![]() ![]() The result of this first iteration of the Gauss-Seidel Method is We then solve for x 2 (1) in the second equation, using the new value of x 1 (1) = 0.750, and find thatįinally, we solve for x 3 (1) in the third equation, using the new values of x 1 (1) = 0.750 and x 2 (1) = 1.750, and find that We first solve for x 1 (1) in the first equation and find that ![]() Let us be clear about how we solve this system. ![]() To compare our results from the two methods, we again choose x (0) = (0, 0, 0). Let's apply the Gauss-Seidel Method to the system from Example 1:Īt each step, given the current values x 1 ( k), x 2 ( k), x 3 ( k), we solve for x 1 ( k+1), x 2 ( k+1), x 3 ( k+1) in This technique is called the Gauss-Seidel Method - even though, as noted by Gil Strang in his Introduction to Applied Mathematics, Gauss didn’t know about it and Seidel didn’t recommend it. So x 1 ( k+1) is found as in Jacobi's Method, but in finding x 2 ( k+1), instead of using the old value of x 1 ( k) and the old values x 3 ( k), …, x n ( k), we now use the new value x 1 ( k+1) and the old values x 3 ( k), …, x n ( k), and similarly for finding x 3 ( k+1), …, x n ( k+1). Where the true solution is x = ( x 1, x 2, …, x n), if x 1 ( k+1) is a better approximation to the true value of x 1 than x 1 ( k) is, then it would make sense that once we have found the new value x 1 ( k+1) to use it (rather than the old value x 1 ( k)) in finding x 2 ( k+1), …, x n ( k+1). Let us take Jacobi’s Method one step further. ![]()
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