First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 2. SOLUTIONS OF EQUATIONS: ROOT-FINDING 81
end
println("Did not converge. The last estimate is p=$pzero.")
Out[1]: fixedpt (generic function with 1 method)
Let’s find the fixed-point of g(x) =x where g(x) =(2x 2 +1) 1/3 , with p 0 =1. We
studied this problem in Example 43 where we found that 23 iterations guarantee an
estimate accurate to within 10 −4 . We set ɛ =10 −4 ,andN =30, in the above code.
In [2]: fixedpt(x->(2x^2+1)^(1/3.),1,10^-4.,30)
p is 2.205472095330031 and the iteration number is 19
The exact value of the fixed-point, equivalently the root of x 3 −2x 2 −1, is 2.20556943.
Then the exact error is:
In [3]: 2.205472095330031-2.20556943
Out[3]: -9.733466996930673e-5
A take home message and a word of caution:
• The exact error, |p n − p|, is guaranteed to be less than 10 −4 after 23 iterations from
Corollary 42, but as we observed in this example, this could happen before 23 iterations.
• The stopping criterion used in the code is based on |p n − p n−1 |,not|p n − p|, so the
iteration number that makes these quantities less than a tolerance ɛ will not be the
same in general.
Theorem 44. Assume p is a solution of g(x) =x, and suppose g(x) is continuously differentiable
in some interval about p with |g ′ (p)| < 1. Then the fixed-point iteration converges to p,
provided p 0 is chosen sufficiently close to p. Moreover, the convergence is linear if g ′ (p) ≠0.
Proof. Since g ′ is continuous and |g ′ (p)| < 1, there exists an interval I = [p − ɛ, p + ɛ]
such that |g ′ (x)| ≤ k for all x ∈ I, for some k < 1. Then, from Remark 39, weknow
|g(x) − g(y)| ≤k|x − y| for all x, y ∈ I. Next, we argue that g(x) ∈ I if x ∈ I. Indeed, if
|x − p|