:

  d.u - c.v + b - a = u^2 / (u + A) = v^2 / (v + B)

The latter pair are equivalent to:

  B.((2A/u + 1)^2 - 1) = A.((2B/v + 1)^2 - 1)

So taking:

  2A/u, 2B/v = s, s.t

gives either s = 0 or a linear equation in s which implies
the following general parametrization of u, v in terms of t:

  u = t.v = -A.(A.t^2 - B)/(A.t - B)

and plugging that into "d.u - c.v ... = " gives the quartic:

  (A.t^2 - B)^2 + (A.c.t - B.d + b - a).(t - 1).(A.t^2 - B)

         - B.(b - a).(t - 1)^2 = 0

This is a homogenous quadratic in A.t^2 - B and t - 1.
But unfortunately if A.c != 0 there is a pesky t in the
middle coefficient, and this means it can't be solved
directly as a pair of quadratic steps.

(If you decide to use the preceding, it would be a good idea
to check the algebra!

Cheers

---------------------------------------------------------------------------
John R Ramsden (jr@adslate.com)
---------------------------------------------------------------------------
"Eternity is a long time, especially towards the end."
  Woody Allen

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