First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 2. SOLUTIONS OF EQUATIONS: ROOT-FINDING 59
We have, from the proof of the previous theorem (see (2.3)) : |p n −p| ≤ 1 (b−a). Therefore,
2 n
we can make |p n − p| ≤10 −L , by choosing n large enough so that the upper bound 1 (b − a)
2 n
is less than 10 −L :
( )
1
b − a
2 (b − a) ≤ n 10−L ⇒ n ≥ log 2 .
10 −L
Example 30. Determine the number of iterations necessary to solve f(x) =x 5 +2x 3 − 5x −
2=0with accuracy 10 −4 ,a=0,b=2.
Solution. Since n ≥ log 2
( 2
10 −4 )
=4log2 10+1=14.3, the number of required iterations is
15.
Exercise 2.2-1: Find the root of f(x) =x 3 +4x 2 − 10 using the bisection method,
with the following specifications:
a) Modify the Julia code for the bisection method so that the only stopping criterion
is whether f(p) =0(remove the other criterion from the code). Also, add a print
statement to the code, so that every time a new p is computed, Julia prints the value
of p and the iteration number.
b) Find the number of iterations N necessary to obtain an accuracy of 10 −4 for the root,
using the theoretical results of Section 2.2. (The function f(x) has one real root in
(1, 2), so set a =1,b=2.)
c) Run the code using the value for N obtained in part (b) to compute p 1 ,p 2 , ..., p N (set
a =1,b=2in the modified Julia code).
d) The actual root, correct to six digits, is p =1.36523. Find the absolute error when p N
is used to approximate the actual root, that is, find |p − p N |. Compare this error, with
the upper bound for the error used in part (b).
Exercise 2.2-2: Find an approximation to 25 1/3 correct to within 10 −5 using the bisection
algorithm, following the steps below:
a) First express the problem as f(x) =0with p =25 1/3 the root.
b) Find an interval (a, b) that contains the root, using Intermediate Value Theorem.
c) Determine, analytically, the number of iterates necessary to obtain the accuracy of
10 −5 .