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L9_NonlinearSystems

# L9_NonlinearSystems - MA2213 Lecture 9 Nonlinear Systems...

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MA2213 Lecture 9 Nonlinear Systems

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Midterm Test Results
Topics Roots of One Nonlinear Equation in One Variable Secant Method pages 90-97 Roots of Nonlinear Systems (n Equations, n Variables) Newton’s Method pages 352-360 Calculus Review : Intermediate Value Theorem, Newton’s Method pages 79-89 Mean Value Theorems for Derivatives and Integrals Applications to Eigenvalue-Eigenvector Calculation Applications to Optimization

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Mean Value Theorem for Derivatives 2 Theorem A.4 p. 494 Let x y = y x . b)) ((a, C f b]), C([a, f 1 a) - (b (c) f ) ( f ) ( f ' = - a b ) 0 , 0 ( ) 0 , ( 1 x There there is at least one point b) (a, c such that 5 . 0 ) 0 2 ( ) 5 . 0 ( 2 0 ) 2 ( ' - = = - y y
Newton’s Method ) ( f x y = Newton’s method is based on approximating the graph of y = f(x) with a tangent line and on then using a root of this straight line as an approximation to the root of f(x) )) ( f , ( 0 0 x x y x S x x x / ) ( f 0 0 1 - = ) ( f ) ( ) ( f 0 0 0 ' x x x x y + - = ) 0 , 0 ( ) 0 , ( 1 x ) ( f 0 ' x S =

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Error of Newton’s Method Newton’s iteration for finding a root ]) , ([ C f 2 b a of α Mean Value Theorem ξ 5 between )) ( f ( / ) ( f ' 1 n n n n x x x x - = + )) ( f ( / ) ( f ' 1 n n n n x x x x + - = - + α α )) ( f ( / )] ( f ) ( f ( ) ( f ) [( ' ' n n n n x x x x - - - = α α )) ( f ( / )] ( f ) ( ) ( f ) [( ' ' ' n n n n x x x x ξ α α - - - = the error satisfies α and n x )) ( f ( / )] ( f ) ( f [ ) ( ' ' ' n n n x x x ξ α - - = Mean Value Theorem η 5 between ξ and n x )) ( f ( / ) ( f ) )( ( ' ' ' n n n x x x η ξ α - - = | ) ( f | min / | ) ( f | max , | ) ( | | ) ( | ' ' ' 2 1 x x B x B x B n n - - - α α Question Compare B with estimate in slides 33,34 Lect 1
MATLAB for Newton’s Method )) ( ( f / )) ( ( f ) ( ) 1 ( ' n x n x n x n x - = + MATLAB implementation of formula 3.27 on page 91 Start with one estimate ) 1 ( x >> x(1)=2; f(1) = x(1)^6-x(1)-1 >> for n = 1:10 S = 6*x(n)^5 – 1; x(n+1) = x(n) – f(n) / S; f(n+1) = x(n+1)^6 – x(n) – 1; end For n = 1:nmax end >> x' ans = 2.00000000000000 1.68062827225131 1.43073898823906 1.25497095610944 1.16153843277331 1.13635327417051 1.13473052834363 1.13472413850022 1.13472413840152 1.13472413840152 Example 3.3.1 pages 91-92

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Secant Method ) ( f x y = is based on approximating the graph of y = f(x) with a secant line and on then using a root of this straight line as an approximation to the root of f(x) )) ( f , ( 0 0 x x y x S x x x / ) ( f 1 1 2 - = ) ( f ) ( 1 1 x x x S y + - = ) 0 , 0 ( ) 0 , ( 2 x 0 1 0 1 ) ( f ) ( f x x x x S - - = )) ( f , ( 1 1 x x
Error of Secant Method )) ( f 2 /( ) ( f ) ( ) ( ' ' ' 1 1 n n n n n x x x ς ξ α α α - + - - - = - It can be shown, using methods from calculus that we used to derive error bounds for Newton’s method, that the sequence of estimates computed using the secant method satisfy equation 3.28 on page 92 where 0 ) ( f = α where n ς is between n x and 1 + n x and n ξ is between the largest and smallest of 1 , , + n n x x α The analysis on page 92 and Problem 8 on pages 96-97 shows, using the growth of the Fibonacci sequence, that 2 / ) 5 1 ( , | | | | 1 + = - - + r x c x r n n α α and c is a constant

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