First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 2. SOLUTIONS OF EQUATIONS: ROOT-FINDING 56
Step 1 : To start, we need to pick an interval [a, b] that contains the root, that is, f(a)f(b) < 0.
From the plot, it is clear that [0, 4] is a possible choice. In the next few steps, we
will be working with a sequence of intervals. For convenience, let’s label them as
[a, b] =[a 1 ,b 1 ], [a 2 ,b 2 ], [a 3 ,b 3 ], etc. Our first interval is then [a 1 ,b 1 ]=[0, 4]. Next we
find the midpoint of the interval, p 1 =4/2 =2, and use it to obtain two subintervals
[0, 2] and [2, 4]. Only one of them contains the root, and that is [2, 4].
Step 2: From the previous step, our current interval is [a 2 ,b 2 ]=[2, 4]. We find the midpoint 1
p 2 = 2+4
2
=3, and form the subintervals [2, 3], [3, 4]. The one that contains the root is
[3, 4].
Step 3: We have [a 3 ,b 3 ]=[3, 4]. The midpoint is p 3 =3.5. We are now pretty close to the
root visually, and we stop the calculations!
In this simple example, we did not consider
• Stopping criteria
• It’s possible that the stopping criterion is not satisfied in a reasonable amount of time.
We need a maximum number of iterations we are willing to run the code.
Remark 27. 1. A numerically more stable formula to compute the midpoint is a + b−a
2
(see Example 20).
2. There is a convenient stopping criterion for the bisection method that was not mentioned
before. One can stop when the interval [a, b] at step n is such that |a − b|