roots2 - - x - 8 3x/2 3 f(x) = 5*ln x + 2 - x - 8 3x/2 3...

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Week 4 - Bisection method - Newton's method
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Bisection Overview: 1. Take the midpoint 2. See if f(mid) is positive or negative 3. Throw away old point with same sign
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f(x) = 5*ln x + 2 - x - 8 3x/2 3
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f(x) = 5*ln x + 2 - x - 8 3x/2 3
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f(x) = 5*ln x + 2 - x - 8 3x/2 3
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f(x) = 5*ln x + 2 - x - 8 3x/2 3
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f(x) = 5*ln x + 2 - x - 8 3x/2 3
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Bisection Example: f(x) = 5*ln x + 2 - x - 8 3x/2 3 f(0.1) = -18.4 f(10) = 31771.5 f(5.05) = 62.0 f(2.575) = -5.8 f(3.8125) = -4.1 f(4.4312) = 12.6 f(4.1219) = 1.7 f(3.9672) = -1.7 f(4.0446) = -0.1 f(4.0832) = 0.7
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Newton's Overview: 1. Find derivative at current point 2. Follow derivative to x-axis 3. New point is above/below step 2 x = x - f(x ) f'(x ) n+1 n n n
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f(x) = 5*ln x + 2 - x - 8 3x/2 3
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f(x) = 5*ln x + 2
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Unformatted text preview: - x - 8 3x/2 3 f(x) = 5*ln x + 2 - x - 8 3x/2 3 f(x) = 5*ln x + 2 - x - 8 3x/2 3 f(x) = 5*ln x + 2 - x - 8 3x/2 3 Newton's Example: f(x) = 5*ln x + 2 - x - 8 3x/2 3 f'(x) = 5/x +3/2 *ln 2 * 2 - 3x 3x/2 2 Newton's Example: f(x) = 5*ln x + 2 - x - 8 3x/2 3 f'(x) = 5/x +3/2 *ln 2 * 2 - 3x 3x/2 2 x = 10 x = 9.1 x = 8.1 x = 7.2 x = 6.4 x = 5.7 x = 5.0 x = 4.5 x = 4.2 1 2 3 4 8 5 7 6 Newton's Example: f(x) = |x| f'(x) = |x| - x = x - 1 f(x ) f'(x ) = -x Other Problems? Newton's Problems:- Derivative is zero- You don't always converge Convergence Newton's: Quadratic = x 2 Bisection: Linear = x...
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This note was uploaded on 10/21/2011 for the course CSCI 2031 taught by Professor Meyer during the Spring '08 term at Minnesota.

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roots2 - - x - 8 3x/2 3 f(x) = 5*ln x + 2 - x - 8 3x/2 3...

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