Example 7.1.1. \(\sqrt{2}\) via Bisection.
For the sake of having a definite problem to work with suppose we want to compute a decimal approximation to \(\sqrt{2}\text{.}\) The Bisection Method works like this: First pick two numbers, one less than \(\sqrt{2}\text{,}\) and one greater than \(\sqrt{2}\text{,}\) In this example we’ll choose \(1\) and \(2\text{.}\)
Next, we take the midpoint of the interval \([1,2]\) as our first approximation to \(\sqrt{2}\text{.}\) We’ll call this \(r_1\text{.}\) The midpoint is the average of the endpoints so in this example \(r_1 =
\frac{1+2}{2}=\frac{3}{2}\text{.}\) Since \(\sqrt{2}\) and \(\frac32\) are both in the bracketing interval \([1,2]\) we see that the distance between \(\sqrt{2}\) and \(\frac32\) is less than \(1\text{,}\) the length of the interval.
Now \(\sqrt{2}\) must either be in the interval \([1,3/2]\) or \([3/2,2]\text{.}\) We need to decide which one. Since \(2\lt\frac94=\left(\frac32\right)^2\) we see that \(\sqrt{2}\lt\frac32\) so that \(\sqrt{2}\) must be in the interval \([1,3/2]\text{.}\)
We take our next approximation to be the midpoint of the (smaller) interval \([1,3/2]\text{.}\) Thus, \(r_2=\frac{1+\frac32}{2}= \frac54=1.25\text{.}\) Since \(\left(\frac54\right)^2\le2\) (confirm this) we have \(\sqrt{2}\) bracketed by
\begin{equation*}
\frac54\le\sqrt{2} \le\frac32.
\end{equation*}
This is really the whole idea. If we are approximating a number, \(\alpha\text{,}\) we begin by bracketing \(\alpha\) between two known numbers, say \(a\) and \(b\text{.}\) We take the average of these, \(r_1=\frac{a+b}{2}\text{,}\) as our first approximation of \(\alpha\text{.}\) We know that \(r_1\) is within \(\abs{b-a}\) (the length of the interval \([a,b]\)) of \(\alpha\text{.}\) If this is sufficiently accurate we use \(r_1\) as our approximation.
If not we determine if \(\alpha\) is in the first half–interval, \(\left[a, \frac{a+b}{2}\right]\) or the second, \(\left[\frac{a+b}{2},b\right]\text{,}\) and repeat the process, finding a new approximation \(r_2\) in an interval half the length of the first.
The Bisection Method generates a sequence of approximations, \(r_1, r_2, r_3, \ldots\) of the root we seek. In our example we have \(r_1=\frac32, r_2=\frac54\) and so on. At each step the new approximation is the midpoint of an interval whose length is one–half of the length of the previous interval. So our approximations can be made as close to the target as we would like.
