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62      2 Nonlinear algebraic systems



                   Iterative methods and the use of Taylor series

                   The techniques that we will use are iterative. We start with some initial guess x [0]  of the
                                                                                  [2]
                                                                              [1]
                   solution, and apply an algorithm to refine this guess to generate a sequence x , x ,... that
                   hopefully converges to a solution x s with f (x s ) = 0; i.e.,

                                                    [k]
                                               lim  x  − x s  = 0                      (2.5)

                                              k→∞
                   We first consider Newton’s method, an iterative technique that is based on the use of Taylor
                   series expansions. As Taylor series are used extensively in numerical mathematics, we
                   briefly review their use.
                     Let us say that we have some function f (x) that we wish to represent as a polynomial in
                   the vicinity of some point x 0 ,

                                          ∞
                                          	           m
                                   f (x) =   a m (x − x 0 )  ∀ x in |x − x 0 |<        (2.6)
                                          m=0
                   Given only information of the function at x 0 – the value of the function itself plus the
                   values of all of its derivatives (we assume derivatives of f (x) to all orders exist at x 0 ), what
                   coefficients a 0 ,a 1 ,... should we use to match the polynomial to f (x)? First, we see that
                                            ∞                ∞

                                                        m           m
                                    f (x 0 ) =  a m (x 0 − x 0 ) =  a m (0) = a 0      (2.7)
                                           m=0              m=0
                   so that
                                                       ∞
                                                      	            m
                                         f (x) = f (x 0 ) +  a m (x − x 0 )            (2.8)
                                                      m=1
                   We differentiate to obtain


                                   ∞                         ∞

                           df                                         m−1
                                =     ma m (x − x 0 ) m−1    = a 1 +  ma m (0)  = a 1  (2.9)
                           dx
                              x 0  m=1              x 0     m=2
                   Taking yet another derivative yields

                      2       ∞                                ∞

                     d f                                                      m−2
                           =     m(m − 1)a m (x − x 0 ) m−2    = 2a 2 +  m(m − 1)a m (0)  (2.10)
                     dx  2
                         x 0  m=2                              m=3
                                                     x 0
                   Continuing this process, we find that if all derivatives to infinite order of f (x) exist at x 0 ,
                   we may represent the function as the infinite series
                                                     2                 3

                                   df             1 d f         2   1 d f         3
                      f (x) = f (x 0 ) +     (x − x 0 ) +     (x − x 0 ) +     (x − x 0 ) +· · ·
                                   dx             2! dx  2          3! dx  3
                                      x 0               x 0               x 0
                                                                                      (2.11)
                   If we only wish to approximate the function in the vicinity of x 0 , we can truncate the series
                   at order n:

                                             n
                                                   m
                                            	   1 d f          m
                                      f (x) =            (x − x 0 ) + R n (x)         (2.12)
                                                    m
                                            m=0  m! dx
                                                       x 0
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