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This is the solution to the “bonus problem” from Monday: Bonus question: Consider f (x) = 1 − 21 (x − 1)2 . This has a fixed point of x = 1. Find an interval [a, b] containing 1 such that any number in that interval is a “good guess” for the fixed point iteration method (i.e. f (. . . (f (f (x0 )))) ≈ 1 for x0 in the interval). Hint: Use the idea of the proof of the criterion for attracting fixed points. Solution. Rearrange this to get 1 |f (x) − 1| = |x − 1|2 (*) 2 Suppose we do fixed point iteration starting with a guess of x0 . Then x1 = f (x0 ), etc. Then x0 is a good guess if the sequence {xn } converges to the actual fixed point 1. That is, we’re trying to make sure the error |f (xn ) − 1| goes to zero. Start by plugging x = x0 into (*): |f (x0 ) − 1| = 21 |x0 − 1|2 (**) Now plug x = x1 into (*): |f (x1 ) − 1| = 12 |x1 − 1|2 But by (**), we have an expression for |x1 − 1| = |f (x0 ) − 1|: = 1 2 1 2 |x0 − 1|2 2 = 1 3 |x0 2 − 1|4 (***) Now plug x = x2 into (*): |f (x2 ) − 1| = 12 |x2 − 1|2 But by (***), we have an expression for |x2 − 1| = |f (x1 ) − 1|: = 1 2 1 3 |x0 2 − 1|4 2 = 1 7 |x0 2 − 1|8 Hopefully you can see the pattern now: 2n+1 −1 n+1 |f (xn ) − 1| = 12 |x0 − 1|2 2n+1 −1 = 12 |x0 − 1| · |x0 − 1| Remember, the goal is for |f (xn ) − 1| → 0 as n → ∞. Since |x0 − 1| is just some fixed n+1 quantity, we need the exponential sequence ( 12 |x0 − 1|)2 −1 to go to zero, and that only happens if the base of the exponent, 21 |x0 − 1|, is < 1. That is, everything works out when |x0 − 1| < 2. That means our interval is −1 < x0 < 3: if x0 is chosen in this range, and we build a sequence xn as prescribed by the fixed point iteration method, then the errors |f (xn ) − 1| go to zero, which means the xn ’s converge to the true fixed point 1. 1