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Example 3

This example is derived from example 2. We notice that in the three
functions of example 2 the second degree terms of are for all
functions . Thus by subtracting the first function to
the second and third we get a linear system in . This system is
solved and the value of are substituted in the first
function. We get thus a system of one equation in the unknown
(see section 15.1.2).
The roots of this equation are 0,-0.806783438.
The test program is `Test_Solve_General2`.
The `IntervalFunction` is written as:
INTERVAL_VECTOR IntervalTestFunction (int l1,int l2,INTERVAL_VECTOR & in)
{
INTERVAL_VECTOR xx(1);
xx(1)=11092.0+(-25912.0+(19660.0-4840.0*Cos(in(1)))*Cos(in(1)))*Cos(in(1))+(
-508.0+(3788.0-1600.0*Cos(in(1)))*Cos(in(1)))*Sin(in(1));
return xx;
}

This program is implemented under the name
`Test_Solve_General2`.
With `epsilonf`=0 and `epsilon`=0.001
we get the solution
intervals, using 32 boxes:

for whatever order.
If we use `epsilon`=0 and `epsilonf`=0.1 we get,
using 50 boxes:

In both cases the solution intervals contain the roots of the
equation.

Jean-Pierre Merlet
2012-12-20