# Lec3 - Solving nonlinear equations Bracketing methods Basic...

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Solving nonlinear equations Bracketing methods

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Lecture 3 2 Basic Problem How do you find the solution to an equation such as ? 2 x exx = + Step 1: Recast equation in the form: Step 2: Plot f ( x ) () 0 f x = 2 0 x −= () f x -2 0 2 -2 -1 0 1 2 solution x f ( x ) a t 1 . 2 f xx ⇒= Graphical method Graphical method is … not precise … but it • provides good initial guesses for more precise methods
Lecture 3 3 Methods for finding roots of equations z Graphical not precise provides good initial guesses for more precise methods helps avoid pitfalls z Bracketing Starts with 2 inputs that bracket solution Usually converges Slow z Open Starts with 1 input Converges very quickly But often does not converge x f ( x ) Oops – no solution

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Lecture 3 4 Bracketing 1: Bisection method -2 0 2 -2 -1 0 1 2 Step 1: Bracket solution x L x U f (x L ) Choose 2 points x L and x U so that x L < x r < x U . A simple test for this is: f ( x L ) f ( x U ) < 0 ? f (x U ) x r = ?
Lecture 3 5 Bracketing 1: Bisection method -2 0 2 -2 -1 0 1 2 Step 1: Bracket solution x L x U f (x L ) f (x U ) Choose 2 points x L and x U so that x L < x r < x U . A simple test for this is: f ( x L ) f ( x U ) < 0 ? Step 2: Choose new trial root 1 2 () RL U xx x = + x R Calculate f ( x R ) f (x R )

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Lecture 3 6 Bracketing 1: Bisection method -2 0 2 -2 -1 0 1 2 Step 1: Bracket solution x L x U f (x L ) f (x U ) Choose 2 points x L and x U so that x L < x r < x U . A simple test for this is: f ( x L ) f ( x U ) < 0 ? Step 2: Choose new trail root 1 2 () RL U xx x = + (a) If (b) If (c) If ()0 L RU R f xf x x x <= L RR f x == DONE x R f (x R )
Lecture 3 7 Bracketing 1: Bisection method -2 0 2 -2 -1 0 1 2 Step 1: Bracket solution x L f (x L ) f (x U ) Choose 2 points x L and x U so that x L < x r < x U .

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## This note was uploaded on 12/29/2011 for the course CHE 132b taught by Professor Ceweb during the Fall '09 term at UCSB.

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Lec3 - Solving nonlinear equations Bracketing methods Basic...

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