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Unformatted text preview: Chapter 2 Iterative Methods 2.1 Introduction In this section, we will consider three different iterative methods for solving a sets of equations. First, we consider a series of examples to illustrate iterative methods. To construct an iterative method, we try and re arrange the system of equations such that we gen erate a sequence. 2.1.1 Single Equation Example 2.1.1: Let us consider the equation f ( x ) = x + e − x − 2 = 0 . (2.1) When solving an equation such as (2.1) for α y=2x y=ex α 2 1 2 where f ( α ) = 0, 0 < α < 2, we can generate a sequence { x ( k ) } ∞ k =0 from some initial value (guess) x (0) by rewriting the equation as x = 2 − e − x , i.e. by computing x ( k +1) = 2 − e − x ( k ) from some x (0) . If the series converges, it will converge to the solution. For example, let us consider x (0) = 1 and x (0) = − 1: 21 k x ( k ) x ( k ) 1.01.0 1 1.632120.71828 2 1.804490.05091 3 1.83544 0.947776 4 1.84046 1.61240 5 1.84126 1.80059 6 1.84138 1.83480 7 1.84140 1.84124 7 1.84141 1.84138 9 . . . . . . In this example, both sequences appear to converge to a value close to the root α = 1 . 84141 where < α < 2. Hence, we have constructed a simple algorithm for solving an equation and it appears to be a robust iterative method. However, (2.1) has two solutions: a positive root at 1.84141 and a negative root at 1.14619. Why do we only find one root? If f ( x ) = 0 has a solution x = α then x ( k + 1 ) = g ( x ( k ) ) will converge to α , provided  g ′ ( α )  < 1 and x ( ) is suitably chosen. The condition  g ′ ( α )  < 1 is a necessary condition . In the above example, g ( x ) = 2 − e − x and g ′ ( x ) = e − x , and  g ′ ( x )  < 1 if x > . So this method can be used to find the positive root of (2.1). However, it will never converge to the negative root. Hence, this kind of approach will not always converge to a solution. 2.1.2 Linear Systems Let us adopt the same approach for a linear system. Example 2.1.2: 22 Consider the following set of linear equations: 10 x 1 + x 2 = 12 x 1 + 10 x 2 = 21 Let us rewrite these equations as x 1 = (12 − x 2 ) / 10 x 2 = (21 − x 1 ) / 10 . Thus, we can use the following: x ( k +1) 1 = 1 . 2 − x ( k ) 2 / 10 x ( k +1) 2 = 2 . 1 − x ( k ) 1 / 10 , to generate a sequence of vectors x ( k ) = ( x ( k ) 1 , x ( k ) 2 ) T from some starting vector, x (0) . If x (0) = parenleftbigg parenrightbigg , then x (0) = parenleftbigg parenrightbigg , x (1) = parenleftbigg 1 . 2 2 . 1 parenrightbigg , x (2) = parenleftbigg . 99 1 . 98 parenrightbigg , x (3) = parenleftbigg 1 . 002 2 . 001 parenrightbigg , . . . where x ( k ) → parenleftbigg 1 2 parenrightbigg as k → ∞ , which is indeed the correct answer. So we have generated a convergent sequence....
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 Spring '09
 Dr.Sharpey
 Numerical Analysis, Equations, Sets, Gauss–Seidel method, Jacobi method, Iterative method, Jacobi iterative method

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