dstefan
03-22-2008, 01:33 PM
Chapter 8, Problem 6:
Let F : {\mathbb R}^3 \to {\mathbb R}^3 given by F(x) ~=~ \left( \begin{array}{c} x_1^3 + 2 x_1 x_2 + x_3^2 - x_2 x_3 + 9 \\ 2 x_1^2 + 2 x_1 x_2^2 + x_2^3 x_3^2 - x_2^2 x_3 - 2 \\ x_1 x_2 x_3 + x_1^3 - x_3^2 - x_1 x_2^2 - 4 \end{array} \right). The approximate gradient \Delta_c F(x) = \left( \Delta_{c,j} F_i(x) \right)_{i,j=1:n} of F(x) is computed using central difference approximations, i.e., \Delta_{c,j} F_i(x) ~=~ \frac{F_i(x+h e_j) - F_i(x-h e_j)}{2h}, ~~j=1:n, where e_j is a vector with all entries equal to 0 with the exception of the j-th entry, which is equal to 1.
(i) Solve F(x) = 0 using the approximate Newton's algorithm by substituting \Delta_c F(x_{old}) for \Delta F(x_{old}). Use h = 10^{-6}, tol_{consec} = 10^{-6}, and tol_{approx} = 10^{-9}, and two different initial guesses: x_0 ~=~ \left( \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right) and x_0 ~=~ \left( \begin{array}{c} 2 \\ 2 \\ 2 \end{array} \right).
(ii) Compare these results to those obtained in section 8.3.2 corresponding to the approximate Newton's method with forward finite difference approximations for \Delta F(x).
Solution:
a
Let F : {\mathbb R}^3 \to {\mathbb R}^3 given by F(x) ~=~ \left( \begin{array}{c} x_1^3 + 2 x_1 x_2 + x_3^2 - x_2 x_3 + 9 \\ 2 x_1^2 + 2 x_1 x_2^2 + x_2^3 x_3^2 - x_2^2 x_3 - 2 \\ x_1 x_2 x_3 + x_1^3 - x_3^2 - x_1 x_2^2 - 4 \end{array} \right). The approximate gradient \Delta_c F(x) = \left( \Delta_{c,j} F_i(x) \right)_{i,j=1:n} of F(x) is computed using central difference approximations, i.e., \Delta_{c,j} F_i(x) ~=~ \frac{F_i(x+h e_j) - F_i(x-h e_j)}{2h}, ~~j=1:n, where e_j is a vector with all entries equal to 0 with the exception of the j-th entry, which is equal to 1.
(i) Solve F(x) = 0 using the approximate Newton's algorithm by substituting \Delta_c F(x_{old}) for \Delta F(x_{old}). Use h = 10^{-6}, tol_{consec} = 10^{-6}, and tol_{approx} = 10^{-9}, and two different initial guesses: x_0 ~=~ \left( \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right) and x_0 ~=~ \left( \begin{array}{c} 2 \\ 2 \\ 2 \end{array} \right).
(ii) Compare these results to those obtained in section 8.3.2 corresponding to the approximate Newton's method with forward finite difference approximations for \Delta F(x).
Solution:
a