       Re: Solve[] for equations?

• To: mathgroup at smc.vnet.net
• Subject: [mg32042] Re: Solve[] for equations?
• From: atelesforos at hotmail.com (Orestis Vantzos)
• Date: Thu, 20 Dec 2001 03:42:04 -0500 (EST)
• References: <9v7857\$s6n\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Your problem is fascinating!
I found a simpler way to reduce the equations, taking advantage of the

The r[i], i=1...4, can be thought of as the solutions of the
polynomial
(x-r)(x-r)(x-r)(x-r)=0
This polynomial can be written as
x^4-Px^3+Px^2-Px+P=0 ,
where the P[i] are the symmetric polynomials of {r,..,r}.
So, instead of the r[i] we can solve for the P[i], and use the
polynomial to find the r[i], (or store them as Root objects).

<<Algebra`SymmetricPolynomials`

The original equations can be rewritten as:
t[i]==r[i](P-r[i])/P, i=1..4 were P is the 1st symmetric
polynomial r+r+r+r.

Turn this into a rule
tRule= Table[t[i]->r[i](P-r)/P,{i,4}]

Now we can produce equations for the P[i], using the following method:
a) Pick a symmetric expression of the t[i]
b) Use the tRule to convert it into an expression of the r[i] and
P.
c) Use the SymmetricReduction function, to convert both expressions to
expressions of the symmetric polynomials T[i] and P[i] for t and r
repsectively, and return the equation.

The function which creates the equations is this:
symRed[expr_] := First[SymmetricReduction[expr, Array[t,
{4}],Array[T,{4}]]] ==
First[SymmetricReduction[expr /. tRules,Array[r, {4}],Array[P,{4}]]]
The T[i] are the coefficients of the 4-th degree polynomial with roots
t[i].

It can be easily shon that the equation that corresponds to the square
of the sum of the t[i], depends only on P and P, hence P can
be found directly:
symRed[Sum[t[i],{i,4}]^2] --> T^2==4P^2/P^2
Of the two solutions, the useful one is the P==TP/2.

I have experimented with many forms, but it seems tht one can not
actualy get a set of equations that can be solved for all
P,P,P.
I can however solve for P and P.
The following set of equations come from symRed-ing the 3rd symmetric
polynomial of t[i], the sum of the form
t(t+t+t)+t(t+t+t)+... and the expression
t^2(t+t+t)+t^2(t+t+t)+...
Out=
3       2   1     2
{P  (-P  + - P  P T +
2

3
P P (P + T) - P  T) == 0,

2                                1     2     2
2 P  T == 2 P P + 4 P + - P  T ,
2

2                   2
P  T T - 3 P  T ==

2
3 P                  1
------- - 3 P P - - P P T - P T +
P                   2

1     2     3
- P  T }
4

We solve for P and P, and using them we assemble the polynomial,
whose solutions are the r[i].

The end product of this procedure are two 4-th degree polynomials,
whose solutions are the r[i]. The polynomials have the T[i] as
parameters, which can be calculated directly from the t[i].
Unfortunately (?) they also have the P as parameter. So before you
can get the solutions you have to provide their sum.
Preliminary numerical analysis of the solution (well, I only found the
whole trick yesterday), indicates that the values of P that give 4
real roots lie in an open interval, and that is only for one of the
polynomials. Which one seems to depend on the signs of the t[i].
Simply fascinating...
Complex solutions are available for any value of P.
Orestis

```

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