APPM 1350 Review Solutions 3

# 1 n n n n 10 0 10 a set up a newtons method algorithm

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Unformatted text preview: gral, thus, "✓ ◆ #Z ✓ ◆9 ⇣ n ⌘9 1 1 19 2 x10 lim + + ··· + = x9 dx = n!1 n n n n 10 0 10. (a) Set up a Newton’s method algorithm to approximate (b) Now, starting with x1 = p 5 Z 1 x9 dx, 0 1 = 1/10 0 2. 1, ﬁnd x2 , the second iteration approximation of p 5 2. Solution: p (a) Let x = 5 2, then this implies x5 +2 = 0, and we wish to ﬁnd x, that is, we wish to use Newton’s Method to approximate a root of the equation x5 + 2 = 0, so let f (x) = x5 + 2 then f 0 (x) = 5x4 and so f ( xn ) x5 + 2 5x5 (x5 + 2) 4x 5 2 n n xn+1 = xn = xn =n = n4 f 0 ( xn ) 5x 4 5x4 5x n n n so the resulting algorithm is xn+1 = (b) Starting with x1 = 4x5 2 n . 5x4 n 1 we have x2 = 4x 5 2 42 1 = = 4 5( 1)4 5x 1 6 5 11. Find a positive number such that the sum of the number and its reciprocal is as small as possible. Solution: We wish to ﬁnd a number x > 0 such that the sum S = x + 1/x is minimized. Note that x2 1 S 0 (x) = 1 1/x2 = and so the critical points are x = ±1. Note...
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## This note was uploaded on 02/04/2014 for the course APPM 1350 taught by Professor Segur,harv during the Fall '07 term at Colorado.

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