4.6.1-eg-1 - -. 3333 , rounded to four decimal places....

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Example Use Newton’s Method to estimate the real solution of x 3 + 3 x + 1 = 0 , starting from the initial estimate x 0 = 0. Find x 1 and x 2 . Solution: Recall that the iterates in Newton’s method are given by x n +1 = x n - f ( x n ) f 0 ( x n ) , n = 0 , 1 , 2 ,..., starting from an initial estimate x 0 of the desired root. For this example we have f ( x ) = x 3 + 3 x + 1 , where f 0 ( x ) = 3 x 2 + 3 , and we are given x 0 = 0 . Then x 1 = x 0 - f ( x 0 ) f 0 ( x 0 ) = 0 - 0 3 + 3 · 0 + 1 3 · 0 2 + 3 = - 1 3 . =
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Unformatted text preview: -. 3333 , rounded to four decimal places. Further, x 2 = x 1-f ( x 1 ) f ( x 1 ) =-. 3333-(-. 3333) 3 + 3(-. 3333) + 1 3(-. 3333) 2 + 3 . =-. 3222 , also rounded to four decimal places. Thus, the rst two iterates from Newtons Method are x 1 =-. 3333 , and x 2 =-. 3222 ....
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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.

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