hw2solns

hw2solns - 2) 10 = 1 / 1024 < 10-3 . However | p...

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Math 151A HW2 Solutions Will Feldman 1. For part (c) use the error estimate for bisection method | x n - x * | ≤ 2 - n | a - b | where ( a,b ) is the initial interval and x n = ( a n + b n ) / 2 where ( a n ,b n ) is the interval at the n th stage. So if you want error less than 10 - k choose n so that 2 - n | a - b | < 10 - k or, rearranging, n > ( k log 10 + log | a - b | ) / log 2. 2. Use your code from (1). 3. Use your code from (1). 4. Use your code from (1) and also the bound on the number of iterations derived in (1). 5. Let f ( x ) = (1 - x ) 10 . p = 1 is the root of f , let p n = 1 + 1 /n be a sequence of approximations of p . Note that f ( p n ) = (1 /n ) 10 which is decreasing in n and (1 /
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Unformatted text preview: 2) 10 = 1 / 1024 &lt; 10-3 . However | p n-p | = 1 /n &lt; 10-3 requires that n &gt; 10 3 . Notice that we are in a situation where the residual is very small while the error is relatively quite large. As we discussed in section this is because the function has 9 derivatives which are zero at the root, so the residual decreases like the error to the tenth power near the root. 6. Do some algebra. Note that there are many dierent ways we can convert the root nding problem into a xed point problem. 1...
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This note was uploaded on 01/30/2012 for the course MATH 131a taught by Professor Hitrik during the Spring '08 term at UCLA.

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