First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 2. SOLUTIONS OF EQUATIONS: ROOT-FINDING 58
In [3]: x=1.319671630859375;
x^5+2x^3-5x-2
Out[3]: 0.000627945623044468
Let’s see what happens if N is set too small and the method does not converge.
In [4]: bisection(x -> x^5+2x^3-5x-2,0,2,10^(-4.),5)
Method did not converge. The last iteration gives 1.3125 with
function value -0.14562511444091797
Theorem 28. Suppose that f ∈ C 0 [a, b] and f(a)f(b) < 0. The bisection method generates
a sequence {p n } approximating a zero p of f(x) with
|p n − p| ≤ b − a ,n≥ 1.
2n Proof. Let the sequences {a n } and {b n } denote the left-end and right-end points of the
subintervals generated by the bisection method. Since at each step the interval is halved, we
have
b n − a n = 1 2 (b n−1 − a n−1 ).
By mathematical induction, we get
b n − a n = 1 2 (b n−1 − a n−1 )= 1 2 2 (b n−2 − a n−2 )=... = 1
2 n−1 (b 1 − a 1 ).
Therefore b n − a n = 1
2 n−1 (b − a). Observe that
and thus |p n − p| →0 as n →∞.
|p n − p| ≤ 1 2 (b n − a n )= 1 (b − a) (2.3)
2n Corollary 29. The bisection method has linear convergence.
Proof. The bisection method does not satisfy (2.1) for any C