exam1s - r ) n , The iterations are 0.01 0.00972183...

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Numerical Analysis/Exam #1/October 6, 2004 Solutions 1. (a) f (0) = - 1 < 0, f (2) = 9 > 0, and f is continuous, so the Intermedi- ate Value Theorem tells us that f has a root in (0 , 2). (b) a b m e 0 2 1 1 0 1 0.5 0.5 0 0.5 0.25 0.25 0.25 0.5 0.375 0.125 0.25 0.375 0.3125 0.0625 Final estimate is 0 . 3125. (c) We have to solve b - a 2 n < 10 - 5 , giving n > 17 . 6096, so n 18. 2. g ( x ) = x - f ( x ) f 0 ( x ) = x - x 2 + 3 x - 1 2 x + 3 , iterations are 1 0.4 0.305263 0.302777 0.302776 0.302776 . 3. The iterating function is g ( x ) = x - f ( x ) f 0 ( x ) = x - (1 + r ) n (1 - kr ) - 1 n (1 + r ) n - 1 (1 - kr ) - k (1 +
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Unformatted text preview: r ) n , The iterations are 0.01 0.00972183 0.00968964 0.00968925 0.00968925 . 4. L ( x ) = 1 2 ( x 2-3 x + 2) L 1 ( x ) =-( x 2-2 x ) L 2 ( x ) = 1 2 ( x 2-x ) L ( x ) = x 2-x + 2 f (1 . 8) ' L (1 . 8) = 3 . 44 5. L 1 ( x ) =-x ( x-2) , L 1 (3 / 2) = 3 / 4 L 2 ( x ) = 1 2 x ( x-1) , L 2 (3 / 2) = 3 / 8 L (3 / 2) = 3 8 1 + 3 2 ! = 0 . 69976...
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