Problem2.6

# Problem2.6 - 2. By induction, the distance between g n ( x...

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Convergence to the 2 Using Newton’s Method James Keesling Theorem 0.1. Let f ( x ) = x 2 - 2 = 0 and suppose that x 0 > 0 . Let g ( x ) = x - x 2 - 2 2 x be the Newton function for f ( x ) . Let x 0 > 0 be any positive number. Then { g n ( x 0 ) } n =1 converges to 2 . Proof. Suppose that x 0 > 0. Then either x 0 > 2, x 0 = 2, or x 0 < 2. If the middle case, then g n ( x 0 ) = 2 for all n and we have convergence to 2. If we have the ﬁrst case, then for all c between 2 and x 0 , g 0 ( c ) = 1 2 - 1 c 2 . Thus we have 0 < g 0 ( c ) < 1 2 since 0 < 1 c 2 < 1 2 . Now use the Mean Value Theorem over the interval [ 2 ,x 0 ]. The theorem states that there is a c between 2 and x 0 such that g 0 ( c ) = g ( x 0 ) - g ( 2) x 0 - 2 . Since g 0 ( c ) > 0 for this c as argued above and g ( 2) = 2, we must have the g ( x 0 ) and x 0 are on the same side of 2. That is, g ( x 0 ) > 2 as well as x 0 . Furthermore, since 0 < g 0 ( c ) < 1 2 we must have that ± ± ± ± ± g ( x 0 ) - 2 x 0 - 2 ± ± ± ± ± < 1 2 . 1

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This says that the distance between g ( x 0 ) and 2 is less than half that of the distance between x 0 and
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Unformatted text preview: 2. By induction, the distance between g n ( x ) and 2 is less than 1 2 n of the distance between x and 2. Thus, in this case as well, g n ( x ) 2 as n . The last case to consider is x < 2. In this case for any c between x and 2 we must have g ( c ) < 0. Using the Mean Value Theorem again, this implies that g ( x ) would be greater than 2. Thus, all the elements of the sequence { g n ( x ) } n =1 are greater than 2 and the previous case applies to show that g n ( x ) 2 as n . A similar argument shows that for any x < 0, g n ( x ) - 2 as n . 2...
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## This note was uploaded on 11/12/2011 for the course STA 3032 taught by Professor Kyung during the Summer '08 term at University of Florida.

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Problem2.6 - 2. By induction, the distance between g n ( x...

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