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222    NUMERICAL DIFFERENTIATION/ INTEGRATION
           ASCII data file named “xy.dat”, we can use the routine “diff()”toget the
           divided difference, which is similar to the derivative of a continuous function.
            >>load xy.dat %input the contents of ’xy.dat’ as a matrix named xy
            >>dydx = diff(xy(:,2))./diff(xy(:,1)); dydx’ %divided difference
             dydx =  2.0000   0.50000  2.0000

                       f(x k )             f(x k+1 ) − f(x k )  f(x k+1 ) − f(x k )
                x k             x k+1 − x k
                                                           D k =
           k  xy(:,1)  xy(:,2)  diff(xy(:,1))  diff(xy(:,2))       x k+1 − x k
           1    −1       2          1             2                 2
           2     0       4          2             1                1/2
           3     2       5        −1            −2                  2
           4     1       3



           5.5  NUMERICAL INTEGRATION AND QUADRATURE

           The general form of numerical integration of a function f(x) over some interval
           [a, b] is a weighted sum of the function values at a finite number (N + 1) of
           sample points (nodes), referred to as ‘quadrature’:

                 b          N

                         ∼
                 f(x) dx =    w k f(x k )  with a = x 0 <x 1 < ·· · <x N = b  (5.5.1)
               a
                           k=0
           Here, the sample points are equally spaced for the midpoint rule, the trapezoidal
           rule, and Simpson’s rule, while they are chosen to be zeros of certain polynomials
           for Gaussian quadrature.
              Figure 5.3 shows the integrations over two segments by the midpoint rule,
           the trapezoidal rule, and Simpson’s rule, which are referred to as Newton–Cotes
           formulas for being based on the approximate polynomial and are implemented
           by the following formulas.
                                 x k+1

                                            ∼
              	midpoint rule
       f(x) dx = hf mk                      (5.5.2)
                                x k
                                                                     x k + x k+1
                               with h = x k+1 − x k ,  f mk = f(x mk ), x mk =
                                                                         2
                                              h
                                 x k+1

                                            ∼
            	trapezoidal rule
      f(x) dx =  (f k + f k+1 )            (5.5.3)
                                              2
                                x k
                               with h = x k+1 − x k ,  f k = f(x k )

                                 x k+1        h
                                            ∼
             	Simpson’s rule
       f(x) dx =  (f k−1 + 4f k + f k+1 )   (5.5.4)
                                              3
                                x k−1
                                           x k+1 − x k−1
                                   with h =
                                                2
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