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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