# 4.6.5-eg - 4 15 is x = 2 . Then x 1 = x-f ( x ) f ( x ) =...

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Example Use Newton’s method to estimate 4 15, displaying the two iterates x 1 and x 2 . Solution: We need to ﬁnd a function f for which 4 15 is a root, preferably something that is easy to diﬀerentiate (like a polynomial). Since we’re seeking x = 4 15 , (1) raising both sides of (1) to the 4th power leads to x 4 = 15 , or x 4 - 15 = 0 . We therefore select the polynomial f ( x ) = x 4 - 15 , and note that f 0 ( x ) = 4 x 3 . In order to start the Newton iteration, x n +1 = x n - f ( x n ) f 0 ( x n ) , n = 0 , 1 , 2 ,..., an initial estimate x 0 is needed, where x 0 should be suﬃciently close to the desired root. We know that 4 16 = 2 , so a reasonable initial estimate of

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Unformatted text preview: 4 15 is x = 2 . Then x 1 = x-f ( x ) f ( x ) = 2-2 4-15 4 2 3 = 2-1 32 = 63 32 . = 1 . 9688 , rounded to four decimal places. Further, x 2 = x 1-f ( x 1 ) f ( x 1 ) = 1 . 9688-1 . 9688 4-15 4 1 . 9688 3 . = 1 . 9680 , also rounded to four decimal places. Thus, the rst two iterates from Newtons Method are x 1 = 1 . 9688 , x 2 = 1 . 9680 . (Note that the second iterate equals 4 15, when rounded to four decimal places.) 2...
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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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4.6.5-eg - 4 15 is x = 2 . Then x 1 = x-f ( x ) f ( x ) =...

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