Evaluation using Interval Arithmetic

The problem was to evaluate the expression
f(x) = (((1.013x**2 - 5.262)x - 0.1732)x + 0.8389)x - 1.912 using four decimal digit interval arithmetic with x = [2.278,2.280].

The sequence of operations is:

x^2              contained in [ 5.189,5.199]
1.013x^2         contained in [ 5.256,5.267]
1.013x^2 - 5.262 contained in [-0.006,0.005]
* x              contained in [-0.01368,0.01140]   (4 SIGNIFICANT figs.)
-0.01732         contained in [-0.03100,-0.005920]
*x               contained in [-0.07068,-0.01348]
+0.8389          contained in [ 0.7682, 0.8255]
*x               contained in [ 1.749, 1.881]
-1.912           contained in [-0.1630,-0.03100]

This function is rather ill-conditioned. If the program


!  The following program uses machine rounding, rather than
!  four digit rounding

program problem_3_b_with_interval
   use interval_arithmetic
   implicit none

   type(interval) :: x, value

      call simini
      x = interval(2.278d0,2.280d0)

      value = x**2
      write(*,'(1x,a15,2(1x,d14.6))') "^2:", value
      value = 1.013d0*value
      write(*,'(1x,a15,2(1x,d14.6))') "*1.013:", value
      value = value-5.262d0
      write(*,'(1x,a15,2(1x,d14.6))') "-5.262:", value
      value = value*x
      write(*,'(1x,a15,2(1x,d14.6))') "*x:", value
      value = value-0.01732d0
      write(*,'(1x,a15,2(1x,d14.6))') "-0.01732:", value
      value = value*x 
      write(*,'(1x,a15,2(1x,d14.6))') "*x:", value
      value = value+ 0.8389d0
      write(*,'(1x,a15,2(1x,d14.6))') "+0.8389:", value
      value = value*x 
      write(*,'(1x,a15,2(1x,d14.6))') "*x:", value
      value = value -1.912d0
      write(*,'(1x,a15,2(1x,d14.6))') "-1.912:", value

end program problem_3_b_with_interval

is run, the following output is produced:


             ^2:   0.518928D+01   0.519840D+01
         *1.013:   0.525674D+01   0.526598D+01
         -5.262:  -0.525531D-02   0.397920D-02
             *x:  -0.119821D-01   0.907258D-02
       -0.01732:  -0.293021D-01  -0.824742D-02
             *x:  -0.668088D-01  -0.187876D-01
        +0.8389:   0.772091D+00   0.820112D+00
             *x:   0.175882D+01   0.186986D+01
         -1.912:  -0.153176D+00  -0.421438D-01

If, instead of x = [2.278,2.280], x=[2.279,2.279] is used in the program, and a d24.16 format is used, the following output is produced:

             ^2:   0.5193840999999999D+01   0.5193841000000000D+01
         *1.013:   0.5261360932999998D+01   0.5261360933000000D+01
         -5.262:  -0.6390670000016030D-03  -0.6390669999998266D-03
             *x:  -0.1456433693003653D-02  -0.1456433692999605D-02
       -0.01732:  -0.1877643369300365D-01  -0.1877643369299960D-01
             *x:  -0.4279149238635533D-01  -0.4279149238634609D-01
        +0.8389:   0.7961085076136446D+00   0.7961085076136539D+00
             *x:   0.1814331288851496D+01   0.1814331288851517D+01
         -1.912:  -0.9766871114850395D-01  -0.9766871114848263D-01

One concludes that the values are sensitive to x.