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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)
This program is implemented under the name
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