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