hw3sol - Math 3110 Homework 3 Solutions Exercise 4.1.1 (i)...

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Math 3110 Homework 3 Solutions Exercise 4.1.1 (i) Claim. If a n L and b n M , then a n + b n L + M . Proof. Let ± > 0 and let d n = a n - L and e n = b n - M denote the error terms for the given sequences. Since a n L and b n M , then by the error-form principle, d n 0 and e n 0. So there exist p and q such that for all n p , | d n | < ± 2 , and for all n q , | e n | < ± 2 . Let N = max { p, q } . Then for all n N , | d n + e n | ≤ | d n | + | e n | < ± 2 + ± 2 = ± Hence, d n + e n 0. But d n + e n = a n + b n - ( L + M ), so by the error-form principle, we have a n + b n L + M . (ii) Claim. If a n L , then ca n cL . Proof. Let ± > 0 and denote the error term of the given sequence by e n = a n - L . Notice that since | c | +1 > 0, then ± | c | +1 > 0. Since a n L , then by the error-form principle, e n 0. So there is some P such that for all n P , | e n | < ± | c | +1 . Then for all n P , | ce n | = | c || e n | < | c | ± | c | + 1 < ±. Hence, ce n 0. But ce n = ca n - cL , so by the error-form principle, we have ca n cL . 1
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Exercise 4.3.3 Using Newton’s method to find M , the unique positive zero of f ( x ) = x 2 + x - 1: (i) We give the recursion formula for a n +1 in terms of a n . Since a n +1 = a n - f ( a n ) f 0 ( a n ) , then a n +1 = a n - a n 2 + a n - 1 2 a n + 1 = 2 a n 2 + a n - a n 2 - a n + 1 2 a n + 1 = a n 2 + 1 2 a n + 1 . (ii)
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hw3sol - Math 3110 Homework 3 Solutions Exercise 4.1.1 (i)...

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